MOLECULAR THERMODYNAMICS
John D. Simon
University Science Bouks Sausalito, Califurnid
lllrivrrsity Science Books
Preface
551) (;,11c I:~vcKodd S , ~ u \ ~ l ~CA t o . 94965
xi
Acknowledgments
x i ii
CHAPTER 1 / The Energy Levels ot Atorns and Production manager: Susunna Tadlock Manuscript ed~tor:Ann McCuire Designer: Robert Ishi Illustrator: Johtt Choi Compositor: Eigent)pe Printer & Binder: Edwards Brothers. Inc T h ~ sbook is prlnted on acid-free paper Copyright 01999 by Un~versityScience Books
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Reproduction or translation o f any part o f 1111swork beyond th,~t~ ~ c r n i ~ thy ted Sect~on107 or 108 of the 1976 U n ~ t e dStates Copyr~ghtAct wlthout the pcrlnissron of thc copyr~ghrowner IS unlawfill. Kequcsts for p u r n i ~ s s l ~orn further inlornlation should be addresaed to the Perni~sstonsDepa~tmcnt, Univers~tyScience Books.
Library o f Congress Cataloging-in-Publ~cationData
M A T H C H A P T EAR / Numerical Llethods
McoJarr~e,Donald A. (Donald Allen) Molecular thermodynamics I Donald A. McQuarr~c,John D.Simon. p, cm. Includes bibliographical references and index. ISHN 1-891389-05-X I ' I ' l i ~ ~ r ~ ~ i ~ ~ t l y n aI .mSimon, i c s John D. (John Douglas), 195711
1111<.
IOO(I 01) ~ o . I A I \ l i \.t 1 \'I+) ,I, .'I
98-48543 CIP
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M o l e c ~ ~ l c sI
1-1. Tllcs Eltactrunic Fnergy of Atoln~ctIyclrogt.11 is (>LI~IIIIIL(YI 2 1-2. Tlic Allowcd Energies o f d System arc, Ol)t,~~r~eel Iron1 tlie Sc lirotlinger L c ( t r , t l t ~ ~ t t 1 , 1-3. Atonis Have Translationdl Energy 111A t l d ~ t i o nto F l c ~I r o n ~ cEnergy 7 1-4. The Vil~ratlorlalMotion ot a D i a t o n i ~ cMoleculc C,rn t)cx hlc~tleledIly 1' I I.IIIIIIII~II Ov i I l , ~ l o r 9 1-5. 1 lit, I Ilcq!,y I (>vtsls(il ( ] L I J I I ~ L I I ~ ~ - M ~ ~ I I IcI ~~r IlIi~~ I( C O\( ~ ~~l ~~~ cl l ~ ~Art' t(lr I l l ~ ( t , I ;) \.\.1111 1' = 0. I. 2. . . . 1 .$ 1-0. I I I ~I, l . t r ~ ~ r o r iO n j 1 il1,itor Accourils for the Infrarc,cl Sptrc Irunl (11 '1 1)1,1toni11. 1111. 1-1 h111lts1 1-7. 1 lit, I ) I \ ~ o I , I I I ~ 1I ~rlcrfiy <~nci the Ground-State Election~ct n c r ~ \o i a U ~ ~ i t c ~ l n ~ c I I I I K I I I I I I = I , I / 16 I-I!. 1111. I I I ( , I ~ \ .I (.\.I,I\ 111 ,I K ~ g i tKot,~tc~r l Are c , = ~ ' J ( . t I I ).l?I IH 1.9. Ill(.\ ' 1 1 1 1 . 1 1 1 ~ 1 1 1 ~ 01 I'II~~.IIIIIIIII hlo1t.c i~Ic,\Art, Rcy,reselit(~tl I)y Xr~rni,ilMotlc> -12 1-10. Ill,. l i h ~ ~ , i l ~ r\IIIV , ~ ~ .IIIIII~ tl 01 $ 1 I ' I I I V ~ I I ~ I ~MOICY ~ I I iilcl 1)(~11(~1111\ (111011111(, MOIII(.III~01 5 I I I I I t I I h111lc~ LIIC111 t l ~ t K , ~ g ~ ~ l - l < oI I~I~III~III<-O\( t~to~ 11l~it111 ,~\~III~~\II~I,~~I~II 1-11. rli(x I o(3rgy01 ( ,In IN' \Vr~tten.I\ t = c ,,,,,,\ + ~ 5 , < > , - + t \ , l > ~ t27 t,l
30
Problems 4 0
CHAPTER 2 / The Properties of Gases
49
2-1. All C a w s Belidvc Itlrally li They Are Sufficiently Dilute
4')
2-2. Tlirl V J I , der \V~'ils Eq~rdtlonand the Krtll~c-h-KwongEqir,ition Arc Ex,ln~[~lt,s of Two-f'ardmeter tcluat~onso i St,~tc 54 ill1 2-3. A TLII>I~:Equ,it~onot State Call [ k s c r ~ l ) cI311fti the c;,~itaous ;ind I I C I L I IStlit(,i ~ 2-4. T11c' VJII dcr l'V,>.ils E c l ~ ~ a ~ant1 i o n tlit' Kcell~cliti\\.orig Erlu,it~on C)I~tsytilt, l .I\\ of C'orrcxsptrn(lingSt,itc>s 67 2-5. Sccoriti t'irial ( . o c ~ i i i c ~ e ~Cdn ~ t , Bc Ilsccl III D c t c ~ r l ~ i i rIlnr ~ l ( ~ r l i i r ~1l 1i 1~~ 1r1 I ' I I ~ I , I I I I . I ~ ~1 ~
I-(,. I ( I ~ I ~ I II I ~ II \ ~ , 1,111 ~ ,4.h IC I \ ~ uI*)~II,II ~ ( ~ 1111, I . ~ ~ c i *( \ it~ t ! l r ~ I ) ~ ~ tot iIo~ n IrC" ~Tt>rriii r i lhr! 2
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M A T H C H A P T EBR/ I'ruI7ahility and Stalislics
9;
Prrrtdr~ms lIll
C H A P T E R3 / Tho Roltzrndnn Factor and Partition Functivns
I 05
3-1. Tlw Bollzlnann Factor Is Cjne of tht- Mo+t Irriport,~ntQudntilies in the Phvslcal 5r iences 106
3-2. T t ~ Protwbililv r Tlwt a System In an tnst,mk>lr. Is i n thr. Stdtr! wilh Energy E , ( ! V . V ) IC I ~ ~ , ~ ~ ~ ~ >10 I I(,~ C'lY'I" I I I w ~' n r 106 3 4 . i.2.t-I'r~
l>j 1 I 0 3.1. I l ~ e H t ~C.q).~r t irv ,it Ct>rl+r;lr)lVolulne Is ihe lbnlperat~lreLlerivativr nf thr Avel ace Ener~t 1 14 '1-5. i'r:r- ('.iri Fxprrqr lhr Prrswrr In Tellns of a Partition Fun(-tirlri I1 h 3.6. rI.r, l',)~ tillon I - ~ l l ~ r t nf ~ o,In5ystrrii of Ir~rlc*l>mtlt~~jl, Di~tinjiuisl)ahleMolecules I\ i t i t , Prrlrlucr ni hlolecular Partition IKunctiuns 1: 19 3-7. 1114. !',~rtitiori Funr tior) rlf a Syslenl of Indepe~tdent,lndlstinguishable Atorr~srlr .kl<~lecules (:3n 1Jsu;ljly R t : LZ.'ritfcrlA S Iq(V. . I ' ) ] ? ' ! N ! 120 3-0. ,A Mnlerul~rPaltition Function Carl H t - Dr.tr>rnposcrlinlo Partitior) runctlons irrr r,3rli I ~ P ~ ni P frwrlnm P 12 i Problelns l ?n
M A T H C H A P T ECR/ Series and I-irnits 135 Problems 13U
CHAPTER 4 1 Partition Functions and Ideal Cases
I 43
4.1. Tlrc Tra~~sl;ll~onal Partitlnn runtt~unof an /\tom in ;l Monatrmir Itir!dl <;a!, Is ~ 2 , - r ~ ~ ~ k l , 1 / l ~ ' j143 3"V 4-2. XZost Aiorrlc Arc ir> tl1r Ground Eleclrunlc State at Room Temperature 1,15 3-3. 1 hr 1-ner);r. uf ,I n i . ~ f r > r r ~hlrilvr ir ulr?L i r > R r h Al)llroxin)d~cdas a Suln rd 5t.pardrr Tcrnls 1 4Cl 4.1. Mo+IMolc~cult!~ Arc it) lhc Ground Vit>ral~oi>al State at Room Ternperatur~ 1 5 2 4-5, hh~sthlr,lr>cult.+ hrr. in Exr i t t h t l Rot;lfinr>,>lS1;Hcs dl Ordinary Telnperati~res 155 4-6. Kot,ttional I'arlltlriri I unctions Conhin J Syrrlnlrtry Numt~rr 158 4-7. -1 t i t > L'ihr,itirjri,ll I',irlilion Ful)clifir) of a Polvatornjc Molecule Is a t'roduct uf H~rmonic ClsrlI1~tort'art~t~on Funr tion5 for fat-h Nrlrrrl;il Corlrdirlate 160 4-U. The iorln oi tlie Kot~tionalI'artitiun Fun~tionni a Prilyatt)rrlir Mt>iucu!c Dcpcr)ds upon ! l ~ r ?5h,lpe of the Molecule 163 4-9.( . - r ~ l c u ~ l 1 3 t rMolr>r ~rl I tr.nl C;ll);lc III(!F Arc 111 Very C ~ o Agreement d wlth l-xl~rrinient,jbI I=IIA I Oh
f'roht~ms I h(j w
CHAPTE5 R / The First Law of Thermodynamics
185
5-1. I? Con~n~on Type of LVrjrk I S P~~SSIITP - VoI~nie \,VRrk 1 95 5-2. iVmk anrl Hrmt Arr Not Slal(. Funr-t~on~, 1)11t Ent'lg) I< a Stat(. Ft~nctlcm I tli< 5-3. Ihr. kirst L.IW of Tli~~rr~iodyti,>r~lir + S~p5thr! trlrlrgy Is Clotr! Fuuc l i r ) ~ ~1 ' ) L 5-4. Ari .Arli~ti.ltic:Proc-th\s Ic A Pror thcsill l.L'liir h hrj Encbrgv ,is t-lc~tI $ Tr,>r>dvrrr.d 1 \ I I 5-5. Ihe Icmljc~rdtl~rt. rlf A Gas Ilvr rr.,i\t-+ in ,i Kt>vt,rsit>lr-htli,~l).itir.txp,~n\ior~ IYh 5-6. iVrirk anrl H t w H a w . ,>Silriplt. n-lol(.c-ul,~rI~~tt:rlir~.l.>titjr> 198 5-7. -1l1r.Erltllalpy C h r l r ) ~I$ ~ .EtjudI 1 0 tllf Erlergy Trar>sierrcrlas t leal In ;l Corlstzrll-t'rcs~:~-r I'rorcss Irivrllving Only !'-\I Wt>rk 2 0 0 5-0. Heat Capacity 1s a l'ath tunr-tion 1112 5-9. Relatlve Cnthall~lesC-an I<e Ut,tern,ined from Hcat (.al~.ititv I),lf,> anrl Ht..)f< o i Transillon 205 5-111. Ert~hatpyChdrlgcs tot Cl~e11~1caI Eq~lati~tis Are Addltive 207 5-1 1. tivats o i Rt.dctir>ri!,Carl Br C,llr ulalftl (rotn T,d)ul~lcdI-lmls of Fortndllon L 111 5-12. The Tenlperature Dependence of A, I I Is Liven in lerrn+ r i l th*. Hth.~t t s Prorllfr ts 21 7 Capxities rjf the R r ~ c f ~ r >,irirl Problems 220
M A T H C H A P T EER/ The Binomial Distribution and Stirling's Approxim at 'ron 229
CHAPTER 6 / Entropy and the Second Law of Therrnorlynamics
;I I -
6-1. The Change ol C n e ~ pAlone Is Not Skrffiritwt tn I ) ~ t ~ r ~ nthr> ~ r iI)~rtvr-tir>l~ c* of a Sllorltarleotrs Process 2 3 7 6-2. hlr>r)rqrril~briurn Isolated 5vsten)s Evolve In a IJlrection T h ~ Increase< t I huir Ui!,orrlcr L 39 6-3. Irrllikt- ryw%, Er>trrjpyI s ,I St~fcFur,t.litin 241 6-4. 1 IJP S~rurldI aw of I htbrrnotlyn~rrlir+ 5t~tc.sT h ~ !ht. t Er11roi)yr i l r l r > Icr>lr~lvtl Sy~l(~r11 Irlcreases AS a Result of a Spont~neuusl'rur-e- 2.15 6-5. The Mosl ralnuus Equation of Statistic-a1Therrnndrnam~c-sI+S = L,,111 11' 2-4(? 6-6. !Ve MI AS^ Alw~ysUevise J Revers1ble t'ruc-es5 trj (-;ltr ulatv tntropy (-.han~r.s 2 5 <
6-7. Thermodyiian~icsGives Us Insight into the Conversion of tleat into i,ilr>rk '58 6-8. Fr~frol>v T,I~I R r Exl)rcssctl in Tcrnls nf ,I Parlillon Function 2GO 6-9. The Molecular Furmula S = k, In W Is Analrycll~stn thtl 1 ttc~rmrldyridrnirForrlil~l.~ d S = S q t c h ; l 263 Prt~filcms L h4
C H A P T E7R / Entropy and the Third Law of.Thermodynarnics
2 71
7-1. Entropv Increases with Increasing leml~erature 273 7-2. The Thlrd L a w of Therlnodynaln~csS ~ v That s tlie Entropb ui a Perfect (. ryst,il Is Zero ,lt 0 K 2 7 5 7-3. Al,<S -- Alp H ; 7iP at Ph,isr Tr~nsilirjrl 2 7 7 7-4. [ 1 hirrt I.,iw th Thrrrriot~yr>,~~riir.~ Asscrts Th,it I',. 0 ,>5 I , 0 I -'r< 7-5. I'rx tic-,11Al>\t)lulc tritropics C,rn Bc Dc~t!rrr~ir~cd Calorirnt,tr~r, 1 1 1 ~ .) 7-6. rJrLltl i e a1 Al~srjlu~e Enlropies ot C , ~ s eCan s Be C,llcul~tcdt i c Ilrl 1'.u 1 1 1 1 ,r t ~I ~ I I 114 I ~111% 7-7. TIrc* V,iluc$ th 5tdr)ddrd EoIropic~.DFPCI! L I ~ O I > ,%lr>l
':> I
7-8. Iht. 511t" 1"l5( 1)1)1( t ~ l t r [ ~ i ) of ~ !as tc!w Sutlsr.lnces 110 Not Agrthc w ~ l hthe ( ,tkrrirnt.lric Enfropit<< 28H 7-9. 51.1rlddr(l Entropics Can R? Used trl Calculate Entropy Changes o i Chcnucal Kt,ns 289 Problems 29CI
CHAPTER8 /
Helmholtz
and Gibbs Energies 301
.
8-1 1 hv 51gn01 Ihe Hr.lmholt7 Energy Ch,~nge[>c.lerrn~ncsthe Direction of d 5prjrllane01ls i 1'rotr.s~In a Systcm a1 Cnristhnt Volume ~ n rTernp~ralure 301 8-2. 1 tic C~t>t>s Energy I3eterrnincs the Ilircction o i a Sporjtan~!usProcr!ss for a System at Constdn! Presslire 2nd rclrnperature 304 8-3. Mdxwell H
CHAPTER 9 / Phase Equilibria
-44s
9-1. A Ph3w Diagrdnl S~rnrn~~rizes thc Sol~d-Liq~l~d-Gds D e h ~ v ~ of o r a Substance 350 9-2. I hr C~bbsEnergy t j i a 5ut,stancr Has a Close Connrbction to Its Phase Diagram ,157 9-3. The Chemical Pott~ntialsof a Pure Sul)slan~t? in Two Phases in Equilibrium Are Erlual 159 9-4. The Cldus~usXlal~evrcm Eytrdtic~nLives the Vdpor Prrswre ui a Substdrlce As a F~rnct~or~ oiTcnlper~turt: 365 9-5. ( Iir'llric .il P o t i ~ r l ~ Car) i ~ ~1 l 3 rvaludlerl ~ 1ri)ln ,I P,rrtit~r>n Furlct~on 369 Problems 373
C H A P T E 11 R /
5 i ~ I u t i r ) n sII:
5r)li(l -I.icli~itiSolutlol~>
-1
1 1 - 1 . L%kII>[,lul~onsof S(111rls [)h+srllvr>rl in I icl~uds 4 11-2. 1 he Ar-tivity of n Nr,nvol.itllc Soluttb (-an Ht. I ~ l > t , ~ l r ~ ri r~ trrl~ r tht' i Vdl)trr l'rc>++clr(, ,,I GI
!
Solvent 443 11-3. Colligativr. Prof)crt~~s Are Sr,lrrtir,n Prr>pc.rtirhcThat Dcpthnri (.h~lyIlprln tllc NI.,I , s . Den5ily ui Sol~~tc Particles 448 . 11-4. Osrntjt~cPrcssurc C J Be ~ Used to Detr.r~ninth~he Muler uldr M;tssc-+ot l ' t ) l y ~ n < ~:, ~ 11-5. Solutions of E1e~trolytr.sAre Nonltleal dt Rel~livelyLobv Cunt r-~~lralions4; 1 11-6. The ilc:byr-tliikr.l Theory Gives an Exat I txpress~onut In y+ Lor Vt.rp D~lute Soltrtions 459 11-7. The Mean Spherical Approxinlatir>[lIs An Extcns~orio i thc Urhye-Hucktd Tht~o~; tlj klighrr Concentrdtion~ 36 1 Problems 4415
CHAPTER 1 2 1 Chemical Equilibrium 477 12-1. Chenllcal Eqtlilibrium Results when tkrc Cit)bs Frrcrgy Is a Mllunium with Itc.qrt , ' to the Extent of Kthdctiorl 477 12-2. An Equilibrium Constant IF a Functiol~o i Temp~.raturrOnly 481 12-3. Standard Gibbs Energies nf Fornidt~oriCan Be Chcd to Calr date Equ~libr~ur~~ Conslants 484 12-4. A Plot t ) i the Cil~kxEnergy of d Kcldctiurl hlixture Against the Extent of Kt~dcticiI Is a M~nimurnat Equiltbrium 106 12-5. The Rdt~oof tlie Reaction Quotit-nt to the t q u ~ l ~ b r Const;lnt i ~ ~ m K)ctern~ine> ihc rjirer-tion in whlr:h a Kc;lct~c>rl Will Prncced 488 22-6. lhc Sign oi A,G And Not T h ~ of t A < G ' Determin4.s the D~rectinn r r f Redctiorl 5finntant:ily 490 12-7. The Variation u i an Equilillri~lmC.onsl,lnt with rr!rnprhraturr!Is Given 1)). th~. Vdn't tiofi Equation 491 ( , I l'ortit~oriiuoclion., 4')i 12-0. We C d r l C-alculdtr Equilibrium Crlnstarits In 1. 12-9, hlnlec-ular I'art~tionFuriclions ~ n Reldtcd d -1 hcrmrjdy~urnictldta Are txtun+ivcly rL~t>uldlecl499 12-10, Eq~~ii~tlrlunl Corrstant+lor Hcdl C;dses Are kxljres5r!rl in Tc~m!,ui l ' l ~ ( t ~ d l Fugar ities 50h 12-11 . Tlit~r~nurlynarntc tqu~lihriuqnConstants ,%retxpreswrl in Terms of Ac r~vit~cs 12-12. The Use of Activities Makc5 a Sig~~ifirdnt nlffrrtlrice in 5mlubiiity C~lr.ul;rtrrin\ Involving lunic Sllecic~ 5 12 Prohlerns 51 5 )rl
CHAPTER10 / Solutions 1: Liquid-Liquid Solutions 387 10-1. Pa~tldlMolar Quantiti~sAre lnlportarrt Thermcldynamic Properties of Solutions 387 10-2. I h r Lihbl-Duhtam Equation Rdalr!s lhe Chnnge In the Chvm~calPotentidl of Onr Crmrponent of a 5olution to the Change in thc Chemir dl Poteriiial of lhe 0thr.r 390 1U-3. At Fr~l~iiibrlurn. lhe Chcmical I'r,lcnt~al r ) i Each Cornpunt-nt Has the S d r r ~Valur ~ in Each I ' h ~ cIn Which the Colnponr.n~r\ppcars 393 10-4. Thc Cornprmc.nts of an ideal Sol~rtionObey Rauult's Law fur All Cunt cntraticlrlr J9.1 10-5. hlo5t Solutron5 are Nut Irleal 4 ~ 1 10-6. The Gibt)s-Cl\~h(.rrEqtlation Reldtthsthe Vdpor Prthssuresof the l w n Components o i a Volatile Binary 5olution 1115 10-7. The Ghntral Thermorlynamir quantity ior Nonldedl Solutions is the Activity 410 10-8. Activrties Must Be Cdlc u1att:ci rv~thKcspect to Slandarrl States 413 10-9. We Can C-rjrl\truc~a M o l uldr ~ Mrjrlcl r l i Non-idtldl Solution5 414 Problems 426
CHAPTER1 3 / Thermodynamics of Electrochemical C e l l s
i:')
13-1. A n Elpt ~rochcnli~dl Cell Prurluces dn Elcctrir. Uurrcnt a< (he Result o i d I-I~V;!II<.
Kedcllori 524 13-2. Hali (Yells Tan Be Cla~sitiedinto Varirlus iypes 53 1 13-3. A Cell D~agranlIs Used to Ktlprestlnt ~ r Elec:trocht!~n~c,~l i Cell 5 < 3 13-4. Th(: Ciblls Cncrgy Changr. of a Cell Reaction Is Dirtlclly Kclatr-rlto ihe t i r , ~1 1 , Fort e uf the Ccl l 5 16 13-5. The Stand~rdfrrrf of an E1t.c-lroc.ticmii,d Ccll Cdn Be: F~~urlrl t ~ vE\ttl~jloll~l I( 13-4. lt'e Carl .4ssign Values td E' lo Sirlgle Electrodes 541 13-7. Clec.tror he~nicalCtllls CAI] Be User1 10Ilrrerrrrine Actrvity Cr)r.iiir h r m r l t \ . ,
.,.
.
1 { - H . 1I1.i r r o , t i ~ ~ ~.II r Mt>.lwrtbinr>r~t< i ~ t (:an Iic Uscd to Deternlirlc Valu~.sof A r H , ~ r ~ A r lr X of ( t.11 Kr..~tlion5 557 1 J-9. Soluhrli ty Prtirluc-t!, C.I Hv Ilt!tt!rmintd with Flectrorhem~calCells 553 1 J-10. Thc Cljssoc~atiotlCurlsf,>r?t+r > i L i c ~ Arhk k (-dn He Iletermined with Efectrochenlical Cells 556 13-11. W e Can Assign Tht.rn~udynamicValues to lndiv~dualIons in Solution 554 13-12. Rattcries ant1 Furl C(!lls Art, I l ~ v i r e s That Use Cl~emicalReactiorm to Pror1ut:r. Elrrtrir rurrcrlt., i h 5 Prr~hlcms 5h8
CHAPTER 1 4 / Nonequilibrium Thermodynamics
581
14-1. Entropy Is A l w ~ p sProdutc~din a Spontdneous Process 582 14-2. Eritroyiv Always ln~reasesWhen There I s a Iblalcrial Flow from a Region uf Higher ( hcm~calPotrrl~iallo 2 Rt.~ir>n o f I . ~ ~ wChernical cr Potential 505 14-3. h\;lrly Flux-Fort-(*Kclat~onsAre Linear 5RD 14-4. l~hr,C)ns~jier Reciprocal Kclalior>~ Say Thnl I., = 1. 591 14-5. Various El~c~rokinrticQudntities Are Kelated th;0nsagcr Reciprocal Krldtions 593 14-6. The Clnsager Rec~procalRrlatir>r~s Art. RA+CYI(In thp Principle of Detailed U~lance 597 14-7. Elccfroclwrr~iral I'rittwtials Play the Role of the Chernicdl PotcnfidI for Charged Skstr!rr~ri r i rlifit-r~ntPhases 602 14-I<. The hlagnilurh r > i thv I irll~idlunrt~onPotential Depends Upc~r)Trdn\lir>rt Numbers 6 U i 14-9. The Liqtrid Jut~ction Poic>ritidlIs i,Vt,ll Approximated by the I fendersorl Equdtion 6TO 14-10. Thp Flux-Force Rchtiorls Ir)r C(>ntinunusSystems Involve Cradienls of Thcrrnr>tlynarr!ir Qu~ntit~es Rather Than Differerlccs 61 7 14-1 1. :I Slt*,~rlvState 15 a State of Minimum Entropy Produt.tir)n 622 14-12. TIw Cl,~n$rjnrff-Pr~gugine Inequality Applies to Systfms That 110 Not Necessarily t l.avt? I inr,.~r Flu,: Force Relations 025 P r o t ~ l ~ l lt,,!i{ ~\
hv
/ \ r > \ t ~ t l r c trr
thc Nunierical Problems
639
This book has cvolved from ou~.physiualchemistry tcxt, Pli~siculChrrnirtry, A MoIt~r,u h r Approach. I n (hat book, we emphasize a molecular approach to the l c a c h i ~ ~01-g physical chcmistry. Consequently, unlike most other physical chcmistry books, ours discusses the principles of quantum nlechar~icsfirst and thert uses these ideas i r l thc subscqucnt clevelopment o f thermodyrtamics. We l'ollr,w that same approach i n this lext. Mo1err~lu1Ilrcr~noriyna~rlics. Marly of thc chaplers irt this b~lclkare siniilar to those i n our physical chenii\trjr book. hut we have added new material to sevclal chapters attd have three c n ~ i r e l yIlew chnpters. l'he first chapter, The kncrgy Levclx of A t o m and Molecules, is new and presents resulls of quantum mechanics that arc needed i t ) Inter chapters. Only a fcw quantu~n-mechanicalresults are rcally needed Students learn to lhink in tcrnm ol'discrete clcctn)nic energy lcvels i n general c h c m i s t ~and organic chemistp. We begin our discussion o f thermodynamics ill Chapter 2, where we discuss the properties o f gases. We introduce the Boltzmann factc~rand partition functiuns i r i Chaptcr 3 and the11 use the results of that chapter i n Chapter 4 to ualculate energies and heat capacities of ideal gases. Note that we do this beforc we introduce the First 1,aw c ~Thermodynamics f i n Chaptcr 5. This approach works perfectly wcll heci~use we treat only rneuhat~icalproperties (pressure, cnerpy, and heat capacity) that 5tndcnt.s havc encountered i n previoux chetriistry and physics courscs. This approach all(~wsus to immediately givc a mnlecular intcrpretalion to the three laws of thermodynt~rnics and l o nlnrty thermodynamic relations. The molecular interpretation of entropy is an obvious example, but eve11thc vr:ncepts o f work and heat it1 the First L a w of 'I'her~ttndynamic'; have a nice, physical. molecular interpretation i n tcrlns c ~energy f levcls and their populaiions. In Chapters 5 through 8. we in~roducethe principal Ihermndynamic state func~iuns,giving them a molecolar interpretiitio~~ throughoul. I t ) Chaptcr 9. we examine nnc-component phase equilibria artd introduce chemical potential. The next two chapters concern solutionx: C'hnptcr 10 discusses solutions of l w o liquids and Chap1t.r II , solutions oi'sulids dissnlvcd i n liquids. Following Chapter 12. on chen~iciil z r electmchemiciil cells. 'I'hc finill chitptr~'sctvc.s equilihriu~n,we present a new c h ; ~ p ~ on as an itltroduction 10 notlequilihrium lherrrlodynamica, or irreve~'siblctIicnnodyr~;i[nic~ This chapter is a unique feature of our book hcvau.se Sew pedagopical i n ~ r t d u c t i n nare ~ available to this topic, itlthnugh it finds exletlsive application i n biopl1ysic.c and hiology. As i n our physical chemistry book, we have included a nutnher 01-so-called MathChapten, uhich arc short reviews of the mathematical topics used i r ~suhsequcnt
c h u p c r . 'l'hc tivc Mirthl'haptcrb AI-cNUIIIC~IL'III Method\. I'rohahili~y and Stilt iatlcs, Scrics und I.imira. Partial Uiffcwntiation, a11d Thc Binomral Distributiotl and Stirling's Appronirn;liio;~.In cach one, thc discussions art: brief, elementary, and self'contained. At'tcr reading cach MathChapler and doing the problems, a student will be able to focus 1111 lhc subsequent physical chemical lnatcrial rather than having to cope simultanct)usly with the physical che~rlistryand the mathema~ics.We believe this fcature greatly enhances the pedagogy of our text. A n important feature of the text ih the inclusion of about I 0 to 12 wclrkctl example.+ in each chapter. In addition. cacl~chapter provides sotne 60 problems, and soluticlns to he numerical l~rnhlcmsare given in the back of the book. Some prohlemc exlend the material in the chapters and introduce tiew topics that are somewhat more advanced. They have becn written in such a way as to leiid the student step hy step hrough the material. Tuday's sludents are comfortablc with compulers. In the past few yeal,s, we have seen homework assignmcnts turned in for which students used programs such as MathCad and Marhcmatica to solve problems. rather than pcncil and paper. Data oblained in laboratory courses rtrz now gri~phedand fit to functions using programs such as Excel, L o ~ u s123, and Kalciciagraph. Almost all sluderits have access to prsonal compulers, and a mudern course it1 the phyhical scicnces should encourage students to take advantage of thesc trcmcndous resources. As a result, wc have written i1 number of our pmblelns with the use of computers in mind. For example, the hrst MnthChaptcr introduces the Newton-Raphson melhod for sulvirlg higher-ordcr algchr~icequatiuns and tra~~scedental equatiuns numerically. We scc n o reason any Iclnger 10 limit calculations in a physical chemistry coursc to solving quadra~icequations and orhcr artitic-ial examples. Students should graph data, explore expressions that fit experimental dala, and plot fu~lctionsthat describe physlcal khavior. The understanding of physiral concepts i s greatly enhanced by exploring the propenits of real data. Such exercises remove the abstractness of many theories and enable students to appreciate the mahematics of physical chcmistry so thal (hey can describe nnd predict the physical behavior of chemical systems. Keeping in mind that our purpose is io teach the next gerlcritlon 01. chemists, the qua~lrities,units, and symhols used in this text are those presented i n (he 1993 lntcrnational tJnion o l Pure and Applied Chcrnistry (IUPAC) publication, Qltunririps. I/nir.~,und Sj~mbolsin Phy~irzr/Chemi,l-rq. by Ian Mills et al. (Blackwell Scientitic Puhlrcatiotjs, Oxforti). Our decirion to follow the 1UPAC' recornmendation5 means t h ~ some ~ l of thc syrnhols, units, and standdd slates presented in this h w k may differ frt~mthose used in the liieraturc and oldcr tcxthooks and may Lw utlfa~niliarto some instrt~ctors.in Fnmc irl\tances, we look some time ourselves to comc to grips with the new nutalion and ullils. but ~hcreis. indccd, an underlying lugic 10 their use, and we f o ~ ~ the t ~ deffort worthwl~ile.
Many peuple have contributed to the writing and prtxluction of this book. Wc thank our ccrlleagucs Paul Barbara, James T. Hynes, Veronica Vaida, Juhn Crowell, Andy Kurn~nel,Robert Continetti, Amit Sinhn. Juhn Wearc, John Whcclcr. Ki1r13;iltlrirlpv. Jack Kyte. Sill Trtjgler. and Jim Ely for slirnulating discussions on thc topics th;lt sh(luld be included in a rntldenl physical chcmistry cuurqe. and our students B;trry Holding. Peij~tnCong, Ruben Dunn, Scott Fcller. Susan k r z u t , Jcff Grei~thuu
MOLECULAR THERMODYNAMICS
theory: thry ertil rc\nh\ 111thcrn~utlyr~;~n~icsnr)l h:lscd on ony atornlc or m~)lecul;~r ; I I ~i r l d c p ~ ~ d u(I!r l ~ ;itrl~l~ic 311dt~lolecular~nrjdels.The development of thermndynanli c b ;~l{tng tllcse lincs is cnllcd rloaricnl th~rmotlynnmirs.This character oT classical thcr~nodynamicsis hoth a strength and a weakness. We can bc assured that classical tl~crrnrdyr~atnic results will never need to be modified as our knowlcdge of atomic and molccul;~r\ll.ucture itnprnves, hut classical thermodynatnics gives us o~llya limited incight at thc n~ulr!cul;~r level. With the develnpnient of atonlic and muiecular theorics in the late 1800s and early 1900s. tlic~n~udynamics wns given a molecular interpretation, or a tnolecular has~s. 'l'his field i~ called .rmristic,rrlt~~ern~uciyr~nmirs hccause it relates avcr;iges of ~nuleclllar p r o p t i e s l o ~nrtcroscnpicrtlermodyriamic pruperties such as temperature or pFCshUE. The tnatrnill ill Chiipters 3 and 4 i r actually an elementary treatment uf statistical thcrmody~~a~nius. Many of thc resul~5of slatistical tl~ermodynaniicsdepend upotl thc mr~lecu!nrrl~odelsused. so these rcsl~llsare not as solidly based ~5 are those c~fclns.cical rhcl-mcdynnmics. Ncvcrlheless, the intuitive advantage of having a ~nulecularpiuh~rc of certain quantities or processes is very convenient. Consequently, irl ou~~deuelopnient of thermodynamics in this bonk, we will use a mixture of classical and statistical thermodyi~n~nius, evcn though this approach will cost u< some of the rigor of thc 1-ccult5. For 11sto use a statistical ther~nudyna~nic approach, we must use n few qtrantilrtl ~ncchanicalresults f(lr alorns and tnolccules. It is not at all necessary to be an e x p r t i n quiintum mechanics to w e these results. I n this chapter, we will discuss the quantum rllcchanicnl energies 01-a~omsand nlolccules and relate them to atomic and rn(llecu1ar \ ~ ~ w t n ) x o pEvery y collcge general chetnistry course treais thc energy lcvels of atonilc Ilytll,clpcnand it< associated spzclrun~in sornc detail, so we begin with a discussion of thl\ lopic in the first section. We then discuss niultielectron aloms, the vibrational and i c > t ; ~ t ~ t l i i : icncrgief l nf diatomic mt~leculcs,and pdyatornlc rnulectllcs.
1-1.
I lit. 1Ir.c
i t o ~ l i cEnergy
of Atomic Hydrogen Is Quantized
' I ~ I II ~I ~ I:I ;II \V ~ . lc..i~nc.d III pcncral cl~emistryand organic chemistry that the energies of III(. c.lci-trrlr~\111 ;ilr~ln..;jnd molecules arc quantized; that is, they are rcstricred to only L
crl;l~rlthrc Ic+lrv : ~ l ~ cti)r \ L ~ K ; L I T I thc ~ ~ ~ energy , of an electron in a hydrogen atom is t h hrtlr~ul;~ ~
c1\1-11 0)
-2.00 -
,
F I G U R E 1.1
1-he :~llrnvedelrclrt~riiccncrglcs trl I: hyd~a~f:i,n nroul. 'l'heenergirsarc ~ I V C I I>y I rhc I'r>~-liull:i t ,, --2.17864 x 10 '",'II'. whcrc ttlc qrj;ullurlr nulrlbcr n 1. 2. 7 , . . .. Notc thnr the vc~tical axis i~label hy sm/lO-lhJ. Thi\ nt,pafir>r~IIic.arl5 that the tlirnencionlcc:, nurllbcrs on that :lxis :,rc cnrrgicv divided hy l0-ld J. We will u\e [ h ~ c notation to label colunins in tablcs and axc? in figures bccause of its unambiguous nature and algebraic ccr~menience.
-
-2. 8,1
'.-
-
Notc that F , e, P , . . . hecause of (he ncgativc sign in Eqr~ation1.1. The stilte of 7 ~ r oerlergy occul-v when 1 1 -+ w in Equation 1.1. In this state, the prototi and thc electroll arc so far apart that they do not attract cacli other at all. so wc take their intrraction energy to he zero. At cloacr distances (snlnller values 01' I ! ) , the proton and the clccrron atlract each othcr heciu~seof their opposite charses. i\ statc that has negative energy i s Inore stable rhan nnc thal has Leru energy. The a l l o ~ e dellcrgy states yiven by ['quation I . 1 are called srarir?nay stittcr. I'hc statc of lowesl energy (n = 1 ) is called the gmtmtl srutu. The othcr states itre called rxr,iitd .rrarcs; the state with n = 2 is called thcjirar excited stntr. that with n = 3 is ci11Ie~l the .rrr-o~~de.~rit~.dedt~tte, and so on. When an electron makes a transition from one sti~tic)rlnrystntc to another, i t emits or abhurhs electrntnagnetiu rildii~tior~. Wc piclure clectrurnagnctic radiatiurl as consisting 01- packets of cnergy called phoronr. wht~se cncrgy is equal to h v . where v is the ftequcncy of- the radiation and h is the I'lnnck constant ~h = 6.6261 x 10." J.s). Consider a transitinn frtlrn the state rt = 1 tn n = 2 (see Figure 1.1 ). Because e? > F , . cncrgy in the lor111of a photon must he sbsnrhcd. and we kave e, = F , -th v, .+: by cotlscrvation ot ene1,gy. or h 11, = e, - e l . For thc 2 4 I transition, conservation c ~ fenergy gives F~ = xI hvz +,. or Iru141 = s,- - E I ' Notice thal h v I + Z= k v I , , ; the freque~~cy depends unly upon the magnitude of the difference between the energies of the two states. The general result &,
+
..
WIICIV 1111. c~11:111111111 11~1111l1t.1. 1 1 . I \ I I + \ I I 11 I C 10 ~ till: integer values 1, 2 , 3, .. The units of rnergy 111 kl~cItlrctrl;~l~rl~~,~l S!\lr.ll~ (11 I! I I I ~~;~l\hr~vi;ited \ SI) fronl tlie French Systkrne 1nter11ation:llc11' l l ~ l i t c;IT\. ~ I I I ~ I ~ L -~lr\ipll:~tviI ~. f>y I. Clnc ,jt111lci s equal to the kinetic encl-gy of a mahs 01 lwr~kilcrp~;lrn\Ilnr\111gwith ;I qwcd 01-one nlcter per second. If you remember that kinet~ce n c y y I S ] rrr r*'. lllcrr y t r ~c;lrl %ccth;if I J = 1 kg.m2.s- The clcctrunic cnergy of a hydrorct~;1111111 p1vi.11 I1y ~ : ~ I ~ I : 1I II I iqI Iplc~ttcd I~ ill Figure 1.1.
where Afi j~ tlie (positive) d i f f c i c ~ ~ in c e rhe energy nf the two slates invnlvcd in tl:e tmnsitiori, is billcd the BoI!~-jrr~qrtt~nr:y motditior~. Figure 1.2 shows tllc oh.cerved emission spectrum of atotnic hydrogrn irl the vihihle i111tlnear ul~ravioletregion of the elcctrornagnetic spcctru~n.Note that the lincs
Chapter 1 / The Energy Levels of
I
Alnrns and Molecules
E X A M P L E 1-2 Expre\s the alloucd electrt~nicencrpiec or atomic hydrogen given hg Equation 1.1 in unrh oicrrl ' instead ofja~ulcs.
I
SO1 L'TI O N . The conversion fromi joule\ to c~n-Iis given hy Equatio11 1.5.
Thus. the encrgies are II
-F,,
/c~n-l
1-3, 4toms Have Tran~l.?iionalFncrgy rn Addiiirjn
IO
Flectrunic Energy
A wavc function has the physical interpretation that + ' ( x ) d x i s the probability that the par~iuleis located between x and x J.r. A wave function i s the most tnn~plete description c~fa particle that is pussihle, and is said to specify the state of the panicle. Quanlum mechanics (and the Schrijdingcr cquatiun) forces us to ahiindun our intuitive notion lhat we can locate a particle a.c wcll as we wish and replace that notinn wit11 a prubilhilisriu i~~tcrpreration inatesd. It turrls out that for the types of particles that w e work with in everyday life, the prcjhahilities are sn
+
1-3. Atoms Have Translational Encrgy in Addition to Efcctronic Energy
1-2. The Allowud Energies o f a System Are Obtained from the
Schriidinger Equation 7 lie exprr.s.cio11for ttle allowcd clcctrunic encrgies of atomic hydrogen given by bui1tiun I . I as rvcll as the allowcd energies of any other atom or molecule arc obtained tiom t l ~ eSchriidinger eqi~atinn,which is thc ccntral equation of quanti~mmechanics. We will not have to solve thc Scllriidinger cquntion nur use it at all ill this book, but wc wilt r~ccdto use .colnc of the sc~lutionsto the Schriidinger equation for atoms and molecules. Conscqucntly, w e should discuss briefly the nature of the Schriidinger equaliun. For a qingle partiulr in one dirnension. the Schriidinger equation takes the fumi
wherc TI
(called h har) = h/27r, IPI i< thr trl;t>s oi. the pilfliulu. .t rlcnotrc rhc Iocnriorj
of the particle along the x axi.c, V ( x ) is Ithe prjtct~ti;~l rllclpy of tllu pilrticlc. > i \ ;HI allowed cncrgy. and $(x) is the wave func~inn01- 111cpitr~iulc.W l i r ~yc>u ~ "\olvc" thc Sclirfidirlgrr equatiot~for a particle iti a pnfliculiir p o t t ~ ~ t i\:I~ l t. I. yo11get :I ~ r of t w ; l ~ cfunctions, I/I~(x), and a set u l r o c r g i c ~ .F , , , ;~s.ct,ciiilctlw l ~ h1l1r u.l\r I r ~ r ~ t , t ~ a ~ i ~ \ . where rr (nftcn ari in~egedlabels the wavc funclion and lllc clicrgy t:rrr cx;ul~plc.~ l l c r.levtronic energy uf a hydrogen atom is given by Equ:ilior~ 1 . 1 ; ~ r ~ ~r l h cr>rrc~~n)ritlin~ c wavc futlction.c arc thc hydrogen aton~icorbitals, 1s. 2s. 211. . . .
In the tu-o prcvious sections, we U F C ~ttie allowed electronic energies of atomic hydroger~to illustrate the idcas of a sti~lionarystate. the line spectrum associa~rdwith transitirms hctween stationary statcs, wave f11nction.s~degeneracies, and othcr quijrituln concepts. We uxed atomic hytlrupen as an cxatr~plehcuausc ntn~r~ic hydrcyen. its a ~ o m i corbitals, and its emission spzctrum arc discussed in nI1 gcncriil c h r ~ n ~ s t r y courses. Atoms and ~nolecules.however. have other types ot cncrgie\ id :id(lilion In electrorlic energies. We discuss these othcr ~nntle:.of-cncrgy it1 this ;~rld~h' t0Ilowi11g sections of this chapter. J,etqs consider atorns first. In ;~dditir)n10 I I : I V I I ~ ' I~C C ~ ~cricrgy, ~ I I ~ Ciltulns have kitletit (translational) energy. The allowcii ~ c t n \ l ; ~ t ~ rc)r ~ ~; ~r pl io1'a c s p;lrticlc arc fnuild by solving the Schrtidinger equatirrt~of' thc p;~riiclcof Irl:i\h trt cont~nedto some region in spacc. If we consider the n r ~ c - t l i ~ t ~ r ~ ~ch; iI o\ ~r fljr. ,~ ; ~4rijl)lictty. I then wc have a particle cot~strained10 the one-ditr~cr~\io~~;d ~r~tcrr.:tl 0 -. t .: ( I . The allowed energies in this case arc given by
w ~ t h:In : ~ \ \ t ~ r . i i tlcxcnrracy. ~~rd K,, = 1. (When g,, = I , we SiIY that the state is nmir l r ~ ~ g r ~ r ~ r ~I~ 'I'li<. r r r r ~I .I ~ L - C ri~rl~unsional version of this system is n parlicle vonfincd tu a ~ l u (' I~I I I ~ ( * I I \ ~ ~ I VIIIUIIIG ~I;I~ V , rather than n om-dimensional interval. jf we take the vulut~lcio tw ;I ~ L ' c I ; I ~ ~parallelepiped II~~I~ of sidesu. 6 , and r, then the alluwed encrgirs arc ~ I W I Ity
In Equation 1-9, the dlree qu;lnturn numbers indcpc~~dclltly takc o n thc lrltcpcr values 1.2, .... Equation 1.9 gives lllc kinetic (tt-ans1;rtionul) encrgics c)f-;u~y11;rrtic)~ot' rrli~ccP J I . Atrims also have electrrmiu energy, bul lhcre is no simple fortnula lor rllc clcclmnic er~ergicsu l atotns other than hydrugct~.Ncvert1des.c. the electrunic c ~ l c r p c sof ilt~)lt~h and their ions. which are deicrrrlined by a t ~ m i cspectroscopy, arz well ~ a h u l ; ~ t e dThc . standard tabulaliun of a ~ o m i cctlcrgies appean in the US Government t'rinting Oftice puhlica~iun.A l r i w l i t . E~rer.gyLr,~*tl.r,hy Charlotte E. Monre. Chemists uslially refer to t t a s r tables as "Moore's tables." Table 1.2 lists the energy-level datn for tllc first few levels nf atonllc .;odium, whose ground-\tale electron configuri~tiut~ is ls22s'2ph3s'. The Iirst column in Table 1.2 shows the outer-shell electron configuration o l h e Icvel; h e second column lists the degenerncy of the level; and the third column lisls the energy of thc state it1 cni-'. We will see that the energies o f t h e exciled electronic stales usually lie su rnuch higher than the ground-state energy that we du not have to consider them it] thc thermotlynamic c n l c u l a ~ i o r ~that s we will carry out in Ialer chapters.
SO1 I J T I 0 N : 'I'hc ground-htatu cnrgy in Table 1.3 i c tnkcn to he 0.000 c n ~'.. To c~>riil)~tc ~ h crc\ult~of Fquatiun 1.6 with rlle dilra in Table 1.3, wc must tuhc the ihii~rctlcc~ ~ W C C+I ,I2nd t , , 111 Lluatton 1.6. Thuu, w e have
The arni~llr l ~ c c r c p n c yhc~wccnthecc valucr ~ I I I L I I I I O ~ C ~ I TI ~ h l e1.3 sten1 frnni the
magnetic clfccl:. ~ncntionedahlvc.
1.3 The tirst few electronic TABLE
statcs of
atomic hydrogen."
T A B L E 1.2
Data fmrn "Mwre's lablcs", liali~igthe dcgencrocies and cnergich (in cm ' j of the firct Fcw states of atomic stwiiu~n.
--
Electron coniiguratir,n
Degeneracy
Energytcm
'
Electron contigurstion -
Is
-
EIIL'~~Y/CIII
2
O (HX)
2
82 258.917 82 258.932 82 259.272
2
ZP
4
3~
2
07 442.198
3s
2 4 0
97 492.20X 07 44?..lllh 97 492.342
?(I
Pnnlunp Ofhce.
,-
2~ 2h
,111, 3ri
"F~rlniC.E. Mc,r,rc. Alvrrlrr f:!zr,x\
Degc~~eracy
--
Lw~-ri$. Krill Bur. Std. CIW.No. 467 (LS G o ~ ~ c m r r l c r ~ l
Wu\hl~ip,t#~n. I)(-. 1'1191
E X A M P L E 1-3
Table 1.3 shrws ~ h hrct c few rlettrrrrllc <1:1tc9r d , ~ ~ o mhydrogen ic accordi~lgt r l blr~ore's tablcs. Comparc the ralu& of the crlcrgles In this table with tltose given by the expression fur E,, io Equatlon 1.6. Notc that ~tlccncrgies in Tablcs 1.2 and 1.3 occur as sels uf closely cpuccd values. T h ~ hst>-c:~llcd hliluting results from rnagrletic elTects associated ed,ith the spin rlS thc clectron I ~ , I I. I I C I I ~ I I~rlcludedin Equation I.h.
'Fmm C.E. Mrlrrw, Alumic h p r g y h v d i , Ndll. Office. W ~ h h l n ~ t u lIK'. i. IWY I.
Bur. Sid
C'lrc.
No. 467 (LmS Gr>bcrnu~cr~l Plintlng
1-4. The Vibrational Motion of a Diatomic Molecule Can Re
Modeled by a Harmonic Oscillator In the previo~~s scctiun, wc discussed the trarlstntio~ni~l and c l e c l r o ~ ~cnergies ic of atrmls. The expression for the trarlslational encrgy ul' a diatomic molccule i s thc same as t h a ~
(.tlq)lrr L ! I he Lnergy Lcvcls oi Atoms and tviolcr ules
5 0 L U T I C I N : Recall that the u n i h r i k arc N . I ~ I -i~~ldrhat '. I N z I kg rn s = mass rime\ accelernliun). Thercforc, thc unlts of V itre given by
' Clorcc
1-5. The Energy Levels of a Quantum-Mechanical Harmonic Oscillator Are E , , = h v ( v {) with v = 0, 1, 2 , . . .
+
Whun thc ScItriHlirrgcr cquation for ;l one-dirncnsiond harrr~or~ic nscillalur is snlvcd. physicull y wcll-hchlrvcd w:~vcL'unctionscan he obtaincd only il- the energy is reslricted ~llr q u u ~ ~ t i l rvrilura d
Rvfo~r:wc disuuch thc qu;~r~(u~~i-~nc.ctiilriic'ill lrcuilllrnt irf u ltur~nr~niu rlbcillulirr, wc should discuss how grnxi an approx~~nirirm it i s fur a vibraling diritonl~crnc~lcuulc. The inlemuclear potential li)r a di;lt~,~nic n~r)lcculuis illus~rarcdby ~ h csolirl line in Figurc 1.5. Nuticc thal thu curvc rlxcs ~ c p l yto thc left (I( the ~ninirnurii,iriilicating the rlll7iculty r)T pushirig the two tluclc~closer tugether. Thc curve to the righi s ~ d uf e the rquilibriu~npoaitiun rihe* iritially hut eventually levels off. 'I'hc potential cncrgy at large separations is cssenlially the bond energy. Thc dashed line shows the potel~tial j k ( K - kc)' associated with Hooke's law. Although the harmonic-oscillator potential may appear to be a terrible approximation lo the experi~nentalCUI-ue, nore thal i~ is, i~ldced,a go(d approxinlation in the rcgion of the minimum. It turns out that this region is the physically important region lor many n~olcculcsat room temperalure. Although t h t harmonic oscillator unrealistically allows the displacement to vary froin - x to +,m, ihese large displacements produce poler~lialenergies that arc so large thal they do IIOT oftcn nccllr in priotiue. The harrnor~icoscillator will hr a guod approximation for vlhratiuns with snlall arnp1itude.c (Problem 1-24).
where
Each encrgy level of a harmonic oscillatc~ris nondegenrate. In uther words, the degeneracy, gL,,i s equal to one for all values of u. The allowed energies of a harmonic oscillator are plotted in Figure 1.6. Note that the encrgy levels are equally &paced,with a separation ku. Note also that the energy of the grclund statc, thc slate with I J = 0, i s
i h v and is not rero as the Iowesl classical energy is. Thc energy of this lowest energy Icvcl is called the :em-pint twcrgy of the harnmonic c>scilhkor,and the fact that it is not zero i s stric~lya quantum-mechanical resull.
F I G U R E 1.6
l'llc cllcrpy Icvcl+~ r ;lI~l~~;tr~~~~~n-~i~ccha~~rc.al Ililr~uri~~ic rnr ~llaklr. w F I C U K E 1.5 A colnpiwson of the harmonic-oscilla~t>r putentin1 (kx2/2; dashed line) with ~ h ccon~plelc ~ntcrnuclearpote~ltial(solid line) of a dia~umicmolecule. The harmonic oscilla~urpotential is a satislactory approxllnation at small displacernunt~.
E X A M P L E 1-5 (iivcn th:!~ttnc v ~ l u rul . ~ h ciorct.vt,n\~an~ rlitl?1s 575 N-m-I, calculate the iurda~nc~~t;~l vihri~liolli~l frcqucncy of H,.
Chapter 1 / The Energv Levels of Aroms and Molecules SOLUTION: According to Equatio~~ 1.19, v = ( k / p ) [ ' 2 / 2Thc ~ , reduced mass of H, is given by (Equation 1.13)
1 6 . Thc tiii~mon~c C)~rillnla~r Ar r rlrlnt, for
thc Infrared Sperirun~or ,I rli,rt#rm~c Mvlccule
accord with cxperirrlenl. and this linc i s called (hejundurnetltd vibra~ionrriii-eq~~rni-?'. Fur diatomic ~ n t ~ l e c u l these c ~ , lilies clccur at around l o 1' Hz 10 10'" [IT, which i s irl thc infrared regioti. Eqtralion 1.22 or 1 23 enables us lo deterlninc force corlslanls it-
the furtdilrnental vibrational frcqucncy is known. Fur cxa~nple,for H ' ~ c I ( ~ ) , 2886 cni and so, according 10 Equillion 1.23. the force constant o ~ H ' ' c I ( ~ is ) > ,,\;
'
is
arid s o
-481 kg-s ' = 3 X I N
1-6. The Harmonic Oscillator Accounts for the Infrared Spectrum of a Diatomic Molecule
111
'
Forcc cr)nst;lrlts for din~orrncr~lolcculc\:Irth 1y11ic:lfly~illbc orilcr 01 l r l ' N .In I . Tahlc 1 ..lliqts (he iurldanitntal vih~:ilitrri;~I I r t + t ~ ~ ~ rr o ~ tt; v~ it ~ r~\ constnrlts ~. ~ : ~ l (Scction 1-7). bond lengths, dcpcr~cr:icir\. ;illcI clcctru111c.cricrpich 01' illc g r ~ j u t i slales ~l ot. some diatomic molecules.
A diatomic molecule can make a transition from one vibntio~~al energy state to another by absorbing ur emiiting electromagnetic radiation whose observed frequency satisfies the Bvhr frcqucncy condition T A B L E 1.4
11tunls out 1hn1the hartnonic-oscillator model allows transitions only between adjacent cncrgy stitcs. \(I that we havc the candirion that Av = &I. Such a condition is called I:
I.UIE. kor absorption to occur, ALI=
srlectiotz
+ 1 and so
Thus. rlte observed lreyuency oT thr. radialion absorbed is
or, i n aauenu~nhcrs,
xlicre thc tilde indiuatca that thc units arc c m I . Furthermore, because successive cllcrgy statcF of a harmor~icoscillator are separated by the same energy, As is ihe same lor all allowed transitions, so this rntxlel predicts that the spectrum consists of just one line whoce frequency is given by Equation 1.22. This prediction is in good
The turldattlctital vibrational Ireql~cnc~c\ (I,). ~ah;rtiorl,tlwn%lulll% (fib, h~ltl Icl~gllih(K?j. degeneracies, and equilibrium ellergie\ I I>< I 17f tlw prarlr~ltl%~uic\ 111 wrii~cJ~iirt~rliic ~nolcculrc Thcse purunieterc were ohtailled ~II~III ;i Y:II ~ r . l y t11 WIIIILC* :hilt1 11t1 11111 rcI)rt\Cnt lhc 11!ost accurate d u e s hecau~ethcy wcrc r~hr:~lr~ctl u l ~ t l r~~l ~ rr t .~ 1 ' 1 rtrl;btrjr t t : ~ t r i ~ o~~~\ ~+t ~ , ill:~Ior approuimatiun. Molecule
:/CI~I-'
~/CIII
H,/~I~II Ikprr\cr:rcy
I)?/kJ IIIOI I .-.
1-7. The
1
EXAMPLE
nl\sr~cial~on Fntmrgy and
thc
(;round-Stale Elcttrorilc LrIcrRy ui d [ I I ~ ~ ~h10lt.r OI~ 1111, IIC
I
1-6
Llse the data in Tahlc 1.3to calculate lhr value ( I ( the force cul~stanlr)1 ' Y ~ " i ~ l g ) . SOL U - IICj N : The force crlnskint is given by
The rcduced rrlass of
is
' 9 ~ ' Y ~
(38.964 alnu)(3X.464 amu) (1.661 x 10 P = (2)(3R.964amu)
Lh
kg.ilrl~~i I )
w :' I",
'I#
The hnmionic-oscillii~or S L ' I ~ U ~ ~nllu I ~ I sirya ~ not osily 111;ti A t ) = 11, hut that the thc rtu)lucl~lcvihratc5. 'Thus, the dilx~lemomen1 of the ~rlrhcculc11lus1ch;lrIpc harmonic-oscillator rrlodcl prcllicts 1 1 1 ~ 1li('l(g) ;thsorhs In thc infrared but N,(g) does not, which is in good agreement with cnprrllilcnt. Tliere are, indeed, deviaticlns from the liaminnic-oscillatclr niodzl, bur thcy arc fairly small, and we can systematically irltroducc corrcctiorls and exten\ions to ;kccount for them (see Problelns 1-31 through 1-33),
1-7. The Dissociation Energy and the Ground-Statc Electronic Energy of a Diatomic Moleculc Are Rclated by D, = U, h u / 2
+
k'ipure 1.5 ahowed the pu~cntialenergy ul'a riiatornic molecule as a function of the internucle~rseparation, R. This potential cnergy that the two nuciei experience results from the distribution of the cleclrons around thc twu nuciei and is called the electronic ?nergy u l thc nioleculc. The complete curve can he calculated from the Schriidinger equation by holding the nuclei a fixed dis~anceK apart and then solving for the distribution nf electrnns and the correspondit~genergy. Cnlculainns l ~ k ethik must be llotlc on a conlputer, hut they have been cnmed nut for a number of ~noleculcs.The in~crnuulei~r separiit icln at the minimum of the curve in Figure 1.5 is the eyuilibriuln i1ltcl.nuclear scparaiioll. RL,, or thc bond length of the mulecule. Figure 1.7 s h n ~ , ttlz s pr)tent~al-energycurves of thc ground and first excrted elccironic statcs ol a tyllical d i ; ~ t o m ~molecule. c Note that the bond length ill the excited electron~cstate is not ncc5ssarily equal to the bond length in the gruund electnlnic stale. As for alorns, usually the energy of the first excited electro~iicstate of adiatomic lnolecule lies so much higher than the ground electronic state that we need nut consider it in thc themullynamic cillculrt~ionsthat we carry out in later chapters.
I),,
;
1, v, 2
F I G U R E 1.7 The grt)und-alateand llrct cxc~tcd-stateclcclrr)r~~c rnclgy ti! a ~ I ~ I I C ! I I I IIIOICCIIIC IC ~ I O I I C L I ilpalrl.;l the lnlernuclcar scparatlml. H, showing that he equil~briu~~~ grou~~d-htatc clcclronkc trlcrgy, /Ir. and the dishmiation cncrgy, I>(,,am rcbtcd by De = I),, + h v / 2 .
According to Figures 1.5 and 1.7, the elecironic energy of the grtlund elccrronic state ~ ( R to S zero as the internuclear separation increases. Energies must alwnys be reckoned relative 10 sonic (arbitrary) rcrO of encrgy. In this caw, we arc tithing tlic zero of the gnlund-state electronic energy of a dialoniic n~c~lecule to bc the sey,u;)tcd constituent atoms in their ground electronic states. Consequently, thr. gro~lnd-sli~te elcclronic cnergy of the molecule at its etjuilibriuni ~nterr~r~clear sep;iriition IS - I ) , , . n.c shown in Figure 1.7. For exa~nple,lhr valuc of DLfor H,(g) is 458 kJ.fl101 I , which m e w s that thc ground-state electronic encrgy of Hl(g) ak its cquilihriu~nintcmu~lenr separatic~n(74.1 prn) is 458 kJ.rnol-l below that of two vrpararcd hydn)gen atoms in their ground electrur~icstate (Is). or thal the elcctroc 158 k J . i n ~ l - ~ . This is not the energy required to dissocinte H,(g) into , . ....; .I:LIZ hycllr)gzri atoms, howcver. Recall that Ihz Iowcst v~hratio~lal cncrgp ktatc. hdk .I ~clro-l>olnter~e~,gy s,, = h v / 2 . Consequznlly, thc energy required clihsocia~zH, rg) liorn it3 ground slare into two ground-state hydl.c>gcn ;ktoIrl:, ic I ) , - I r l ] / ? . which wc. dc.\~gn;htuh y I),, 'I'l~us. wc WI-ilein at1 cquatlon
Givcn that v = 4401 C I I ~' fair Il,(gl (i'ron~L b l c 1-41,we see that D,,, the dissociatior~ energy of Hz(g) is 458 kJ-IIIOII - 2h kl.tl101 I = 432 kJ.rnol-I.
Chapkr 1 i The Encr~yI.ev~Is of Atoms and Molerules
I EXAMPLE 1-7 Givcn that the d~ssocintionenergy of F,(p) is 154 kl ,moi and that the vihraliorlol frequency i, 892 cln I. ralculatc rhe value oi the ground-atale clcctro~~ic energy.
'
S C l I l l T l Oh': We are givct~values or FI-o~mEquation 1.24. we have
freq~lcncyof vrG,[cycles per second, then ~ h cvelocities of the two lnasscs are u , = 2 r R II J ~ , , and v, = 2rr R,um,,. We define the anguhr v ~ l o c iw~, ,by o (radians per second) = 27r v,,,, so w c write I ) , and vz as u, = R, cd and a, = K,a). The kinetic cnergy of the r i ~ i rrot:ttor l i\
13,and 5 and wish tocalc~llntethe value of De.
L
whcrc I, ttir. jtlranltbrltr l j irtr.riro, i s p t v l - i ~ hy
1-8. The Energy Levels of a Rigid Rotator Arrg
F,,
-
whrrc
h'.l(.l t 1 ) / 2 1
sccticll~wc will cli\cucs :I sirllplr rilcnlcl M o l c c u l r ~rotate as wcll ;IS vihrii~c,and i n for :I rotatirlg diaton~iurrlo~cculc.'The ~lioclclconcists 01' two puini II;I>\CI nt I i111Cl 111, a1 6xcd distance5 K , and R, l'rolr~ thcir ccrltcr ul' iii;isc (ct. f:igurc I .H} neuausc the distance b e t w c c ~thc ~ two Iiussu\ IS liacd, thi5 mtxlcl is rrI'Urrc~Ito ils the rigid-n)lr~ror nmdrdl. Even though a diatolnic mr,lecule vibrates as i t nllatcs, the vibrational amplitude is sn~nllcoll~prurd&ill) thc bond length. s o considering the bond length fixed is a good ;~l>p~-nximutioli. A rigid rotntor rotates around its center of mass, which is given by the condition In, R, = nt,K, (see Figure 1.8). IF the molccule rotates about its center of mass at a
-
H I t K, ( ~ h ciixed
\cpalation of the two
niassesj and p is tllc rrdu~erl
t11(1.1.1
Whcr~we sulve the Schrudinger cquatiorl for a rigid rotator, the e a p w s s i n ~for ~ the allowed energics comes out to be
fil = -J(J
E,
21
+ 1)
J = O h I. 2. . . .
( 1 .2X)
Once again. we ()blain a set of discrete c n e g y levels. In addition to the allowed eiicrgies given by Equation 1.28, we also find thal each energy Icwl has a degeneracy g, given by
'l'hc seleu~ionrule for il rigid rtltatur says that trarisitinns are allowcii clnlp fro111 adjacent rtates or that A J = +I.I n addition tn the rzquiretrlcllt that A J = fI , the moleculc nus st also possess s pcrlnoncnt dipole mornrrlt to :thsorb eleclrnmagnctic radiation. Thus, we say that HCl(g) has a pure rotntional s l x c t s ~ ~hut ~ t Nl(g) ~, does nol. In the case of absorption of electronlagnetic radiation, the molccule g w s from n state with a quantum nrirnher .I to one wit11 .I I. 1-hc cnel-py d~llcrcncc,then, is
+
TI'
h'
I
4n'I
= --(.I+\)=
-(.I t 1 )
U s i r ~ gthe Bohr frequency ct~tltlit~rm At- = 1; I!, the I'req~~uncies at which thc ahsorption transitin~~s ouc111-;!re C
F I G U R E 1.8
'1-wrr tllncic., Jn, and r t l , shown rr)Ulir~gabout their center of Inass
I tr The. Fncrc) I t ~ v c l br,l
'l'l~coornrnon practice
ill
.j
K~jil~ K+)l;rlur l Art.
a , --A' 1 IJ : L!/II
ttticrowilve ~prctrriscupyis to write tqu;iliurt I .3 1 as
is cnlltd the romtiond ron.rtunt of thc rnolecllle. Alscl, the rrancitiort frcylrrrr~y13 o i ~ c n expressed in terms of wave numbers (urn-') rather thil11 hertz (Hz). If we usc ihc rclaticttl P = r/r, thcn Equalion 1.32 becorncs
whcre B is thc n~talionalconstant cxprzsscd in uni~col wavc rtu~~~lrvr\
mc)~lelprcdiuth that the Frcwn tither Equatio11 1.32 i)r I .3-1. wc KC th;~ttl~c~rigtrl-ro~i~tor ~nicrrjwavespeurrunl uc a diatulnic moleculr cortsists a scries of equally spaccd lines with a \epiuatinn o f 2 3 H~ or 2 8 crn I , a\ shuwn in Figure 1.9. Let's use Equation 1.32 to calculate thc values nf the ahsorotiun frequencies of H"cI(~) tising the equilihnurn bond length given in Table 1.4 ( 1 27.5 pni). Recause I = K:, we must tirst calculatr Lhe vnlric rjl' lhz rcduczd Inass ot H~'CI. "1
and
(1.008 arnu)(34.97 amu) -1.008 amu -t31.97 antu- ( l . h h l x 10 "kg.amu
I)
F I G U R E 1.9 The cnergy lcvels nttd ahcotption transi~ionhoI a rip~drotalor. l'he ubsorphon t~-aris~~ionh r)cclll belwecn adjacent Icvcls, so the ahsorpliun spe5trum shuwn helow the energy lcvels consisis uf a series of equally hpoced lines. The quantity H i\ / t / 8 r r 2 c - I (Fquatlrln I .35).
<(I
transitions of dintn~nicrnolzcules occur irl the rtlicrr)wavc region, and the dircci study of rotational transitions in ~nolcculcs1%called ttlil,ron,uime,sprrtrosr,oj)y. From thc xeparatiort hc~wrcnthe ahaogicin Ircquericics in a r ( l ~ a ~ i u lcpcctrurh ~al rjf a (lialutnic rrlulccule. we c;in de~erminrthc ~rr>t;~t~r)n;~l constiillr arld hence the Iunmerit ofitlcrtia o f t & 111olec~1~. Furtherrntrrc, ~ L . C ; I I I C C -I -- 11h':. n.hrrc RLis the inrcmuclcnr distartcc or bond lrrtgth, we cxn dctel-r~tittc~ l t chotlrl Icltpth. I'hih ~>lricedurc i\ illurtl-ated irt Exarnple 1-8
C
As .I lakes on the valrlcs 0, 1. 2, . . ., rl takes on the virlues 6.343 x 10" HI, 12.686 x 10" Hz, 19.029 x 10" Hz. and so on. By I-ekrrirtg to F~gureI . 1 1 in Problem 1-1, wc see that these kecluc~~cies lie in the rr~icrowaveregion. Consequently, rutational
EXAMPLE 1 4 To :I g t ~ ) d~ I ~ ~ [ o x Itllc I I t~t ~~ i I~ Ir ~t ~I'~~wctruul w~ , ~~ ~ . ~ of . "N~~"CI(E c.ol!slsl\ ] r,l ;I \rries riTurlually qvaccd blur!, >cp;lratcdhy 0.4322 cm '. Cnluulste lht. bond Icrtglh ot " ~ a " ~ ' l ( ~ )
C h a p t ~ r1 I The Ln~rgyLerrlr c ~ Aromr f
arid M#>l~rules
SOL UTI 0 N : According to buation 1.35, the spacing of the lillcs ill the lnicrowave spcctrutn of "~n"Cl(g) is given by
and so
Solving hiir eq~~stir,i~ f(>rI. we have
The reduced masv r ) l " N ~ " C I is
1 Y The Y~hratrr,r~\ rlf Pnlyatr,mic
Mvlcr 1 1 1 ~ 5Arc
Represented by
Nrrrrri.~!Modes
three coordinates corresponds to translational motion o f the center of Inass ol' the molecule, and so we call thcse three coordinatcs tran.rlationul degrees nffrrcdrirn. ' h o conrdini~tesare rcquired to sprcil'y thc rjricntation of a linear molecule about its ccntcr uf mass and three coordinate\ to spcci t'y thc oricntatirm u l a n o ~ i l i ~ ~molecule car about its center of nlasq. Because rnutiotl alonr thew coordinatcs corresponds to rotatic~nal rno~ion,wc say lhai n linear ~noleculch;i\ two rlt,grt'e.s of~mtr~tionrrlJrePdnm and [hat a nonlinear rilnlcculc h;~sthree dcg~ccsof ~ol;i~iijrj;~l f'rccrlr>tn.Thc remaining coordinatrs (3rr - 5 [or a Lil~carmolecule and 3r1 - (I for ;I ~lulili~~ciw tr~ulcc~ilc) specify thc rclativr positions of the n n~rclci.Bcuairsr rrlotior~ long thrcc c-trc~rrlin:~tch correhpontl< to vibrational motiun, we say that a linriir n~olcculchas 3rr - 5 l,ihrrrtiot~ctlrlrxrrc..r rd' frpedonr and that a norjlincar mulccult: hii* 311 - - h vil~~:i~iijn:~l tlcprcck of trc.t.tlol~i. These rcsults are .s~i~nmnrizcd in Tirblr 1.5.
E X A M P L E 1-9
Deterrr~iricthe 11urnht.rul- various degreev (yi frcctl+,~n111 Il('1. ('0:.111( ). Nll,. and CH,. Using tl~clact that I = p R', wc obtni~~
5 0 1 IJTION:
Problclns 1 3h, I 38, and 1-39 ~ i i olljcr c cxarnple+ trl thc tlctcr~~~i~i:itiu~~ trl hrlr~d 1e11gth~ l'rtun tllicrtrwavc data.
A diatolnic rnoleculc ih not truly a rigid rotatub; bccat~srir s i r ~ ~ l t ; i ~ ~~ihr;itcs. uo~~ly however srn;tll the iunplitude. Cun.sequentlg, we might expect th:~t a1thr~ugh~ h cr ~ i cruaave speutnlm oL H diatomic lnr~lecuIeconsists nf a scrics of cqually spaccrl lint*. their crparation is nut r,x(~crlycunstant. There is a straightforward prncedurc tu correct fur thc fact t t ~ the t bond is riot exactly rigid (Problem 1 4 1 ) .
1-9. The V i b r a t i o n s of P o l y a t o m i c M o l e c u l e s Are Represented hy
L l ~ ~ d the c r hi~rrlro~~ic-o~cill:~t~~r ; L I I I ) I O ) ~ I I I \ : ~ [ I ( ~IIIL. I ~ . Y ~ ~ > I , : ~~IlIl ~ o Il iInIof~; I~;I~ 1)1iIyatumic n~nlcculecar1 be picturcd as ~ h ~ c ~ l o t i rll u t ~I I ~ , , ~nilvprr~dcn~ harmonic ohcillators, where II+,, is the number of vihr;1tio11;11dcgrccs 01. freedom. Fur example, a water moIecule has three degrees of vihr;i~iol~;i! t ~ c ~ d o(7n r r ~ (7 = 3 x 3 - 6 = 3), and the
N o r r n a l Mndes Poly;~turriicmoleuule.s have translatir~nal,vibrational, rotational, and electronic energics. Thc tral~slatinnalcl~crgyis the satne as that of an atom or i1 diatomic moleuule, and the electronic e,nergies of Iriany pulyalornic molecules are wcll tabulated. We will discuss thc vibrational cl~crgyof polyntnrnic molecules in this secticm i ~ n dtheir rotnti(~nal erieIgy iri the next section. Cunsidcr a lnolecule corltaitlitig r l liuclei. A complete specification of this inolec~ile rn space requires 317 coordil~ates,thl.ce Cartesian coorditintes lor each nucleus. We say that the n-atomic nlolecule has a tutal of 3n d q r e e s offreednm.Of these 3n coordinates, thrcc can hc uscd to specify the center uf mass of the molecule. Motion along these P
TABLE
2.5
'l'he nunibcr of viiriou, tic~rccbuf Crcctiuln of a polyaton~icmolecule contairi~rlgn :Il#>nl\. Degrws (if I'rrctlt~in
--
1,inear
Nunlinear
-
Ch;qitrr I i Ihe Energy Levcls uf Atun~r,lrhtl
Molecules
kihratiunal motioll of a water molecule can bc broken down into the thrcc cIi;~r;~cterlstiv vibrational motions ahuwn below.
T A B L E 1.6
l'hc flndntllcntal vihrxt~r>nal ~requenciesand the rotatiunal consr;ujts r h f home pol!~atnnllc nrolrcule~.7.k n~ilrthcrsin pmenlheser indicate the degeneracy uT the ilnrrrlal mrhdc. Thesc prnnttttrlr wcrc t~blilil~e~l frt)111 il variety of sources and do not reprt'scnt Ihc most xcuruto
valtlr>~TVUIIMI
Sylnmetric slrclch 3 5 8 6 crn
'
Asymmeiric s l r t t c l ~ 3725 cru
'
t l c ~ r d ~ rl t li i~r l l t r n ISVS rln I
C'O!
N?cl
CH,
These three charactcristic vibritional modeh of a w;iivr 11lulcc18luibrc cxul~~plva of normul modes. The arrows in the above drawillgc iudicatc thc rtirrciic~n111alcuch atrlln rnnvcs in the nurmal ~rlude.Bccausu thc norrr~;tltrltnI~'\i i ~ ta h ~ I I ~ C I ) C I I ~ ~tli~r~tlt)nic CII~ uscillntors, the vihrationi~lenergy of .I ~roly,ltr>~r~lr. ruulrr,ttlc ir p t v c ~by ~
II~\L*;I~
lu~car \phcrical I U ~ '
2 43Tb I3 i 3. blj,$t2) 222.1. 1270. 594).8(21
0.390 0.41 0 5.27
?Ht)N, I5 1 S(3). 1{K12(3). l 3(l(Y3)
CCI,
spher1c;il ~clp 458.7.215.5(?). 77X 4 11. 312.N( 11
NH,
symnlrlric (up
CH,CI
where v, is thc vihratic~nalI r c i l u r ~ ~01~ yrllc jth I I ~ ) I I I I 11lralc. ~~~ 'l'hc\u frcquvncies are indicated ior the normal ttltdcies clt a walcr rtlulcculc htiown ithovc. I:.acl~type of ~x)lyatr>mic moleculc has n+,,charactcristic riorlt~i~l ~utnlcs.f41r ('0:.for c x a ~ t l p lthere ~, ;Ire i0ur nul'rnal modes ( 3 n - 5 = 3 x 3 - - 5 - 4):
t ~ h l ~ i li~ntlcr i ~ d the rig~d-rotatorharmonic-osci1latr)r opproximaliu~l.
~ wtrc P ~
H20 So?
(1 1157
3336.945.2. 330?(21.
1619I2) syn~me~ric top 2968, 1355, 729.8. 3U44(2), 1487(2), 1015 ( 2 ) asym~letrictop 3725. 3586, 1545 asymmetric top
1 153, 1362, 521.3
h.?(l
'1.45
9-45
5 04
0.343
11 443
27 0
14.6
2.03
0.344
t)
71
0.243
-1-1 0, The Rotational Spectrum of a Polyatomic Molc.cule Depends Upon the Moments of Inertia of the Molecule The rocation;rl ~notlunof a I l n e ~ polqatornic r ~x~olec~rlc (such as C'O, ur HCN) ic gwcn hy the same expression as thal for a diatomic mulecule. Thus,
'l'he ~lnrlnalmode shown h v e at ihe Far right indicates vihl-.lt~or~;~l lno~ior)pcrpcndicul;~rto t t ~ cpage. Note that the hcnding mode i b doubly rlcgcrlcr;ltu; t11u twtrlling motiuns in thc plane of the page and prpmdivular 10 it art: thc halnc I ~ I O I I ~l ~I ~i occur t~ in ditkrent directions. A selev~ionrule for vibratitic~nalabsorption spectroscopy is th;it t h ~tllllcllc . Ii\onicnt ot-ihe molecule Inu.st vary during the motion ofthc normal mode. When rhls I > .+(I. thc normal modeissaid to hc itlfmrcdactive. Ott~erwise,it is infruredit~nt,lir'c,. Nurc 111;tt the dipole Inotnellr changes during the [notion of ail three nurmal modcs o f a wdtcl-~nolcclllc (showr~above), so all three normal modes of H,O arc infrared active. Thclcfrlrc, ;ill thrcc line\ are observed i r ~its infrdrrd spectrum. Nnte that there i s IIO ch;il~gcit1 tliliole ItluIrleIlt during the sym~netricstrctch ofCO,, huwever, so this mode is lnlrared ~r~;~r.tivc. Tht other modes arc infrara active, but the bending Inode 1s doubly dcgcncr~tu.s o il leads m only one infrared line. Tablc l .ti list.+ihe characteristic vibrational l'requel~cics of the nornlal nlodcs of some pulyatornic molcculcs. The numbers in parc~~thcces ii~dicatethe degeneracies of thc modes.
with g , = 2 J -1- 1 , whcre I is the moment of-incrila. In this case. hovievcr, I is giwn by
where m, is the mass of the j t h atom and ma\s of the moleculc.
EXAMPLE 1-10 Given that K,,, = 1M.h pm n r d #, thc
mnlnent OK 111enia.A. and
6.
t, -
,= 115 -1
(,ni
r<,,,1s its distance from thc centCr ol'
1
4
I~ I " ~ " 'N. L ~ ~ I L ' I I ~t ~l Il c~\illlle5 O f
('h~per1 / Ihe Lne~gyLevels of Atoms and Molcculcs
2U
wlicrc h = h/2x and IIS the mornenl o f ~ n e r l i auf the nwlecule. For a diatorn~crigid rotalor. I = p R:, where p = rn,rn,/(m, t. rn?) 1s the reduced mass. and Rt IS the c q u ~ l ~ b r i u hnnd m length. The energy IS communly expressed in unlts OK c t ~ i .u~d E q i ~ a l ~ o1.42 n wntleri as
w11cl.c fi = h / 8 r r Z ~isr the rr>tatir,nalct>nspantexpressed i r ulrl ~ I . Koiiltiotial lrallsitions occur in thc ~nicrowavtand far infiraced region uf the apzctrurn and yiclrl illformation ahnut thr molnent of inertia and hence the equilibrium hunrl Icrlgth\ ot nlolecules (Example 1-8). The vibration;~lmolion oKa di;11~111ic 111ol~c11lt+ i\ I~~(uIcIc(I xs i t I~;I~III~!I~c owillatorb whose energy is given by
1-1 1. The tncwy oi ,h hlolccule Irl the Rigid-Rut.llor Harmoni( -Oscillator A[l[rrt)ximat~on
To conclude h i s chapter, we prcsenl a sununary of thc energy Icvclx of aturns and lnuleculcs thal we w i l l use i n later chapters.
E clcr
= see l'ahlcs I.I and 1.3, for cxalnple
Diatomic molecules
where p is the reduced rnass of thc rnolcc~~lc, A is thc Iorcc ctlrihtunt, al~tt11 is the natural vibratio~~al frequency of the moleuulc. V~hr;~t IOII:LIIri~nhit~tlrih i ~ c i In ~ rthe intrared region uK the spectrum ;md car1 yicld int'cu-~il;~~ioti ; ~ l n > thc \ ~ t ftlrcc ct)l1sfitnfs of n~olcculzs(Example 1-h). As for atr,m.c, lherz is no analylic 'xplcccltrll 101tht. c l i - t ~ l r t r rlll-rpic* ~ u ~ ~ ol d ~ : ~ ~ t r r ~ i i c but thcy arc wcll tahulil~ctl f.'~)rIIII>\~ (..Iw>. wc IICC~I ( ~ r i l yItic rllcrgy and ~r~c>lecliles, the dcgzneri~cyclf- the proulld c l r ~ l ~ ' u l\l;llr* l~c The cncrgy of a ~ n ~ l y ; ~ t oIII~>IL~L.IIIL. r l l ~ ~ 111 lhtnI I ~ I OrtrI,tI(Ir II;LIIINIII~~. ~ h ~ . ~ l l ; ill)~ttlr ~I-LLXIIII:I~~OI! i\ ;IIUI g i w r ~tjy I ~ . ~ ~I 4I1. I'[~ tic ~ t rL i ~~ t ~; ~I l i ~ ~t ~i cor ~gI> ~ ytllc i ~ l\;11iic as lhat or a11 i j l o i l ~or I~~;I~~IIIIC III(~IC.CIIIC .l'hc ~t~l.ili(lli;hIclicrcy ~IL~~K'III!.~ ~ I [ U I~~ IL ' \Ii;i~w c ~ tlic f molecule. tior ;I 1111cii1. IHI~~;II{>IIIIC IIIOI~CIAII~, tllc r t ~ l ; ~ l i oI'IICIX~ r ~ ~ ~ l I\ \;t1111- ;I\ ~ t m tof a diaton~ic~ncll~.culc, w h c ~ cthc rllr)tllcllr r j l IIIC~II;I 1h p 1 1 ~ 1Ily 1 I:L~I~,~IILII\ I IH. For a nonlinear polyalorlllc ~liolrck~lu. lllc rol;ilio~l;~l cncrgy tlclwclcI\ uInrn w lirllichr lhc molcculc is a sphel-ical 1c1p (all Ihrcc rn0111c11ts01' l~lurlirtcqual), ;I xylnlllclrlc hrll Ilwcl r l f the three molnentx (11 irlcrtl;~cqu;~l). or LII; asytt~lt~cl~,iu tup (all ihrcc riiomcflts o f inerlia differ en^). In the harmonic-cl\cillatl,t. a l ~ l ~ ~ c j x i ~ n athe ~ i crihl-ational j~i+ r n o ~ i c ~of-a n poly;~lorn~c molecule decomposes into 3 n - 5 (Ilnear) ur 3n -- 6 (nonlinear) nor~nalniodcs, cach of which acts as all indepcncicnt 11:~rrllonicc~sciIlntorThe vihratir,nal energy is given
hy
where
I I ~ ,is ,
the number of vibralioni~ldegrees uf- Ireedom.
(Equal~or~ I .2X)
( E q u ~ ~ i o1.18) n (Table 1.4)
Chapter 1 IThe Energy Levels of Atom
1-4. Calculale the vnIuc o f the energy of a phntr111 for n u~avelengthof I00 pln (;thout one atomic r:~diusj. 1-5. C':~lc~~l;~lc Ihc ~ I L I I I I ~ ~ of JIIIOIIIIICill :I Z.(HJ-rnJlight pulse a1 (a) 1 .Uh prn. (b) 517 rlrn. and (c) 2hh r11r1
(Equation 1.9) 1-6. A h C t i u ~ ~ ~ - 111arr l e l ~ (uwil n ill s i ~ ~ w r l n ~ rsc~nncrs) kct emits Iight at 632.8 t~lri.Calculate 111c Krcqucncy r ~ IIII~ l l i p l ~ tWhnt i\ thr cllrrxy of' llrc photon generatud by this lasrr'? ' I > ,h
F v ,
Chv,(z!, + f)
each
vJ = 0, 1, 2, ...
(Equation 1.36)
1-1
1-7. 'I hv J M ~ W C 011111411 I of' II I:~rcr I\ I~IC:I\II~C~ 111 1111111 ut wath ( W J . whcrc m e watt is cquul 11) t~ihc~lnalcpcr wad ( 1 W = I J , h I ) . Illlw 111:lrlyphutoilh ~ I P crnitrc~ll)er ~ e ~ obty ~ad I (HI I~IW nlrrljgcn I ~ ~ u r r " ' l wilvclr~~gth l~c crnlltcd by a ~ ~ i t r o j i I:~\cr c r ~ i?337 rrrll.
1-8. Il+rILloill~onI .I 10 ualculatc {he vnluc uf the ioni7atiun enelgy r,f a hydrrlgen alolri in itq ~rllllllr!rlct'~ruriic'dutc
Problems
t -9. A llrir
1-1. Kadialinn i n the ultmviulet regiorl o f the clectro~nagr!tlicspcclrum is usually dcscribcd In ~ givcn d in nanonieters (IO-~ m). Calculi~tethe val~iesOC I,, i,. lcrms 01- wavelcngih. 1. a ~ is and F for ultra%ioIcl radiation with h = 200 nil), and compare your results with tl~rrrcin k i g ~ ~ I. r cI I.
1-10, A pruut~tl-htatrhydrogen atorrl absorhs a photon of light that has a wavelength ryf 97.2 nnl. 11 t t ~ r r givcx l oft' a photon that has a u.avelength elf 4x6 nrn. Wh:~t i s t t ~ cfinal
II~ tllc t.yr~~:~r! scries of hydrogen has a wavclcngth o f 1.026 x 10 IIII~III~Irllr-rpy Icvr-l ol'thc clcctrt~n.
'
In, F i r ~ d[he
1-11. A conllrltmly user1 non-S1 urlil denergy is an electron volt (eV). which is the energy tllal an rlcctron picks up when it passcs tt1rr)ugh a potcr~tialdifference oCr)ne vdr. Givcn [hat :I joule is a coulomb tittle\ a vnlt, show tl~at1 eV = l.M)22 x 10 !'IJ. 1-12. Using the recult of the prt.\iuu\ prublcln, calculate thc value of the ioni/:atic>n energy oC n hytlrugel~x o l n i r units ~ uf elcctnm vrllts. 1-13. IJcing the data in Figure 1.2. show that a plot of frequency ayainyt I / n 2 is linear.
1-14. '1-hcrc is another serics (IT lines ill thc emicsion spcclrurn o f aton~ichydrogen i n The near infrnrcd region callcd the Pi1chen scries. This serics rehullu from n -t 3 Iral~sitions. Calculate the wavelength o f the ~ e c v n dline i n the Paschcrl series, and show that thir l i t ~ e Ties i n thc near infrared region; [hat is. in thc inirared region near the visible region
F I G U R E 1.11 'l'hc rcgionr o i electromngnetic radiation.
1-2. Kadlalirln i n the infrarcd r e g i u ~is~often cxprerscd In I ~r115I (11 w.lw ~rt~r~lhcra, i = I/ A , A typical value of i.III this rcpiun i s IU' cm Calculate 1l1cvalurr i7t I,, i , ;nirl I for rildiil~i(>n
'.
wit11 I.= 10' ctrl
'. :ind crjmparc l o u r rr-ct~llcwith 111usc 111 1:1gt11t'I I t
1-3. Pa\t thc ir~lrul-cdregion. 111 the direction ot luwer encrgic\. 1s rluc t ~ ~ i t ~ r n rrcgiqrn : ~ v t ~ In t h ~ sregion, radialir)~~ is uhually charactcril.ed hg its frequcr~cy. c.il,~c\\ctl III urlirs oC n~egahertz(MH71,where thc unit, hertz (H7),is one cycle per sevlrntl. A ~ypicalrt~iurowave frequency is 2.0 x 1 f l MH7. Calculate the values of C, A, ant1 6 l i b 1 Illi, t;idr:~lirln, and cr)mp;irc your results with tliuse i n Figure I.11. 18,
1-15. Consider all elcctnln i n a I s nto~nicorbital i n a hydrogen atoln. The averagc dislunce o f this cleclrrln from the pmton i s g i v c ~by~ u, = .Irr~,fi'/rn~e'. whcrc co i s the perrllittivity o f free space, TI = h/2la, where h is the Planck constant, m e i s the inass oC an electroll, md F is the protonic charge. Calctrl;~tcthe value o f (I,, which is callcd the Rohr rurliu.~,in units o f picorrieLer\
1-16. Equatiun 1.8 has a nice phy\ical itncrprt.latiun based upon tlic i d c i ~o r a de Brtylic wavelenpth. Rccull that niovrllg plrticles havc an iwsociatcd de Broglie wavclcngth given by E. = h/ntv. where rrf I S the mars of thc parlicle and c is ils \peed. Show that i f we n%surne thal tlnly shndlnp de Rroglic waves can fit in the interval O to u. then I, = 2 u l n K r w usc h = 12/11118 and E = irnf:' ro dcrivc Equitiun 1.8.
12
(
h.~pler1 ! Ihr, Er~evbl evrl5 oi 4tomr anrl hltr1t.r
1-1 7. Tlie nlotion of rnacrt>ccopicphnicles is gorerncd by Newton'c equnriun of can be written in the 1i)rnl
drs
I I I ~ I ~ I ~ which HI.
wllzre x ( t ) is ihe pusition of the mahs In and f (s) is the Force acting on the partlcie. Fiwdtiun I i h a differential equation whose solu1ir)n ylres x ( l ) , the tralectrjry of the mass. I n ihe case of a harmrlnic oscillator, rn is the wduccd Inass and ,f( x ) = - k x (Iioclke's law). so that Ncwlrm'x eyLlatlon is
wl~cre11 IS thc reduced Illass of an electron arid proion. c is the prolutlic charge, E,, I, Ihe pcrmittir~tyof Cree hpauc, and h is the Planck constant. IJs~ngthe valuus given in khe inhide f in terrncofn2. Cnn!pafe your rehult with Equation I I . Irtrl~lcover. CHICI~IXIC Ihc v a l u c ~ 6"
1-24. It) ~ t ~ 111artrlc111, i h wc ~nvc\tljwic11111 h:lrnion~c riscill~torpoteniial as i l ~ eleading temi in n 'I'ayl~lrcxpanrluli ol Ihc achlal ~ntcrllu~lcar plentla!, V(H), about its cqu~llhriurnposiric~n. K . A c c u n l i ~ ~tog l'rrrhlrni ( ' 1). Ihc tirht icw kcrnls in thih cxpanqicln are
d'x
p-- , - -k. dt2 -
Show lhal ~ ( r = ) A cos2xvr suli+hes this equaticln i C v = ( 1 / 2 x ) ( k / ~ ) ' ~T' .h ~ result s is ralld only for a aacruscopic nscilliltrlr, called a clas.~icolhrrrmvniu usrillntor.
1-18. 'l'he kinetic energy of a clas~icalharmonic rlscillalur is
Usillg Equalio~ls1.11 and 1.12, shuw that
If R is always close to Kt,then R - He ia alwayr >111illl. ( ' ~ I I W ~ U C I I I I ~ . I ~ rCe r ~ ~on~ tht h right side of Etjualion 1 gel brnaller and smaller. The Krs~Icrrn in 1;yuatlon 1 i c a ctlllstanl and depends upun where wc chorw thc 7ero u i energy. It ~h crmvcnicnt 10 chocl\c thc /.cro ol~energybuch that V ( R w )equals zero wd to relaw V ( X ) I r ) lhih col>vt.nticm.Ihplain why the linear knn in the displacement vanishes in Equulio~~ 1. (Nutc rhdr - d V / r l N ir ttlu hircc
lrlturprel this rehull phyhlvally
acling betweerl the two nuclei.) Denote R - y by x , ( d ' ~ / r l ~ ~ )by , =k , and ~ ( r l 3 ! ' / t l ~ ' ) , = , ,hq y to wrirc Equation I as
1-19. Thc solution Ftlr ~ h clahzical c h:~rlilun~c ozci1latr)r ~cn ( I ) = A c o b 2rr PI (Equation 1.1 1). Shrlw that the ~iicplacclncl~~ ~rscill:~lclr httwrtn + A and - A with a frequency v cycle.s-'. !What is the pritd of the oscillat~tu~&. Ihi11I\. 11rm l u 1 1Ir ~ unc cyclc? 1-20. From Prr)hlcn~1--1'1, we \ct. tt~:it ~ h cIWI I I I ~r ~ :It ~ ~ , I ~ I I I rltnitla)t~ I I I I I C ~h r = I / v . The iirrrage r ) i the k l ~ ~ c tenercy is uvcr orlc cycle. 15 glvcn I,y
Stww that { K )= E / 2 where E 1s thc total encrgy Sllriw ~ 1 5 t thdt l (V) instantaneous potential energy is given by
-
t / 2 . wllcre the
Argue (hat if we restrict otlrselres to slnall dlsplncements, then s will be s m d l anti wc can n g l e c l thc terms beyond thc qundratic term 111 t..quatlon 2, showing that the gencrdl potential cllergy function V { R j can bsappruxitnated hy u harniuri~c-obcillatorpolcntial. Wc can co~lsiderC O ~ I - ~ L ~ ~ CorI Iextcrlsionc IS o f the h a r r ~ ~ c ~ ~ ~ i ~ - ~ l s cniudcl i l l a t cby > r thc higher-ur,der ,i,m!c. terms in Equntirm 2. These terlns art. caller! ur~hurtrloffic An analytic cxprzssion thal is a gcxd npproxi~llation to ~ I inlrrlnoleculnr I po~enlial energy c~irvt.15 B Mnr.vc put~r~tiul
Illterpret he result (Kj= { V } .
1-21. Calculate the value of the reduced mass of an electron in a hydrtrgcrr , I I ( ~ I I I 'I'ilkc thc macses of the eleclrrjn and protor1 to bc 9.113390 x 10-" kg and l.h72021 x I0 kg, wcpectlvely. What ic tllc percent difference k t w c e n t h ~ sresuit and thc rcrl I ~ I : I \ ~(11- iltl clcct~,on'!
''
C
1-22, ljuanlurn rricchanicq gireh that the electrorilc c n e g y of a hydrogcl~atom
15
and $ are pwameters that dclmul uptw the ~ntilecule.The pamlncter DLis the where ground-st& clec1rr)nic energy of the rnuleculc 1ne:lrured from thc minimum of I : { R ) , and fl is a rnedsule (>I khc curvature o f I 1 ( H ) at 11s minirn~lnl.Drrivc a reliltiul~k l w c e i ~[he force constanl and the parameter!, LlC and p . Giren tll:it /Ir = 7.3 1 x 10 lLJ~.rriolzculu-'. B = 0.0181 pn-'. ant1 R,, = 127.5 prn ior t ICI(g), calculute the forcc cun>Pallt of HCl(g). Plol thc hlvrhe potenlial for lICI(g), and plot thecorrcspt)nding harnlonic ncclllat~irpotcrlti.~l un the same graph (cf. Figure 1.5).
I:hapter 1 / The Encr~yL~cvelsuf Atclrns and Molecule$ 1.7 The vibrational spxtrum of ~ ~ ' C l ( g ) . TABLE
1-25. Uw the refult o f Problcm 1-24 and Equation 1.23 to show that
;obx/cnl-l H arrnonic oscillatur
Given thnl i, = 2886 ctll-l nrld DL= 440.2 k l - m o l ' for 1 1 ' ' ~ t ( p ) ,calcul:~tctllc vnluc of P . rompare your r e ~ u lwith t that ill Ploble1111-24 1-26. Cnrry out thc Taylor erpanqicln o i rhc Murhc p o ~ c n ~ iin a l Problen~1-24 through terms in (X - K:)'.1;rpresu y in Equnriull 2 ol Ilrt,l>lc~rl1-24 in tcrnms o f IIc and P .
1-27. I t lurrl!, r)ut that thc solution 11I'rhr Schrlnlinger equation for the Morse potential (ProbIcul 1-24) call he expressed ua
Transition
--
(I -. 1 (fundanlental) 0 + 2 (fin1 overtone) 0 -. 3 (second overtone) [) -. 4 (third c>vertonc) O + 5 (fourlh ovcnone)
C(,/cm
'
2885.9 5(10H 0
1 = 2X85.90~ 2885.9 5771.8
8.741 0
H657.7
10 413.1
1 1 543 b
13 396.5
14 429.5
whcrr
Given that fi = 2x86 rrn and 4 = 440.2 kJ. ~ n o l - 'for H ~ ' C I ( ~calculate ), the value\ o f I and 1.4.Plut rhc tibrat~unnlcncrgics of 11"Cl(~) for a Murse prllential.
with i~lcrcnsing71. The scJcction rule for ari anhermonic. oscillntr)r 1s that 11I: can havc any ir~tcgralvalue. although the intensities o f tllc A t l = *2. f3, . . . transitions arc rlluch lers ttinrl for the A I I= fI transitinns. Shuw that ifwe recrlgnizc t h a ~most diatort~icmolecules arc i n ihe ground vibrational slate nt room ternpralurc, tllc frequencies of the uhrerved 0 4 II trnnsitions will be given by
'.
1-28. In the infrared spectrum of H"Rr(g), there is an intense linc at 263U ccln U x 1hc hnr~nonic-osciilatr,rapprt>ximatiunt o calculale ihe valucs of thc forcc con\t;ittt t ~I11"13rlp) f ;jnd rhc p r i o t i (>rvihrafion ur H ' " R ~ ( ~ ) . 1-29. 1-he force conslant o f ' " ~ r "~ r ( g ~h ) 2 4 ) P4.m-I. Use thc harrllrlr~ic-oucillatorilpproximalion l o ca1cul;lle tllc vulucs rjf thc t i ~ n d i ~ ~ r l vihritiun:~l c ~ ~ ~ : ~ ! f~t.rlut.~~<,y i l l ~ c Ihc l ICIII-~X)I~I~ cncrgy (ill 1i)uIrc per [II~IIC) (11 "'15?'13r(~),
1-30. In Ihe I;~r-ir~lr;~rcd SIICC~I~UIII r j I ' " K ' ~ C I ( ~ I~CIC ). i~an inlcli~cline at 27H 11 1.111 the v;tl~~c\ a~rthc.f+il,c<.I r r r l \ l . i l l l .IIII~tllc peritnl ut vihrntion ul " ' K ' ' c I ( ~ ) .
('i~l~~rl.iI~
1-31. T t u ~ rlar, uc. 11:j~c.~rc.,itt~ltllc vthrntit>n:!l 111c~rion o f a diatomic rnoleculc by rncaris or a h:i~-~~~t~r~ic-r,\c.ill:ltr,r rilrjdcl. Wc. >;IN. i n Scctiut\ 1 4, however, that the internuclear polenlinl cr~crgy1 5r1l11 ;I \inll>lthII.II.L~N>~:L 1 ~ 1 1ili nlorc like that illuctrnted i11 F1gul.c 1.5. The har~nor~ico l ~ l {he y quurlratic term i n [he Taylur exp;tnsiun occilla!r)r :~pl,ra~xirir;ilicrnc r j r ~ \ i \ t \ (~I-kccplng o t V ( H I [ w c I'IO~~L'III t 1.1 I.:111r1 1 1 p r c r i l ~thilt ~ \ lhcre will be only one line in the vihraticlnal 1;1~)1.11111c.11ri1l dala shr)rr, thcrc is, tndced, one dominant spcclrwrl ul ;I di.itc)nlit II~~IIc*~~~IIL* line ccnl1t.d ~ t l cjrrrrrlrr~rrt~r~rtrl~ hul :I]\(> 11rat\r>f wcilkcr i ~ ~ t e ~ ~: ~salmost ti t y inlegral rnulliples of ~IIC lundo~ncl~ritl. 'I hrrc lulcx ;uc c;~llctlrrLmrJr/rrnc.v (hee Table 1.7). I f the anharmc~nictcrlli+ ill t-'all~:~llr>r~ 1 19 arc lakcn into account, thcn a quantum lriccharlicnl cnluulatia~rlp r c 5
Curve l i t Equat~on2 to the e x p c r i r n c ~ ~data t ~ l i n Table 1.7 to find the optinlum ral~ie!,ur ij and i-fi. llse Equation 2 to calculate the value!, (h~ h cobserved freqt~cncicsant1 cr)mpuru yuur rcsillth with ~ h expcrilnc~~tnl c d;ita.
-
'
for ' ' N ~ ' ~ F (calculate ~), the v:~lucs of 1-32. ( ;wen tll;it si 52h.10 cln arid isi : :1.4 CIY I ~ L .t r t . q ~ i r ~ ~ ~ o ,f ithc c \ l i r ~ollrl l wcr1ntl v i h r : ~ t ~ o ~uverl(Jncs ~nl (\cc I'ruhle~ii 1-3 1 I. 1-33. Ihc lurldi~~nt.~l~;~l linc ill rhc itilrilrctl 5FK.ctrurrl 111 l ' ~ ' l ' O ( ~clcrurs ) :it 2143 0 crn I . n l ~ d i L farr ''(-l''O(F) the 11rrt Iwcrrorlc o~-<-llr\ ; ~ t4 2 ~ (1~ ~0r 1 1 ' , ('i~lct~!:ircIIIC L,;IIIIC\ il ZIIILI ( U C C I)rot?lc111 I -3 1 ). 1-34. Tllc Morse putentin1 is presented in I ' r n h l t ~I ~24. ~ ( i i v c ~thiit ~ I>< = H 75 x 10 l u ~ . t i ~ oI . l F r 1556 cm-', and Re = 12U.7 pm for 0,. plot a Ma,rse pc~tcr,ti:ll IOI. 0, Plot the cor1.rspontling kurnonic-oscilla~(~r potcntirtl on the same graph.
1-35. Show that the momcnt of inertia for a rigid diatomic rotator can be written as I= @ K:. where Xr = R, -t R, (the fixed separation of the two masses). HI and R? arc the divlanccs o f t t ~ ctwo malise+fro111t l ~ ccenter o f rrioss. n ~ J I~ i\dthe wduccd mass. 1-36. I n [he tar-infrared cpxtrum o f H ' ' U ~ ( ~ ) ,there is n scries of Iinec sepnratcd by 16 72 crii - ! . Calculutc thc values or [he rllonlent uf inertia n t ~ dthe internurlenr separation in H'"B~(€).
where i is cnllcd the cmltrrm~onrt~~~v c-rrr~rtrrr~r 'l'l~e:~l~harli~rll~ic corrccrion i n Equalion 1 is r n ~ ~ smaller ch than the harmonic Ierm hec:~usci<< I. Shuw lhal tt~clcvcls arc not equally spaced as they are li)r a harmr)ri~cr)scillalur and, in bcl, ttial tlltir scl~araliondecreases
1-37. Given :hat i' = 2330 cm-I and that D, = 78 715 cr11-I for N,(g), calculate the \ d u e of I I c .
Chaptcr 1 / The Lnelgy Levels ofAturns anrl Moleculer
1 -3U. The .I = O to J = 1 trarlsltiorl Ibr carbon monoxide 1 1 2 ~ 1 f ' ~ ( cxcur!, g)] IV MHz. Calculale the value o f thc bond l c ~ i g r hi n carhon rnonoxidc.
at I ,153 x
Curve t i t Equatiol~2 to thc experimental data above and hnd ttle r)plimum v a l ~ ~ eofs B and D.Compare the predic~ionsof thc rcsulting Equatirln 2 with the c.r~rirnerltn1dara
1-39. 'l'hc Inlcrowave spectrum of " ' ~ " ' ~ ( g jcotisials ul- u series uf lines wl~oscspacing is allnust constant at 3633 MHI C'alculate the bond lcngth ~ T ' ~ K ' ~ ' I ( ~ ) . 1-40. Assuming thc rotutinn uT a diiltomic moleculc i n ttlc J = 10 slate may he approxinlatcd
~) w i n g classical ~ncchnnics,calculate how Inally revolutions per hecond " N ~ ' ~ C I (makes = 10 rotatiorrnl stale. The rotntic>nal cnnxtallt of " ~ a ' ~ C l ( ic ~ )6503 Mllz.
i n the J
1-41. The rigirl-rotator 1ntdc1predlclc th:~tthe line!, ill thc rotaiionul spctrum of a diatomic rr~olt.culeshould he equally spaced The following rable lists sorne o f the uhcerved lines in thc rowtiunal hpectmm o f H ' ~ U I ( ~ )
C C ~ = , ~Z B ( J
Tranhi~ion
fi,,,h/~~~~
A , c m
B
+ 1)
1-43. Given that
= 8.4hS c ~ n ;~rttlI ) I1 I)IXtIJI\ C I I ~ I l;)r t i " ' l l r ( g ~ ,C~IIUUIA~C thc imquencyc)flhcJ=O-Jz1.J 1 * I -?.J-:2 . I - 3 . --.J=6-J=7 tra~~sitions ill thc rutat!unal hpcIrtIriI (11 1t l U l i r l p ~
= 111.243 crn
- .-
3+4
83.03
82.72
1-44. neter~ninethe nulnbcr ut *,lrliulr t l c ~ ~ r *I! c \ trccthr~~b trt N:, (':ti,,
21.07
4+5
103 40
104 I 0
124.311
6+ 7
145.0'3
7-8
1h5.51
C': ti,, arid
1-45. nulerniine llle tr>tal 11umhc1i ) t 116rr11r,1l ~ r ~ i 0 1~ 1 rkII~#,~~IIIII r 111 J1('N. ('I>,.SO,. Sl,,,, .III~~
20.211
5-h
(':ti,,
C,H,.
(CH,),CO.
124.011
20.73
1-46. Using the dala i n Tahle I.tj, calc111.11rtllc \uluc III t l r rrrrb ykurlr 1hrilliun:11C . I I C . I ~ ~
143.7h
~ j t,I
water molecule.
20.4K 16.5 44
1-47. Using the data in Tablc 1.h. ralculiitc ~ h cr.ill~c+>I IIIC trrur methane n~olevule.
186.12
1-48. Givcn that K,, = 112.8 ~ I Iand I Hh,, I I N .I
20.35 H
,-+
9
185.86
tllc 1llolIIeIlt r d inertia ant1
211.52 9-
10
206.80
206.38
11
226.50
227.48
I. c , ~ hb~lrtctllc rdluc, or'
i.
t ~ , ~~ l! T ~ T I I < 1j1rI) ~ 111~111~ ~ I C I Ir ~ I : i l ~ v>c~ ~ ; 1 r ~ t i o l l mcnsiorl, inleracling thr(111ghrl p t r t ~ , ~ lI)I.II ( A , , 1 1 = 1 iI I , I I;I\~II 111.111tn- I011(.-11 t111gII~R~II 111~ j l h parllcle ( . -K 0 t / , \VII.LI 1.1% I, itair ' is = -(aL'/?I r , 'I.ahrm 111:11 1, Newtcln'x erluatlnns fur 111, ,irltl rrr . .dl<.
fi
The d~ffcrcnceslisled in the third column clearly ahow that the lines are ~ ~c ox ~tc t l yequally spaced as thc r~gid-rolalorapproximati011 predicts. The discrepancy can be resolved by r c a l i ~ i n gthat a chenlical bond is nr)l Lruly rigid. As rhc ~nulcculcrr)talec more energetically (lllcrcasirig J ), Ihc centrifugal fu~,cccause+ the boild to 'irretcI1 slightly. I f Ihis !,mull effect i\ t ~ k c nInto :iccouril. Ihcn [he energy i s gwcn by
whzl-e fi is called Ihc cmlrijic,~rrldr.rlor11mrrcorrtttrof. Show [ha[ the frequencies due to I + J -iLtrtr;l~l\irir>ns are givcn by ahsorptio~~
I~II+ It* "N"N'?
VI~~I,~~I(I!LII C I I C I , 01 ~ ~ :I
1-49. I n this prohlem, we -*ill \ec I~rrx~ h ba)rlrr(lt r u r l rctlitctl ril.l\\ .ul\cr ~wrul-:~lly whsll discus~ingthe irilrractiun of Iwr> parllt Ir.. ( ' a l ~ l r ~ lwrr t ~ r Iimurwr, m, mil r r l , , ill unr d i -
20.12
10-
prllll
ot thc Solve fur
A,
al~rlA, 10 ~IM,IIII
Chapler 1 / The Energy Levels or Arorns and Molecules
Show that Newton's equations i n Ihece cvordinales are d ? ~m m d 2 x
M
dr-
"$
dr2
A
Dx
' +-aU M
d' x - m nt, d 2 x 'lt'
MATHCHAPTER
au
=--
-
dl2 -
DS
Now arid thew In(?ctluatint~sto hnd
lulerl~crhi.; rehull. NOW divide thc lirs! cqu:ltio~~ hy m ,and thc F ~ C O I hy I ~ 111! :lnd '.uh1rilct ohlain
I(1
d'x dl:
/.I--- =
JL7 ax
--
whcrc ,A is the reduced mass. Interpret (his result and discusc how Ihe original IWII-hxly has been reduced to two one-hdy prnblerns. ~mhlutl~
You learned in high school that n quadratic equation r r I : given by the so-called quadratic fornlula:
+ hk
t
(,
= 0 h;~stwo nlols,
Thu\, the 1wo value, o l x (called mots) that satisfy the equation xZ t 3.t
2
: 1)
ijrc
Al~hougI~ Ihcl-c al-c ggcerrll furmul;~cfor the ror>t\ 01 cubic ;111dqu;~rticr t j \ ~ ; l l i o ~th~.) ~\, vcry inuunve~iientto use, ant1 t u ~l~el'n~orc, f tIlc~cilrc nr) f'r>r~llul:l\lur ty t~:~lio~ic or' ~ h tifth c degree or higher. Unll)rtunatzly, in pri~c!ice we cncourltcr ~ u c rtlu;~tirlr~s t ~ freque~ltlyand must lcarn to dcal with thcm. Fortunately, wirh Ihc advent 01- 1l;lnd raluulatun and personal compulers. h e t\uitlerical ~ o l ~ ~ofl jI ) ~I ~ nI ~ I ~ ( ~ I I I ~ C< II II I I : I ~ ~ O ~ S and other t y ~ ot' s equi~~iur~s. such as x - cus T = 0, is routine. A l t h ~ ) u r llhc\c ~ ;III~ other equations can hc snlvcd by "brute force" trial and error, much more o r ~ ; l n l ~ e d procedures ran arrive at an answer lo almost any desired degree ui'accu~auy.1'crh;lps thc mosl widely known procedure is the Newton-Raphsnn ~ncthod,which is hcsi illustr~ted by a figure. Figure A.1 shows a I'unv~ionf ( x ) plol~edagainst .r. The s u l u ~ i c l10~ ~the equation ,f ( x ) = 0 is denotcd hy x * . The idca hchind thc Newton-Rnphcon method is to guess ;in initial value of .x (call it x,,) "sufficiently close" to x * , and rlr;~wlhr IarlgerH to the curve f ( x ) at x,, as shown in Figure A. 1 . Very or~eri,the extension the Iilrlgcrlt line through:he horizontal axis will fie closer to x* [hall docs x,,. iVc dcl~otctl11svaIue of .r by x, and rcpcat thc plnccss using x, to get a new value uf x,, which will lie even clr~serto x*. By repeating this process (called iteration) we can approach x* to essentiully any desired degree of accuracy. :ilC
3')
T A B L E A.1
Ihe results 01- the :II~~IIC~II~+)II o E the Ncwto~~-Kaphhr)rl rllcthod f ' ( . x ) = 4 . r 1 - ~ . 7 2 r ' fK.721 - 2 . 1 8 = 0 . tl . -.
x"
0
0.251HI
1
0 3442
?
11.355Y
1
015hl 0 35hl
4
f (1,;) -4.825 -4 855 -6.281 1.703
x 10
.
'
x ICI-'
Ihc >olutic>nuK ttlc etlualio~~
f'c-T,,) 5 . 1 10
3.139
x 10-I x 10
1~
'
4.033 4.031
F I G U R E A.1
A grilphical il1ustr;ition 01 lhc N ~ w ~ r r t l - K ; ~ l ~111rth~u1 h\cl~~
Wc can use Figure A.1 to derivc a convenient formula for the itzrativc valucs 01'x. The slope of f ( x ) at x,,, f '(x),), i.c given by
,
EKAMPLE A-l 111( ' ~ l i l ~ 2~, tWl' ~l~
1 1 '%tlI%c 1 1 1 L ~' U ~ ~C Cl ~ ~ l b l ~ l ~ ~ l l
Uhc thc N t ! w r ~ r i ~ . K a ~ r tilcthtd h ~ i ) ~ ~ 1~ llrltt flrc rcill rtwh t r I 1111, ctlit,rtltrrl tigureh.
Itr
Ilrc rrg~tllicanl
Solving this cqurlticln for x,,, gives f ( x ) = x'
which is the ~terativeformula for thc Newtun-Kaphct~nmethod. As an application of this brinula, consider the chemical equation
whosc relared equilibrium constant is 2.18 ; ~ ta certairl lempcrature. (Chemical equilibrium is discussed in Chaptcr 12, but we'rc simply using the algebraic equation below as an example at this p i n t . ) 11' 1 .OO atm i>l'NOCl(glis intruduccd into a rei~ctionvcsscl, then at equilibrium PK,,l = 1.00 - 2.r. Ph0 = 2x. and PC,, = r; thusc pressures satisfy the equilibrium-constar~texpression
+ 3x' + 3x - 1 =
By inspcction. a solutlo11lies trelwcco O and 1. 1Jsl11gx0 = 0 5 rcs11Itsi l l Ihc it>IIow111:! table:
The answer to tiire t.iglliticant figures is A = 0 25992. Nihk I~:II1 (Y,,) is s i g ~ l ~ f l ~ f i n t l y cmaIlcr at each s~cp,as ir should be as we approach thc valuc, of r t h a ~satistiec 1 ( r ) = 0,bur [hat j'(q1ducs not vary spprcciahly Tho same khar,ior can he >ern in W l c A. I.
which we write as As powerful as it is, the Newtun-Kaphson method drlcs not always work: when it il is trtwious the mcthod is working, ;I~IL~ whcn it doesll't W O I - ~ , i t irlay tje even m o l t obviul~z. spectacular failure is prrlvidrd by thc eqo:tliort / ' ( r) - r I." = 0. for xhict~.r,= 0. 11-we b c ~ i nwith .w0 = I , we will ohtairi r , = -2. .r2 = 1 4.1, -. - 8 , and su on. Figure A.2 shows why tile neth hod rs h i l i n g to convclgc. Thc lrjrssagc here is that you shoillrl always plot f(x) first to gel all idca c ~ f where the relevant moth
does work,
Because u i the stoichiometry of tht: reaction eyuation, the value of-x w e arc seeking nu st he between 0 and 0.5, so l e r ' ~chnose 0.250 as our initial guess (.you,). Table A.1 .shows thc results of using Equalion A.I. Nutice that we have converged tu threc cignilicant figurcs in just three steps.
then
Thc t u ~ ~ c t iF(.r) o ~ ~ ir so~netimescallcd thr ;~n~ulcri.;nive of f ( x ) . 11- therc i s nu v l c l i ~ r l ~ l : ~function ~-y F ( x ) whose derivutivc i\ j [ r ), w r m y II~;II111eiritegral of f (.r) c i ~ n r ~hr c ~cv;lluatcd t :inalyticdly. By elclr~e~~f;iry 1'tlrlc.tion, wc Illcan a lunctinn that rat1 hc cxprc*xcd
S I; fir~itccomhinntinr~n l p o I y r ~ o ~ ~ nI~-igrjr~c>r~~~-tr.ic, ;~l. cxpc~rlcntii~l. ;ltld h ~ ~ n r i ~ l i lu~lctions. ~r~ic. 11 Iurli* ~ > uthat l numerous integrals cannot hc cv;du;ltcd ~ r i : ~ l y t ~ c . iA ~ lp:miualnrly ly. Irrlpmunt cxiit~rpleof an integral that canllot hc cv;~lu:ltcrl i t 1 tcrrlls 111. s l t . r ~ ~ c ~ r ~ ~ ; ~ r y fulwIit~n\ir
FIGURE A plot
A.2
=
I,',
illuslraling that the Newton-Raphson method fails in this case.
lie and to see ~ h athe l lunc~iondoes nut havc any peculiar proprlies. You should do P~t,blc~n\,%-I In A-8 m bccolnc proficient with the New~on-Kaphsonmethod. 'l'here arc :ilso nulncricnl method\ lo evaluate integrals. You learned in ualculus that an integral is the area between a curve and the hori7.ontal axis (:uea under a curve) hetween the integration limits, so that the vnlue c~f
i s givell
by the shaded area in Figure A.3. Rccall a fundamental theorem uf calculus,
which says that if
I ~ u u ~ ~A..1 r r wrvrs n (r>dclincrhc(nonclementary) I ~ l r ~ c t()1{ i u1~) ~'I'tr~v;~luc nf 4t.t)for iuiy YUIIK. tnt I is ~ ~ V by ~ the I I area undcr thc curve ,/ ( , I ) - t. " hurl, rc z I) to 11 = x. Irl'\ ~otlhidcrl h nxli-c ~ general casc givcrl hy Erlu;ltit)~iA ? r ~ itlc r rhadcd :lre;i i l l i.'ipnrc A 1. We cun approximate this area in ;I ~ ~ u l l l h(I!r rwily\. f:lr\l Jlvldr rhc i n t c r v ~ l
,
( ~ J , / I ) iritar n cqurrtly spaced suhinterr;;ils u , - : I , , , r t : II . , , , ida ith wit11I,,, = (1 rlrld 11" h. Wc w i l l Ict h = u, - u, l i ~ ,Ir x 1 ) . I. . . . rr t I:ipurv A.4 ultrlws II ~rla#ftifirlblirm(1'one of thc suhiritcrvals. h;~y illc i r , , i r , , ~ r ~ h i l ~ c r r .011c i l l . wily 10 under thc curvc 1s to cotirlcct t l ~ cpc~irl~s !'(I, 1 a r ~ d((rc,, ) hy $: Ilpprruisialr ~ h arca c ~ t l l r i p hIIIK ~ 11% SJIOWR i n Figure A.4. The area uridcr the 5tmight line ; ~ p p r u x i r ~ ~ a t i ( ~ r ~ iu / t u l Itr tlrr itticrvill is the sum of the xrce of The reutanglc [A f ( u , ) ] and the area af
.-
.
,
,
Ihr ~ r t n r y k1 Ihl I( l r , + ,) - j ( u , ) l ) . Using this app~nximatior~ for all intervals, thr total r m a ur~clrrr l b r curvc fmrn u = (1 to u = b is given by the sum
.
N~)tt-1h;~1thv ~ . r ~ ' l l ~ ~ in ~ iEquation c ~ i f s A.4 go as 1, 2. 2, . . . 2. I.Equatiun A.4 is for n = 10 or so and on a personal con~puter ;Ig \prc:~(lclicc~lirr I;~rpervalues of n. The approximation to rl~eintegwl given hy uci~~ I
FIGURE A.3 I ' l ~ eintegri41 of f ( u ) f r t m n tr) h is ~ i v e nh\: rhe shadcd arca
(hat with n = 100, the result lor Simpson's rule differs from the "acccpted" value by only one unit in the eighih decimal place. The error for Si~npsoll'srulc goes as h" c t i ~ n p x e dwith h' for the triipc7c1idal approxiniatic~n.In fact, if 2V.f iia the lagest value of i f [ " ( u ) ) in the interval ( ( I , h ) , then the crrur is (at mua! M (b - u ) h 4 / 180. Prohlerns .4-4 to A-1 2 illustrale the use (I!' thc tmpzzwdal approximi~rioriand Si~npstln'srule.
EXAMPLE A-2 Onc Ihcclry (fro111I lchyc) ul' I ~ I Crln)lur t FIGURE
A.4
An iliustnltion of the arra oilhe j
~ c cap;tclty ~t r)f a mo~iatr~inic crystal piye!,
+ 1st subinterval fur the t1-qe7oidilI ilppro~ititiitit)li
l a g e s t value of I f " ( u ) l in the interval (a, h), then thc crror is ot most M ( b - rr)h2/ 12.1 'l'able A.2 shows the valucs of
whcrc Kirihe 1i111lurgua cc~rlhlunlIl( 4 I4 I h: ' ,rrtcll )irr~tl(-r,,, thcllchye ternpcrutt~rc is u puralrlcicr c h u r r r t c r ~ ~ rbf t ~ cthc 1.ryci~I111w ~ u h s t i l r i (~~ ~I V. C I Ilhit 1-),, = DOV K for ctrpwr, c:~lculaicihc rtlolar hcat upaclty ul'cuppcr ill 7' = 103 K. $ 0 1 0 II O N ,
h r n = 10(h = O . t ) , n = 100(h = O.Ol),andn = 1000(h =0.001).The"accepted" value (using more sophisticated numerical integration methods) is 0.746824 13. to eight decimal places. We can d c v e l ( ~ pa rtiore accurate numerical il~tcgrationruutinc by approximating j(11) in Figure A.4 by something olher than a straight line. If we approximate f (u) by a quadratic l'tunctinn, we have Simp~o>~l'.r rule. whose fnrmula is
AI I' = 103 K, thc h u h s 1111~gr~l 1 0 cvilluulc ~lu~twrl~.ally 15
(e*
- 1)'
rlr
IJsing the tlapzzr>idnl approximalio~l(Equatiun A.5) and Simpcu~l's rule (Fquution Ah), we Iind the follouring values of I : II
It
i,, (~rape~oidal)
I," (S~mpsun'\rule)
in Nule that the coel'ficieilts go as 1, 4. 2, 4, 2, 4, . . . , 4, 2. 4, 1. We write Equaiion A.6 hccnuse S i ~ n p s u n ' srulc requires rhac there k a n even number 01-interirnls. Tahlc A . 2 show\ the values oT $( I ) in Equation A.5 for n = 10, 100. and 1000. Notc
T A B L E A.2 The application c ~ l - t t~,.~lx./uiti:~l h~ :qq~rc>x~riiation (Equatiurl A.4) and Simpson's rille (Equalicm A.b) to the evaluatiorl of @ ( I ) pvcn t>y 1Lqu:llion A.5 The exact value tr) eight decimal places I S 0.74682413.
-.
I7
11
10
0.1
109
iwo
-
I,, I
I I ~ I ~ / . ~ ~ I L ~I!,, ~ (~ ~) ~ 1 1 1 ~ 5 0l t1l ~l ~~ )9
. .
.
.--
..
I 1 7.1(r218{Kl
I I fJOH?4'1-1
0.01
Il.lJ(1HlH~Hl
11
o.w~
[I
7-1(>~2.107
7.111H?.l I J
o 7,1t1~2-I I \
+,r(', = l h . 5 J . n ~ u l ' , K-' , ~nagreeme111H ith the cxperirricntal value
Although the Newton-Raphson nicthod atid Simpson'r, rule can he irriplemcntrd casily on a spreadsheet, therc are a nunilxr of ensy-to-use nu~neriualsuftwarc plickagrs wch as MnthUud, Krxleidagruph, Mclthenlutiru. or Maple that can he i ~ w dto evaluate thc roots o f algchraic equations and integrals by eken inorc hophisticateti n u ~ n e r ~ c a l methods.
This inlegral can be evaluated analytically; it is given by tan-'(1). which is cqunl to x/4, so I = 1).7R539816to eight decimal places.
Problems A-1. Solve the eqtrati#>t)ur betweell U and 1 .
+ 2 r 4 -t41 = 5 to tour significal~tiigurcs ior the rlMlt that lie';
A-10. Evaluate In2 to six dccimal place!. by vvaluating
In2=/;'f
A-2. IJve lhe Newtun-Raphscln rnethd lo derive the iterative formula
What must n be to assure six-digit accuracy?
A-I 1. IJsc Simpsun's rule to evaiuale for Ihe value o f *h Thiv formula was discovered by a Babylonian mathematician rrlorc that) !lKMJ yc:lr\ :)go.Use this tor~nulato cvaluate f i to live significant figure,. A-3. Ckc thc Y\lcwrnl~-R;~ph?rw n~etllndto snlvc thecquatlon fipurc>.
t,-'
+ ( . ~ / 5=) 1 to Coursignificant
and cornpale your rcsult with the exact value, &/2. A-1 2. Use Simpson's rule to evaluate
A-4. ronridcr the chemical reaction described hy the equation
at 3tH) K. II 1-00 atln of CfI,,(g) and ll,O(g) are introduccd into a rcaction vesseI, the prccaurch at equilibrium nhry the equation
10
six duci~nalplaces. The exact value is rsd/l 5.
A-1 3. Use a nu~nericdsoftware package such as MafhCud, Kaleidagmph, or Muflrcrncrfi~ato evaluak the integral
for value:, u i u between0.200 and (1.300and show that S has o maxirnurn value at a
A-5. In Chapter 2, we w ~ l solve l 11icrul>lc cquntlon
= 0.27 1.
A-14. Use Si~npcon'srule to evaluate the integral (.we Fyuation 2.31)
( A % 'I h ~ t! 12.1 - l = lJ Usc the Nexvtrrn-Knpllson 111cthodto f i t ~ t fhc l o111yrt:ll rrwf ot thi+ctpiltion to five significant figui-ec
A-6. Solve the equnlinn .r'
-
71- + 1 = (1 (or ;1l1 rl~rcru T i i \ rr*rlr I,, faiur. dccirilal places
A-7. I n Example 2-3 we will solve the cl~hlcctluiitltirl
Ucc tllc htw~orl-Raphsonmelhrd lr) find the rtwl rnrhi< r011.1114111 111:11
I \ 111.~11 \'
0 I
A-8. In Sertir)n 2-3 we will sol%ethe cubic e q u a t i o ~ ~
USCthc N C W ~ O I I - K ~ ~~ncthod I I S O ~to I show [hat the three rrjuts t o hi\ rqu;ttlcrrl urc 11.11707>. 0.07897. and 0.2167. C
A-9. Use hc trape~oidalapproximation and Simpson's rule to cvaluatc
for T * = 2 . W and 3.W. The accepted values are -0.6276 and -U.1 152, respectively
CHAPTER
/
The Properties of Cases
Wc kgin clur study of thermodynamics with the propertics of ga5e.c. First, we will Jiwusr the ideal-gas equatinn and then some extensions of this equation. of which !llc vim Jcr Wuitls cquaiion is the most famous. Although the van der Waals equation ucuounlb in p;~r!for deviations Tmm ideal-gas hehimior, a nlrlrc cya~ernaticiind accurate uppn~uchIS to k15c il so-callcd virial expansion, which is itn expression tur thc pressure ut'u guh UR B p ~ l y n i ) ~ t ~Ini arhc l density. We will relate the cr,efficients in this polynt~niial trl ttr energy 01' intcracliun k t w c z n the molecules o f the gas. This relation will take ir+ ~ I I ~ u I , disc.us.\irm r)f how lnolecules interact with o n e another. Wc will sce that ilcvtalic~rlst'rorn idci~l-gnak h a v i o r teach us a great deal about molecular intcrxlions.
2-1. All
(;,IWS
Ht-I,,~vcIdcally If They Arc Sufficiently Dilute
It a pur Ir c u l Y l ~ , ~ r ~ rl~lutc ~ ! l y t h a ~its c~)n\titucritm o l t c t ~ l c sare s o f i r apart h l n e a c h c~thcrtnrrllc rlvrrupc 111~1w t ciln Ignore lhcir jnttr;~ctit)ns,it obeys the cquation of state
Johannes Diderik van der Waals was born in Lciden. the Netherlands, OII Nove111brr 23. 1837, and died in 1923. B e c a ~ ~ shee had not learnd Latin and Greek. he was at lirrt not able to continue wilh university studies and so worked a? a schr~11~eacherin a scco~iriaryschool. Aitcr pashape of new leg~slation.Iiowcver, van dcr Wials rjbtained ail cxcr~lptionlruln the unlbersity requiremen~s111 clacsical larlguageh n11d defended hls docturnl disscWatior1 ot I.eydr11 of the Ll~liirerhity111 1873. In his discertation. he proposed an t.xplsnatir)n 01- the crinlin~~ity p \ e o u s and liquid phases and thc ]~henr)~r~unon of the critlc:iI point. as well as a derivatiori of a 11twequatirln of s1;ite 111 gases, now callcd the ran der iVxals cquatirin A tcw years latcr. llc p r t ~ p b ~the d law of corr~spondil~g states, whlzh reduces thc propertics ot' all gabcs tu rrne common dcnomiwtor. Although his d~srertationwar writtcn in Ijutch, his wnrk quickly camc tu the aucrltion IIC Maxwell, who published a r w i e w o i it ill English in the British journal Ndiure in 1875 and so brolighk thc work to the ancntion of a luuch hrrwdcr alirliencc. In 1876, van der U'aals waq appointed the iirst Prufe~sorof Physics at the newly crcatecj Ilnlvcrsity u T Anwterdam. The University bccanic a c e ~ ~ tior e r both thetxetical and expcrimenral rcscarch un Buids, largely through van der Waals' inflilence. \'a11 der Waals was awardcd rhr Nohcl Prize lur physics in 1910 "for thc work r,n thc cquation uf state for gases and liquids."
II wc tllvldc. t n > t lhides ~ ol'lhis equation hy n , we obtain
w l ~ c r tv = V / n i s the molar volume. We will always indicate a nrolitr quantity hy dr;iwi~igi~ l i ~ i eabovc the symbol. Either of Equatiul~s2. I , h ~ n i l i a even r t o high school htl~ti~tits, IS called the idmi-gus eqac~tionof lar re. Eyuations 2. I are called i i ~ equi~lion i ul >t;ttc becausc they scrve as a relatiun hctween the pressure, volume, and tcniperature r l t ' the gas fur a given quantity of gab. A gas that obeys Equat~crns17.1 is callccl i i n idtit1 g i i c , o r the gas is ?aid to bchave ideally. The distinction between V and 7 illustra~esan important ul~iiraclerof the quat~tihes or the variables used to describe macroscopic systems. These qunntities are of twri types,
49
5U
(
t1,qitr.r ? i Thc Propcrries elf (idres
c21lIcd cxtcnsive quantities and intensive quantities. F~~rrtsivr q u o ~ ~ r i r i ror . ~exrri~sive , vririr:blrs, are directly proportional to the size of a system. V r > l u ~ ~mass. ~ c . and energy are exilmples of extensive quantities. Itrtetrsive q~rantitir.r, or. irlrr,r~sil~r ~.~~ri(~hle.v, do not depend upon thc si7c of the system. Pressure, temperature, and detikity ;Ire cxamplec of intc~lsivequar~tities.If we dikidt anextensivequantity hy thc riur~~hcr ot p;~rticlcsor ihe ~~urnhcrnf nloles iri a system, we obtain all intcnsive quantity. Forun:i~i~l~lc., V ( r l i l i ' ) is ;in cxte~iai~c quantily but (rlrn'.mol ' ) is an intcnsive quatitlty. Dislinguishinp htwtlen 11ic p r o l s r t i c ~of e x t c ~ ~ s i and r ~ e intenqive rluanlities is often irnportarlt i ti desc~-ihi~lg cf~emicalsysiems. The reason Equations 2.1 are encuuntered so frcqueti~lyin chc~iiisirycilurscs is thrtl all gascs nhcy Equatio~~s 2.1, as h ~ n gas thcy are sufficiently tlilu~c.Atiy ir~cliviiluiil charactcri~ticsof the gas, such as thc shape or size o l i ~molecu1c.s s or huw thc rti~>lrcuIcs interacr with each othcr, are lost in Equations 2.1. Jn a setlse, lhesc cqu;~tion\am a crlnirnon dct~ornitlilturfor all gaqcs. Experimenlatly, most gascs s;t~i\l'yI+li~:~tit,rir2 , I l o approxirnatcly 1%) at one altn and O'C. tquations 2.1 require us to discuss the system of units (SI) ;~drq~tcd t ~ yihc Iritcrnational I!nion of Pure and Applied Clicmistry (IUPAC). For cx:~~nlllv, i~ltl~rrt~ph thc SI 1111it01- \ , ~ I U I T I C is 111' (meters cubed), thc unit L (liter). which is dctirictl ilk cxr~ctly I rlrli' ( d e r i r n e ~ ecubed), r~ is an acceptable unit of volume in the Il!PAT' cycic~lb.'Ibr Sl unit of pressure is a pnscal (Pa). which is equal to onc newlor1 p r \~ll~:irrr i i ~ t ~ r (PJ= N.m = kg,m-' .s-'). ~ e c a lthat l a newton is the SI unit o f i'orcc. ~u wr w1.c ~ t i ; t ~ pressure i s a ~ K per P unit area. I'ressure can be mcasured experin~ent;~lly hy ol>ccrvi~tg how high a column of liquid (usually mcrcury) is supported by the gas. I!'III i c rl~r 1 1 1 : ~ ~ of the liquid and 8 is the gravitational acceleration constant, the pressurc is givrn b y
2-1.
/\I1
(;~ct.r I h . h ~ v cldcatlv IiThey h r c 511fflt ~t.~nlly Dilurr
Strictly speaking, new textbooks should use the IUPAC-suggested SI units, but thc units of pressure are particularly prohle~nalic.Although a pascal is the SI unit of pressure and will prubitbly sce incrc;~singu w , the alniosphere will t~ndouhtlycontinuc to he widely urcd. One ~ , i ~ ~ r o . t(at~il) p I ~ ~i~~defined r ~ ~ as 1.01325 x 10' Pa = 10 1.325 kPa. lone atmosphere uscd lo he dchned ;IS lhc prcssure that supports a 76.0-cni colunirl of mcrcury (see Example 2.1).1 Notu that nllc kPn i s ;~pproximately1% of an atmosphere. One iitrr~osphcreused tt) he thc c~:ind;~rd 01- preshure, ill the sense that t:~bul;lted prnpcrtiea nf substances werc preser~tctli1t c~ric:LIIII.Will1 the change to SI unils. Ihe standard is now one bar, which is equal to 10' [';I. c>r0.1 Mt'i~.Thc relation bctween ban and atmospheres is I atm = 1.01325 bar. Onc otlicr c ~ > ~ t ~ i s ~uscd o n l unil y uf Ilrcssure is a torr, which is the prcssure that supports ;I I .(M)-lu~ii coluirln <>I' mcrcury. Thus 1 iurr : :( 1 1760) attn. Because we arr experiencing ;I tr;11ihitio11period hclwccn the wirlespread use of atm and tnrr on the OIIC hand and the l'uturc u\r ut h;lr and k I 4 OII thc other halid, students of physical chemisiry must bc prolicien~it1 both \el:. or plcacurt' units. The re1ation.s betweell the various units of pressure are collectctl ill 'rahlc 2.1 Of the three quantities, volurne. prcssure, and tcmperature, ternpcr;~turei s the most difficult to cunceptualize. We will present a molecular i~~rcrpretaliorl of ternprature later, but here we will give an operational definition. The fundamental tempcraturc scale is based upon the ideal-gas law, Equationh 2.1. Specifically, wc detine T to he
'
where A is the babe area nf the colunln, p is the dcnsity of the fluid, and h is thc hcr~lli ut- the ctllumn. The gravitatjonal acceleration constant is equal to 9.8067 111 \ .'. Ilr 980.67 cn1.s '. Nntc that the area canccls nut in Equation 2.2.
because all gaqes behave ideally ill the limit u l P + 0. Tile unit of temperature is the kelvin, which is dcnoted K . Note that we do not use a degree synthnl whr.11tl~c temperature is expressed in kelviil. Rccuuse P and t'cannot takc on neeativc v:tIucr. lhe luwest llossible \,al~ieofthe temperature is L! K. 'rrmperatures a
E X A M P L E 2-1
Calculale the p r e w r e cxencd by a 76 0 0 - c r n column of mercury. Take the density (I[ Inercury to bc I? 5% g cnl '.
T A B L E 2.1 Various units lljr expressing prcssure.
'
1 pawal (Pa) = I N . niL' = I kg,mA pascal i s equal to N .rtj-'
or
k p - m - s-', so the prescure in pascals is
1 a~mosphere(atnl) = 1.01325 x 105Pa = I .C113?5 bar
= 101.325kR = 1013.25 lnbnr =
760 turr
1 har = lo5 Pa = 0.1 MPa
, S-'
Chapter 2 1 The Prupertirs [if
(;ASH
2-1.
All Cases Uehavr ldcally If They Arc Sufiic~cntlyD~lule
To establish the unit of kelvin, the triple point of water has been assigned the temperature of 173.16 K. (We will learn about the pnlperties of a "triple point" in Chapter 9. Fur our prescnt purposes, i t is sulficient to know that the triple point of' a substatice corresponds t o an equilibriu~nsystem that c0ntain.c gas, l i c l ~ ~ i d and . wlid.) We now have a definition for 0 K and 273.16 K. A kelvin is thcn defined aa 11273. l h of ttie temperature of the triple point of watcr These definitiuns of 0 K ;ind 271. Ih K generille a linear temperature sci~le. Figure 2.1 plots cxlxrilnental -V versuh T for Ar(g) at dill~rrvrilI)rv**ktrcs. As cxllccted from uur drfitlition of thc tcmperature scale, the extrap1~xatit)i~ (II thrar dutu shows that 1' i O a.s 'V + U. The kelvin scalc i s related t o lhc corlirnorily uwd C'clhi~h~c:IIL. t)y
Figure 2.2 shows PV data plotted against P for several gascs at T = 273.15 K. A l l the data plotted extnpnlale t o PV = 22.414 L.attn as P + 0, where the gases certainly behave idcally. Therefor-2, we can write
We w i l l use the lower caw t Il>r l' it110 ihc (rppcr cast 7' for K . Note lilsrt t h a ~tbc degree symbol ( ' 1 i s associ;itcd with valucs 01. rhc tctnprralure in tllc I'clsius s c ~ l c . Equatic~n2.4 l e l h US ihat O K = 773. IS I.. t)r ttiitt (1 (' = 273.1 5 K . Htxitk~acof thc gcncral use r>l C i r i I;thor;ttur~c\. ii higuiiic;~r~carriuirtit of thcrrlltxlyu;~u~~c data are tabuli~tedfor suhstanccs at 0 C (273.15 K ) and 25'C' (2YH. 15 K):this latlur valuc 1s
whcrr wc hwc uacrl (tic h;rclttiut I h,m' = I N , n i = 1 J. Because of the change of rhc lrtundunl 01 prehhirrc I r t ~ t ruttnosphrrcb ~ ti) hirs, i t is also convenient to know the vuluc ctf H in u n ~ (11, h I..bur. 111t1l I .K I . Ilring ttlc k ~ c thi~t t I atm = 1.01325 har, we
Using the fact that I atm = 1 .O 1325 x 10' Pa and that 1 L = 10
' nl', we have
~ C lhal C
conurnonly called "rtnjrn tcmperaturc." 11' we lneacure at 273.15 K for any gas at a sul~ticienilylow prcswrt. that ils hehawor is ideal, then
PV
22.251 (I
_ I
I
0.2
0.4
0.6
(1.8
I
P / arm T/K FIGURE 2.1 1* Expcrir~lcntalmolar volume!, (urdid lirich) o f Ar(g) an-eplotted as a function of TIK at 0.040 atm, U.020 attn. and 0.010 alm. ,411 three pressures extrapolate to Ihe origin (da.rhc,dlines).
F I G U R E 2.2 A plot of experinlcnpl values of
PF
P for H,(g, tclrj\cc\). Nl4
and CO,(g) (circles) at T = 273.15 K. The dala fur PV = 22.414 L.atrn as P + O (ideal behavior).
(rlia~~lo~~dh),
all thrcc p;i..cr cutr;~lu)latrIIIa value of
T A B L E 2.2 'I'he values of the molar gas constant R in
variuur unit?.
H = 8.3145J rnol I.K.] = 0.083145 drn'.bar,mol-l . K - ' = 83.145 cni'.bar.r~lol-l.K-I = U.082058 L.atm.nlu1-' .K-I = 82.058 cni'.a~rn.rno~-l .K-'
2-2. The van der Waals Equation and the Rcdlich-Kwong Equation Are Examples of Two-Parameter Equations clf Stdtc The idral-gas equation is valid for all gasc5 at sufficiently low pressures. As the presstlrc o r 1 n gjvcn quantity nf gas is ir~creased,however, deviations iinm the ideal-gas e q u a ~ i r ~ n appear 'fl~cccdcviatinr~scan be displayed graphically by plotting P ~ / R as T a fullctio~l of pressli~t,as stiuwri iri Figure 2.3. The qurtntitj P V / R ?is- callcd the cclmn~)ressihility #u(.torand is denoted hy Z . Notc that % = I under all co~ldiiionsfur an ideal gas. For I-calgnccs. % = 1 at Inw pressures, but Jevia1ion.c frurn ideal behaviol.(Z# I ) arc sec11 as the pressure increases. The extent of the deviations from ideal behavior al a given pressure depends upon the tempcraturc and the nature of the gas: The closer the as is to [he point at which ir begins to liqueiy, lhe l a g e r the deviations from ideal behavior wrlf hc. Figure 2.4 shows Z plotted against P for methane at various temperatures. Note that Z dips below unity at lower temperatures but lies above unity at higher telnper2lurt.s. At lower temperature5 the molecules are moving less rapidly, and so
P l bar F I G U R E 2.4
The cnmpressihility factor o f methane versus pressure ;it various temperature\. I h i h 11gurt shuwf that the eltect of molecular attraction bccomes ~ E inlponant S al highcr lclllpl.r:llurr\
are more influenced hy their attractive lurccs. - Because of these attractive forces, thc molecules arc dri1~11 topcther, rhus i~lakingVmallcss than Vidri,,,which in turn causes Z to t x less than unity. A si~nilareffect can be seen in Figure 2.3: (he oirlcr ol. the curves sl~owsthar the ctfecl of molcculi~rattractions are in the order CH, > N, > He at 300 K. At higher letnperaturea, the niolccules arc moving rapidly enough thar their attraction is much s~nallerthan k,T (which wc will see ill Chapter 4 is a measure of their tbcrmal energy). The n~nleculesare influenced primarily hy their repulsive I-orces I. at higher temperatures. which lend to make I '-, 1',,lt4,1, and so Z Our picture of an ideal gas views the molecules ;I.; rnr~vi~lg it~dcpenderlllyt~l~c;lch other. not experiencing any intermolecular intrrnu~ion.;,P'igurcc 2.3 and 2.4 allow rhnt this picrure fails a1 high pressures, and thar the altlacrivc ;trld rcpi~l\ivcintermulecular interactions tnust be taken into account. Many equation\ c.xtentl the idcrtl-gas equi1lio11 to account for the intermolecular intcl-actions. 1'crh;lpc thc most well known is the vat? der Wi~alscquntiun,
Ideal gas
?OI1
U
3011
I'
M 1101) 800 I000 /11;1r
2.3 11plot of P ~ / R Tversus I' fur one mole of hcliurii, t\ilrt>pen,and methane at 30U K.n i s tifurc shows thnl the ideal-gas equation. fur w h ~ c hI ' V / H ~=' 1 , is "(71 ralid at high pressure. FIGURE
p
where designates moiar vt,lulllr. N r l ~ i cIII;SI ~ l:quutir,n 2.5 reduces to the idealgas equation when 7 i~ large. i~ Inuhr. Thc con.clants cr and h it1 Equatin~l2.5 are called vrm der 1N1ul.r c ~ o t ~ . r ; l t ~ l ~whtrcu .\-, vi~lucsdepend upon the particular gas (Table 2.3). We will see in Suction 2-7 thai the value u l a reflecls how stnlngly the mol~culcsof' u $35 altr;wl C:ICII other and the value of h reflects the size of the n~oleoules. Lct's use F.quation 2.5 10 calculate the pressure (in bars) cxened by I .(XI mol of CH,(g) thxt occtlpicc ;I 250-mL cuntai~lerat O"C. From Table 2.3, we find thal
;,\
.
57
2 - 2 I l l e van dcr r;\/,l,ll> Lquat~unand the Redlich-Kwrlg Lquat~on TABLE
2.3
At h ~ g hprc\surcs, thc lirst tcrm in Equation 2.6 dorninatcs because b l ~ l a l lilnd , ut low prebhurcs the sucond tcrm dtllninntes.
Fall der Waals constanlu for vanou\ substance!. Species
a/dmh.bar-~nol-'
a / d n ~ ~ ~ a t r n - m o l - ~b/dln7.mul
Heliur~~
0.U3.1598
0.034145
0.023733
Neon Arg011
0.21 666 1.3483
CI.21382 1.3307
0.017383 0.U3 1x30
Krypton Hydrogcrr Nirrngcn
2.2836
2.2537
0.24646 I ,3661
0.24324 1.3483
(3.038650 0.0266hS
1.3820 1.3734
1.3639
Oxygen
Carbon ~nonoxide
C'arbori dirjxide Arn~nut~is hlerhane Ethane Ethene Prrjpane Buhne 2-Methyl propane
Pentune Benzene
3.6551
1.4541
4-6112
3.6073 42181 2.2725 5.5088 4.5509
4.3919
9.2601
4.3114-1
2.3026
5.5Kl8
I' - b hectlrr~es
13 888
13.706
13.328 19.124 18.876
13.153 18.874
18.629
EXAMPLE 2-2 IJw ~ h van r tkr Wuith C~~U;III<)II
ttj
c;~lcul;ttc [he molar vt~lumeoi clhn~leilt 300 K a113
0.038577 0.03 1860
0 039523 0.0428 16 0.037847 U.043057 0.064143 0.058 199 U.(W0494
Ihe Ncwum Ihphwm Wrl pwe ur
0.11641 0.1 16.15 0.14510 0.1 1974
WJKW V, LAW.
-
bw units Ialr r,llnrrlllcllur. 'fhc ~tlcitlgas w l u t of V i c w b w R T / P 0,1211. d I , ul k1.1uw 0 111 I r11o1 ' a!, our inilia1 gutas. In {hi!. wc & h a i n 3
a = 2.3026 dm6bar-mu].' and b = 0.043067 d n ~ ~ . m u lfor - ' melhanc. If we divide Equation 2.5 by V - h and solve for P , we ohtain 1 IK r n ( r t t r h r r b t r I vrlw br 1) (11I 1. IIIUI~ I '1'11~C ; I ~ C U ~ ~ I ~ Oof I ~ prehburc preceding this rnilrnihr pawl ttw Imk t t l ~ 4 4~ ~~h c~vi1111r11r t ~ In th~xenumpk: show [hat the van der Wads rtlttunalrl, u IIIIF1 1 u ~111 r IUIMIC thi~ri ttlc ~tkal-gakequation, is nut panicul;uIy acctirnle. U r u 111 ~,.LII > l r u ~ t l y~ h I$KIC ~ t #trcI T ~ ~ ~i lI~C~ u r aequiltl~ll'i le of SMC.
\
I By comparisnn, the ideal-gas equation predicts that P = 90.8 bar. The prediction o f the van der Wdals equation is in much better agreement with the exprirnental value of 78.6 bar than is the ideal-gas equiltion. The vim dcr Waals equation qualitatively gives the hehavior .chowri it1 Figures 2.3 ntld 2.4. We can rewritc Eqni~tion2.5 i n thc form
Iu{, ~ ~ t t l ~r.l,rt~rcly rr ~111qilr t+(IuUllllnh 01' state thiit are ~ n u c hrtlorc accurate and ~I-.~.I~II 1t1~1r1 ~ I I C \ it11 LII'I, W:I,I~'~ cquiltio~lare the Kcdliclr-k'wor~g~qrrution
t 1 t . 1 1 ~r 1tit11r
I'
K 'I'
r/ p
- -
u
v(vtp)+flcV-B)
58
2-2. Thc rCandcr LVFqllnlinn and the Rcdl~ch-KflungEq~f,ilir)ri T A B L E 2.4
S O L U T I O N : SubstituteT = 300K,P = 20Uatrn. A = 97.539drn6. a t m mol-I, K'," and 8 = 0.045153 drn7.rr~ol-lillto Equation 2.9. r r l (>blain
'I'hc Rcdlich-Kwong equation parameters for various substances. ~/dm"har.mth 2,K1!2
Spec ics
~ / d r n ~ , a t m o l - ' , ~ ~ : ~R/dln'.lntrl-'
--A.
I Irliurn Neon Argc~r~ Krqpturi tlydro~r~~ Nilrngc~~
0hygfl1 Carhnn monoxidc ritrhrln rlic>xidc An~rnoni;~
lfethrri~e Etl~anc Elhcne Prr>p;~~~c Hutnnc 2-Meihy l propane
Penta~~e Urnzcnc
0.079905 1.363 1 lh.7Rh 13.S7h 1.4333 15.551 17411 17.21)8 64.597 87.808 32.205 9K.Rl l 78512 183.02 240.1 h 272.73 419.97 453.32
0.078860 1.4439 16.566 33.137 1.4145 IS 348 17.183
16.983 63.752 86.660 31.784 97.539 77.486
180.63 286.37 269.17
414.48 447.39
0.0 I 0450
0.01 21WU U.02201>2 O.OZh780 0.0 18482 0.026738 0.022082 0.027394
0.029677 0.026232 0.029850 0.WS153 0.040339 0.U62723 0.08I)h8 0.0807 15 0.10057 0.082996
where we have supprc~sedtllc u-~ ~ ifor t s convenience Sulving thir crluation by thc Newtun-Raphson method gives I' = 1) 075U dlr~'.lnol I. cr,rnp:~rcd nit11 the rail d ~ , r W a a l ~I-erultof V = 0.04h dn~'.mol-l and the experimental recult of 0.07 1 d l n l -r11r)l ' (see Example 2-2). The prediction of the Redlicti-Kwtlng cquation i\ ne:~rlyquantitnlivc, ulllike the van der Waals equotlci~~. u'h~cllpredicts s value (IT th:!t 1s alwut ?Or, luo Iargc.
1:ipurc 2.3 colnpares experimental pressurc vcrsus density d a ~ for a ethane at 400 K with t h r predictions of the various equations uf state it~trnduccdin this chilplcr. Note that the Kcdlich-Kwong and Peng-Robinson equations are nearly yuantitntivc. whcreas [tic von d c r Waals equation fails colnpletely a1 pressures greatcr than 2(H) bar. One of t l ~ ci r ~ ~ l l ~ c \ \ ifeatures vc of the Redlich-Kwnng a l ~ dPcng-Robinson equation'; i \ Ihat ~ t ~ itrc c y 11ci1rlyquantitative in regions uhcrc the gas liquefies. For ex;~rnple.Figure 2.6 rlli,w\ r~ri.h\urc vcrhtts dcnsity datn f ( ~ethane r at 305.33 R. whcre it liquefies at around 4 0 h;~:. 'l'ltch t ~ r i ~ t t n t region al in the hgurc rcprcscnts liquid and vapor in cquilihriu~n will1 P I I C ~rrlhcr. Noic th;~t the I'el~g-Rnhinwn equatiorl is better in the l i q ~ ~ i d - v a p o r rcgirlri 1w1 that lhc KcdIich-Kwor~gequatior~is better at high pressuEs. The van dcr Wuuls cquotirtn ir not shown hcc;~usci t gives negative values of the pressure under thew cc~t~diliiuis.
where A. R , ct, and b. itre prtlamelers Ihi11 depend upon the gas. The vi~luesnf A and R i n Ihe Rcdlich-Kwong equation are listed in Table 2.4 for a variety of suhstances. 'I'11epi~rillnctclLY it1 the Peng-Robinson equation i s a sumrrvhat complicated function of lernpcraturc, F(I ~ t b:i l l not t a b u l a ~ evtllucs of a and 8. Equations 2.7 end 2.8. like the van der Waals equaliun ( E x a ~ r ~ p2-2), l e can be w r i ~ t e nas cubic equations in V .F o r exitmple, the KedIich-Kwong cqtin1iun hecomes (Prublern 2-26)
Problem 2-28 has you show that the Pcng-Robinson equation ol' statc is also a cubic equation in 7.
EXAMPLE
2-3
U s c thc Redlich-Kuo~~g equation to cillculute the molar volume of etl~arieat 300 K and 100 ntni.
t l ~ . ~ I U2.5 l
tixywrn~~c~~l;~l pic,w~revcrsus density data for ethane at 400 K (sol~dlinc) is compared r i t h Iht. cri the van der Waals equation (dot-dashed line), the Redlich-Kwong equation (long prctiuc~~c~~i\ tie\l~crIIinc). :~ritithe Pcng-Robinsan equation (short dashed line).
CICURE 2.6 Thc txperimen~alpreshure versus den\ity data (solid I~ne)for ethane at 305.33 K IS compared wilt1 he prediction!. of the Redlich-Kwong equation (long dashed line) and the Peng-Robinson equation (short dashed line). 'The liquid and vapor phascs are In equllibnurn in the hori~ontal
,
!I
region.
0 !
l o /l . . l b l ' b l
Although Figures 1.5 and 2.6 show compitri~unsclnly for ethane, the conclusions as to the relative accuracies of the equations art. genernl. In gcncral, the Rcdlich-Kwong cqilation is superior at high preasulrs, whcrcas the Peng-Kobinson equation is superior i n the liquid-vapor rcgion. In fact, these two equ;ltions of stale have been '-con~trucred" so that this is SO. There i r e mtlre sophisticated cquatiol~so f state (sorne containing Inore lhan 10 pnrnmeterh!) that call reproduce the experiniental dat;~tu a high degree o f accuracy over a large range o C pressure, den~ity,and tcniprature.
2-3. A Cubic Equation of State Can Describe Both the Gaseous
..
L ---J
li~pcrilntntnipressure-volume isc~thermsof curtkln d~un~dc around i t b critical tempcmture. MI V) ( ' I'O~III> (;. A, 0, and L are discussed $11 11ict c x ~
lincs in Figure ? 7 15 unllnl the rht#~Drmtr c,un.r. trrcuuw ally p o i ~ l within t this curvc comspc~ndsto InlrilJ unrl p n ct*rln~lnb in r q i ~ ~ l i h r i u with r n ciucl~other, At any pairit o n or o~ilcitlcthi\ ctbrvr, cmly t n e p h w 1% prcwnr I:trr ca:ltr~l~lc. ;it lwlir~t(i 111tJ1~ligtiw wc havc {rldy i t p,rr plhnr I[we lum kfurt ul ( i :illti crllliltlc\\ thr, G;I\ iltorlg llrc I.t.? C' 1hoth~1.111. l ~ t l l t l t wall l hrnt u1rpc.r w l ~ r l wc l ICIKIII ~ L Ch r ~ r ~ / ( ~ r ~ 11111t ; t l;!I 1njjri1 A . 'l'hr plccclil,c w i l l rclt1r1111t,rtnncmr IU wc r.ta~lrnrctlic Eikr 111 l u o l ; ~vr>li~rnc ~ 0. 1 1 ,111u1 (poilil A ) i t r Ilittutl rbl 111u1nt VIIIIIIIH.r l l ~ ~ p p r l x ~ l ~ 1107 h i ~ lI~ t111u1 y ( ~ n ) i t lI)). l Ahtr
'
rt.;~clllrlglrrllil 1). It& ~ncnburcr m r c u r r rlhiirllly w ~ t ;li ~ lunhcr dv~rc;i*c 111 vo~tltlle. t)cc.lu\c wc Iltrw llrlvr all Ilqutd UIHL I ~ vtrlt~rt~c F q d ;t I I ~ I~'1l.lrlg~'h ~ I ~ VCI-y littlc wllll I11I'CIIIII'
and Liquid States
;I-,ihr 1~111lk'rututc ~ ~ l i ~ r r qowirrrl : ~ u ' ~t h L,I ~ 11lc;tl tc~ul~ur;ttill.r.lhe horiIIIIC'I L ~ ~ M P~Hl i rdl i h a p p r ut the c v i l i ~ : ~ t~ I f l l l x r ~ t i At l r ~Ihlh , point, the nletlls-
Nirlr ih,it lrllllill
A reniark~blefeature of equntions o f state that can be written as cubic equalions i n V is that they describe both the gaseous arld the liquid rcgions o f a substance. To understand this fcaturc, wc start by discussing some exprimentally determined pluts of- P as i t fur~ctlunut V at constdnt T , which are cotnlr~onlycalled isotlrrrms (i.50 = cunstant). Figure 2.7 .chows cxl?crimentnI P versus V isotherms fbr carbon dinxidc. The isother111sshow11are iri the neighborhtxxl o f the critical tcnq>craturc, 'I, which 1s the ternperiturt: ahove which a ga5 cannot be liquefied, regardless of the pressulr. Thc critical prcssure. t', and the critical vulurne. are the corresponding prc.c.culc and thc molar volurpe ;I( the r,rilic,~rl pnir~r.Fur exrtmplc, for carlmn dioxide, 1: = 31M.1 1K {3O.YY C), q = 72.9 iltln, and = 0.094 L..lnol I. Notc t t ~ a t~ h r i.sotherlns in Figurc 2.7 flattrrl 0111 SI; 7' + TLtrum above and that tllcle arc h~rri70111ul regions when T is less than Tc. In the horizontal rcgions, gas and liquid coexirt 11) ecluilihrium with each other. The daahcd curve connecting the ends uf the h o r i ~ o l i ~ d l
8.
vL
lr~urrli ~ l t cIiqiritl illnl 11- vnprr tll\;l~~r;irs:tnd thur,c i.\ 110 ~lihtinctionbctwcun Iltluld ;tilt1 p.1~. ltw rut lrti,c tclirltbll dl~ill)lsilrh,1110 t JIG p ? r ;i11i1 Ilquld phases both l ~ ; t \ r i l r r w ~ r ~ t~cc t l u c ill 1 tlrthtly Wi. u 111 t l ~ r ~ ,~ i~ l i \cc~ \ 11ic;tl lx!irll i~rll)r)rc det;til in t 'I1.111lcr0 I +yurt .' H \ ~ ~ I H .\11111l;tt 'I I . ~ ( I ~ ~ C I I T \h~ \ thc' Y~III dcr Wad\ cqulirio~liltld the RedlichKuilli~ rqurhllrlrt Nd~tbrci1t:~i the twi) quiltions of state give fairly similar plots. The ~MIIIOLI\ hurpr ~ l ~ i : u i ~ttw c t l1' -r 1; rcsult from the approximate naturc of these cquat l t r l l c r r l ~ t l l t ct.apilrc1 2.') ?rhi)w~ a single van der Waals or Redlich-Kwurlg isotherm lor I - 1 : . '1-tic curvc. ( ; A l l ia the curve that would bc observed cxpenmer~tallyupon ~~crlillirca\lng the g i b . The hurizontal line D A is drawn so that the iilras of ihe Ioc)p hrlth ;ihcivt and below DA are equal. (This so-called .Moxwrll equrrl-urcrt
>-'I
A
(
ulj~rF<(u,>titm of S~dtcClili 111,511 rilw Roth lhr (
: l r l ~2nd ~ (Liquid ~ ~ ~51,1t('~ ~
is consistent with the fact that the van der Waals equation call bc wrilten as a cubic polynomial in the (tnolar) volunle (see Exalnple 2-2). The volume corrrspondi~~g to point D is the tnolar volurrje o f thc liquid, the volume corresponding to p i n t A is the molar vulurne of the vi~pori r cquilihriur~~ ~ with the liquid. and the third root, lying between A and D i s spurinur. At 142.69 K and 35.UO Gtln. argon exists ;IS two phases ill cquilihri~lrnwith each (111111' : i i r l c 1 thc dcnsitiw 01. thc liquid ;rnd v;IIwr ~ ~ : I S L I S ;II-c 22,491 rnol.L and 5.29 1 tllul.I . I , respectively. 1,ct's sec wh;it thc van dcr W,I:I~~cyuntion predicih ill this case. As we saw i n Exarrjple 7-2, wc can write ~ l j cvat1 dcr W,r;ils equation as (1 (I
0 2
V/
L. rnol
0.4
'
0.0
0.2 0.4 V / t.+~nv~-'
I
FIGURE 2.8 Prcrsure-vnlunle ihutherms uf carhun dioxide ar-oundits critical temperamre, au calculated from (a) the V~UI drr Waals cquation (Equaliol~2.5) and (b) thc Redlich-Kwong equation (Equalion 2.7).
U s i t ~ gthc values of ri and h irorrl Table F4~1ati1,n2.10 becorncs
7.3. T = 142.69 K. and P = 35.(XI atm.
whcn.. for ~(~nveniencc. we have supresscd the units of the coefliciellt~.Thr three nwtr ot tlli* cyuation arc (Problem 2-22) 0.07073 l . . r n t ~ l - ~0.07897 , L,rnr)l-I. and 11.2 l h7 I.-11101 I . The smallest nlot represents the molar volumc of liquid argon, and the I ; ~ r g r ~II~I\OIIL.LIII:I~~~I~.C ;ire tiii~,ly nccul-ale. and the Pcng-Robitlsot~cquation i s about IU'X r ~ i { t ;~L,L,I~I;+IC r~ 111 this Iiqtrid region 'I'ln. ~n'irit('.I3,i t 1 I.'ipurc 2.7 is the critical poirli, whcrc T = TL. P = ft.and IIlr IHIIIII ( ' l',i\ :111 i ~ ~ l l ~ poir~t. ~ t i o:iltd ~ l 50 1' ,
'
-',
'
FIGURE 2.9 A typical van der Wmls pressure-volume isotherm at a telnlnpcralure less than the critical lernpcrature. The I~ori/.ontalliric has bee11drawn so that areas of the loop above and bclow are equal.
Ihc coexisting vnpur and the point I) represcllts the liquid. The line DL represents the change of volume of the liquid with increasing pressure. The steepncss o f [his line results from lllc relaiivc incomprcssibili~yof the liquid. The segmenl AR is n metastable region correspo~~dirlg to the supel-hcnted vapor, and the segment CD curresponds to the supel-cooled l i q u i d . _ n e scgtrlent I3C i s a region in which ( ; ) P / ~ T > )0., This conriitiorl signifies an unrtabfe ~ g i n t which ~, i s not observed for equilihrium syxtelns. Figure 2.9 shaws that wc can obtain three values of the volumc along the line DA 1111. a given pressure if the temperature is Icss than the criticnl telnpcrature. This result
Wr g .In u\r 1I)r.cr I w t c ~ ~ ~ r ~ t l i t v10> ri l \t . ~ { ~ r t ~ Ihr ~ i t l~.ritic:tl c conciants in terms o f a and b ( l ' r ~ ~ h l r ?t t ~? ? ) ,411rnhrcr HI*). to (11% f111\. I I O WI\~(I)Vwrite L-~ the. van der Wxals 1 . I!). tqil.ltla>rr :I\ ;r c\rtbr~rtlu.ltlurl In 1 ' , I.
Ilcrl~p ;I c ~ i h c r~ i l \ l . ~ t ~ u11~h;$\ l . tttrcc rotrth. For 7' :7;. only one of these roots is real (the o ~ h cIWO ~ ;KC.c c l t ~ ~ ~ l c;arjd n ] .fur 7' ,. 7; ; ~ n dP % all three rnots are red. A t T = thc\c 1llrt.c. IIHI~, Iilrrgc info orlc. ;uld c t we ~ can writc Equation 2.10 a5 (7- Vc}*= 0, or
q,
q,
64
Chapter L / Tlw I'rul,crtics of GasT A B L E 2.5
If wc colnpare this equation with Equation 2.10 at the critical point. we have
The expcri~nentalcrilicul constants of variou, subsranceh. .-
3Vk = h
Eliminate
HT
+2
9 '
-2
3Vc
a = -,
9.
and
-I
V =
9
ab
(2.12)
-
q betwcen thc second two uf these to ohtiiin Helium
and then substitute this result into the third of Equations 2.12 t o obtain p =
U
(?.13h)
27b2
and last. substitute Equations 2.13r1 and 2.13b into the first of Equaiiuas ? I !ta~~rhluin
5.195U
Neon Argon Krypton Hydmgen
{ha in a
150.95
49.288
48.h33
U.07.530
210.55
56.61 8
O.U9?20 0.06500 O.WO10
(1.03 110
126.20
34.000
Oxygen
154.58
50.427
50.768
0.07hJO
( ' i ~ r h ort~onclxidc ~~
132.85 4 1h.(l
34.935
31.478
79.9 1
78.87
0.093 1U 0.1237
72.877
o.~l(wi
Nitrogen
--
U I U I ~
Iclbrr l-~lrn
~~
The Iolluwing example shows III;II Ihr. V;II~ (lcr W;I;I~\ rtltlitf lull i111dt h KC~II~CII-KU.~)II~ ~ equation make iln interccling pretl~c,l~c!~l ;rhoui t l ~ vv;tlut of I:-v*/ H 7 : .
11 (15780
55.878 12.670 33.555
cwbun ~ {pi
2.2452 26.208
32.938
Ihlrwl~tc The critical constants of a nu~nbcrof suhstancca are givc.11In '1'11t~Ic 25 The values of the crilical constanis in terrw of ttic I)uraltlclcr\ A v ~ Hd Redlich-Kwong equation can he deterrn~ncdIn ;I\IIIII~.~I I~C~IOII ' f h I~~IYI)IC~IH.~' hit morc invnlved. and thc rcsults arc ( P r o h I c ~2t ~ 1 7 )
2.2750
26.555
44.415
1% Il41Irw !M r l h y l ~ l p r r l'r~ilnrr II'.ll/rlu
12.838
3 r ~14 ,
73.843
M7 126
220 55 II I rl 4 5 UHll
41s.lo luo.l3 M8.W
#t. JJ W.U. 42s. I& I$ Mb9 .MI 79
2 17 hh O I n4 45 3 7 0
0.05595 o.071-st1 11 IFltl0()
48 '114 4
48 1177
I1 I4HO
ru 761
!I 1?'n1
4t.rrt 17.w
41 u21
7
1121 ul (1 lwl
Yllla JlWJ 79n
1g024
02hU1
3J2(11
LIl(M0 t j 251~1
4n I31
EXAMPLE 2 4 CaIc~llarcthc r:itiu i f LJL/KYLtur Ihc van tlcr Wdals equariun anti the Kcdlich-Kwong equation.
5 0 L U T I ON: Multiplying Equation 2.13b by 2.13u and dividing by R t ~ o n2.130 gives
lime:,
Equa-
Srlnilarly, the Redlich-Kwong equatic~ngives
1
I v
Equations 2.15 and 2.2 predict that R7: should bc the same value for all substances but that the numerical valuer, dil't'er slightly f i x the two approxinlaie equalions uf &talc. The cxptrimenlal values of give11 in Table 2.5 show tirat neither
!~T/R<
cvc/
t r l ~IYIP 1% qumtltuive. T k corrrspading value for RT for the PcngUI~IIMII P I J U ~ I L(1 ~ ~O,M)1)740(PrrMcm 2-28), which is closer to the experi~~~ental \,lllbr\ I h u t ~c~thcr181 IIW vdt1c* sivtn by the van der Waals equation or the RdlicliKHINI)~ V~I~PII~MI NIHP.IIIWCYC~, ~ t i i all ~ t threc cquatiuns of state do predict a curlstitnt \ ,111hr l41t 1: V4/ I I T . II~LII Ihc cxlwrirr~ent;~l data i n Table 2.5 (how that this valuc is ~ i ~ t l c rfruly tt ctrllntullt 'flliq ot)wrvation is an example ot' thc iaw oT corresponding \t.ttrTr, W(~IL h r a p 11u1l lttu ~)nqxrlicsof all gases are the same if n c compare t l ~ e ~ n i~lltlct~ h r;attw r t u ~ l i l ~ t l u lrl~r l a t i v eto their critical point. \VC w i l l discus\ the lib& of t , t r ~ i v rrr~iltnw l~ r l ; h l C ~ Innlc lhc~roughlyi n the next section. Allh<~trph ut. hlnc writtcn and I n terms 01-o and b in Equarior~s2.13 rlr i n I~.I In\ (11 A + l r d H I!) l i r l u ; ~ ~ ~ o2.14, n s i n practice these constitnls arc usually evaluated i n ~ c r ~ trt n \r ~ l w r t r ~ ~ c crilical r ~ l ; ~ l constants. Because ihere arc three critical conataIits and clnl y t * ~ )r~ull~i;lIHc I t ) r citch equation of stdte, therc is honle ambiguity i r l doinp b o . For c\,r~nl>lt+. wc ctlulti u\c t~.quntiuns2.13a and 2. 1.38 to evaluatc ' I and h 111tcrnic 01 C~ILIIIO~~
vL,<,
<
h6
Chaplrr Z i Ihe I1rr>perties rrf Case
and $, crr use another pair of equatic~nq.Because we ilsc Equatiol~s2.13h and 2.131- to obtain
27(~1;)* u=-.and 64P
and
6=
are known more accurately,
R7; -Re
L e t ' ~.stalt with the van der Waals cquittion, which wc can write i n an il~teres~irlg and practical fonn by suhstituting t11r \rct)l~d{I!. F;qu;~tiori:, 2.12 l'i)r a and Equalion 1 . 1 3 ~ 1 forb inti) Erluatioo 2.5:
I,ikewisc, frrlnl Equatiuns 2.14. rve obtain t l ~ cRedlich-Kwnp cor~stontc. R?,~r:?
A -. U.32748----
and
B = n.08hMO
R 7:
P,
-
2 4 . The van der W a d s Equation and the Redlich-Kwong Equation Obey thc Law o f Corrcsponrling States
(2.18)
I:
Thc vari del- Waals and Redlich-Kwollg constatlts in Tahlr:. 2..1
;III~2.4
h;lvr hcen Divide thrnugh by I: and
obtained in 1hi.c w:~y.
E X A M P L E 2-5 l l ~ ct11c critical-cr)nstant dala in Table 2.5 to evaltiatc Ihe im;ltltltr W,I;II~clrrlslatlta fnr
I.:to gct -
where we havc used Equation 2.15 for I: V , . Nrnv IIIII~MIII~L' ttlc n'iftrr.c,rfqldrlt11itir.t P, = P/?, = T / y , and ?, = T / T to cjhl:~irilllc V;I dcr W;l;~lscquiltic~r~ wrillen in terms nf' rcduced quantities:
ethane.
vR
'I
27(0.083 h.~r.lllul K ')'(.!(l5.34 K]' =-- 145 din' --tA(18.7 14 bar)
and
Equation 2.19 i< retrlarkable in that thcrc ;Ire IN) IIU;III~~~IL'\ i n thi\ r(111;hllon ~IIJI ul' any pa~ticulnrgas; 11 is ;I ullivct\;tl cil\l;lliol~101 (111 g;iccs. I t says, Ibr example, that the value o f 1; will hi. rhc c;i111c ti11 ill1 F;lrr< ;I! Illc s;lrllc v ~ l u e sof C', atld T,. k t ' s ct~nsiderCO,(g) ;IIILIN , ( p ~t'ur I', = 71) ;III~1, = 1.5. According to Equation 3.19, PR = 0.196 wherl i;,- 11)1) :ind 7; = 1.5. Usirlg thc values of the cri~icalconstants given i r i Tahlc 2 5 , wc t i t i t i tll;~l lllr rcduccci clu;jrl~i~ies P, = O.lgh, V, = 20.0, and T, = 1.5 corrccpo~it! 10 /: ,, 11 3 ;a1111 = 14.5 bar, are characteristic
(I).OK3145 cllri' bar-rnol I K - j(305.34 K ) h= R(48.714h.~r) = O.Ob51.W
dm',111ol-I
-
V = 1.9 L.nl(rl ', and ?,,, = 45h K anti I,, I',, = 0.M .illn : h.hh hdr. I'N: = 1.8 J,.lnol I, and T,? = 189 K. 'These two gasc\ 111\(1crt h r w crl~ltlrtionrare xilid 10 bc at correspur~dingstates (same values (jt- PR, : i l l t i 7; ). Accordil~gt o the van der Waals equation, thcsc quantities are relatcd by t:cjual~r>r~ 2.19. \(I Hquatir~n2.19 is an example of the Inw nfcnrrespc~ndingstates, that ;rll p;iaes have the same - properties if they we compared at corresponding condition\ (salnc viiluibs r ~ l ' P,. V,, and T,)
,,,!
E X A M P L E 2-6 Usc t l ~ ccritica1-constan1data i n Table 2.5 10 evnluate A and 8 , the Redlich-Knong const,lllts Tor ethane SOLUTION:
(U.UX31.15 drrh'-har,~nol .K-')'(305.34 K)':' ----48.714 bar = 98.831 dtnh-hnr.111ol - K ' ' = ~ 97.539 dln6.a1Tn n l ~ l - ' . ~ ' ! ~
I
v,.
A = 0.42748 -
(0.083 145 dm7,bar-mol- -K-')(305.34 K) B = 0.OAh640--. 48.714 bar
E X A M P L E 2-7 Express the Kedlich-Kworlgequation i n tsrrns u i reduced qual~lilirq.
S(1I I J T I O N : Equalions 2.18 show that
Chapter 2 ! The Propert~rGr ~Gases f
Substituting these fquivalencies inlo Equation 2.7 givcs
r l i v i h the nurneratrjr and thc demn~inatorof the first term on the right bide by the second by to get
vL2
i/t and
14. The v.m rler bVaal;ialsEqudllnll and the Hedl~ch-KwongEquat~onObcy thr I .w of (:urreqiunrIing 5 t d t h
Equations 2.10 arid 2.2 I express Z as a universal function of 7, and T,, or o f any other two reduced quantities. such as P, and T,. Althc~ughthese equalions can he used to illustrdte the law ot corresponding states, they are based on approxiinate eq~~ntions oi' state. Nzvenhcless, the law US corresponding slates i s vlilid for a grcut variety of gi15e.c. Figure 2.10 shows rxpcrimzntal data for Z plotted against P, at u a r i o ~ ~vnlues s of I , for 10 gases. Note that the data fur all 10 gnses f i l l OII the came cli~vcs,thus i l l u s t m t ~ n ~ the law of corresponding states in a more g r i ~ c rway ~ l that1 either Equatiot~2.20 or 2.2 1. Much more extensive graphs are available, particularly in the engineering litcralurr, and are of great use in pmctiual applications.
-
Divide both sides by {,nl~duce the Fact that 1: V L / R 1 ; = 1/3 in the cccond tcrm tr, gel
Finally, multiply and divide the numeralor of the first lerm on the righl side by Tc to obtain
Thus, we hee that the Redlich-Kwong equdtion also obegs a law of corresponding
states.
Thc cr>mpressibility factur, Z , asswiatzd with the van der Waals equation also obeys the law of corresponding states. To dclnt~nslratzthis point, we start with Equation 2.6 and substitute the second of Equations 7.12 fur a and Equation 2.13b for b to get
pm , d,'
x
Nbtrogen
Melhane
(I,?
Now use Equation 2.15 for pLVL i n the secnntl tcrm and introduce reduced variables to
.r,= 1 .ou
n-Bulurle
lrnpcnlnnz
Ethylcnc
A
rr-Hcptarle Cerhun dioxide
Propane
m
Water
a Ethanc
i.
o s c
get
Similarly, the cunipressibiljty lilctor for the R~dlich-Kwongequation is {Prt~hlern 2-30)
F I G U R E 2.10
,
AII illustration of the law of corresponding stale$. The compressibility faclur, Z, is plcltled ugdinsr the reduced pressure, P R ,uf each of the 10 indicated gahes. Each curve rcpresenls a ~ I V C I reduced I tcmperatuw. Note that fur a given reduced renlperaturc, aH 10 gasec KaI1 on Ihc aamc curve because reduced quantitiec are uscd.
f,q
Chapter 2 E X A M P L E 2-8 l i s e bigure 2.10 to cstirrlalc lhc ~rlcil:~rvr~lunlcof' :lrnlilcrni;l
1 Thc Properties of Caw5
1 ,IT? 15 ( ' :tr\tl
4(A) har
5 0 L U T I 0 N: Ucing [he critici~l-con%~:~~~i d:ita 111 7bhlc 2 3, wc l i r t t l Ihat 'I; = 1.20 T h e rmllu and P, = 3.59. F i g ~ ~2r c10 chilwx thar 7 ?- 0 . M ) ur~tlcrtllcrc ~,crntlil~rmr volu~neis
Tl~cIilw o t ' c t > r r ~ * s l > c ~st;lttns ~ l d ~ II~IS ~ ~ g;I I~~c'c. ~)hy\iculiliturprvtatio~~. Ally tcnlpcrii1ui-c sc;~lc wc urc to dc\crjhc ;I F;IS i\ ~ i ~ c c c w r iarbitrary. ly Even the Kelvin .cuale, with its futlJarllc~~t;~l fcro 1e111pcra1ul-c, is arl~itraryin the sense that the size of n degrcc 011the KClvi~1 scale is arhilrary. 'I'hus, the numerical value we assign to the temperature is rnca~linglcsras far as the gas i~ uuncemed. A gas docs "ki~ow"its ctidual temperaturc, and theretnre i s '-itware" u f its temperaturc rrlurivr to its cnliual temperature or its rcduced relrlperature, T, = T / ?;. Similarly, pressure and vulunie scales arc ilnposed hg u ~ but , the reduced pressure and the reduced volume arc quantities t h a ~art. uf significance to a particular gas. Thus, any gas that has a certain reduced temperature, pressure, and volume w i l l behave i n the sanic mallner as another gas under the same conditions.
2-5. Secorld Virial Coefficients Can Be Used to Determine
Intermolecular Potentials Thc most fundatnel~taleqt~ationo f state, i n thc sense that i t has h e most xou~ldtheorctical foundation, i s the virinl equurion clj'srrrrr.The virial equ~tionof state expresses the compressibility factor as a pulynomial i n 1
/v:
The coefficients in this expression art: functirms of temperature only and arc called vrrrrxl corb/"fit.imrs.In part~vular,B,,(T) ir called the second virinl ro~@riclrtt,B,,(T) thc t h l ~ d and , so nn We w i l l see later that other propenleh such as energy and entropy can he exprcwcd as polyriorrli:~ls i n I/V.and generally these reIalions arc called viriul rbxpunhion.\. We can also cxpress the cornpress~b~Ilty factor as a polynomial in
P
2 5.
5t.t r b r ~ r l ' i r ~ a l Cric,ft~cienrs Can
Bc I!rcrl to nctr.rm~nt*Ir~lt.rrr~r,lrtu l a r t'rller~ri;ls
F ~ i ~ a t i o2.23 l i i s also caIlcd a virial expansion or a virial equation of state. The virial cc~.fficicnlsH,,, (7') ;ind R , , , ( T ) :rrc wlijtcd by (Prnblc~n2-36 )
Nutr ill llq(~;~!~nib 2.22 or 2.2 1 Ih:r~ % - + I SI; 7 becomes large or as P becomes LIII~~II, jirrt 11% 11 shrrultl. 'luhlc 2.h gives ;in idc;~of the magnitudes oi the terms i r l t$uatii~li 2,2? UA II IUIK.I~IL 01prthh~ln.I'or ;hrgrlii it! 25 C. Nrjticc that even at lo() b i ~ r ~ip thr lirmt lhrtc lrrnlr nre sunicicnl f i r c ~ ~ l c u l u t i %,
I A A l t 2.6 'l'l~c~ , t ~ t ~ I r i h i01'~ tI ~~aC!~I+I ~ r ~tcw Icrril\ III t l ~ ck~rialtxpdnhlnn c>f1,tiquatiorl 2.22, for argon a1 23 C'.
I + B'','T' . =_..+ I'
ZI1y\T'+ rcrnaining terms
-
v-
'I'hc sccnnd virial coefficient is the most important virial cocfticient heciiuse it reilccts the first deviation from idcality as the pressure oT the gab is itlcrcascd (or the volurijc is dccrt.;tscdj. As such, i t i s the most easily ~lieasurctlvirial crxffic~entand i s wcll lahulatrd I-or many giihus. Auvtwtling to f.qu:ltion 2.23, il ciin be deterlr~ined c x p c r i ~ ~ j c r ~ t iL'lon~ ~ l l y lhc slopc of ;I plut of % ;~K;LII~lccul;ir interiictions. Consider two inleracting rnulecules as ~ h t l 11 n 111 1.1g11rc2 1.1. 't'hc intcravtion of the two molecules dcpcnds upon the disti~nce twtwccr~tllcir c.cnlcr\, r , and upon their c~rienlations.Because the molccule~are ruaverage out, so for simplicity we assume that the laling, thcir ~ i r l c r ~ t ~ t l opartially ns inlcr;~cl~orl tlrpctids only u p n r . This approximation t l ~ r n sout to hc satisfactory f i ~ r
FICUUt
0.998
1
(1
I
0.U2
0.04
I
I
I
0.06
0 08
0.10
2.13
TWOliilcraullng l ~ m wriiolecules. Gencrully, thc interrnrdecula interaction bcrween h a rnolcculcx cLcycdx upm Ihc Jtrtuncx herwccn their cenlers (r) and upon their orientations (1)' , It:, 1 1 d $1
P / bar FIGURE
A
2.11
plot of Z Venus P at luw pressure? for NH,(g) at O'C, I W C , and 200°C. The alopes of
the lines :ire equal lo B , , ( T ) / R T according tn Equations 2.23 and 2.24. The respectice slopes give B1,(O C) = -0.345 dm'.mol-'. B I , (100'C) = -0 142 drnJ.mol I. and B1,(200 C ) = - O 075 dr~~'.mrll-'.
whcrc W, is (tic A c I I @ ~ cttnatml ~ und k , i s thc Hnlumann constant, which i s equal El divided by the Avtbgadro constant. Note that B I , ( T ) = O it' u ( 1 . ) -- 0; IIItlhcr wdn, rhcm arc ntl Jcviatii)ns l'rotn idcal hchavior if- there arc I ~ O illlr+r~ntrltculur ~r~lcrwlii~~ln. l!quatitrli 2.25 h t w a lhxl cmc u ( r ) l a Lrowll, i~1s ;I *il!l~,lc Inaitcl' 11) c;llculatc HI,. (7')aa rl l u ~ i c l l i ~ofn ~empcmture,or cr~rlvcncly.14) dctc.rmine u ( r ) I!' &2C. ( T ) IS known. [ti priric~plc,r r ( r ) can k c~~lculutcd I'rorri quantutn rncch;~nics, hut this is ;I dillic~bllcn~iipututior~al pnlhlclt~.11Gun be 5hijw11, h
for l a r g ~v i i l l l c ~(*I'r. In lhih cxprcsaicln, r., is a conhtant whose value depends up011 the p:inic.ular ~ntcr:hciingn~olcculcs.'l'hc nugativc sign in Expresslun 2.26 indicates th;~t thc iwt) ~ n r ~ l c c i l lnttruct t\ ciucl~t)thcr. This attraction is what causes suhs~ancesto cuntlcl~rr;1t r u l l ~ ~ , l c yl i iluw l i c r ~ i l l c r a t u i ~There r. i s no known exact expressinl~like 2 . 3 tor \1u:111rlt\tiitlc~>.I>ut 11 rtnlal hc ~ ) l ' albrm thal reflects the repulsion t h a t uccurs u hcn t HI) ~III~ICCIIIP\ a ~ ~ p t o i ~rl(lstiy. cIi Uhually, we assume that
FIGURE 2.12 The second virial ctwtticient t l , , ( T ) of several gases plotted ugainst temperature. Note that H,,,(T) i s negative at low temperalures and increases with tempzrature up to apornt, where i t passus through a shallow lnaximuni (observable herc only for helium).
many molecules, especially if they are not very polar. If we I c l u ( r ) be the potential energy o f two molecules separated by a distance r , the relation between the second virial coefficient B,,(1') and u ( r ) IS given hy
for' \rilall v;lluca nt r . In Equation 2.27, n i s an inleger, often taken to be 12, and u,, is a colihi;bnl wllo\t' V ~ U dcpends C upon the two molecuIcs. A n ~ntcrliic>lcculur potential that embudies the long-range (rlttraclive) behavior 01I:c)u,~lli~rl 1 2 0 :tr~llthe short-range (repulsive) behavior of-Equatio~l2.27 is simply the s ~ l l iul l I~IC IWO. It' wc take n to be 12, then
I$uatlon 2.28 is uiually written in the form
Chaprcr 1
!
Prtq,eri~t.& oi Case.;
Tlir.
tvhcrc r12 = 4m.I' and c, = 4 ~ r r Equation ~. 2.29, which i s c;~Iledthe LPn,lnd-Jones potetlri{d. i s platted in Figure 2.14. The lwo paratneters in the Lennxrd-Jorics potcnri;il I w e thr ii)llowir~gphysical irlterpretation:e is the depth of the p t e n t i a l well atirl cr is ttie distal~cea! which u ( r ) = O (Figure 2.14). As such, s i s a Ine;thurc (11 Ilow slrtlngly thc molecules attract each (Ither, and cr is a measure o f the s i ~ of e thc rrjolcculcs. 7htsc Lrttrzuni-Jorlr~~s paranterr/-.Yare tabulated for a number c ~ molevulcc f i n 'I';~t)lv2 . 7 .
I
E X A M P L E 2-9 Shnw that theinin~~nulr~r)lthe 1-cnnard-Jo~ne.;pa>tcrlti;rI txi,urr ;II r,,<,m F.valuarc r r k ) at r ,,,,,,.
2'
'rj
= I
I?",
SOL U II 0 N: To find rmln. IIC dilTel-e~ltiiltvE:qu:~~ir>ri 2.21)
Thu< A i s the dcpth of the potential well, rclntive to the infinite separatiun
T A B L E 2.7
Lennard-Jonesparanleters, s and 0 ,for various substances. {r/k,)lK
Spcuies
-. . - - .-.-- -- .. ...-..-.--
olpm
(2;co'~,,/3j/cm~.rnol.-'
t lc
i(I.22
256
2 1.2
Nc
15 h
275
26.2
AI
I !I1
74 1
50.0
Kr
I(rl
183
70.9
Xc
2 2'1
1IK> 20 3 3 7U 358
86.')
]II
37.0
N!
95.1
0:
I IX
C'U
ICW
CC),
189
Ct4
152
CIi, C?H4
149
CI H,
243
C~HR C(CH,),
242 232
199
31 7 63.V
57 .o
376 449 47U 378 452 395 5M 744
67.0
114.2 131.0 68.1 1 16.5
77.7
226.3 519.4
If we substitute the Lennard-Junes potet~tialinto Equation 2.25, ae obtain Equation 2.30 may Icluk cumplicated, but it ran he simplified. We first define o reduced
temperature T' by T * = k , T / € and l e I~- / l r =
whul,tH3,(7-'1
f I G V R E 2.14 A plot o t i ~ ( r )= /~ 4 [(:)I'
w
-
( f ) 6 ] versus
T/D lor the
Lennvd-Jones palentiri Note i b t the
d c p ~ hof rhe putcntial well i s c and that u(r) = 0 at r/n = 1.
-
I
try g u ~
H,,~I'~/(!~r~'~~/.~~.1~~1;1tioti2.J1showsthatthereducedsecond
~IIW>II only the reduced temperature, T '. The integral viriill ~ . ~ w l - f i ~/I,:, ~ i t ~1.'r ~1. lIIC~WIIII\ , in I Q u ; ~ t i t ~2r ~11 IIIII\I tw TV:IIII:IIC~ ~ i ~ ~ ~ r ~(Mathchapter ~ r i ~ n l l y Anl for each value of 7 ' * . I'.ut~-rl\~it+ t ; ~ t ~ l t .111 \ It:, I I' I vcr\u\ 7" are available. I,qu:ltlulr 2 . 1 I i\ ; ~ ~ l t r ~r tx~; ~c ~r i l p lof c the law of corresponding states. If we take cnpcl ~lllc.r~t;~l \:lluc\ 111 H : , I 1 ' ) - divide them by 2rsu3~ , / 3 and , then pIot thc data vrrcllc I' X ,I I/ r , t l ~ r u s i ~lor l ~ nll gases will fall on one curve. Figure 2.15 shows rucll :I l ~ l t r t ' 0 1 5 i a gi~scs-Conversely, a plot such as the one in Figure 2.15 (or better yrl. l~ulllc.r!c;~l 1:ihlcc) c:u~he used to evaluate B , , ( T ) for any gas.
2 6. London Oispers~onForr es Arc Otten the
Largest Cuntributkin lo thcr
y,cdl at pressures such that the contrihulion r,l
and the ideal-gas value cwfticient i s ncgligihle.
I
k' ,,b -I
I
E X A M P L E 2-10 Estilnnte H,,{T)for N,(g]at 0 C'.
--
-
5CILLJTLON: 'L'dhle 2.7givcs f / k , 95.1 K and 2 r r t r ' ~ , / . l hB 0 c.111' ~ r l c , l Nllg). 'Thus. 7' = 2.87. and Figure 2.15 givcs Or, (7")s - 0.2 Therefore,
lor
If wc had used numerical tablcs fol- H ; , ( T ' ) iristcad of Figure 2.15, we would hare uhkniried B;,.(Y'') = -0.16, or U?,. (1')= - 10 cln3.mol--'. I
I
The value of Ll,,.(T) has a slrriple intcrpretrttic~n.Consider Equalion 2.23 under uondiriorls whcrc we ran rgnoIe thc terms in P' and higher PV k,.(T) = I H2,,(7')P= 1 +KT' RT
+
By ~riultiplyingthn~ughby K T - / P and usirlg
TI,%,= R T / P , we can rewrite this
equation in the Form
-
H,, = V
KT.
I'
-
-
\I,,,
I
,F
(O.lln.7145 ~.r l . ' , h -: a.,mol --~ ')(300.0 K) . .. . I har
= 24.31 drn'-l~~ol ' - 24 04 tlll~'.nirll '
= -0.b7drll' mol
= - -hlO c n l ' , lnul
'
Although we have been discussing calculating EL,(7') in t e r n s of the LennirrdJones potential, in practice it's the other way m u l i d : Ler~nard-Jonespanttleters are usually detennined from experimental values o f Dz,(T). T h ~ dcturminatiun s i< usually made through trial and error using tables of B , * , ( T h )'l'he . valucs o l lhc Ennard-Jotlcs para~netersin Table 2.7 were determined frum cxperimcntitl second virial c o c t l ~ u ~ t . r ~ ~ d a u . Because the swond virial coefficient rcflccts the initial deviutluns from idcal behavior, which are caused by intermolecul~interactions, experimental F -V-7' k i l a turn out to bc a rich source of information concerning i ~ ~ t e m ~ o l e c uintzr~cliol~s. l:~r Once Lcnnard-Jones parameters have been determind, they can he uccd I ~ ualculi~tr I many other fluid properties such as viscosity, lherrnal col~ductivity,ttriits o!' vi~p~riiation, ;rrltl various crystal properties.
2-6. London Dispersion Forces Are Often the Largest Contribution t o the r P 6 Term in the Lennard-JonesPotential In the previous secrion, we used the Lennard-Joncs pr)tc1111;1l(l.:rli~;ttiui~2.29) to represent the inter~nc~leculnr potential between rnvlccl~les. I'hu r- '' tcnrl accuunts fclr the repulsion at short distances, and thc r-"ern1 nucclullt\ t t r ~thc u ~ r a c ~ i oilli l larger disw~~ccs. The actual fr,m of the rcpulsivc terrrl ic no1 well r.\~;ll>llshcd, but the Idcpende~rceof the allractive term is. 911 1hi5 \cc.tiorl. wc will dihc.i~\\three cc~ntributitlns 10 the r - h attraction and comparc thcir ~elativcI I I I I I O I I.IIIC,L,. Conbidcr twu dtpuliir tnoleculcs, nhc>+e~li(xllc.~rrrrrn~.r)l\ ~trc11 ilnd p , . 'The rtltcraction of thc\e dlpolcs del)cn~lsrllxln hclu lllcy :LIP orlcr~~cd with rehpcct tu cach otl~cr. The energy will v;try Iron1 rcpulsrvc, W ~ I C I Ir t ~ u pill-c cjricnted Itcad-to-head as sh(>wrt
,
Thus. we see that I 4 , ( T ) represent< the difference betwccn the actual value nf
the third virial
E X A M P L E 2-11 Thc molar volulrn: c>f 1sc7hulancat 300.0 K and onc bar i.; 24.31 drn'.rn~l-~. kstir~late the valuc of H ! , lrrr i\ohu~uncut 31Ml.O K. 5 0 1 . I J T I ( 1 N . 'I hc itlcitl-g;~\rll<>li~r v~~lurt~c at 3(K) O K and o ~ l bar c
F I G U R E 2.15 A plot of the reduced secolid virial corffioient 8 ; , . ( T X )= B , , { T ' ) / { ~ ~ u ~ N (rtllid , / ~linc) ) ~ T' = k,T/c. Expenmenpal data of six gahev (nrgull, nitrogen. again\[ the w d u c temperature oxygen, 'arbo~t dioxide, and sulfur hex,~Rur>ride) are alho piolted. 'lhis pto~ib another illurlralir)n of the law uf correspontling \tale$.
77
lern,
2 h.
I.r)rirlc)ri
T l i \ [ ~ t . r h ~ c > rFur<:?\ ~ Arr
Illlrn t
h 1.1rgrsl ~ (:untrihr~tiwi tu the rFhTcrm
E X A M P L E 2-13 Calct~l:~le the value r)f the cr~fficientc ~ rf
' i n Fquarion 2.33 at 300 K for twn I!CI(EI
~r~chcculcc. '1';tblc 2.H lists the dipole Illolilcnts o f v:iriou\ ~~ioleculcu. 5 ( 11 11 Il r > N , According to Table 2.8. 11, = p : = 3.44 x 10 "' C - m . Therefore.
ill
FICURE
2.lh
Turn[$ pcrrt1:lilrnt d~pcjlc\orrclllctl
[ill t~c;~rl.~o-l~catl and (h) 11e:ld-to-fail.Tile head-to-tail urictltaliol~is cllrrgctici~llyI:~vtll.;rt?lc (c) A rr~oleculcwith a pcrnlanenr dipole Illolrlent will induce a dipole moment In a neighhrlring molecule. (d) The instantaneous dipole-dipule correlation strr)wn herc is what leudr to a Lundon attraction between all atoms and mvlecules
in Figi~~-e 2. I b i ~to attractive. when they art: oriented hcad-10-tail (Figtlrc 2.16b). Both ~riolcculesrotate in thc gas phase, and if we were to avtrnge both dipoles randomly ovcr their oricntn~iuns,the dipole-dipole interactions would average out to 7cro. Because diikrcnt orientations have difl-ercnt energies, they do no1 occur to equal extents. Clearly. thc lower-cnergy hcad-to-tail orientation is favored over the repulsive hend-tubrad nrierilation. If we take into account the energy of the oricrltation, then the overall a v c r a p interaction between the two molecules result$ in an attractive r - 6tenn of the form
n~ol-'-K~-' ) (31111K)( I 1 13 (G8 .3140 2 J2-~10:' ~ rnrjl-'
.- -
--
.,
.
-.
--
x l l i ' ~ ' r ' . k gm - ' j 2
'1111\ ~lu~r~rrlcrtl rck1111111ily YCL'III exceedingly small, but remerltber that wwearc cnl. C I I I I I ~ I I I ~ t 'id,, , ( { rJ. At :I \cp;1r;ition of 3(X) pm, ~ , , - ~ (isr )equal to -2.5 x 10 J . cc~lnlxlrrdwith u 1hcrr11:tlcncrgy (kl,7-)of 4.1 x 10 'I 3 at 300 K.
;i perrtlilnent dipole morrietlt. Equar ion 2.33 rcyuircs [hilt holh ~ncllcu~ilcs Even il- one nlr~leculcdocs 1101 h:tvc ;I pcrnxjncnt dipole nloment, thc one withnut a permnrletlt dipole Inonierll wlll have ;I dipcjlc molncnl induced by the other. A dipole moment can be induced in n niolcculc that docs t~othavc a permanent d i p l e motnerlt bccause all atoms and molecules are pulrrrizmhlr. When at1 amrn or a rljolecule i n ~ e r ; ~ c ~ s with an electric Lirld, the (negat~ve)electrons are displaced in one direction slid thc (positive) nuclei are displaced in thc opposite direction, as illustrated ill Figure 2.1612. This chargc separation with its associated dipole rnolrient, is proportional to the strength of the electric field, and if we designale the induced dipole monlent by IL,,,~", and the electric tield by E,we huve that I L ~ M~ E~. Thc ~ , proportionality ~ constant, whiclt n c dcnntc by 0,is called the yolarizubilir~,so we have the defining expression
l'hc u n i t < of k; ;irt v.111', so ltie units ot'cu in Equation 2.34 are C - m / ~ - m - I= C .rn"-v I . Wc C;III put CY into rnorc trrinsparent u11it.cby using thc fact that energy = (charge)'/4rrr,, (di.;tariccl, whicl! it1 SI ulli~sg i v n E X A M P L E 2-12 Show thal rhe u1111c of tllc right qidc r l i Equotlnn 2.33 are energy.
Similarly, trom cIc.clroclalic~~, wc Ii;~vcIh;~t SC) LLJT I O N :
The unit? ot
11
are C.tn (chargc x \eparntiun), and so we have
-'
Equating these lwo cxprc.c.cions l0r joulcs gives C . V = c 2 . m / 4 n ~ , , ,or C.V = (4rtn)In. Now wc .;uh,~itulc ihic rczult irito the above units fur ur ( ( C - ~ ? - V - 'lu ) gt
T h u ~u-e , scc tt1;tt r v / 4 r r ~ ,has , units of m 4 .The rluilntity ur/4~r~,,, which is sometimes referred lo ;I.; Ihc /~r~lrtri:nhility~,olumc,has unils of volu~ne.The easier it is for thc
1 6 . Londun Dispersion Furces Are Often the Largest C t m t r ~ b u l i otu~the ~ r
dccrric field to deform the atomic or molecular charge distribution, the grcdter is the pularizability The polarjzability c~fan atom or a molecule is proportional to its size (notc the unjls of a/47r~").or 10 its number of electrons. This trend can be seen in T a h l ~2.8, which lists the polanzahility volumes of some atoms and molecules.
SOLUTION : Thc two tcrnih Uhing the dara i11 Table 2.8.
111 Equation
Term
2.35 are the s:lnrl: to1 identical molecules
T A B L E 2.8
The diprde moment (lc), the pnlarizahilrty volunie (u/4nsU), and the iunization energies (I) of various atoms and molecules. -
Species u/lO-"Cm He
Nc AI~r Xe
NI
Note that this rcsull is abut 30% of the re9ult we ohtalned i r ~krarrlyle 2-13 lcjr - rnlc4, d(r).
(u/3n~,)/lU-~~rn' 1/10-~~1
-.
0 0
0.21
U
n
1-63 2.48
0 0
4.0 1 1 77
2,243 1 .%I1 2.40h Z.o(I.1
0.39
CH4 C?H,
0
1.60
O
4,43
I K1(1
C,H, CO
0.W 0.40
1754
COI
0
6.31 107 2 (>3
HC'I
2 f>l
HI
3 44 1.47
*Hi
5.iK)
2 2<
H,O
(1.14
1.57
S,.$?
1
3.93'1 3.454 2.525
I ~ J fqui~litrr~* I ~ 2.33 and 2.35 equal zert, whcn neither niulecule tins 11 pcrnlanent tltpcilt mtttrrcnt.'1')ru third contnbutiun to the r "term m Equation 2.29 is nonzcro evcn if h ~ nw~lcculcb h ilrc nclnpnlar. This contribution was first calculated hy the G c m ~ a r ~ ~ . i c n t ~I;ril# b t I . o ~ t ~ l in o l ~Ic).10using quantum nlechanics and is now called a 1;)1xrio11
drrlrrr,ticm rrIfri~t,rir~rt. Although thih ilttr;~ctionis a a~rictlyqudnttlrn-mechanical effiut. II Ir~tilk~ t w l l10 IIIC I o l I t ~ w i ~r ~ . tg~ t ~ i ~ l oil*ccl ~ i l y cl;tssical picturc. C'onsidcr lwu atoms ub shown in I:~purc 2 . l h l ~ C P ; I ~ U I hy U ~ il dihtsncc t - . The electrons on one etom do nut cnrl~plc~cly hlucld Ihc high prsitrvc ch;rrgc (111 111c~ruclctlsFrom thc electrons o n thc olhcr illoi11. HUCUU~C I ~ IINIIUC'HIF C is ~ n ) l i l r j ~ ; ~the t ) l ~~, l c ~ l r o n w;)i7c ic f~~nctlon call dibti)rt ;Ihit IO iunhcr hlwur thc ir11c.r-actioncncrgy. If wc avcrijgc (his elcctrol~ic allruclion quanrurn itluchar~ic~ally, wc oht;1111;In ;tttr;lctivc fcrin [hat varies as r ". The c x x t yuantunt-r~tcch;tx~ical calcu);rri~)~~ i\sor~lcwh;~t complic;~lcd.but ;HI ;ipprclxlmale iorin ot' the Lirtul waul1 rs
2.!44
2 t(YI 2 {L# 3 I (r(4 I h?H 2 t120
We now return lo the dipole-induced dipole interaction hhl)wrt in I:~gurc 2. Ih. Because the induced dipole moment is always in a head-to- ail oricr~tatloriwith r c s p c t to the perinanent dipole momenl, the interacljon is a1way.c itttr~ctiec.and i \ ~ I V ~ hy I I
The
term represents a permanent dipole moment in molecule 1 and ;ill inducctl dipole moment in molccule 2. and the second represents the oppusitc s~tuation. first
C
I
EXAMPLE 2-14 Calculalc tile wlue of thc cneflicicnt (I[
I r-6
for ~,,~~,(r) for two HCl(g) mc>lcuule>.
where I, is the io~~izalion energy of alurn or rr)oleculc j . Nnle th;~tEquatlon 2.3h duch not involve a permanent dipole moment and that the interaction energy is propurtlonal tu thc pruduc~of the polari7ability volumes. Thus, the irnportuncc ~ I ~ I I , , , , ~ ( I -increases ) with lhc sires oi ~ h uloms c or jnolecules, and, in fict, is often the duininar~tcoritrihution l o the Fb ter~nin t y u a ~ i 2.29. o~~
E X A M P L E 2-15
Colc~llutcthe value of the cmflicient of r " for u,,,,,cr) Ibr two 11rItg3 rr~c>luculc.b.
Chapter 2 I The Properties of
82
Cases
This quantity is about six times greater than -r6u,,(r) and 20 times greater than - r ' ~ , ~ ~ , , ~Similar ~ ~ ( rcalculations ). show that the disperison term is significantly larger than either the dipole-dipole term or the dipole-induced dipole term except for very polar molecules such as NH,, H,O, and HCN. The total contribution to the r-6 term in the Lennard-Jones potential is given by the sum of Equations 2.33, 2.35, and 2.36, giving C C U(r) = I Z- A rI2 r6
2-7. The van der Waals Constants Can Be Written in Terms of Molecular Parameters
This potential represents hard spheres of diameter 0. Equation 2.38 depicts the repulsive region as varying infinitely steeply rather than as r-". As simplistic as this potential may seem, it does account for the finite size of molecules, which turns out to be the dominating feature in determining the structure of liquids and solids. Its obvious deficiency is the lack of any attractive term. At high temperatures, however, meaning high with respect to & / k , in the Lennard-Jones potential, the molecules are traveling with enough energy that the attractive potential is significantly "washed out," so the hard-sphere potential is useful under these conditions. The second virial coefficient is easy to evaluate for the hard sphere potential. Substituting Equation 2.38 into Equation 2.25 gives
with (Problem 2-53)
for identical atoms or molecules.
2-7. The van der Waals Constants Can Be Written i n Terms of Molecular Parameters Although the Lennard-Jones potential is fairly realistic, it is also difficult to use. For example, the second virial coefficient (Example 2-10) must be evaluated numerically and one must resort to numerical tables to calculate the properties of gases. Consequently, intermolecular potentials that can be evaluated analytically are often used to estimate the properties of gases. The simplest of these potentials is the so-called hard-sphere potential (Figure 2.17a), whose mathematical form is
which is equal to four times the volume of N , spheres. (Remember that a is the diameter of the spheren.) Thus, the hard-sphere second virial coefficient is independent of temperature. Note that the high-temperature limit of the second virial coefficients shown in Figuren 2.12 and 2.15 is fairly constant. The curves actually go through a slight maximum because molecules are not really "hard." Another nimple potential u d fairly often is the square-well potential (Figure 2.17b):
The parameter E is the depth of the well and (A - I)a is its width. This potential provides an attractive region, as crude a. it is. The second virial coefficient can be evaluated analytically for the square-well potential
.
F I G U R E 2.17 (a) A schematic illustration of a hard-sphere potential and (b) a square-well potential. The parameter u is the diameter of the molecules, E is the depth of the attractive well, and (A - 1)u is the width of the well.
Note that Equation 2.41 reduces to Equation 2.39 when A = 1 or E = 0, there being no attractive well in either case. Figure 2.18 shows Equation 2.41 compared with
83
2-7. The van der Waals Constants Can Be Written in Terms of Molecular Parameters
Comparing this result with Equation 2.22, we see that
for the van der Waals equation. We will now derive a similar result from Equation 2.25 and interpret a and b in terms of molecular parameters. The intermolecular potential that we will use is a hybrid of the hard-sphere potential and the Lennard-Jones potential
We substitute this potential into Equation 2.25 to obtain F I G U R E 2.18
A comparison of the square-well second virial coefticient for nitrogen. The square-well parameters for nitrogen are a = 327.7 pm. ~ l k = , 95.2 K, and A = 1.58. The solid circles
represent experimental data.
B2,(T) = -2n NA
I'
(-
l)r2dr - 2n NA
In the second integral, we assume that r6/k,Trb (MathChapter C)
experimental data for nitrogen. The agreement is amazingly good, but the square-well potential does have three adjustable parameters. We will finish this chapter with a discussion of the second virial coefficients for the three cubic equations of state introduced in Section 2-2. First, we write the van der Waals equation in the form
I"
<< I
[ec~''~Tr6 - l]r2dr
and use the expansion for ex
x2
eX=I+x+-+... 2! and keep only the first two terms to obtain
Comparing this result with Equation 2.43 gives We now use the binomial expansion of 1 /(1
to write Equation 2.42 as (letting x = b/V)
-x
) (MathChapter C),
2n Njc, 3u3
a=-
and
2 n u 3N, b=3
Thus, we see that a is directly proportional to c,, the coc.lfiL.icntof r-h in the intermolecular potential, and that b is equal to four times the volume of the molecules. From a molecular point of view, the van der Waals equation is based on an intermolecular potential that is a hard-sphere potential at small distances and a weak attractive potential (such that c6/kBTr6<< 1) at larger distances. In a similar fashion (Problem 2-55), the second virial coefticient for the RedlichKwong equation is
and the second v h a l coefficient for the Peng-Robinson equation is (Problem 2-56)
86
Chapter 2
/
The Properties of Gases
The second virial coefficient from the van der Waals equation and the Peng-Robinson equation have the same functional form, but they have different numerical values because the values of the constants are different. Also, the parameter a is a function of temperature in the Peng-Robinson equation.
Problems 2-1 1. It takes 0.3625 g of nitrogen to fill a glass container at 298.2 K and 0.0100 bar pressure.
It takes 0.9175 g of an unknown homonuclear diatomic gas to fill the same bulb under the same conditions. What is this gas? 2-12. Calculate the value of the molar gas constant in units of dm3. t o r r . ~ -.mol-' '
Problems 2-1. In an issue of the journal Science a few years ago, a research group discussed experi-
ments in which they determined the structure of cesium iodide crystals at a pressure of 302 gigapascals (GPa). How many atmospheres and bars is this pressure? 2-2. In meteorology, pressures are expressed in units of millibars (mbar). Convert 985 mbar to
torr and to atmospheres. 2-3. Calculate the value of the pressure (in atm) exerted by a 33.9-foot column of water. Take the density of water to be 1.00 g.mL-I.
2-4. At which temperature are the Celsius and Farenheit temperature scales equal? 2-5. A travel guide says that to convert Celsius temperatures to Farenheit temperatures, double
the Celsius temperature and add 30. Comment on this recipe.
2-13. Use the van der Waals equation to plot the compressibility factor, 2,against P for methane for T = 180 K, 189 K, 190 K, 200 K, and 250 K. Hint: Calculate Z as a function
of p and P as a function of V ,and then plot Z versus
P.
2-14. Use the Redlich-Kwong equation to plot the compressibility factor, 2,against P fdr methane for T = 180 K, 189 K, 190 K, 200 K, and 250 K. Hint: Calculate Z as a function of and P as a function of V,and then plot Z versus P. 2-15. Use both the van der Waals and the Redlich-Kwong equations to calculate the molar
volume of CO at 200 K and 1000 bar. Compare your result to the result you would get using the ideal-gas equation. The experimental value is 0.04009 L.mol
'.
2-1 6. Compare the pressures given by (a) the ideal-gas equation, (h) the van der Waals equation,
(c) the Redlich-Kwong equation, and (4) the Peng-Robinson equation for propane at 400 K and p = 10.62 m ~ l . d m - ~ .The experimental value is 400 bar. Take a = 9.6938 ~ ~ . b a r . m oand l - ~0 = 0.05632 L.mol-' for the Peng-Robinson equation.
2-6. Research in surface science is carried out using ultra-high vacuum chambers that can
2-1 7. Use the van der Waals equation and the Redlich-Kwong equation to calculate the value
sustain pressures as low as lo-'' ton: How many molecules are there in a 1.00-cm3volume inside such an apparatus at 298 K? What is the corresponding molar volume V at this pressure and temperature?
of the pressure of one mole of ethane at 400.0 K confined to a volume of 83.26 cm'. The experimental value is 400 bar.
2-7. Use the following data for an unknown gas at 300 K to determine the molecular mass of the gas.
2-1 8. Use the van der Waals equation and the Redlich-Kwong equation to calculate the mo-
lar density of one mole of methane at 500 K and 500 bar. The experimental value is 10.06 mol. L-.'. 2-19. Use the Redlich-Kwong equation to calculate the pressure of methane at 200 K and a
density of 27.41 mol .L-I. The experimental value is 1600 bar. What does the van der Waals equation give? 2-8. Recall from general chemistry that Dalton's law of partial pressures says that each gas in
a mixture of ideal gases acts as if the other gases were not present. Use this fact to show that the partial pressure exerted by each gas is given by
2-20. The pressure of propane versus density at 400 K can be fit by the expression
where P, is the partial pressure of the jth gas and yj is its mole fraction. 2-9. A mixture of H2(g) and N,(g) has a density of 0.216 g.L-' at 300 K and 500 tom. What is the mole fraction composition of the mixture?
for 0 5 p / m o l . ~ - ' ( 12.3. Use the van der Waals equation and the Redlich-Kwong equation to calculate the pressure for p = 0 mo1.L-' up to 12.3 mo1.L.'. Plot your results. How do they compare to the above expression?
2-10. One liter of N2(g) at 2.1 bar and two liters of Ar(g) at 3.4 bar are mixed in a 4.0-L flask
2-21. The Peng-Robinson equation is often superior to the Redlich-Kwong equation for tem-
to form an ideal-gas mixture. Calculate the value of the tinal pressure of the mixture if the initial and final temperature of the gases are the same. Repeat this calculation if the initial temperatures of the N,(g) and Ar(g) are 304 K and 402 K, respectively, and the final temperature of the mixture is 377 K. (Assume ideal-gas behavior.)
peratures near the critical temperature. Use these two equations to calculate the pressure a density of 22.0 mo1.L-' at 280 K [the critical temperature of C02(g) is of 304.2 K]. Use a = 4.192 b a r ~ ~ . m o and l - ~ = 0.02665 L.mol-' for the Peng-Robinson equation.
at
Chapter 2 / The Propertis aof Cases
2-22. Show that the van der Waals equation for argnn at T = 142.69 K and P = 35.00 aim can be writken as
where, for convenience, we have supreshed the units in the coefficients. Use the NewtonRaphson method (MathChapter A ) tu find the three roots to this equation, and calculate the values of the density of liquid and vapor in equilibrium with each other under these conditions. 2-23. Use the Redlich-Kwong equation and the Peng-Robinson equation to calsulate the densities of the coexisting argon liquid and vapor phases at 142.69 K and 35.00 atm. Use the Redlich-Kwong constants given in F ~ b l e2.4 and take a = 1.4915 a t r n . ~ ~ . m o l - and ' p = 0.01981 L.mol-' for the Pmg-Robinson equation.
2-24. Butane liquid and vapor cmxiht a1 370.0 K und 14.35 bar. The dcnsities of the liquld and j:ap>r phascs are 8.128 mu1 . L-' and 0.63 13 ~nol. L- respectively. Use the van der Waals equation, the Redlich-Kwong equation, and rhe Peng-Robinson equation to calculate these densities. Take a = 16.44 bar. L2.rnol-' and = 0.07245 ~.rncll.-'Ibr tllc Pzng-Roblnson rq~ratioo.
'.
Problems
Divide this equation by
vC3 and let B/T= x to get
Solve this cubic equation by the Newton-Raphson method (MathChapter A) to obtilin x = 0.25992, or
B
(4)
= 0.259927~
Now substitute this result and Quation 4 into Equation 3 to obtain
2-27. Use the results of the pmiwa pmblcm lo derive Equations 2.14. 2-28. Write the Pcng-Rubinson cquetim aa a cubic p,lynomial equation in 7 (wilh lhc eelefficient of 7'qua1 tu one), d cornpoue it with (7 = O at the critical point tcl
- vC)'
obr~in
2+25. Another way to obtain expressions for the van der Waals constants in terms of critical and ( a 2 ~ / 3 ~ ' )q,u a 1 to zero at the critical point. Why parameters is tu set ;ire thcse quantities equal to z k t ) at the critical 'pnint? Show that this procedure leads to
(a~/av),
Equations 2.12 and 2.13.
2-26. Show that the Rcdlich-Kwong quation can be written in the form (We write q because u & p d a upon the temperatuw.) Now eliminate q/l: betwecn to ubtain Equatimr 2 and 3, md rhen we Equation 1 for Now compare this equat~onwith
(v- y)' = 0 to get A
BRT B?
Solve this cqulrlirm uning Ihe Newton-Raphaon method to obtain
and
Note that Equation 1 gives
w
Now solve obtain
Eqwa~lon3 for A and substitute the result and Fquation
Subslitute ~hihrerult and Equalion L into Equation 2 to obtain
4 into Equation 2
10
Last,uuc Equqion I b show that
Chapter 2 1 The Propertie uf Gases
2-29. l . w k up the boiling points or the gases listed in Table 2.5 and plot these values versus the critical tclnpcratures 7;. 1s there ally correlation? Prvpose a reasoll to justify your conclusionr frnm thc plol.
2-30. Show that the compressibility factor Z for the Redlich-Kwong equation can be written nq i n Equalion 2.21.
2-35. Use Figure 2.10 to estimate the molar volume of CO at 200 K and 180 h.An accurate experimenta1 value is 78.3 cmi.mol-I. 2-36. Show that B,,(T) = RT B,,(T)(see Equation 2.24).
2-37. Use the following data For NH,(g)
at 273
K to determine B , , ( T ) at 273 K.
2-31. IJse the following dala Fr~rethane and argon at T, = 1.64 to illustrate the law of c m sponding st~tcsby pluiting Z against
vR.
-
Ethane ( T = 500 K) ~/~.rnol-'
Atgun (T = 247 K) v/~.rnol-'
Plbm
U.500
2.00 10.00 20.00 40.00 60.00 XO.00
100.0 12fl.0 1MI.O 2(X).O
240.0 ~OWI
350.0 400.0 450.0 500.0 600.0 700.0
Platm
83.076 20.723 4.105 2.028
2-38. The density of oxygen as a function of pressure at 273.15 K is listed below. Platm
0.9907
0.h46 1 0.4750 0.3734 0.3116H
1 0.2500
0.500
C1.750n
1.0000
-
Use the data to determine BB,,(T) of oxygen. Takc the atomic mass uf uxygn to be 15.9994 and the value of the rnt~larp s constant to be 8,31451 J.K..' .rnol-' 0.0820578 dm'.ntm-K ' .rnul-I.
0.2265
0.1819 0.1548 0.1303 0.1 175 1). 1085 0.1019 0.09676 0.08937 0.08421
2-39. Show that the Lennard-Jones potential can be written as
where r' is the value of r at which u ( r ) is a minimum.
2-40. Uning the Lcnnd-Jones parameters given in Table 2.7, compare the depth of a typical knnard-Jonca pdential to the strength of a covalent bond. 2 4 t . Compare ihc ttnnerd-Jones potentials of H,(g)and O,(g) by plotting both on the same ormph,
2-32. Use the dm;' in Problem 2-31 to illustrate the law of mrrespnding states by plotting Z iignirist P,.
2-33. Use the data in ProbIem 2.31 to test h e quantitative reliability of the van der Waals equation by comparing a plot of Z versus from Equation 2.20 to a qirniler plot o f
&
ve.
242. Use hc &la in Tables 2.5 and 2.7 to show that roughly & / k = 0.75 T and b, = 0.7 Thus, critical wmslanlr can be used as rough, first estimates of r and b,, (= 2 n ~ , n ' / 3 ) .
243. Pmvt thml rht nwond virinl coefficient calculaled from a general interrnolpcular potential of the Iwrm
the data.
I I ( ~ )=
2-34. Use the data in ~rohlt.; 2.31 to test the quantitative reliability of the Redlich-Kwong equation by cornparitig a plot of Z versus frum Equation 2.21 to a similar plot of the
&
dava.
(energy paramelcr) x f
r distance parameter
rigornusly obeys the law ~Tcorrespondingstates. Does the Lennard-Jones potential satisfy
this condition?
Chapter 2 / The Properties of Gases
2-44. Use the folluwing data for argon at 300.0 K to determine the value of B,,. The accepted value is - 15.05 cm3.1nol-I.
2-49, Listed b l o w and xenon.
ilre
experimental secund virial coefficient data fur w o n , kryptoo,
TIK
2-45. Using F i p t e 2.15 and the Lennard-Joms for CH,(g) at 0°C.
2.7,
g h lh
Argon
Krypton
Xenon
Btv(r)
2-46. Show that B,,(T) ubeyb the Iuw uf ~uwrerponding a k h rw r rpvlrrd afied value of A (in other words, if dl molecub hd d~ un# vJw dA&
wllb
2-47. Using the Lennard-Jones parametem in Table 2.7. shmv thrr Ihp b Im l dr#rl cofficient data satisfy rhe law of corresponding suwa.
Argon TIK
khsm
Nitrogen
Bzv(T) /1w3dm3-mol-'
TIK
Bzv(T) /lo-' dm3.mol"
TrK
Blv(T) /10 drn'.mol-'
'
U l c h e Lennd-Jones parameters in Table 2.7 to plot B;,(1"), the reduced second virial eoowitnt. versus T*, the d u c e d temperalure, to illustrate the law of corresponding
mm. IdA Uw rhs cri~icaltemperaturesand the critical molar volumes of argon, krypton, and xenon ID
a#*
(wh Irw
of m s p o n d i n g states with the data given in Prohlcn~
-
241. Bvdu#o 8;,(T') In b u t i o n 2.3 I numerically from T' = I .OO to 10.0 using a packagstl ~ u c has Mathcad or Mathematics. Compare the reduced dIn* weoad virld cwmcienl dda from Problem 2-49 and B;,(Tm)by plotting them all on the
&.
1.51. S h m llur Ihc unlb ofIf# tight mi& 2-53. Shuwr rhrt t 2-48.-In Section 2 4 , we expressed the van der Wads equation in reduced units by dividing P, V ,and T by their critical values. This suggests we can write the second virial coefficient in reduced form by dividing &(T) by and T by T (instead of 27rNAe3/3and ~ / ask we did in Section 2-5). Reduce the second virial crdlicient data given in the previous prublem by using the values of and 1; in Table 2.5 and show that the reduced data satisfy the law of corresponding states.
vc
vc
k
01
h u ~ r i o n2.35
;w Energy
aum of B g l u l i ~ 2.33. r 2.35. and 2.3h gives Equation 2.37.
2-54, C o m p the valurx or thc caeMcicnt uf r "or LnnurJ-Jonc~prnmetcn piven in Tabk 2.7.
N,(g) using Equation 2.37 and the
Chapter 2 / The Properties of Gases
and
MATHCHAPTER
for the Redlich-Kwong equalion.
B
PROBABILITY A N D STATISTICS
2-56. Show that the socond and third virial coefficients of the Peng-Robinson equation are
Rz,(T) = B
a --
RT
and
-
2-57. The squnrc-well parameters for krypton are E / k = 136.5 K, o = 3 2 7 . 8 p m , d A 1.68. Plot R,,(T) against T and compm your results with the data given in R'obkm 249.
2-58. The coefticient o l thermal expansion ru is defined an
2-59. The isothermalcompressibility K is defined as
In many of the following chapters, we will deal wilh prt~hahilitydistributions, avcrage values, and standarddcviations. Consequently, we take a few pages here to discuss some basic ideas of probability and shnw h ( ~ wto calculate average quantities in gencr:~l. Consider some experiment, such as thc tossing of a coin or the rolling r ~ fa die, that has n possible outcomes, each with probability pj. where j = 1, 2, . . ., n. If the experiment is repeated indefinitely, we intuitively expect that
Show that where N, i s the number of times that the outcome j occurs and N is the total number N, p, must satisfy the condition of rcpelilions of the experiment. Because 0 5 N j i i ~an r ideal gas.
When p, = I , wc say the event j is a cenainty and when pj = 0, we say it is impossible. In addition. because
we have the noprrlization condition.
Ma~hChaplerB/ P K O B A B I L I T Y A N D S T A T I S T I C S
Equation B.3 means that the probability that some event occurs is a certainty. Suppose now that some numberx, is associated with the outcome j. Then we define the avemge of x or the meun of x tit be
where in the bst tern we have used the expanded notation p(x,). m c a n l ~dm prokbility of realizing the number x j . We will denote an average of a q m d w by the quantity in angular brackets.
Another quantity of importance is
The quantity (x2) i s called the second moment of the distribution [p,)and is analogous to the moment of inertia.
'
I
EXAMPLE 6-1 Suppose we are givtn
the following data:
I
E X A M P L E B-2 Calculate the second moment of the data given in Example B-1. SOL UTiO N: Using Equation B.5, we have
N& from Examples B-1 and B-2 d Ihu we will prow below. Calculate the average value ofx.
SOL UT I 0N : Using Equation B.4, we have
It is helpful to interpret a probability distribution like pj as a distribution of a uni~mass alr~ngthe x axis in a discrete manner such that pi is the fraction of mass located at the point x,. This interpretation is shown in Figure B. I . According to this interpretation, the average value ofx is the center of mass of this system.
B.l The discrete probability fwquency function or probability density, p ( x ) . FIGURE
that (x2)f { x ) ' . This nonequality is a general
A mWly m m interesting quantity than (x2}is the second centml momcnr, or
Ilw -,
defined by
w, a
b & lWYiM W ¬e rhe squnre mt of the quantity in Equatio~lB.6 by u,,wbleb b h-rd deviation. From the summation in Equation B.6, we cm # Uut of will k Lupc if x, i s likely 10 differ from ( x ) , because in that m e (x, ( x ) ) md IO (1, (%))I will bt large for the significant values of p ) . On the other h i d , a,' will bs uarll If&,b likely to differ from (x), or if the x, cluster around {A ) . because thm (x, b))' will k small for the significant values of p,. Thus, we see that either Ihe d a m a tb stnndard &vialion is a measure of the spread of the distribution n b # l i~ mew. Equation 6.6 whow~thuo: i~a sum of positive terms, and so 0: 2 0. Furthermr~re,
-
-
-
The first term here is jusl (x') (cf. Equation B.5). To evaluate the second and third terms, WP need to realize that (x), the average of x j , is just a number and so can be factored out of the summations. leaving a summation of the form 1x j pj in the second tenn and p, in the third term. The summation 1xjp, is (1)by definition and the
Mathchapter B / P R O B A B I L I T Y A N D S T A T I S T I C S
Mathchapter I3 I P R O B A B I L I T Y A N D S T A T l S T l C S
summation C p, i s uni~ykcause of normalization (Equation B.3). Putting all this together, we find that
EXAMPLE k 3
Perhaps the simplest continuous distribution is the so-called uniform distribution, where
p ( x ) E constant = A =0 Becnusc rrf 2 0. we see that (x2) 2 (xj2.A consideration of Equation B,6 fihowa t b t = (x2)only whcn xi = ( x ) with a prubability of one. a caw thU h M really probabilistic because the event j occurs on every trial. So Tar we have considered only discrete distributions, but continuwr d # t d b u h al-e also itnportant in physical chemistry. It i s convenient to use the unit #ms Cansider a unit mass to be distributed continuously along the x mi#, interval on the x axis. We define the linear mass density p ( x ) by
Show thu A mu8i equal I/(b
nt2= 0 or ( x ) '
- a). Evaluate (x), ( x 2 ) ,.:rc
SOLUTION: h s e p ( x ) muu
b-a
+
Prob(x, x
+ d x ) = p(x)dx
and q for this distributiot~.
mmnalid,
1 p ( x ) ~ -
x dx. By analogy, then. where d m is the fractinn OK the mass lying bttwten x we say that the probability that some quantity x, nwh u the polition of a p t i c l e in a box. lies between x and x f d x i s
a 5x 5b otherwise
asX5b
The mean of x is given by
03.9)
and that
and the second moment of x by
In the mass analogy, ProhIa 5 x 5 h } is the fraction of mass that lies in the interval n 5 x 5 b. The normalization condition i s
h s t , the variance is given by Equation B.6, and so
(B. 11) and the standard deviation is
Following Equations B.4 through B.6, we have the &finitions
!
EXAMPLE B-4 The most curnmonly occurring and most imponant continuous probability distribution is the Gaussian distribution, given by
Find c. { x ) .
4 and 4.
I
MathChapterB I PROBABILITY A N D STATISTICS SOL U TI 0 N : The constant c is determined by normalization:
I i you look in a table of integrals (for example, Tha CRC Stutzdard Mnrhtmnticcil Tables or The CRC Handbook of Chemistry and Physics, CRC Press), you won't 6nd tile above integral. However, you will find the integral
MathChapterB /PROBABILITY A N D STATISTICS The mean of x is given by
The niin Equation B.18 is plotted in Figure B.2(b). Notice lhnt this graph i~ antisymmclrk about t k venlcal axis and that the area un one side of the vertical axis cancels t k carrmponding uea on he other side. A funclion that ha the mathematical propeny Ihu f ( x ) r -/(-x) i s called an oddJiincrion. For an tldd function,
The reason that you won't find the integral with the Limts (-ma m) i s illustrated In Figure B.2(a), where e-""' is plotted against r . Note that the graph is symmetric about the vertical axis, so that the corresponding areas on the two sides of the axis are equal. A function that has the mathematical property that f (x) = f [-x) and is called an rvenfuncrion. For an even function
(I)I( Z w d ] - l n
1:
l#-'at*'dx = 0
If we recognize that p ( x ) = ce-'21'"2 is an even function and use Equation B. 16, then we find that
orc = 1 / ( 2 ~ a ' ) ' / ~ .
cm hk Imdin integral tables, and so
klu~ (x) = 0, : a
= ( x 2 } , and so vI is given by
The standard deviation of a normal distribution is the parameter that appears in the exponential. The standard notation fur a normalized Gaussian distribution function is
F I G U R E B.2 la) function f (x) = e '-is an even function. se -'is an crdd function, f ( x ) = f (-x).
+
-
f (x) = J ( - x ) . (b) The function f (x) =
Mathchapter H / P R O B A B I L I T Y A N D S T A T I S T l C S
102
Figure B.3 showx Equation B.21 for various values of 5.Note that the curves become narrower and taller for smaller values of q . A more generd version of a Gaussian distribution i s
8-2. Calculate the variance asswiafed with the probability distribution given in h b l e m B-I. The necessary integral is (CKCiab1e.r)
8-3.Using the probability distributiun given in Problem B-1, calculate the probability that the This expression looks like those in Figure B.3 except that the curves are centered at x = {x) rdther than x = 0. A Gaussian distribution is one of the most important and commonly used probability distributions in all of science.
particle will be found between 0 and a / 2 . The necessary integral is given in Problem H-I.
B 4 . Prove explicitly that
by breaking the integral from -m to w inlo one from -m to 0 and another from 0 to m. Now let z = - x in the first integral and z = .r ill the second to prove the above relal~un.
B-5. By using the procedure in Problem B 4 ,show explicitly that
B-6. According to the kinetic theory of gaws, the molecules in a gas travel at various speeds, and that the probability that a molecule has a speed ktween v and u + du is given by
F I G U R E B.3 A plot of a Gaussian distribution, p ( x ) , (Equation B.21) for three values of 5 The dotted curve corresp)nds to crc = 2, the soIid curve to oG= I , and the dash-dottad curve to ux = 0.5.
where rn is the mars of the particle, k, is the Boltzmann constant (the molar gas constant H divided by the Avogadro constant), and T is the Kelvin temperature. The probability distribution of tnolecubr speeds is called the Maxwell-Boltzmann distribution. First show that p ( v ) is normalized, and then determine the averige speed as a Function of te~npc~lture. The necessary integrals are (CRC tables)
Problems B-1. Consider a particle to be constrained to lie along a one-dimensional segment 0 to a. Quantum mechanics tells us that the panicle is found to lie between x and x dx is given
+
and
by
where n! is n faclorial, urn! = n(n
= 1 , 2, 3. . . .. Firs1 shuw that p { x ) i s normalized. Now calculate the average psition of the panicle along the line segment. The inlegtala that you need are (Th CRC Handhlr~kr!F Chernistv and Physics or Th@CRC Standard Mathematical Tobles, CRC whcrc rr
Prefs)
- t)(n
- 2) . . . (I).
B-7. Use the Maxwell-Boltzmann dishbution in Problem EM tqdetemine the average kinelic energy of a gas-phase molecule as a function of temperature. The necessary integral is given in Problem R-6.
The Boltzmann Factor and Partition Functions
Ludwig Bo~tzmannwas born in Vienna, Austria. on February 20, 1S44, and died in 1906. In 1867. he received his doctorate from the University of aenna, where he studied with Stefan (uf the Stefan-Boltzmann equation). He worked on the kinetic thwry of g ~ and s did experimental work on gases and radiation during his stay there. Although known for his theoretical work, he was an able experimentalist but was handicapped by poor vision. He was an early proponent of the atumic theory, and much of his work involved a study of the atomic theory of matter. In 1869, Boltzmann extended Maxwell's theory of the distribution of energy among colliding gas molecules and gave a new cxpression For this distribution, now known as the Boltzmann factor. In addition, the distribution of the speeds and the energies of gas molecules is now called the Maxwejl-Boltzmann distribution. In 1877, he published his ramous equation, S = k, In W, which expresses the relation between entropy and probability. At the time, the atomic nature of matter was not generally accepted, and Boltzmann's work was criticized by a number of eminent scientists. Unfortunately, Boltzmann did not live to see the atomic thwry and his work corroborated. He had always suffered from depression and committed suicide In 1906 by drowning.
,
In previous chapters, we learned that the energy states of aroms and molecules, and for all systems in fact, are quantized. These allowed energy states are found by solving the Schrijdinger equatiun. A practical question that arises is how the molecules arc distributed over these energy states at a given temperature. For example, we may ask what fraction of the molecules are to be found in the ground vibrational statc, the lint excited vibrational state, and so on. You may hive an intuitive feel that h e populations of excited states increase with increasing temperature, and we will see in this chapter that this is the case. Two central themes of this chapter are the Boltzmann factor and the partition function. The Boltzmann fdc~oris one of the most fundamental and uset'lll quantities of physical chemistry. The Boltzmann factor tells us that if a system has states with energies E,, E,, E,, . . ., the probability pl that the system will be in the state with energy E, depends exponentially on the energy of that state, or
where kB is the Boltzmann constant and T is the kelvin t e m p t u r e . We will derive rhis result in Section 3-2 and then,discussits implications and applications in the remaindcr
of the chapter. The sum of the probabilities must equal I , so the normalization constant for ihe above probability i s l / Q where
The quantity Q i s called a partition function, and we will see that partition func~ior~s play a central role in calculating the properties of any system.For example, we will show that we An calculate the energy, heat capacity, and pressure o f a system in terms of Q. In Chapter 4, we will use partition functions to calculate the heat capacities uf monatomic and polyatomic ideal gases.
105
'
3-1. The Boltzrnann Factor Is One of the Most Important Quantities in the Physical Sciences Thermal reservoir \
Consider some macruscopic system such as a liter of gas, a liter of wqakr, or a kilogram of some solid. Frum a mechanical pc~intof view, such a system can be described by specifying the number of particles, N , the volume, V , and thc forces between the particles. Even though the system conrains on the order of Avugadro's number o f particles, we can still consider its Hamittonian operator and its associated wave functions, which will depend upon the coordinates of all the particles. The Schrijdinger equation for this N-body system is
Thermal
insulation'-
F I G U R E 3.1
where the encrgies depend upon both N and V , which we wilJ emphasize by writing
E,(N, V ) . For the special case of an ideal gas, the total energy E , ( N , V) will simply be a sum of the individud molecular energies,
because the molecules of an idea1 gas are independent of each other. For example, for a monatomic ideal gas in a cubic container with sides of length a, if we ignore the electronic states and fwus only on the translational states, then the ss are just the translational energies given by (Equation 1.45)
An enfemble, or collection. 01- (rnacmscopic) systems in thermal equilibrium with a heat reservoir. The number of systems in the state j [with energy E ( N ,V)]is a,, and the total number of systcms in the ensemble is A. Bccause the ensernbleJisa conceptual construction, we may consider A tu be as Iarge as we want.
energies E l ( N , V ) and E , ( N , V). The relative number of systems in the states with energies E, and E, must depend upon E, and E,, so we write
where a , and a, are the number of systems in the ensemble in states I and 2 and where the functional form o f f is to be determined. Now,because energy is a quantity that must always be referred to a zero of energy, the dependence un El and E, in Equation 3.4 must be of the form
- -h h2 a Z (nf t n5 +n:>
En x R ) n r -
Note tha~EjCN, V) depends upon hr through the number of terms in Equation 3.2 and upon V Ihrough thc fact that a = v ' I 3 in Equation 3.3. For n more gencral system in which the particles interact with each other, the E , ( N , V ) cannot be writtcn as a sum of individuaI particle energies, but we can still consider the set of allowed macroscopic energies V ) } ,at least in principle. What we want to do now is determine the probability that a system will be in the statc j with energy E , ( N . V). To do this, we consider a huge collection of such systems in thermal contact &th an essentially infinite heat bath (called a heat reservoir) at a tcmpenlure T. Each system has the same values crf N, V, and T but is likely to be in a different quan~umstate, consistenl with the values of N and V . Such n collection of systcms is called an ensemble (Figure 3.1). We wit1 denole the number of systems in thr slate j with rnergy E, ( N Y V ) by rzj and the total number of RySkmS in the ensemble by A. We now ask for the relative number of systems of the ensemble that would be frorrnd in each state. As an example, let's fwus on two particular states, I and 2, with
{q(N,
In t h i s way, any arbitrary zero of energy associated with El and E, will cancel. Thus, we have so far
5 = ,f ( E , - E , ) Equation 3.6 must be true for any two energy states, so we can also write I=f(&-E a and , ) a,=f(~,-E,) a,
But
so using Equations 3.6 and 3.7, we find that the function f must satisfy
Chapter 3 1 The Boltzrnann Fnclor and Pdrlilion Functionc
108
The form of the function f that satisfies this equation may not be obvious as first sight,
3-2. The Probability That a System in an Ensemble Is In ll~eState j
If we substitute this resuit hack into Equation 3.1 I , wc obtain
but if you remember that
then we can see hat
f ( E ) = eSE
The ratio q / A is the fraction of systems in our ensemble that will k found in the state j with energy E,. En the limit of large A, which we are certainly able to take because we can make our ensemble as large as we want, a ) / Abecumes a prohahility (Mathchapter B), so Equation 3.12 can be written as
where B is an arbitrary constant (see also Problem 3-2). To verify that this form for f does indeed satisfy Equation 3.8, we substitute this functional form of f (E) into Equation 3.8:
Thus, we find from Equation 3.6 that
There is nothing special abrlut the states 1 and 2, so we can write Equation 3.9 more generally as
The form of this equation implies that both antand an are given by
where pi is the pmbability that a randomly chosen system will be in state j with energy EjIN, V ) . Equation 3.13 is a cenual result of physical chemistry. We customarily let the denominator in this expression be denoted by Q,and if we s p i f i c a l l y includc the dependence of E, on N and V, then we write
Equation 3.13 becomes
We are nrlt quite ready to detenninc B at this point, but liiter we will present severill different arguments to show that
where j represents either stale rn or n and C is a constant.
3-2. The Probability That a System in an Ensemble Js in the State j with Energy E j ( N , V ) Ir Proportional to e - E ~ " 5 V ' i k ~ T Equntion 3.11 has two quantities, C and f3, that wc must determine. Determining C is fairly easy. We sum bolh sides of Equation 3.1 1 over j to obtain
B u t the summation over a] w s t equal A, the total number of systems in the ensemble. Therefore, we have
where k, i s the Boltzmann constant and T is the kelvin temperature. Thus. wc write Equation 3.1 5 as
v;lrl
We will useEquations 3.15 and 3.17 interchangeably E ~ u ~ ~3 I15 I 1s Ljust I ~ as ~ acceptable as Equation 3.1 7. From a theoreticai point of view, b, or I / k,T, often happens to be a more convenjenl quantity to usc than T itself. The quantity Q ( N, V, B ) , or Q ( N , V , T ) , is called thra ~ ~ ( z r i i l ~,#rtnc~ion on (IT the bystcln, and we will see in the next fcw chapters that wc can express all the macroscopic properties of a wstem in terms of Q ( N , V ,b). A t t h ~ spolnt, il may not seem pc~ssiblc to determine all the energy states { E , ( N , V ) ) never mind g ( N , V , B ) , but yclu will learn that we can determine &(A', I.', B ) for a number nf interesting and ~mportant systems.
Uhaptcr 3
/ The Boltzmann Factor and Parlition Functions
3-3. We l'ostulstt: That Ihe Average Ensemble Encvy Is Equal tu the Ot>semltntlrgy ot a Syrlern
We will learn in Chapter 4 hat fur a rnona!omic ideal gas, EXAMPLE 3-2 We will learn in the next chapter that fur the rigid rotator-harmtmic oscillator model of an ideal diatomic gas, the partition function is given by
where where
For a monatomic ideal gas in its electronic ground state, the energy of the system is only in the translational degrees of freedom. Before we substitute Equation 3.22 into Equ~tion3.20, we write In Q for convenience as a sum of terms that involve B and terrns that a r t independent of 8: InQ=NInq-lnN! 3A' 2
In this expression, I is the moment of inertia and v is the fundamental vibrational frequency of the diatomic molecule. Note that q ( V , B ) for a diatomic molc~uleis the same as the expression for q ( V , B ) fora monatomic gas (Equation 3.23, a tnnstational term), except that it is multiplied by a rotational term, 8 r r ' l / h 2 @ , and a vibrational term, e-ph"/2/(1- e-Oh'). The wd5m for this difference will become apparent in Use this partition function to calculate the average energy uf one mole of Section a. a diatomic ideal gas. S 0 L U T 10N : Once again, for convenicnce we write In Q as the hum of terms that
involve 8 and terms that are independent of 8:
3N
--In B + terms involving only N and 2
InQ = Nlnq - l n N !
V
Now we can see more easily that
+ terms nut involving fi Now 3NdInp
N d- l- n- b- N Nhv
dB
atid that (Equation 3.20)
For n moles, N = nNA and kaNA= R , SO (E)= ; ~ R T
This observation leads us to a fundamental postulate of physical chemisiq that the ensemble average of any quantity, as calculatd using the probability distribution of Equation 3.17.is the same as the experimentally observed value of that quantity. If we let the experimentally observed energy of a system be denoted by U, then we have
for one mole of a monatomic ideal gas. (We indicate a molar quantity by w overbar,)
2
3 Nhv U = ( E ) = -Nk,T+Nk,T+-+2 2
dIn(1-e-"") dB
~hve..~~"
For one mole, N = N, and N,k, = R , so
Equation 3.24 has a nice physical interpretation. The first term represents the average translational energy, the second term represent5 the average rutatiunal energy, the third term represents the zero-point vibrational energy, and the founh tenn represents the average vibratihal energy. The fourth term i s negligible ill low temperatures for most gdseb but increases with increasing temperalure as the excited vibrational states become populated.
3-4. The Heat Capacity at Constant Volume Derivative of the Average Energy
Is the Temperature
E X A M P L E 3-3 111 1905, Einstein proposed a simple lnndel for an alomic crystal that can be used to calculate the molar lwat capacity. He piclured an atomic crystal N atoms situated at lattice sites, with each atum vibraling as a three-di~nensiunalharrnorlic oscillator. Recause all the lattice sites are identical, hc further assumed that each atom vibrated with the same frequency. The partition functinn aswiated with this model is (hob-
The constant-volume heat capacity, C , , of a system is defined as
lem 3-20)
C, is then a measure of how the energy of the syslem changes with temperature at constant amount and volume. Consequently, C , can be expressed in terms of Q(N,V , T ) through Equation 3.21. We have seen that = 3 R T / 2 for one mole of a monatomic ideal gas, so The heat capacity
-
C, = ; R
monatomic ideal gas
SOLUTION : The average energy i s given by (Equation 3.20)
For a diatomic ideal gas, we obtain from Equation 3.24
5
2
where v, which i s characteristic uf the particular crystal, is the frequency with which the atoms vibrate about their lattice positions and U , is the sublirnatiun energy at O K, or the energy needed to separate all the atoms frvm one another at 0 K.Calculate the molar heat capacity of an atomic crystal from this partilion function.
diatomic
(ideal ) gas
(3.27)
Figure 3.3 shows h e theoretical (Equation 3.27) versus the experimental molar heat capacity of O,(g) as a function of temperature. The agreement between the two i s seen to be excellent.
NOW that U consists uf three terms: U,. the sublimation energy at 0 R,3 N h 1 ~ / 2 , the zcro-point energy nT N thrcc-dimensional hannonic oscillators; and a term that represents the increase in vibrational energy as the temperature increases. The heat capacity at constant volume is given by
where we have used the fact that
r i c u u E 3.3 l'hc r x p r i l n c ~ l ~and ; ~ l theoretical (Equation 3.27) molar heat capacity of O,(g) from 300 K to I(HWI K 'rhc ~hcorcti~al cunvc(wllid curve) is calculated using hulk = 2240 K.
N = N, and NAk,= R for one mole.
Equation 5-29 contains one adjustable parameter, the vibrational frequency v . Figure 3.4 shows the molar heat capacity of diamond as a function of temperature calculated with u = 2.75 x 1013 s-'. The agreement with e x ~ r i m e n tis wen 10 be fairly good considering the simplicity uf the model.
3-5. We Can Express the Pressure in Terms of a Partllion F u n c l ~ r > ~ ~
Using the facl that the average pressure is given by
we can write
This expression can k written in a more compact form. Let's stan with
F I G U R E 3.4
The vhserved and thoretical (Einste~nmodel) rnulnr heat capacity of diamond as a function of tempemure. The solid curve is calculated using Equation 3.29, and the circles represent expt.ri~ilentnldata.
and differentiatei t with respect to V keeping N and fi fixed:
Comparing this result with the second equality of lquation 3.31 shows lhat It is interesting tu look at the high-temperature limit of Equation 3.29. At high temperatures, h v / k , T is small, so we can use the fact that d I x for small x (Mathchapter C). Thus, Equation 3.29 becomes
+
Thix result predicts that the molar heat capacities of atomic crystals should level off at a value of 3 R = 24.9 J-K-'.mol-' at high temperatures. This prediction i s known as the law of Dulong and Petit, which played an important role in the determination of ittolnic masses in the I XOOs. This prcdictic~nis in good agreement with the data shown in Figure 3.4.
3-5. We Can Express the Pressure in Terms of a Partition Function We will show in Sec~ion54-that the pressure <>famacroscopic system is given by
Just as we equated the ensemble averdge of the energy with the observed energy, we equate the enselnble average pressure with the observed pressure, P = ( P ) . Thus, we see that we can calculate the observed pressure if we know Q ( N , V. 8 j. We can use rhis result to derive the ideal-gas equation of state. Firqt, considel :t monatomic ideal gas. Recall from Equation 3.22 that Q ( N , V , @ ) for a monatomic ideal gas i s given by
where
Let's use this result to calculate the przssurc o f a monatomic idcnl gas. 'To evaluate Equation 3.32, we write out In Q first for convenience:
118
Chapler 3 1 The Boltzmann racror and Partition Functions
Because N and
are fixed in Equation 3.32, we write In Q as
36. The Partition Function of a System of Independent, Distinguishable Molecules Is the Product of Molecular Partition Functions
In Q = N In V t terms in N w d /? only Therefore,
and substituting this rcsult into Equation 3.32 gives us
as you mrght have expected.
Nolice that the ideal-gas cquation results from the fact that In Q = N In Vf terms in N and f l , which comes from the fact that q ( V , T) is directly proportional to V in Equalion 3.22. Exampfc 3-2 shows that q ( V , T ) is directly proportional to V for a diatomlc ideal gas also, and x t ~P V = N k , 7' for a diatomic ideal gas. This is the case for a pc)lyatomic ideal gas as well, so the ideal-gas equation of state results for any ideal gas, rn~lllatc~rnic, diatomic, or polyatomic.
EXAMPLE 3 4 Calculate the equation of state mociated with the p d t i o n function
where
(I
The general results we have derived up to now are valid for arbitrary systems. To apply thcse equations, we need to have the set of eigenvalucs ( E j ( N ,V ) ] for the N body Schrdinger cquation. In general, this is an i~npossibletask. For many important physical systems, huwever, writing the tutal energy of the system as a sum of individual energies is a good approximation. This procedure leads ro a great sirnpliticalion of the partition function and allows us to apply the results with relative ease. First, let's consider a system that consists of independent, distinquishablc par~icles. Although atoms and molecules ;ire certainly not distinguishable in general, they can hc treated ax such in a number of cases. An excellent example is that of a perfcct cryxtal. In a perfect crystal, each atom is conlined to one and only one lattice site, which we could, at Icast in principle, identify by a set of three coordinates. Because each particle, then, is confined to a lattice site and the lattice sites are distinguishable, the particles themselves are distinguishable. We can treat the vibration of each particle about its lattice site as independent to a fairly good apprt~ximation,just as we did for normal modes of polyatomic molecules. We will denote the individual particle energies by ( E ; ] , where the superscrip1 denotes the particle (they are distinguishable), and the subscript denotes the energy state of the particle. In !his case, the total energy of the system E , ( N , V) can be written as
a~lrth are collstents. Can you identify the resulting cquation or slate?
SOLUTION: We usc Equation 3.12 to calculate the equation of shte. First, we evaluate In Q,which gives In Q
=N
ln(V
- N b ) + PUN' -+ terms irl hf and 9 , v
We now differentiate with respect
tc)
and the system partition function bcuomes
only
V , keeping N and constant, to get
Because the particles are distinguishable and independent. we can sum over i , j, k. . . . independently, in which case Q ( N , V , T) can be written as a product o f individual summations (Problem 3-21): and so
Bringing the last tern tu the left sidc and rnulliplying by V
- Nb
gives us
where each of the q (V, T) is given by which is the van der Waals equation.
120
Chapter 3 / The Boltrmann Factor and Part~tinnFunctions
In many cases, the (e,) is a set of molecular energies; thus y ( V , T ) is called a rnolecuhr purrition function. Fquation 3.33 is an important result. It shows that if we can write the total energy as a sum of individual, independent terms, and if the atoms or molecules are dis~rrtguishable,then the system parlition function Q ( N ,V , T ) reduces to a product of molecular partition funclions q ( V , T). Because q ( V , T) requires a knowledge of the allorved energies of only individual atoms or n~olecules,its evahatiun is often feasible, as we will sce For a number of ca.ws in Chaprer 4. H' the energy stater of all the atotns or mo1ecu)e~ are the same (as for a monatomic crystaf), tlien Equatir~n3.33 becomes
Q W , v , T ) = [q(V,T)]"
distinguishable atoms or rnoleculcs
)
0.35)
where
3-7, Thc Part~t~on Flmr 11onof a System of Independent. Ind~,l~nguishabIrAtvrrls ur Molcr ulc5
(note the lack (IT distinguishing supcrscripls, as in Equation 3.33) and the system partition function is
Because the particles are indistinguishablc, we cannot sum over i,j, k , . .. separately as we did in Equation 3.33. To see why, we must consider a fundamental property of all particles. You iemed in general chemistry that the Pauli Excluslun Principle says that no two electrons in an atom can have the same set of four quantum numbers. Anc~therway of saying this is that no two electrons in an atom can be in the same quantum state The Pauli Exclusion Princ~pieis actually more general than the above staiement, and applies to all particles o f spin 112, 312, 512, and so on. Such particles, vallcdJerennrons, have the restriction h a t no two fermions In a system can occupy the same quantum statc. Examples of fermions arc: electrons (spin 1/2j, protons (spin 1/2), and neutrons (spin 112). Particles that have spin of 0,1, 2,. ., called busnns, do not have any restri~rion regarding the occupancy uf individual quantum states. Examples of hosons are alpha particles (spin 0) and photons (spin 1). It turns out that fermions and bosuns constitutu all the known particles in nature. We must recognize the occupancy requirementh 01 fermions and b s o n s when we aKempt to c q out the summation in Equation 3.37. Let's go back now to the summation in Equation 3.37 for the case of fermions. Because no two identical fernions can occupy the same single-particle energy slate, terms in which two or more indices are the same cahnol be included in the summation. Therefore, the indices i , j, k , . . . are not independent of une another, \o a direct evaluation o f Q ( N , V, T ) by means: of Equation 3.37 p s e s problems for fenniuns.
.
The Einstein rnidel of atomic crystals (Example 3-3) considers rhe atoms to be fixed at lattice sites, so Equation 3.35 should be applicable to that model. Notice that the partition function of thal model (Equation 3.28) can be written in the form of Equation 3.35 if we let u, = U J N he the sublimation energy per atom at O K, in which
case we have
EXAMPLE
3-7. The Partition Function of a System of Independent, Indistinguishable Atoms or Molecules Can Usually Be Written as
3-5
Consider a system of two noninterxting identical fermions, cach of which has states with energies E , , c,, E,, and E,. Enumerate the allowed total energies in the summation in Equation 3.37.
[q(v,T I I ~ I N ! SOI.UTION: Forthissystem
Equaliun 3.35 is ail attractive resuit, but a t o m and lnolecules are, in genera!, not distinguishable; thus the utility of Equation 3.35 i s severely limited. The reduction of a system partition function Q ( N , V, T) to m o l e c u l ~partition functions q ( V, T ) bccornes somewhat more complicated when the inherent indisringuishahility of atnlns and mulccules cannot he ignored. For indistinguishable parlicloa. !he toral entrgy is
-
Of the 16 terms that would wcur in an unrebrricted evaluation of Q, only six are allowed for two identical fermions; these art! ihc tenns with energies
121
Chdpter 3 1 The Boltzmann ractor and Panition t'unctions
l'he six terms in which ihe el are wri tten in reverse order are the same as those above (because the particles are indislinguishable), and the four terms in which the R, are the same are not allowcd (bccau= ttle particles are fermiol~s).
Bost>nsd o nut have he restric~ionthat no two of the samc type can occupy the samc single-particle state, but h e summation in Equation 3.37 is still complicated. To see why, consider a term in Equation 3.37 in which all the indices are the same except for one; for examplc, a tcrm like
123
3-7, The Pnrriticm Funrtiun ul a Syst~mof Ind~-pentleut,lndirringuishableAtoms or hlolecu!m
same state. Although most of the quantum-mechanical systems we have studied have an infinite number of energy states, at any given temperature many of these will not be readily accessible because the energies of these states are much larger than k,T, which is roughly the average energy of a ~nolecule.If, howcver, the number of quantum states with energies less than roughly k,,T is much larger than the numher of particles, then essentially all the tertns in Equation 3.37 will c o n t ~ i nE'S with different indices, and so we can evaluate Q ( N , V ,T) to a good approxilnation hy summing over i,j , k , . . . independcntly in Equation 3.37 and then dividing by N! to get Q ( N 3V . T ) =
independent, indistinguishable
N!
(3.38)
(in reality, these indices might be enormous numbers). Because the panicles are indiswhere tinguishable, the position of the term E , is not important, and we could just as easily t ~ a v e e , , + ~ , + ~ ~ , + ~~, ,~, f~ + ~ , , , o r ~ , , , + ~ , , , + ~ , + ~ , , t ~ ~ ~ + ~ , ~ a n d s o o n . B e cause t h e ~ eterms all represent the same state, such a state should be included only ollce in Equiition 3.37, but an unrestricted summation over all the indices (summing oker I , j , k . . . . independently) in Equation 3.37 would produce R, terms of thiq type The number of translational states alone i s usually sufficient to guarantee that the (the F , can he located in any of the N positions). number of energy states available to any atom or molecule is greater than the n u ~ n k r consider the other extreme in which all the N particles are in different of particles in the system. Therefore, this procedure yields an excellent approximation moleculx states; that is, for example, a system energy of E, E, F, E, -t. . . s,. in many cases. The criterion that the number of available states exceeds the nurnher of Because the particles are indistinguishable. all N! arrangements obtained by permuting particles so that Equation 3.38 can he used is thesc N tcrrns are identical and shauld wcuronly once in Equation 3.37. Yet such terms lvill appear N !limes in an unresrricted summation. Consequently, a direct evaluation of &( N , V , T ) by menns of Equntiun 3.37 poses prublems for busons as well as ferrnions.
c)w
I
E X A M P L E 3-6 Rudo Example 3-5 for bosons
+ + +
+
instcad of fermiolls.
S U L U T I O N : In this case there are 10 allowed terms: the six that are allowed in Example 3-5 and the four in which the F, are the same (bosons do not have the restriction that no two can occupy the same state).
Notc that in every case, the tcrrns in Equatic~n3.37 that causc difficulty are those in which two or more indices are the same. 1f it were not for such terms, we could carry out the summation in Equatinn 3.37 in an unrestricted manner (obtaining [ q ( V , T)]" as in Section 3-6) atid then divide by N! (to obtain [ q ( V ,T ) ] ' / N ! ) to account for the over-counting. For example, if we could ignore terms like a, E , , E, E,, etc. in the evaluation of Q ( 2 , V , T), there would be a total of 12 terms, the six enumerated in Example 3-5 and the six rn which the energies are written in reverse order. By dividing hy 2!, we would obta~nthe correct number alluwed terms. Certainly, if the number of quantum states available to any particle is much greater than the number of panicles, it would be unlikely for any two particles to be in the
+
+
Notice that this criterion is favored hy large particle mass, high telnperature, and low density. Although our discussion at this poinl is limited t o ideal gases (independent, illdistinguishable particles), we show the valucs of ( N /~ ) ( k ~ / X r n k , ~ in ) ~'raahle " 3.1 even for some liquids at their boiling points, just to show that inequality 3.40 is easily satistied in most cases. Note that the exceptional systems include liquid helium and liquid hydrogen (because of their small masses and low temperatures) and electrons in metals (because of their very small mass). These systems arc the prototype examples of quantum systems that must be treated by special methods (which we will not discuss). When Equation 3.38 is vdid, that is, when the numkrofavailable m~lecularatates i s much greater than the number of particles, we say that the particles obey B o l ~ i t n u t ~ n statistics. As Inequality 3.40 indicates, Boltzmann statistics becorncs increasingly valid with increasing temperature. k t ' s test Inequality 7.40 for N,(g) at 2U"C and one bar, Under these conditions,
T A B l E 3.1 The quantity ( N /V ) ( h 2 / 8 r n k , ~ ) 3 tat 2 a pressure of one bar for a number of simple systems.
3-8. A Molecular Partition Function Can Be Decomposed into Partition Functions for Each Degree of Freedom In this section, we will explore the similarity hetwcen a system partitiun function. Equation 3.14, and a molecular partition function, Equation 3.39. We will stan by substituting Equation 3.38 into Equation 3.21 :
System
Liquid heliu~n Gaseous helium Ga%ous helium Gaseous hellurn Liquid hydrogen Gaseous hydrogen Gaseous hydrogen
Liquid neon Gaseous neon
But Equation 3.38 is valid only for independent particles, so
Liquid krypton Elecwns in metals (Na)
( E )= N {E)
-
(3.42)
where ( E ) is the average energy of any one molecule. If we compare Equations 3.41 and 3.42, we see that
and h2 -= 8mk,T
(6.626 x lo-" 4,s)' kgl(1.381 x 10-23 J.K-')(293.2 K)
(8)(4.653 x
We can conclude from this equation that the probability that a molecule is in its jth molecular energy state, x J ,is given by
which is much less than unity.
Let's test Inequality 3.40 for liquid nitrogen at i t s baiting point, - 195.8"C.Expermen tally, the density of N,(l) is 0.808 g . m ~ - at ' its boiling point. Therefore,
N
(28.02molgN?N, ) (6.022mol10")
V = (0.808 g . r n ~ - ' )
I
l
x
Note how similar this equation is to Equation 3.13. Equation 3.44 can be reduced even further if we assume that the energy of n mthecule can be written as
(10' .L)
I m"
and
Because the various energy terms are distinguishable here, we can apply the reasoning behind Equation 3.33 and write
where, for exampk
This, Equation 3.38 is valid, even for liquid nitrogen at its boiling point.
Chapter 3 / Thp Boltzrnann Factor and Partit~unFunLTions
Note that the partition function for a diatomic molecule we used in Example 3-2 was expressed as
1-8. A Mvlecular Partition Function Can Be Decornped into Partition Functruns
and
where EXAMPLE
3-7
Use the partition function for a diatomic molecule giver1 in Example 3-2 to calculate (E"'~).
SOL U T I 0 N : Acctlrding to Example 3-2. we can write
and and so
If we substitute Equations 3.45 and 3.46 into Equalion 3.44, we obtain in agreement
with Equation 3.24.
I
where rr,,,, is the probability that ti molecule is in the ith translational stale, the j t h rutationai stale, the kth vibrational state, and the Ith electronic st*. Now if we sum Equation 3.48 over i (all translational states), j (all rntatiunal states), and 1 (all electronic states), we obtain
I
To this point, we have written partition fi~nctionsas sumrna~ionsover energy strlfrs. Each state is represented by a wave function with an associated energy. Thus, wc write
We will call sets of states that have the sarnc energy, levels. We can write q ( V . T )as a summation over levels by including the degeneracy, g,. of the level:
wliere, as the notation suggests, Chis the probability that a molecule is in its k t h vibrational state. Furthermore, the average vibrational energy of a molecule is given by
In the notation of Equation 3.53, the ternis represcnting a degencratc levcl are rrpculed 3.54, lhey are written once and rnulliplid by #,. For example, we learned in Section 1-8 (Equations 1.28 and 1.29) thul the energy und degeneracy for a linear rigid rotator are gj times, whereaq in Equation
Again, note thc siail:vily-with I$uatit,n 3.2 I .
Of course, we also have the relations and
128
Chaprer 3 !The tloltzmann Factur
rartition Functions
Thus, we can write the rotational parlition function by summing over levels:
Including degeneracies explicitly as in Equatiun 3.54 i s usually more convenient, s o we will use Equation 3.54 rather than Equation 3.53 in later chapters.
Problems
and that
3-7. Generalize thc results of Example 5 1 to the case of a spin-l nucleus. &!ermine the low-ternperdure and high-temperature limits of (E). 3-8. If Nw i s the n u m k r of protons aligned with a magnetic field B2 and No is the numher of protons opposed to llic field, show that
Problems 3-1. How wouid you describe an ensemble whose systems are one-liter containers of water at 25'C:I
+
5-2. Show that Equation 3.8 is eyuiwlcnt to f (x y ) = f ( x ) f (y). In this pmblem, we will prove that f (x) a eU*. First, take the logarithm of the above equation to obtain
= 26.7522 x lo7 rad.T-' .s-' for a proton, cakulatc N,,/Nw its 3 function or temperature for a field strength of 5.0 T. At what kmperature i s No = Nu? Interpret this
Given that y
result physically.
3-9, In Section 3- 3, we derived an expression tor ( E ) for a nlonato~nicdeal gas by a p p l y i ~ ~ g Equatiticln 3.20 to Q(N,V, T) given by Equation 3.22. Apply Equatiun 3.21 to Differentiate both sides with respect ro x (keeping y fixed) to get
to derive the same result. Note that this expression for @ ( N , V. T ) is simply Equation 3.22 with fl replaced by I / k , T .
Now differentiate with respect to y (keeping x fixed) and show that
3-10. A gas absorbed or1 a surface can sometimes b modelled a.i a two-dimensiunal ideal gas. We will learn in Chapter 4 that the partition function uf a two-dime~lsionalideal gas is
For this relation to be true fur all x and y , each side must equal a ctmstant, say a. Show that f (x) a e"'
and
f(y)cr.P'
where A is thc area of the surface. Derive an cxprebsion fur ( f )and comparc your rcsull with the thm-dimensiuml result.
3-3. Show that a,/a, = ~ " ' 4 -" 1 ' implies that a = C@8$ 3-4. Prove to yourself that
C,
=
~ ' " 8
2,:,t.-'4.
3-11. Although we will not do so in this book, it is possible toderlve the parl~tionfunction fur a monatomic vun der Waals gas.
3-5. Show that the partition function in Example 3-1 can be written as
w h m a and b are the van der Waals uvn\tunts. Derive an expression for the energy of a
Ilw lhc fact that d C O S ~X
= sinh x to show that
monatomic VAIIJer Waals gas.
/ ~ X
3-6. IJ,wtrthtr the cxprr*rlr*nfor ( E ) ill Example 3-1 or Lhe one in Pruhlenl 3-5 to show that
3-12. An apprcmirnu~cpurrition function for a gas of hard spheres can be ubtilined from the paflition function 111s monatomic gas by replacing V in Equation 3.22 (and the lotlowing equation) by - b, where b is related to the volume of the N hard spheres. Uerivc errprc~siunsftlr the energy and the pressure of thif system.
3-13. l l e the partition function in Problem 3-10 to calculale the ticat cap:lcity of a ~ w o d~mensionuld e a l gas.
rhapter 3 1 Thc Boltrrnann Fartor and Partition Functiorls
130
3-14. Use the partition function for a monatomic van der Waals gas given in Problem 3-1 1 to calculate the heat capacity of a monalomic van der Waals gas. Compare your result with lhalol a nlonatornic ideal gas. 3-15. Using the partition function given in Example 3-2, show that the pressure of an ideal diak~rnicgas obcys P V = N k , T , juct a
3-16. Show thar if a partition function is of the forn~
and if q ( V . 7') = f ( T ) V [as it dms for a monatomic ideal gas (Equation 3.22) and a diato~nicideal gak (Example 3-2)1, then the ideal-gas equalion o f ftatc rcsults.
3-17 . Use Equation 3.27 and the value ut' i; for 0, given in Table 1.4 to calculate the value of thc molar heat capacily of O,(g) from 3W K to 1000 K (see Figure 3.3).
and that
where U , sirnply represents the zero-of-energy, wherc all
1V
atoms are infinilely separated.
3-21. Show that
by summing over j Rrst and then over i. Now obtain the same result by writing
S as a
product of two separate summations.
3-22. Evaluate
3-18. Show that the heal cnp~citygiven by Equation 3.29 in Examplc 3-3 obeys a law of correspr)nclir~gstales.
3-19. Consider a syslem of indepcndent, distinguishable particles chat have only two quantum states with energy (let e, = 0) and e , . Show that the molar heat capacity of such a system i\ given by
by summing over j first and then over i . Now obtain the same result by writing S as a product or two separate summations.
3-23. How many tcrrns are there in the Following summations? 3
2
a. .T = c c x j l i r=l j=l and that ', plr>tled against BE passes through a maximum value at OF, given by the solution lo P & / 2 = cothpls/2. Use a table of values of cotlix (for cxarnple. the CRc Stor~riitrrlbi(~rl~r~~r(rri~~,II Tuklrs) to show that BF = 2.40.
3-20. Deriring lhe panilion function for an Einstein c r y ~ t a iis not difficult (see Example 3-3). Each 01- [he N atoms of the crystal is aqsumed 10 vibratc independently about its lattice pofition, so that the crystal is pictured as N independent harmonic oscillators, each vibrating in three dircctio~ls.The partition function of a harmonic oscillator is
s=-yc.lYl s = xp-pykk 3
h
3
2
'
2
c.
i-l
,LO
!=I ,=I P=l
3-24. Consider a system of iwu nuninteructing identical fermions, each of which hac states with encrgics E , , E ~ and , E,. How many terms are there in the unrestricted evalualion o i Q(2, V, T ) ' Enumerate the allowed total energies in the summation in Equation 3.37 (see Example 3-5). How many tcrrns occur in Q(2. V . T ) when the ferrnion restriction is taken into account?
3-25. Redo h o b l e m 3-24 for the case of hsons inftead uf rerrniuns. 3-26. Consider a system of three noninterncting identical ierminns. each of which has statcs wilh energies E ~ and , E,. How mally terms are t l l m in the unwslrickd evaluatinn of Q ( 3 . V, T)? Enumerate h e a l l o w d total energies in the summation of Equation 3.37 (scc Exalnple 3-5). How Inany tcrnis occur in Q ( 3 , V. T ) when the fermion rcstrictioii is tillicn inlo account?
This summation is easy lo evaluate if you recognize it as the so-called geometric series (MalhChapter C)
Show that
3-27. Redo Problenl 3-26 for thc care oi bosons inslead of ferrnin~ls. 3-28. Evaluate (A'/ ~ ) ( h ~ / ~ r n k ,(see ~ ) Table " ~ 3.1) for O,(g) at its nnrn~alh i l i n g pr,int, 90.20 K. Usc the idcal-gas equation uF slate to calculate the density of O 1 ( g ) at 'Ml.20 K. 3-29. Evaluate i ~ V ) ( h/2 / 8 m k B ~ ) "(see 2 Table 3.1) fur He(g) at its normal hliting print 4.22 K.Use the ideal-gas equation of state tocalculate the denvily of lIc(p) ul 4.22 K . 3-30. Evaluate ( ~ / ~ j ( h ~ / ~ r n kfor , ~the ) ' electrons :~ in sodium ~nelnlat 298 K. Takc the density of sadiurn to 0.97 g.rnLL1. Compare your result with thc villuc givcn in 'l'ilhlc 3. I.
134
Chal)ler 3 / The Boltzn~.lnnFdcrur arid ILrlitrun Funct~ons
3-42.-The vibrational frequency of NaCl(g) is 159.23 cm-l. Calculate the molar heat capwity, T,, a1 IMKI K. (See Equalion 3.27.) 3-43. Thc energies and dcgencracics of the two lowest electronic siales or aiomic iodine are listcd bclow.
Energy/cm-'
Degeneracy
MATHCHAPTER
c
S E R I E S AND LIMITS
What temperature is required so that 2% of the atoms are in the excited state?
Frequently, we need to investigate the behavior of an equation for small values (or perhaps large vafues) of one of the variables in the equation. For example, we shall show in Chapter 4 that the temperature dependence of the molar heat capacity of a diatomic gas is given by
where R is the molar gas constant and (-Ivi, is a constant that is characteristic of the gas. Suppose now that we wish to determine C, at high temperatures, where 0+,,/ T is small. To do this, we first have to use the fact that P" can be written as the infinite series (i.e., a series containing an unending number of terms)
and then realize that if x is small, then .r2,x 3 , ctc. are even smaller. We can express this result by writing
where 0 (x2)is a brmkkeeping symbol that reminds us we are neglecting terms iiivolving x2 and higher powers of x . If we apply this result to Equation C.1, we have
MathChaptcrC / S E K I E S A N D L I M I T S
Thus, we see that C, takes on a limiting value o f 7 R / 2 at high ternprdtures. I n this Mathchapter, we will review some useful series and apply them to some physical
MathChaprerC IS E K I E S A N D L I M I T S
For the exponential series (Equatiun C.2), we have
problems. One of the must useful scries we will use is the gcometric series: Thus, wc conclude that the exponential series converges for all values o f x In Chapter 4, we encounter the summation
This result can be derived by algebraically dividing 1 by 1 - x, or by the fullowing trick. Consider the finite series (i.e., a series with a finite number of terms)
where v represents the vibrational frequency of a diatomic molecule and the othcr symbols have their usual meanings. We can sum (his series by letting Now multiply S, by x :
in which case we havc
Now notice that
is less than 1, and according to Equalion C.3, S = 1/(1
or that
The quantity
If 1x1 < 1 , then x N t ' + 0 as N + m, so wc recover Equation C.3. Recovering Equation C.3 from Equatlun C.4 brings us to an important point regarding infinite series: Equation C 3 is valid only if 1x1 < 1. It makes no sense at all ~f 1x1 2 I. We say that the infinite series in Equation C.3 converges for 1x1 < 1 and diverges for 1x1 2 I . How can we ~ c l whether l a given infinite series converges or diverges? Thew are a number of so-called convergence tests, but one simple and useful one is the mlio test. To apply the ratio test, wc form the ratic~of the ( n -t 1)th term, u , , + ~to, the nth term, u", and then let 11 hecumc very large:
We say that S has been evatuated in closed form because its numerical evaluation requires only a finite number of steps, in contras~to E q u d o n C.6, which would require an infinite number of steps. A practical questicln that arises is h ( ~ w do we find the infinite series that carrespuncis to a given function. For example, how do we derive Equation C.2? First, assume that the function ,f (x) can be expressed as a power series in x:
x
- I),or
where the c, are to be determined. Then let x = 0 and find that r,, = S(O). Now differentiate once with respect to x
If r < I , the series converges; if r > 1, the series diverges; and if r = 1, the test i s inconclusive. Let's apply this test to the geometric series (Equation C.3).In this case, ~ i ,,,= , xn+' and urn= r",so
w
Thus, we see lhat the scries converges if 1x1 < 1 and diverges if 1x1 z 1. It actuaIly diverges at x = I , but the ralio test dws not tell us that. We would have tu use a more sophisticated convergence test to determine the behavior at x = 1 .
and let x = 0 to find that c, = ( d f / d ~ ) ~ Differentiate =,. again,
and let x = O t o gut c, = (d2f/d~')~=,,/2. DiHi.rcntiatc oncc rrlrlrc,
MathChaprerC / S E R I E S A N D L I M I T S
and let x
= 0 to get c, = (d3f / d ~ ~ ) ~ _ ,The / 3 !general . result is
MathChaprerC / S E R I E S A N D L I M I T S
We can use the series presented here to derive a number of results used thruughc~ut the book. For example, the limit
sinx lim -
x r 0
x
su wc can write
occurs several times. Because this limit gives 010, we could use I'Hbpital's rule, which
tells us that Equation C.9 is called the Maclaurin series of f (x). If we apply Equation C.9 to f ( x ) = e", we find that
,
d sinx dx
sinx
lim -= lim -= lirncosx = I
I+n
x
.-.o
1-0
We could derive the same result by dividing Equation C.10 by x and then letting x + 0. (These two methods are really equivalent. See Problem C-14.) We will do one final example involving series and limits. According to a theory by Debye, the temperature d e ~ n d e n c euf the molar heat capacity of a crystal is given by
Some other important Maclaurin scries, which can be obtained from a straightforward application of Equation C.9 (Problem C-7) w e
In this equation, T is the kelvin tcmpcrature. R is the rnvlar gas constant, and (-1, is a parameter characteristic of the particular crystal. The parameter 0, has units of temperature and is called the Debye temperature of the crystal. We want to determine both the low-temperature and the high-tempcraturc limits of F, ( T ) .In the low-temperature limit, the upper limit o f the integral h o m e s very large. For large values of x, we can neglect 1 compared with ex in the denominator of the integmnd, showing that the integrand goes as x4e-" for large x. But x4e-" + 0 as x 4 co,so the upper limit of the integral can safely be set to m, giving Series C. 10 and C. 1 1 canvcrge for all values ufx, but asindjcated, Series C. 12 converges only for -1 < x 5 1 and Series C.13 converges only for x2 < 1. Note that if n is a positive inleger in Series C. 13, the series truncates. For example, if n = 2 or 3, we have
Whatever the value of the integral here, it is just a constant, so we see that -
and
C , ( T ) + constant
x T~
a
7' 10
This famous result for the low-temperature heat capacity of a crystal is callcd the T' law. The low-temperature heat capacity goes to zero as T'. We will use the T ' law
k q ~ ~ a t i aC.13 n for a positive integer is called the binomial expansion. If n is not a positive integer, the sergs continues indefinilely, md Equation C.13 is callcd the binr~mialseries. Any handbook of mathematical tables will have the Maclaurin series for many functions. Problem C-13 discusses a Taylor series, which is an extension of
a Mnclaurin series.
in Chapter 7. Now let's h k at the high-tempcmture limit. For high temperatures, he upper limit of the integral in Equation C.14 becomes very small. Consequently, during the integration fmm 0 to OD/T, x is always small. Therefore, we can use Equation C.2 for ex, giving
.- .- - - -- -
MathChapterC I S E R I E S A N D L I M I T S
-
lirn C , ( T ) = 9 R
T-tW
-
+ o(x2)idx
wTX ~ [ I
Problems
C-l 0. Evaluate the limit of
as x + 0.
C-1 1. Evaluate the integral
This result is called the Law of Dulong and Petit; the molar heat capacity of a crystal becomes 3 R = 24.9 ~ . ~ - ' . m o l - for ' monatomic crystals at high temperatures. By "high temperatures", we actually mean that T >> (-), which for many substances is less than 1030 K.
1=
1'
it-'
cos'xdx
for small values of a by expanding I in powers of a through quadratic rems. C-12. Prove that the series for sin x converges for all values of x .
C-13. A Maclaurin series is an expansion about the point x = 0. A series of the form
Problems C-1. Calculate the percentage difference between ex and l
+x
for x = 0.0050. 0.0100,
0.0150, . . . ,0.1000.
C-2. Calculate the percentage difierence belween In(] 0.0150, . . . ,0.1000. C-3. Write out the expansion of ( I
+x)ln
f x)
is an expansion about the paint x, and is called a Taylor series.Firsf show that c, = f (x,,). Now differentiate both sides of the above expansion with respect to x and then let x = x,, to show that c, = ( d f / d ~ )Now ~ ~ show ~ . that
and x for x = 0.0050, 0.0100.
through the quadratic term
C-4. Evaluate the series
and so
C-14. h e r on, we will need to sum the series
C-5. Show that
C-6. Evaluate the series
and
-
S2
=
C u2xU "4
C-7. Use Equation C.9 to derive Equations C.10 and C. I 1.
To sum the first one, start with (Equation C.3)
C-8. Show that Equations C.2, C. 10, and C. I 1 are consistent with the relation e'" = cosx f i sinx.
C-9. In Example 3-3, we derived a simple formula fur the inolar heat capacity of a solid based on a model by Einstein:
Differenbate with respect to x and then multiply by x tcr obtain
Using the same approach. show that where H i n Ik mrtlar gas cclnstant and (3, = h v / k , is a constant, called the Einstein Iclnpcrnturr, #hut i h uhurnc~crirticofthe solid. Show that this equation gives the Dulong ;~ntlI'ctil l i m ~ IC', t - - 3HI ut high Icltlpcrillurcs.
Partition Functions and Ideal Gases
In this chapter, we will apply the general results of the preceding chapter to calculate the partition functions and heat capacities of ideal gases. We have shown in Section 3-7 that if the number of available quantum states is much greater than the number of particles, we can write the partition function of the entire system in terms of the individual atomic or molecular partition functions:
This equation is particularly applicable to ideal gases because the rnr~leculesare independent and the densities of gases that behave ideally are low enough that the inequality given by Equation 3.40 is satisfied. We will discuss a monatomic ideal gas first end then diatomic and polyatomic ideal gases.
William Francis Giauque was born un May 12, 1895, in Niagara Falls, Ontario, Canada, to American parents and dicd in 19x2. After working for two years in the laboratory at Hooker Eleclro-Chemical Company in Ni~garaFalls, he entered the University of California at Berkeley with the intent of becoming a chemical engineer. He decided to study chemistry, however, and remained at Berkeley to receive his Ph.1). in chemistry with a minor in physics in 1922. tiis dishertation was on the khavior of materialsal very low temperatures. Upon receiving his Ph.D., Giiluque accepted a faculty position in the College of Chemistry at Berkeley and remained ~ticrufnr the rest uf his life. He made exhaustive and meticulous themochemical studies that cxpk)red the Third Law o f Thermodynamics. Tn particular, his very low terngerature studies of thc cntrlrpier uf substances validated the Third Law. Giauque developed the technique of u~liuhatic&magnetization to achieve low temperatures, He achieved a tempramre uf O.?S K , und othcr rescarch groups subsequently reachod ternperarures as low as 0 . 0 1 4 K using li~:i~lrl~ar*\ twhniquc. Togvthet with his graduate student Hemck Johnston, he spectroscopically i~lcr~ulictl thc iwo t$~iticn{~ unknown oxygen isotopes 17 and 18 in 1929. He was awarded the Nutwl I'rirr for clicrni\rry in 1q49 " h r his cuntributions in the fieldof chemical thermodynamics, prtictllilrly cuilccr1iit1g thc hchuvitrr of suhstnnces at extremely low temperatures.''
4-1. The Translational Partition Function of an Atom in a Monatomic Ideal Gas i s ( 2 j ~ n a k ~ ~ / h ~ ) " ~ 1 . ' The energy of an atom in an ideal monatomic gas can be written as the sum of its translational energy and its electronic energy
so the atomic partition function can be written as
We will evaluate the translational partition functirjn first
Chapter 4 !Partition Funclions and Ideal G s m
The translational energy states in a cubic conlaincr are given by (Equation 1 4 5 ) En", I
.
h2
=
(n:+n:+n;) 8ma2
n I , n y l n z = l , 2, . . .
(4.2)
We substitute Equation 4.2 into qtranh (Equation 3.47) to get
Because e"+h'' = e"ebe",we can write rhe triple surnnlation as a product of three single summations: F I G U R E 4.1 An illi~strdtionof thc approximation of a sulnrnalion C z ,f- by a11 inlegrat. The summation i \ equal to the areas af the mtangles and the integral is equal to the area undcr the curve obtained by letting n be a continuuus variable.
Now, each of these tllrce sirlgle sumniatio~~s is a!ike, hecause each one is simply
so we have that
Thus, we can write Equation 4.3 as
This summation cannot be expressed in terms of any simple analytic function. This situation does not present any difficulty, however, for the following reason. Graphically, a summation such as f R is equal to the sum of the m a s under rectangles of unit width centered at I, 2, 3, .. . and of height f,, f,, f,. . . . as shown in Figure 4.1. IF the heights of successive rectangles differ by a very small amount, the area of the rectangles is essentially equal to the area under the continuous curve obtained by letting the summation index n be a continuous variabte (Figure 4.1). Problem 4-2 helps you prove that the successive terms in the surnmatlctn in Equation 4.4 do indeed d~ffervery lrttle from each other under most uunditjuns. Thus, it i s an excellent approximat~onro replace the summation in EyuaIiun 4.4 by an integration:
where we have written V for a'. Note that q,,,,, is a function of V and T. We can calculate the average translational energy of an ideal-gas atom from this partition function by using Equation 3.51:
z,"=,
In T
in agreement with what
+ terms independent or 1'
we found in Section 3-3.
4-2. Most Atoms Are in the Ground Electronic State at Room Temperature Note that the integral starts at n = 0, whereas the summarion in Equition 4.4 starts at n = 1. For the small valueb: of @h2/8ma2we are considering here, the difference is z a,the above integral becomes negligible (Problem 4-41). If we denote ~ h 2 / 8 m aby (see Mathchapter B)
In this section, we will invesligatc the electronic contrihutiona 10 q ( V , T),It is more convenient to writs the elwtronic partition function as i~ sum over levels rather than a sum elver states (Section 3-8)- so we write
Chapter 4 / Parfitinn r u n r ~ i o r ~JIK! s Ideal LAWS
146
where gci is the degeneracy, and get the energy of the ith eIectronic level. We first fix the arbitrary zero of energy such that = 0; that is, we wit1 measure a11 electronic energies relative to the ground electronic state. The electronic contribution to q can then be written as
T A B L E 4.1 Some atomic encrgy ..
Atom
levefs."
.
Electron configumtiun
Degeneracy g- = 2J -C 1
~ner~~/crn-'
where E ~ i, s the energy of the jth elcctronic level relative to the ground state. Note that is n function of T but not of V. As wc have seen in Chapter I , these p's are typically uf the urder o f tens of thousands ol'wuve numhers.Using the fact that 1.986 x 1 W2' J = 1 cm-' ,the Boltzmann constant in wave numbers is k, = O.6950 cni '.K I. Thus, we see that typically
which i s equal to I0 even for T = 1000 K. Therefore, c-",? in Equation 4.9 typically is :trnund 10-qtilr most atoms at ordinary tetnperatures, so only the fist term in the suill~nutionli)r qclmi s significantly different from zero. There ate surne cases, however, such a q the halogen atoms, for which the first excited state lies only a few hundred wave numbers ahr~vethe ground state, so that several terms in qc,rrare necessary. Even ill these cases, the sum in Equation 4.9 converges very rapidly. As we learned in Chapter 1, the clcctronic energies of atoms and ions aredctcmined hy uromic spectrascupy and are well tabulated. The standard reference, "Moore's tables:' lists the energy levels and energies of many atoms and ions. Table 4.1 lists the lirst few levels for H, He, Li, and F. We can make some general observations tiu~rn t:~bleslike Table 4.1. The first exci~edstates of the noble gas atoms are of order of 10' cm-' or highcr than the ground states; ihe first excited statcs at the atkali metal atoms are of order u l 1 0 4 cni or higher than the ground states: and the first excited stiucs of thc Ii;~logc~i alolns arc only ol'ordcr of 10' cm- higher than the ground statcs. 'I'hus. at ordinary tempertltures, thc electronic partition function of noble gas atoms is essentially unity and that of alkali metal atoms is two, while those for halogen atoms ct~nsistof two terms. Using the data in Tablc 4.1, we can now calculate the fraction of helium atoms in lhc first excited state. This fraction is givcn by
'
"Fmm C.E. Moore, "Atomic Energy Levels" NatI. Bur Std. Cir. 1 467. U.S.Government Rinting OfFrce. Wa~hingtonD.C., 1949
be less than a few hundred cm-' or sn before any population of the excited level is significant.
EXAMPLE 4-1 Using the data in Table 4.1, calculate the fraction of fluorine atoms in the Rrsl exciled state at 300 K, IOOO K,and 2MH)K.
S O L U T I O F : Using the second line of Equation 410 with gel = 4, g,, = 2, and g*, = 6, we have
Ar 3(H) K. p f r : = 770, $0f, 10 '". Even at 3000 K, ji f ; This is typical of ~lic~iohlcc:kwh. 'l'l~cenergy separation between the ground and cxcited levets must
I
Chapter 4 / Partition Funclinns and ldcal Ga-
with s*, = 404.0 cm-' and E
~ = ,
102406.50 em-'. We also have
and
4-3. The Energy uf 3 Diatumic Mulecule Can Bc Approximaled a,
,I
Sum ul Srparate Terms
The first term represents the nvewg kinetic cnergy, and the second lerm represents Ihc average electronic energy (in excess of the ground-state energy). Thc contribution of the electronic degrees of Creedam to the average energy is small at ordinary temperatures. If we ignore the very small cmtribution from the electronic degrees of frmdam, the molar hear capacity at constant votume is given by
Clearly, we can neglect the third term in the denominator off,. The valuc of f2 fur the various temperatures is
The pressure is
Thus, the population of the firs1 excited state i\significant at these temperature.. and so the first two terms of the summation in Equaticln 4.9 nmst be evaluated in determining llc,*L(T). For most atoms and molecules, the first two terms of the electronic partition
Curiction are sufficient, or
At temperatures at which the second term is not negligible with respect to the fin1 term, we must check the possible contribution of higher terms as well. This completes our discussion of the partition function of monatomic ideal gases. In summary, we have
where
qClFf ( T ) = gel
+
+ ...
~ , ~ e ~ ~ @ ' ~ 2
V
+ terms not invulving V )
--Nk,T v
(4.15)
which is the ideaIgas equalion of state. Note that Equation 4.15 resuits because q ( V , T ) i s of the form f ( T ) V ,and only the translational energy of the atoms contributes to thc
pressure. This is expected intuitively, because the pressurc 1s due to bombardment ol' the walls of the container by the atoms and molecules of the gas. In the next few sections, we will treat a diatomic ideal gas, In add~lionto translational md electronic degrees of freedom, diatomic molecules also possess vibrat~onal and rotational degrees of freedom. The general procedure would be tu set up the Schriidinger equation for two nuclei and n electrons and to solve this equation for the set of eignvalues of the diatom~cmolecule. Fortunately, a series of vcry gclcK1 approximations can be used to reduce this complicated two-nuclei, n-elcctron problcnl to a set of simpler problems. The simplest of these appr13ximations i s thc rigid rtaalclrharmonic oscillator approximation, which we described in Chapter I. Mre will he1 up this approximation in the next section and then discuss the vibratimal and rotdtiunal partition functions within this approxitrlation in Sections 4-4 and 3-5.
4-3. The Energy of a Diatomic Molecule Can Be Approximated as a Sum of Separate Terms
Wc can now calculate some of the propeltics o f a monatomic ideal gas. The average 1'111'1,gy is
When treating diatomic or polya~otniumolecules, wc uac thc rigid rt~tatnr-har'moniu oscillator approximation (Chapter 1). In his case, we can write the lute1 energy of tt~c molecule as a sum of its wnslational, rotational. vibrational, and electronic energies:
Chapter 4 1 Partilion Functions and Ideal Gases
4-3. The En~rgyof a Diatomic
Mol~~ule Can Be Approximated a5 a Sum of Separate Terms
As for a monatomic ideal gas, the inequality given by Equation 3.40 is easily satisfied at normal temperatures, and so we can write
Furthermore, Equatiun 4.16 allows us to write
so the partition function of a molecular ideal gas is given by
The translational partition function of a diatomic moIecule is similar to the result we found in Section 4-1 for an atom:
Note that Equatiun 4.20 is essentially the same as Equation 4.6. The electronic partition function will be similar to Equation 4.9. We will discuss the vibrational and rotational contributions to the partition function in the next two sections. Although Equation 4.19 i s not exact, it is often a good approximation. particularly for small molecules. Before we consider q,,, and q\,b,we must choose a zero of energy for the rotational, vibrational, and electronic states. The natural choice for the zero of rotational energy is the J = 0 state, where the rotational energy is zero. In the vibrational case, however, we have two sensible choices. One is to take the zero of vibrational energy to be that of the ground state, and the other is take the zero to be the bottom of the internuclear potential well. In the first case, the energy of the ground vibrational state is zero. and in he second case it is 11~12.We will choose the zero of vibrational energy to be the bottom of the internuclear potential well of the lowest electronic state, so the energy of the ground vibrational state will be h v / 2 . La\t, we tdie the zero o f the electrrmic energy ro be the separated atoms at rest in their grt~undeleclronic states (see Figure 4.2). Recall that the depth of the ground electronic stare po~entialwell is denoted by Dc (4is apositive number: see Section 171,and so the energy of the ground electronic state is eel = -Dc, and the electronic partition function is
where D, and c12 are show11 in Figure 4.2. We also introduced in Section 1-7 a quuntity I),, that is equal to 4 - ikv. As Figure 4.2 shows, Do is the energy difference hctwce~itlic lowcst vibrational state and the disswiated molecule. The quantity Do can hc uwa*uwd spctroscopically, and values of Do and Dc for several diatomic molezules utr yrvcn in l'ahlc 4.2.
F I G U R E 4.2 The ground and firs1 excited electronic states as a function of the inlernuclear separation, illustrating the quantities 4 and Doof the ground state and E?,. The qUntlties 1)* and D,, are related by D* = D, h v / 2 as shown in the figure.
+
T A B L E 4.2
Moiecular cnnstnllls for ~everaldiatomic n~olecules.These parameters were obtained from a variety of sources and do not represent the most accuralc values k a u s e they werc obtained undcr the rigid rotator-harmonic uscillator approximation.
Degencraq of the ground
Molecule
Qv,,IK
HCI HBr Hr
4227 3787
Ntl,
229 133
%
3266
OWtlK
~,/kJ.rnol-'
q / k J ~ r n 0 1 - ~ electronic state
15.02
427.8
12.02 9.25 0.221 0.081
3hZ.h
445.2 377.7
294.7
30H.h
I
71.1
72 1
I
53.5
54 I
1
I
I
4--4. Mosl Muleculrs Are in the Ground Vibratiundl State dl R w m Tcmprdll~rc
4-4.
Most Molecules Are in the Ground Vibrational State at Room Temperature
We can calculate the average vibrational energy fmm qb,,(T)
In this section, we will evaluate the vibra~ionalpart of the partition function of a diatomic rnc~leculeunder the harmonic-oscillator approximation. If we measure the vibrational energy levels relative to the hotturn of the internuclear potential well, the energies arc given by (Equation 1-22)
with v = ( ~ / / A ) ' / ~ / ~where T C , k is the force constant of the molecule and reduced mass. The vibrational partition function q,,, becomes
JX
is itb
Table 4.2 gives (-Iv, for sevcml diatomic molccules. The vibrational conlribution to the molar heat capacity i s
Figure 4.3 shows the vibrational contribution of an ideal diatomic gas to the molar heit1 capacity as a function of temperature. The high temperature lin~itof ?v,vi, is A, and Cv,v,, is one-half of this value at T / O b i b = 0.34.
This summation can be evaluated easily by recognizing it to be a geometric series (Mathchapter C):
with x = FPh" < 1. Thus w e can write
so q,,,(T 1 becomes
F I G U R E 4.3 The vibrational contribution to the molar heat capacity of an ideal diattmic gaq as a funcliol~ol reduced temperarure, T/@v,b.
I Note that this is the vibrational term encountcred in Example 3-2, which presented the p i i t i o n function for the rigid rotator-harmunic oscillator model of an ideal diatomic gas. If we introduce a quantity, (-Iv,, = h v / k , , called the vibmtionul temperature, qv,,(T)can be written as
E X A M P L E 4-2 Calculate the vibrational contribution 10 the molar hcat cap:ioi!y The experimental value is 3.43 I.K-' .rnul-'.
of
N:(g!) at llWW K .
S O I . U T I O N : We use Equarlon 4.26 w~th (-I,,, = 3374 (lahlc 4 ?)
8",/T = 3.374 and
This is one of the rare cases in which y can he summed directly without having to approximate it by an integral, as we did for the translational case in Section 4-1 and will do shortly for the rotational case in Section 4 5 .
The agreement with the e a p e r i ~ l r ~ i l vr~luc ul 1% yultc yrrxl.
1
Ihur,
Chapter 4 1 Partition Functions and Ideal C;ases
An interesting quantity to calculate is the fraction of molecules in various vibtational states. The fraction of molecules in the vth vibrational state is
If we substitute Quation 4.23 into this equation, we obtain
The follnwing exarnplc illusrrates the use uf this equation.
EXAMPLE
I
4-3
Use Equation 4.28 to calculale the fraction of N2(g)mulecules in the v = 0 and v vibra~i~mal states at 300 K.
=1
F I G U R E 4.4 The population of the vibrational levels of Br, (g) at 300 K
S O L U T I O N : We lirst calculate exp(-(+",/T) for 300 K:
Table 4.3 gives the fraction of molecules in excited vibrational states for several diatomic molecules. Therebe, T A B L E 4.3
The fraction of molecules in excited vibrational states at 300 K and 1000 K.
and
eV,K
Gas Notice that essentially all the nitrogen m~leculesare in thc ground vibrational state at 3(X) K.
I],
6332 4227
HCl
N2
3374 3103
x:,
xr+
1.01 x
2.00 x lo-' 1.46 x lo-= 3.43 x lo-' 4.4'1 x I [ ) - I 4.47 x lo-' 7.35 x lo-'
8115
3.22 x 10-' 6.82 x lo-'
I,
308
3.58 x lo-'
4-5, Most Molecules Are in Excited Rotational StaJes at Ordinary Temperatures The energy levels of a rigid rotator are given by (Equation 1-28) F J
or simply
f,,, 1T = 1000 K)
7.59 I()-' 1.30 x
'3
CO
Figure 4.4 shows the pupulation of vibratiunal levels of Br,(g) at 300 K. Notice that most molecules are in the ground vibrational state and that the population of the higher vibrational states decreases exponentially. Bromine has a smaller force constant imd a larger m a s (and hence a smaller value of (-Iv,,) than most diatomic molecules, however (rf. Table 4.2). so the population of excited vibrational states of Br,(g) at a givcn temperature is greater than most other molecules. We can use Equation 4.28 to calculate the fraction of molecules in all excited vibrational states. This quantity is given by fU but because jV= 1, we can write
(T= 300 K)
=~
+ I)
~ I J ( J
21
J=O,
I,
2, . . ,
(4.3ou)
where I i s the moment of inertia of the rotator. Each energy level has a degeneracy of g, = 2J
+1
(4.30b)
Chap~er4 1 Panition Functiuos and Ideal Cases
156
Using Equations 4.30)~and 4.30b, we can writc the rotational partition function of a rigid rotator as
where we sum over levels rather than states by including h e degeneracy expliciLy. For convenience, we introduce a quantity that has units of temperature and is called the
157
&5. Most Molecules Are in Excited Kotat~onal Stales at Ordinary Tcmperalures
For simplicity, we will use only the high-temperature limit, because 8, molecules at room temperature. (See Table 4.2.) The average rotational energy is
<< T tbr most
and the rotatiunal contribution to the molar heat capacity is
ri~tutionaltempruture, 8,:
A diatomic molecule has two rotational degrees of freedom, and each one contributes
where B = h/8n21 (Equation 1.33). Subslituting Equation 4.32 into Equation 4.31 gives
Unlike the harmonic-oscillator panition function, the summation in Equation 4.33 cannot be written in closed form. However, as the data in Table 4.2 will verify, the value of Om,/T is quite small at ordinary temperatures for diatomic moiecules that do not contain hydrogen atoms. For example, 0, for CO(g) is 2.77 K, so 8 , , / T i s about lo-* at room temperature. Just as we were able to approximate the summation in Equation 4.4 very well by an integral because a = Bh2/8maz i s typically small at normal temperatures, we are able to apprnximate the summation in Equation 4.33 by an integral because O,/T is small for most molecules at ordinary temperatures. Therefore, it is an excellent approximation tu write q,!T) as
This integral iseasy toevaluate because ifwelctx = J(J and q, (T) becomes
I
R I 2 to F,,,. We can also calculate the fraction of molecl~lesin the J' rotational level:
I
EXAMPLE 4-4 Use Equation 4.35 to calculate the population of the rotational levels of CO at 300 K. SOL U 'rl 0 N : Using ern[ = 2.77 at 300 K.Therefore,
I
K from Table 4.2, we have that Q,/T = 0.W23
We can present our results in the form of a table:
+ I), thendx = (2J + 1)d.l
Note that this i s the rotational term encountered in Example 3-2, which presented the partition fu-nction_for .... @e rigid rotator-harmonic oscillator model of anideal diatomic gas. This approximation improves as the temperature increases and is called the hightemperature limit. For low temperatures or for molecules with lnpe vulws ol'(-lh,.say H,(g) with Urn= 85.3 K, we can use Equation 4.33 dire~tly.For example, the f rut four terms of Equation 4.33 are sufficient to calculate q,*(T) to within 0.1% for T c 38,,.
Thest rcsullr art plotted in Figure 4.5.
I
J
'
4-6. Rotatiunal Panilion Functions Conlain a Syrnrnctry Nunitwr
Note that this equation is the same as Equation 4.34 for a heteronuclear diatomic molecule except for the factor of 2 in the denominator. This factor comes from the additional symmetry r)f the humonuclcw diatumic molecule; in pwticular, a homonuclear diatomic molecule has two indistinguishable orientations. Thcrc is a two-fold axis of symmetry perpendicular to the internuclear axis. Equations 4.34 and 4.37 van be written as one equation by writing q,", as
F I G U R E 4.5
The
fraction of molecules in thc Jth rotational level for CO'at 300 K.
Contrary to the case for vibrational levels, most molecules are in !he excited rotational levels at ordinary temperatures. We can estimate the most probable value of J by mating Equation 4.35 as if J were continuous and by setting the derivative with respect to I equal to zero to obtain (Problem 4-1 8)
where o = 1 for a heteronuclear diatumic molecule and 2 for a homonuclear diatomic moleculc. The factor u is called the symmetry number of the mr)lecule and reprcsentx the number of indistinguishable orientations of the molecule. Having studied each contribution to the molecular partition function of a diatomic molecule, we can now include the rigid rotator-harmonic oscillator approximation in the partition function of a diatomic molecule to obtain
Remember that this expression requires that Om,<< T, that only the ground electronic bc the separated atoms at rest in their ground electronic states, and that the zero of energy for the vibrational energy is that at the bottom of the internuclear potential well of thc lowest electrunic state. Note that only qm5 is a function of V ,and that this function is of the form f ( T )V, which, as we have seen before, is responsible for the ideal-gas equation of state. state is populated, that the zero of the eleclronic energy is tiken to
This equation gives a value of 7 for CO at 300 K (in agreement with Figure 4.5).
4-6. Rotational Partition Functions Contain a Symmetry Number Although it is not apparent from our derivation of q,(T), Equations 4.33 an9 434 apply only to heteronuclear diatomic molecules. The underlying reason is that e wave function of a born~nucleardiatomic rnolec~lemust possess a certain s y , d with Q respect to the interchange of the two identical nuclei in the molecule. In particular, iC the t w ~ )nuclei have integral spins (bosom), the molecular wave function must remain unchanged under interchange of the two nuclei; if the nuclei have half odd integer spin (fertnions), the molecular wave function must change sign. This symmetry requirement has a profound effect on the population of the rotational energy levels of a homunuclear diatomic molecule, which can be understood only by a careful analysis of the general symmetry properties of the wave function of a diatomic molecule. This analysis is somewhat involved and wili not be done here, but we need the final result. At temperatures such that@, << T, which we Rave seen applies to most molecules at ordinary temperatures, q, for a homonudear diatomic molwule is
EXAMPLE 4-5
Derive an expression For the molar energy Idenii ty each of the terms.
of a diatomic ideal gaq from Equation 4.39.
S O L U T I O N : We startwith Q ( N , V ,T ) =
[ d v ,T)Iw N!
and
Using Q u ~ i o n4.39 fur q ( V , T ) , we have
iterms nut containing T
159
Chapter 4 / Partition Functions and Ideal Gases
and letting H = N, and NAk, = R for one mole,
The first tenn represents the average translational energy ( R T / 2 for each of the three translational degrees of freedurn). the second term represents the average rotational energy ( R T / 2 for each of the lwo rotational degrees of freedom), the third term represents the ?en]-point vibrational energy, the fourth term repreents the average vihritional energy in exccss of the zem-point vibrational energy, and the last tenn reflects the electnmic energy reliltive to the zero of electronic energy that we have chosen, nan~eiythe two sepmted atoms at rest in their ground electronic states.
The heat capacity is obtained by differentiating Equatiun 4.40 with respect to T:
Figure 3.3 presenls a comparison of Equation 4.41 with experimental data for oxygen. The agreement is good and is typical of that found for other properties. The agreement can be improved considerably by including the first corrections to the rigid rotatorharmonic oscillator model. These include effects such as centrifugal distonion and anharmonicity. The consideration of these effects intrduces a new set of molcculur constants, dl of which can be determined spectroscopically and are well tabulated. The use of such additional parameters from spectroscopic data can give calculated values of the heat capacity that are actually more accurate than calorimetric ones.
4-7. The Vibrational Partition Function of a Polyatomic Molecule Is a Product of Harmonic Oscjllator Partition Functions for Each Normal Coordinate
4-7. The Mbrat~onnlPartition Function of a
As for dintofnic ~nolecules,we use a rigid rotator-harmonic oaci1)alnr approximation. This allows us to separate the rotational niotiun from the vibratiunal motion of ihe molecule, so that we can treat each one separately. Boih pmblems are somewhat more complicated for pc~lyatomicmolecules thin for diatomic molecules. Nevertheless, we can write the polyatomir: analog uf Equation 4.19:
In Equation 4.42, qtrM,is given by
where M is the total mass of the molecule. We choose as the zero of energy then atom\ completely separated in their ground electronic states. Thus, the energy of the gruund electronic state i s -De, and then the electronic partition function is
To calculate Q ( N , V ,T) we must investigate q , and qvi,. We learned in Section 1-9 that rhe vibrational motion of a plyatomic molecule can be expressed in terms of normal coordinates. By intmducing normal coordinates, the vibrational motion of a polyatomic molecule can be expressed as a set of indeprndent harmonic oxillators. Consequently, the vibrational energy of a polyatomic molecule can be written as
where v, is the vibrational frequency associated with the jth normal mode and n is the number of vibrational degrees of M o m (3n - 5 for a linear molecule and 3n - 6 far a nonlinear molecule, where n is the number of atoms in the moleculc). Because the normal modes are independent,.
I7,,
e-o",,/?T
The discussicln in Section 4-3 for diatomic molecules applies equally well to polyatomic molecules, and so
4.a =
E",, = As before, the number of translational energy states alone is sufficient to guarantee that
the number of energy states available l o any molecule is much greater than the number of rnoleculcs in the system.
161
Pulyatom~cMolecule
and
,=I
-,.,b,lT
(4.46)
) (4.47)
,=I
Chapter 4 / Partition Functions and Ideal Gases
where O V , ,is a characteristic vibrational temperature defined by
4-8. The Form of the Rotational Panition Functiun o i a Polyatwnic Molecule
For Ovib,,= 3360 K (the symmetric stretch),
hv.
Qv~b.] . =L
k
,
The tvwl vibrational heal capacity at .tK is
Table 4.4 contains values of Ovi,,ifor several polyatomic molecules.
T A B L E 4.4
Values uf the characteristic ruliltional temperatures, the characteristic vibrationat temperatures, D, for the ground state, and the symmetry number, u, for some plyatomic moleculcs. The numbers in parentheses indicate the degenetacy of that mode. Bvlb
co2
33~%,954(2), 1890 5360.5 160,2290 4800, 1360.488q2), 2330(2) 13W+640,1600
HzO
NH, CIO,
so2 N,O NO2 CH, CH,CI CCI,
r
,IK
Molccule B,IK 0.561 40 1,20.9, 134 13.6, 13.6, 8 92 2.50, 0.478, 0.400 2.92,0.495. 0.422 0 3
1650,750. 1960 3200,850(2), 1840 IW, 1080,2330 1 1.5.0.624.0.590 7.54.7.54, 7.54 4170,2180(2), 4320(3), l870(3) 7.32,0.637,0.637 4270, 1950, 1050,4380(2) 2140(2), 1460(2) 0.0823,0.0823,0.0823 660,310(2),1120(3), 45133)
LI,,/kJ-mol-'
o
1596 917.6 1158 378 1063 IIW 928.0 1642 1551
2 2 3 2 2 2 2 12 3
1292
12
1
E X A M P L E 4-6 Calculate the contribution of each normal mnde to the vibrational heat capaciw ofCO, at 4w K.
SOL U T lO N: The values of 8,, are given in Table 4.4. Note that the Bvi, = 954 K anode (bending male) is doubly degenerate. For 8 V ,= j 954 K {the doubly degerate
@)*)ulb,
Note that the contributionfrom each mode decreases as increases. Because B,,h,i is proportional to the frequency of the mode, it requires higher temperatures to excite m d e s with larger values of 6w,,j. The molar vibrational heat capacity from 200 K to 2000 K contributed by each mode is shown in Figure 4.6.
TIK F I C U P E 4.6
The contribution of each normal m d e to the molar vibrational heat capacity of CO,. The curve indicated by triangles corresponds to @v,b,, = 954 K: the curve indicated by squares to Bwib,,= 1890 K;and the curve indicated by circles to, ,@ , = 3360 K. Note that m d e s with smaller values of Ow,,,,,w u, , contribute mwe at a given temperature.
hcnding mode).
4-8. The Form of the Rotational Partition Function of a Polyatomic Molecul8 Depends Upon the Shape of the Molecule In this section, we will discuss the rotational partition functions r ~ l prtlyattrlliic ' tr~i~k. cules. Let's consider a linear polyatomic molecule tirst. In thr rigid-n~lr~tur upprulrl~rution, the energies and degeneracies of a linear plyatolltic tnc~lcculcart the urn nr fm
Chapter 4 / Partition Fun~tiunrand Ideal Ga-
+
a diatomic molecule, E , = J ( J l ) h 2 / 8 x 2 1with J = 0, 1 , 2, . . . and g, = 2 J In this case, the moment of inertia I i s
+ I.
where d, is the distance of the jth nucleus from the center of mass of the molecule. Consequently, the rotational partition function of a linear polyatomic molecule is the same as that of a diatomic molecule, namely,
As before, we have introduced a symmetry number, which is unity for unsymmetrical molecules such as N,O and COS and equal to two for symmetrical molecules such as CO, and C,H,. Recall that the symmetry number is the number of different ways the molecule can be rotated into a configurntion indistinguishable from the original.
I
+8. The Form of the Rotational Partition Function of a Pulyalomic Molecule
Thus, we have the various cases Orot,A =0,,,=+)-),,, spherical top
The quantum-mechanical problem of a spherical top can be solved exactly to give
The rotational partition function is
Foralmostall spherical topmolecules Om,<< T at ordinary temperatures, so we convert the sum in Quation 4.53 to an integral: EXAMPLE 4-7
What is the symmetry number of ammonia. NH,? S O L U T I O N : Ammonia is a vigunal pyramidal molecule and has the three indistin-
guishable orientations shown below looking down the three-fold axis of symmetry.
Note that we have included the symmetry number rr. For 0, << T, the most impomnt values of J are large (Problem 4-26), and so we may neglect I compared with J in the integrand of the above expression for q,to obtain k,
If we let Om,/ T = a, we can write Therefore, the symmetry number is three.
In Section 1-10, we learned that the rotational properties of nonlinear plyatomic rnulecules depend upon the relative magnitudes of their moments of inertia. If all three moments of inertia are equal, the molecule i s called a spherical top. If two of the three are equal, the molecule i s cdled a symmtric fop.If all three are different, the molecule is called an asymmetric lop. Just as we defined a characteristic rotational tempzrature of a diatomic mvle~uleby Equation 4.32, (93)R11 = h2/21k,, we define three characteristic n~tntinnaItemperatures in t e q s of the three moments of inertia according to
or, upon substituting B,/T for a,
The corres~ndingexpressions for a sy~nmeinclop and an asymmetric top are
Chapter 4 1 Partition Functions and Ideal Gases
q,(T) = -
T3
asymmetric top
(4.56)
4-9. Calcula~edMolar Heal Capacities Are in Very Good Agreement with Experimental n a t n
and Cv = 3 + ? -
Notice how Equation 4.56 reduces lo Equation 4.55 when O,a,, = O,,I,, and how both Equatinns 4.55 and 4.56 reduce to Equation 4.54 when Om,,= 8,, = 0.,,, Table 4.4 contains values of a,,,, 0,-,, and B,,,, for several polyatomic molecules. The average molar rotational energy of a nonlinear plyatomic molecule is
Nk,
5
2
+'q+) =
j-l
e-B"w,/T
(1
- e-e",h,!T)2
E X A M P L E 4-8 Calculate the rnular heat capacity of gaseous water at 300 K.
-
501. LIT ION: We use Equation 4.62 with Buib,, = 2290 K. 5160 = 2290 K, (Table 4.4). For 8
K,and 5360 K
or R T / 2 for each rotational degree of freedom, and C,,, = 3 R / 2 .
4-9. Calculated Molar Heat Capacities Are with Experimental Data
in Very Good Agreement
We can now use the results of Sections 4-7 and 4-8 to constructq ( V , T) for polyatomic molecules. For an ideal gas of linear polyaomic molecules, q ( V , T ) is the p d u c t of
Equatiuns 4.43,4.44,4.46,and 4.50:
Slrnilarly F , , / R = 1.00 x 10-"fm O",,, = 5160 K and 5.56 x 5360 K The total molar heat capilcily of water at 300 K I S
(or (-I,,, , =
-
5 = ~ . ( x ) u + O O Z X Z + 1.03 x K
10-5 f5.56 x
= 3.028
The exper~mentalvalue is 3.01 I . Notice that the vibrational degrees uf tre~domcontribute very little totheheat capacity of wakrat300 K. Thecalculatedmdexpenmcntal values at I000 K are 3.948 and 3.952, respect~eb.Figurc 4.7 hhows ihc molar heat capacity of water from 300 K to 1200 K.
The energy is
and the heat capacity i s
For an ideal gas of nonlinear polyatomic molecules,
F I G U R E 4.7
A comparisun of the molar heat capacity of water vaptrr cnlc~alr~cd h t r r u FLuatiim 4.62 :md the experimental value. The experimental data a r t intlicairtl by IIw cbwlc*.
167
Chap~er4 / l'drtition Functions and Ideal Gases
168
lkblc 4.5 gives the vibrational contribution to the molar heat capacity at 300 K for a variety of molecules of different shapes. It can be seen that the vibrational contributions are far from their high-temperature limits and that the agreement between the calculated and experimental values of F , / H is good. A calculatiun for more complicated molecules would show similar agreement between thc calculated values and the experimental data.
Problems ;kgr
4-1. Equation 4.7 shows that {E-) = in three dimensions, and Prublern 4-3 shows 11131 ( E , ~ ~ )= fkerin one dimension and i k , in ~ two dimensions. Show that typical values of translational quantum numbers at room temprature are 0 ( l o q ) for rn = I()-'" kg, a = I dm, and T = 3W K. 4-2. Show that the difference between the successive terms in the sum~nationin Equation 4.4 is very small for m = kg, a = 1 dm,and T = 300 K. Recall trom Prohletn 4 1 that
T A B L E 4.5
Vibrational contributions to the molar heat capacity of some polyatomic molecules at 300 K.
typical values of n are 0(1OY).
-
Vibrational Co~~tributlon Molecule
co2
N2°
*", CH,
H20
Degeneracy
to F,
l8W
I
0 073
3360 454
I
O.O(K)
2
0.458
1840 3200
I 1
0 082 0.003
850
2
0.533
48W 1360
1
0000
1
0.226
4880
2
0.000
2330
2
0.026
4170 2180
1 2
0.000 0.037
4320
3
0.000
W,,,IK
1870
3
0.077
2290 5160
I 1
0.028 0.000
5360
1
O.O(KI
Total
C
R
Total
-
-
C,/R (calc)
C,/K (exptl)
4-3. Show that
in one dirnensitm and that
0.99
3.49
3.46
in two dimensions. Use these results to show that
( E , , ~ , ~has }
a contribution of k , T / 2 tu it!.
total value for each dimension.
1.15
2.65 4.4. Using thc data in Table 1.2, calculate the fraction of sodium alorns in the first excited statc at 300 K, 1000 K, and 2030 K.
0.28
4-5. Using the data in Table 4.1, evaluate the fraction of lithium atoms in tlw tirst cxcited state at 300 K, 1000 K, and 2000 K.
3.28
4-6. Show that cach dimension contributes R / 2 to the molar translalivnal heal capacity 4-7. Using the values of Oviband a,,in Table4.2, calculak the vaues o l De for CO, NO, and K,
0.30
3.30
3.29
4-8. Calculate the characteristic vibrationill tempt.rature 4401 cm-I and
0.03
3.03
ij4
for Hz(g) and D,(g)
(C,,
=
= 3 112 cm-I).
4-9. Plot the vibriltional contribution lo lhe molar heat capacity ofCI,(g) from 250 K to I000 li. 3.111
4-10. Hot the fraction of HCl(g) molecules in the first few vi6~:ional states a1 3(W K and 1W K. \>,>,
,,. \
4-1 1. Calculate the traction of moioculcs in the ground vibrational state and in itl-the excited states at 300 K for each of Ihe molecules in Table 4.2. 4-12. Calculate the value of the chmiclzristic rotalionill temperature +Iwt f<)rH,(g) and D:(g).
(The bond lepgths of H, and D, are 74.16 prn.) The atomic mass of deuterium is 2.014 4-13. The average molar rotational energy of a diatomic n~oleuuleis R7'. Shuw thal typicill values of J are given by J ( J 1) = T/'Im. What are typical values of J for N,(g)
+
at 300 K?
Chapter 4 1
Partilion Funct~unsand Ideal Gases
for a spherical Lop molecule. Show that the correction to replacing Equation 4.53 by an integral is about 1Q for CH, and 0.001$6 for CCi, at 300 K.
4-37. Molecular nitrogen i s heated in an electric KC. The spectroscopically determined relative populations of excited vibrational levels are listed below.
4-28. The N-N and N - 0 bond lengths in the (linear) molecule N,O are 109.8 pm and 121.8 pm, respectively. Calculate the center of mass and the moment of inertia nf ' 4 N ' 4 ~ 'Compare 60. your answer with the value obtained from +Im in Table 4.4. 4-29. NO,(g) is a bent triatomic molecule. The following data determined frum s p e c b scopic measurements are ir, = 1319.7 cm-I, C2 = 749.8 cm-', C3 = 1617.75 crn-', A, = 8.0012 cm-'. B, = 0.43304 crn-I, and Po = 0.4I040 cm-'. Determine the three charac. tcris tic Vibrational temperatures and the characteristic rotational ternperaturcs for each of the principle axes of NO,(g) at 1MlO K.Calculate the value of Id, at 1000 K.
4-30. The experimental heat capacity of NH,(g) can be fit to the empirical formula
over the temperature range 300 K < T c 15(M K. plot C , ( T ) / Rversus T over this range using Equation 4.62 and the mole~ularpwdmeters in Table 4.4, and compare your results with the exprimental curve.
4-31. The experimental heat capacity of SO,(g) can be fit to the empirical formula
over the temperature range 300 K .r T < 1500 K. plat C , ( T ) /R versus T over this range using Equation 4.62 and the molecular parameters in Table 4.4, and compare your results with the experimental curve.
4-32. The experimental heat capacity of CH,(g) can be 6t to the empirical formula
over the temperature range 300 K -z T < 1500 K. Plot P,(T)/R versus T over this range using Equation 4.62 and the molecular parameters in Table 4.4. and compare your resulls with the experimentill curve. 4-33. Show that the moment of inertia oi a diatomic molecule is p ~ : ,where p is Iht reduced niass, and & is the etluilibriu~nbond length. 4-34. Given that Ihc values of and Rlr Hzare 85.3 K and 6332 K. respectively. cillculate ihese quantities for HD and D,. Hinr:Use ihe Born-Oppenheimer approximation.
4-35. Using the result for q m t ( T obbtnined ) in Problern4-14, derivecorrections totbeexpressions ( E m ) = R T and C,-,n = A given in Section 6 5 . Express your result in terms of powers of @,/T. 4-36. Show that the thermodynamic quantirics P and C, are independent of the choice of a rcro of energy.
Is the nitrogen in thenndymrnic equilibrium with respect to vibrational energy? What is the vibrational temperature of the gas? IS this value nexessarily the same a4 the hans!ational temperature? Why or why not? 4-38. Consider a system of independent diatomic molecules constrained to move in plane, that is, a twwdimensional ideal diatomic gas. How many degrees of f d o m Jres a two-dimensional diatomic mulecuie have? Given thal the energy eignvalues of a twodimensiunal rigid rotator are
(where I is the moment of inertia of the molecule) with a degeneracy g , = 2 for all J except J = 0, derive an expression for the rotiltional partitiun function. The vibrational partition function is the same as for a three-dimensional diato~nicgas, Write out
and derive an expression for the average energy of this two-dimensional ideal diatomic gas. 4-39. What molar constant-volume heat capacities would you e x p c t under classical conditions for the following gases: (a) Ne, (b) 02,(c) H,O, (d) CO,, and ( e )CHCI,?
4-40. In this problem, we will derive an expression fur the number of translational energy states with (translational) energy between E and E -+ d t . This expression is essentjally the degeneracy of the state whose energy is
Thc degeneracy is given by the number of ways the integer M = 8mu2€/ha can be written as the sum of the squares of three positive integers. In gcneral, this is an erratic and discontinuous function of M (the number of ways will be i e r o for many values of M),but it becomes smwth for large M ,and we can derive a simple expression for it. Considcr a three-dimensional q a c e spanned by nl, n ",and n:. There is p one-to-one corrcspontiencc hetwoell energy states given by Equation 1 and thc point.; in this nx, n , , n: vpacc ~ i t h coordinates given by pusitive integers. Figure 4.8 shr :, .. . . lirnens~onalversion uf this space. Equation I is an equation for a sphere of radius K = (8mu'~/ h Z ) ' ( zin this spacc 8
We want ttu calculate the number of lanice points that lie a1 some fixed distance from the origin in lhis spm. In general, this is very dimcult, but fur targe R we can proceed as follows. We treat R, or E , as a continuous variable and ask for the number of latrice p l i n t ~
I . .
..
.
a
*
.
. MATHCHAPTER
PARTiAL DIFFERENTIATION
y:::::;:\: .
F I G U R E 4.8
a
*
.
A two-dimensional version of the ( n x ,nyjn z ) spdce, the space with the quantum numbers n*, n y , and nz as axes. Each point corresponds to an energy of a particle in a
.
(twwdimensional) box.
"1
+
h e t w e e and ~ ~ ~E Ac. Tucalculate thisquanlity,it isconvenient tu first calculate the number of lattice points consistent with an energy 5 F . For large E , an excellent approximation can he rimde by equating the number of lattice points consistent with an energy 5 E with the voluitie uf one uctant of a sphere uf radius R. We take ody one octant because n x , ny,and n z are restricted to k positive integers. If we denote the number of such states by @ ( E ) , we can write
The number of states with energy between r: and E
+ AE (At/$ < 1 ) is
Show that
Show that if we takes = 3kBT/2, T = 3(X) K, m = kg, a = 1 dm, and AE to be 0.010s (in other words 1% of c), then W ( E , A E )is 0(1@'). SO-even fur a system as simple m a single particle in a box, the degeneracy can be very large at room tempemure. 4-41. The translational padtion function can be written as a single integral over the energy E if we include the degeneracy
Recall from your course in calculus that the derivative of a function y (x) at some point x is defined as
Physically, d y / d x expresses the variation of y when x is varied. Much of your calculus cuurse was spent in starting with Equation D. I to dcrive formulas for the derivatives of the commonly occurring functions. The function y in Equatirln D.1depends upon only one variable, .t. For the function y ( x ) , x is called the independeni variuble and g. whose value depends upon the valuc of x , is called ihe dependent variable. Functions can dcpend upon more than one variable. For example, we know that the pressure of an ideal gas depends upon the temperature, volume, and number of n~oles through the equation
I n this case, there are three independent variables; the temperature, volume, and amount of gas can be varied independently. The pressure is the dependent variable. We can emphasize this dependency by writing
m
,
T =
D
w ( E ) I - ~ I ~ E ~ ~ E
+
where W ( E ) ~ Eis the numkr of states with energy &tween € and 6 d t . Using the result from the previous problem, s h w that qUm(V, T ) is the same as that given by Equation 4.6.
Experimentally, we may wish to vary only one of the independent variables at time (say the tempcraturej to produce a change in pressure with two of he independent variables fixed (fixed volume and fixed number of moles). To form the derivative of P witb respect to T with n and V held constant, we sinlply refer to Equation D. I and write
( ) R.v
= lim p(n, I' AT+0
+ A T , V ) - P ( n , T ,V ) AT
MathChrrpterD I P A R T I A L D i F F t R E N T I A T I O N
176
We call ( t l P / a T ) m , ,the partial derivative uf P with respect to T, with n and V held cunstant. To actually evaluate this parrial derivative, we simply differentiate P with respect to T in Equaiion D.2, treating n and V as if they were constants. Thus, for an ideal gas
MathChapterD I P A R T I A L D I F F E R E N T I A 1 I O N
We can also fonn another t y p of sccond deriialive, however. For example, we can
form
and we can also form
We can also have
and The above two second derivatives are called cross derivatives, mixed derivatives, or second cross partial derivatives. These derivatives are commvnly written as
I
I
E X A M P L E D-1
Evaluate the two first partial derivatives of P for the van &r Waals equation
We don't indicate which variable is held constant because they differ with each differentiation. Notice that these two cross derivatives are equal (see Equations D.7 and D.8), so that SOLUTION: Inthiscase. P d e p e n d ~ u p o n T a n d ~ ~ s o w e h aPv e= two firs! partial derivatives of P are
P(T,V).~
Thus, the order in which we take the two partial derivatives of P makes no difference in this case. It turns out that cross derivatives are generally equal. and
I
E X A M P L E 0-2 Suppose that
The partial derivatives given by Equations D.5 and D.6are themselves functions so we can form second partial derivatives by differentialing Equations D.5
of T and and D.6:
and
v,
where A, S,and P are functions of T and V . Prove that
SOLU T I O ~ :Take the partial derivative of S with respect to V at constant T:
MathChap~erI3 / P A R T I A L D I F F E R E N T I A T I O N
and the partial derivative of P with rcspect to T at constant V :
MathChap~erEl / P A R T I A L D I F F E R E N T I A T I O N
change in T plus how P changes with
(at constant T) times the infinitesimal change
in iT. and equate the two cross derivatives of A to obtain
EXAMPLE L 3 We can w e Equation D.1 I to estimate the change in pressure when both the temperature and the molar volume are changed slightly. To this end, for finite AT and AT.we write Equation D.1 1 as
The partial derivalives given in Equations D.5 and D.6 indicate how P changes independent variable, keeping the uther one fixed. We often want to knvw how a dependent variable changes with a change in the values of both (or more) of its independent variabies. Using the example P = P ( T , (for one mule), we write with one
A P = P(T
If we add and subtract P ( T ,
+ A T , V -t- AV) - P ( T , 7)
t-AT)to this equation, we obtain
Use this equalion to estimate the change in pressure d o n e mole of an ideal gas if the temperature i s changed from 273.15 K to 274.00K and the volume is changed from 10.00 L to 9.90 L. SOLUTION: Wefirstneed
M~~ltiply the first two terms in brackets hy A T / A T and the second two terms by AV/AV to get
and
so that
Now let A T
+ O and
AV + 0,in which case we have
d~ = AT-.{I lirn [ P ( T
+ AT:;
+ lirn [ P ( T . Y t dr'-+tr
- P(T,
;: -
(D. 10)
The first limit gives ( 8 P l a T ) , (by definition) and the second gives (8PIaV),. so that Equation 13.10gives our desired result:
Equatiun D.t l i s called the total derivative of P. It simply says that the change ill P is given by how I' chailges with T (keeping constant) times the infinitesimal
Incidcntly, in this particularly simple case, wacalculirte the exact change in P fn)m
RT,
RT,
Vl
v,
AP=----.=-
= 3.0J.L-' = 3.0 ~ . d r n - '= 0.030 bar
179
MarhChapterD I P A R T I A L V I F F E H L N T l A T l O N
v).
Equatiltion D.4 gives P as a function uf T and V , or P = P ( T , We can form the total derivative of P by differentiating the right side of Equation D.4 with respect to T and to obtain
Problems
Equation D.12. Equations D.7 and D.8 show that the cross derivatives are equal, they must be for an exact differential.
ah
EXAMPLE I3-l
I
Is
We can see from Example D-1 that Equation D.12 is just Equation D.l I written for the van der Wads equation. Suppose, however, that we are given an arbitrary expression
(D.14)
for d P, say an exact differential?
(D. 1 3)
5 0 L U T 10 N : We evaluate the two derivatives
v)
arid are asked to determine the equation of state P = P ( T , that leads to Equation D.13. In fact, a simpler question is to ask if there even is a function P ( T , ~ ) whose total derivative is given by Equaticln D. 13. How can we tell? If there is such a fi~nctionP ( T , V ) ,then its total derivative is (Equation D.l I )
and
These derivatives are equal and so Equation D.14 represents an exact diflerential. Equation D.14 is the total derivative of F fur the Redlich-Kwung equation of state.
Furthermore, according to Equation D.9, the cross derivatives of a function P ( T ,71, I
I
Exact and inexact differentials play a significant role in physical chemistry. I f dy is an exact dift'erential, then
and
l
dy = yI
must be equal. If we apply this requirement to Equation D. 13, we find that
- y,
(exact differential)
so the integral depends only upon the end points ( I and 2) and not upun the palh from I to 2. This statement is not true for an inexact differentid, however, so
The integral in this case depends not only u v ? h c end pojnts but also upon the path from l to 2.
Thus, we see that the cross-derivatives are not equal, so the expression given by Equation D.13is not the dsrivative of any functiun P ( T , V ) .The differential given by Equation D. 13 i s called a n inrxnct riferenriai. We can obtain an example of an exact dlflerentiul simply hy explicitly differentiating any function P ( T , such as we did for the van der Waals equalion to obtain
v),
Problems 0-1. The isuthclmalcompressibility, K,, of a substance is defined as
182
MathChapterD I P A R T I A L D l F F E R f N T l A T f O N
Obtain an cxpssion for the isothermal compressibility of an ideal gas.
Problems
and given the expression in Proble~nD-7, derive the equation
0-2. Thc coefficient of thermal expansion, u, of a s u b s m e is defined as
Ohtlin an expression for the cmfficient of thermaI expansion of an ideal gas.
0-3.
O.9, Use the expression in Problem D-8 to determine ( d C , / a V ) , for an ideal gas, a van der Waals gas (Equation D.4). and a Redlich-Kwong gas (see Probiem D-5).
Row, that
an exact or inexact differential? for an ideal gas and for a gas whuse equation of state is P = n R T / ( V - n b ) , w k r e b is a constant. This relation is genemlly true and is called ,hreciprr~didentity. Notice that the same variables must be held fixed on buth sides of the identity.
an exact or inexact differential? The quantity CV(T) is simply an arbitrary function of T. What about d x / T ?
0-4. Given that
0.1 2. Rove that where
and that
and k,. rn, and h are constants, determine
U as a function of T
D-5. Show that the total derivative or P for the Redlich-Kwong equation,
where Y = Y ( P , T,n) is an extensive variable.
D-13. Equation 2.5 gives P for the van der Waals equation as a furt~tionof that F expressed as a function of V , T, and n is is given by Equation
V and T. Show
D. 14.
D-6.Shuw explicitty that Now evaluate ( a ~ / a y )from , Equation 2.5 and ( a P / a V j T mfrom n Equatim 1 a b v e and show that (see Problem IT12) fur the Redlich-Kwong equation (Prublem D-5).
0-7. We will derive the following equatirm in Chapter 5:
Evaluate ( a W / a V ) , for an ideal gas, For a van der Waals gas (Quation D.4), and for a Redlich-Kwong gas (Pmblem D-5).
D-14. Refening to Problem &I 3, shcy that
and generally that
0-8. Given that the heat capacity at constant volunle i s defined by
where y and x are intensive variables and y ( x , n, I.') can be written as y(x, V / n ) .
CHAPTER
5
The First Law of Thermodynamics
James Prescott joule was born in Salfwd, near Manchester, England, on December 24, 1818, and djed in 1889. He and hi elder brother were tutored at home by John Dalton, then in his 709.Joule's father was a wealthy brewer, which allowed Joule f m d o m from having to wek employment. Juule conducted his pioneering experiments in lahratories he built at his own expense in his home or in his father's brewery. From 1837 to 1847, ha carried out a sene:. of experiment$ that led to the general law of energy cunservatiun and to the mechanical equivalent of heat. Joule announced all his measurements in a public lecture at St. Ann's Church in Manchester, England and, because his earlier reports had bccn rcjwrcd by the British Association, latcr had his lccture pubiished in the Munch@.$-ter Courier,a new.hpaper for which his brother wrote tnusical critiques. In 1847, he presented his results to the British Association meeting in Oxford, where the 22-year-old William Thomson (later Lord Kelvin) immediately appmiated the importance of Joule's work. Thornson later asked Joule to carry out experiments on theexpansioeofgases. This work led to the discovety of the Joule-Tnomson effect, which demonstrated that a nonideal ggas cools when underguing a free expansion. Joule was elected to the Royal Society in 1850.Later in life, he suffered severe financial losscs, and in 1878 friends obtained a pension Tor him from the government. The SI unit of energy is named in his honor.
As we said in the introduction to Chapter 1, thermodynamics i s the study of the vmous properties and, particularly, the relations between the various properties tlf macroscopic systems in equ~lbrium.For example, thermodynamics tells us the conditions under which a gas will cool upon expansion (an important relation for liquefying gases) and how the vapor pressure of a droplet depends upon the radius of the droplet {an impom~rlt relation in meteorology). Thermodynamics is primar~lyan experimental science and is extensively applied to chemistry, biology, geology, physlcs, environmental sciencc, ar~d engineering. All the results of thermodynumics rest upon three fundamental laws. These laws, appropriately known as the First, Second, and Third Laws of Thermodynamics, summarize a vast body of experimental data on macroscopic systems, and there are absulutely no known exceptions. We shall take up each one of these laws in this and the next two chapters. The First Law of Thermodynamics is essentially a alatcmetlt of rhc law of conservation ofenergy applied to macroscopic systems. To prescnt the First Laal, we must introduce the concepts of work and heat as they arc used in thcrmodynamich. As we will see in the next section, work and heat are modes of energy transfer between a system and its surroundings. We will also see tbrlt although the amount of work and heat involved in a process depend upon huw the process i s carried out, the energy depends only upon the initial and tinal yates and does not depend upon how the final state is reached from the initial state. Later in the chapter we shatl introduce anuther important thermodynamic function, the enthalpy, which we will see arises naturally for processes that are carried out at constant pressure. Finally we w i H introduce standard molar enthalpies of fo~mation,from which we can calculate energy and enrhalpy changes associated wtth chemical reactionh. \>
5-1. A Commgn Type of Work Is Pressure-Volume Work The concepts uf work and heat play impoprlnt roles in thermodynamics. Both work and heat refer to the manner in which energy is transferred between some system of interest
1 85
I
Chapter 5 / The First Law of Thermodyrlnmics
and its surn~undings.By system we mean that part of the world we arc investigating and by surmundings we mean everything else. We define heat, q, to hc the manner of energy transfer that results from a temperature difference between h e system and its sum,undingr. Heat input to a system is considered a positive quantity; heat evolved by a systein isconsidered a negative quantity. We define work, w , to be the transfer of cncrgy between the system of interest and its surroundings as a result of the existence of unbalanced limes between the two. If the energy of the system is increased by the
work, we say that work is done on the syswm by the surroundings, and we take it to b e a positive quantity. On the other hand, if the energy of the system is decreased by the work, we say that the system d w s work on the surroundings, or that work is done i)v tllc system, and we take it to be a negative quantity. A common example of work in phyqical chemistry mcurs during the expansion or compression of a gas as a result of the difference in pressures exerted by the gas and on the gas. An important aspect of work is that it can alMays be related to the raising or lowering of a mass in the surroundings. To see the consequences of this statement, consider the situation in Figure 5.1, where a gas is confined to a cylinder that exerts :r force Mg on the gas. In Figure 5. la, the initial pressure of the gas, P,,i s suficicnt tu I)U'IIthe pisto~lupward, so there are pins holding it in position. Now, remove the pins and allow the gas to lift the mass upward to the new position shown, and let the pmssure of the gas now be If. In this process, the mass M has been raised a distance h, %I, the work done by the system i s
'flie negative sign here is in accord with our convention that work done by a system
But M g / A is the external pressure exerted on the gas and Ah is the change in volume experienced b y the gas, so we have
Note that A V z 0 in an expansion, su w < 0. Ctearly, the external pressure must be tess than the pressure of the inilal state of the gas in order rhnt the expaosiotl txcu~: After the expansion, = Now consider the situation in Figure 5. l b , where the initial pressure of the gas i s less than the external pressure %, = M g / A , so the gas is compressed when the pins are removed. In this case, the mass M i s lowered a distance h, and the work is given by
ex, e.
c,
But now A V < 0, so w > 0. After the compression, we have = positive because work is done on the gas when it is compressed. If is not constant during the expansion, the work is given by
c. The work is
ex,
w=-L v~
LpdV
(5.2)
where the limits on the integral indicate an initial state and a final state; we must have knowledge of how tx, varies with V along the path connecting these two states so we can carry out the integration in Equation 5.2. Equation 5.2 is applicable to either i s constant, Equation 5.2 gives Equation 5.1 expansion or compression, If
ex,
is taken to be a negative quantity. If we divide Mg by A , the area of the piston, and multiply h by A , then we have
E X A M P L E 5-1
Consider an ideal gas that ~ c u p i e sI.Im dm' at a pressure of 2.00 bar. If the gas is compressed isothermally at a constant external pressure, P,, so that the final volume can have? Calculate the work involveti is 0.500 dm3, what is the smallest value using this value of
&,.
cx,
9,
SOL UT ION : For a compression to occur, the value o f must bc at leas1 as large as the final pressure of the gas. Given the inital pressure and.vulume, and the final volume, we can determine the final pressure. The final pressure of the gas is
lnitial s t a t e
Final state
Initial state
Final state
This 1s the sttullest value fx, can be to compress the g a ~ ~ ~ o t h e m afrom l l y 1 00 dm' 5.1 The effect of work is equivalent to the raising or lowering of a mass in the surfoundings. In (a) work is done by the system because the mass is raised, and in (b) work is done on the system kcnuse the mass is Ir~wered.(The system is defined as the gas inside the piston.) FIGURE
to 0.500 dm'. The work
w =
involved using this value o f tX, is
-qn,A V = -(.1.00bar)(-0.500 dm')
= 2.00 dm3.bar ,
= (2.00drn3.bar)(10-'m' .dm ' ) ( I @ Pa.bar '1) = Z(H1Pa m' = 2(X)J
5-2. Work and Heat Are Not State Functions, but Energy Is a Srare Function
Of course, &, can be any value greater than 4.00 bar, so 200 J represents the smallest value of w for the isoohermal compression at constant pressure from a volume of 1.00 dm' to 0.500 dm3.
Figure 5.2 illushates the work involved in Example 5-1. As Equation 5.2 implies, the work is the area under the curve of Punversus V. The smooth curve is an isotherm ( P versus V at constant T) of an ideal gas; Figure 5.2a shows a constant-pressure compression at an external pressure equal to P,, the final pressure of the gas; and Figure 5.2b shows one at an external pressure greater than 4. We see that the work is different for different values of
exl.
5-2. Work and Heat Are Not State Functions, but Energy Is a State Function Work and heat have a property that makes them quite different from energy. To appreciate this difference, we must first discuss what we mean by the state of a system. We say ihat a system is in a definite state when all the variables needed to describe the system completely are defined. For example, the state of une mole of an ideal gas can be described completely by specifying P, 7,and T. In fact, because P, and T are related by PV = R T , any two of these three variables will suffice tu specify the state of the gas. Other systems may require more variables, but usually unly a few will
v,
suffice. A stute function is a property that depends only upon the state of the system, and not upon how the sybtem was brought to that state, or upon the history of the system. Energy is an example of a state function. An important mathe~naticalproperty ' of a state function is that its differential can he integrated in a normal way:
As the notation suggests, the value of AU i s independenr of the path taken betwee11the initial and final states 1 and 2; it depends only upon the initial and final states through AU = U,- U,. Work and heat are not state functions. Fur example, the external prcssurt: used to compress a gas can have any value as long as it is large enough to compress the g s . Consequently, the work done on the gas,
will depend upon the pressure used to compress the gas. The value of tX, must exceed the pressure of the gas to compress it. The minimum work required occurs when &, is just infinitesimally greater than the pressure of the gas at every stage of the compressiun, which means that the gas is essentially in equilibrium during the entire by the pressure of compression. In this special but important case, we can replace and P differ only infinitesimally, the process i s the gas (P) in Equation 5.2. When called a reversible process because the process could be reversed (from compression to expansion) by decreasing the external pressure ~nfinites~majly. Necessarily, a strtctly reversible process would require an infinite time to cany out because the process must be adjusted by an infinitesimal amount at each stage. Nevertheless, a reversible proccss serves as a useful idealized limit. Figure 5.3 shows that a reversible, isothermal compression uf a gas rcrlullcs the minimum possible amount of work. Let w- denote the reversible work. To calculute lalcvfor the compression of an ideal gas isothermally from V, to V,,we use Equation 5.2 with Pexlreplaced by the equilibrium value of the pressure uf the gas, which i~ 11 K? / V for an ideal gas. Therefore,
ex,
FIGURE
c,,
5.2
An illustration of the work involved in an isothermal constant-pressure conlpressiun from V, = 1 .OO dm' to V, = 0.500 dm3t: different values of Pcxl.The smooth curve i s an isotherm ( P vs. V at constant T ) for an ideal gas. In (a) Pcx,is equal to P,, the final pressure of the gas, and in (b) PckI is larger than P, = 4.M)bar, and pins must be used to stop the compression at 1: = 0.5011dm3. Otherwise the gas would be compressed further, until it reaches h e volume thai corresponds to on the isotherm. The work is equal to the area of the Pelt-Vrectangles.
el,
Because V, < V, for compression, we see that w- > 0 as it should be; in other words, we have done work on the gas.
5-2. Work anrl Heat A ~ Not P
State function^, hut Energy Is a State Function
the gas isothermally and reversibly from 1.00 dm3to 0.500 dm3. Comparc both results to the one nbtaincd in Exmurlplc 5-1. S O L U T I O N : Inthetwu-~tilgecumpression,AV=-(1.00-0.667)~lm3inthefirst, step and -(0.hh7 - 0 500) dm' in the second step. Therefore.
We use Equation 5.4 for the reversible procesf
5.3 The work of isotl~crnlnlcnrnprrscion i s the area undcr tllc Pm, versuq V curves shown in tllc figure. The external pressure must exceed ihe pressure of the gas in order to compress it. ?l'hc rl~ir~irnu~ll arnounl r>T work occurs whcn thc cxpansion is carried out reversibly; that is, when txl is just i~~finitesimally greater than the pressure of the gas at every stage of rhe compression. The gray area is the minimum work needed to compress the gas from V , = 1 .IK) dm' tu V, = 0.500 dm3. The constant-pressure compression curves are the same as Ih~j\c:in Figure 5.2.
I I ! , ~= ~
FIGURE
F
E X A M P L E 5-2 Crrn~ideran idea1 gas that occupies 1 .(lodni3 at 2.00 bar. Calculate the work required to cornpresv the gas isothermally to a volunle uT 0.667 dm' at a constant pressure of 3.00 bar filllowed by another isothermal compression to 0.500 dm3 at a constant pressure uf 4.00 bar (Figure 5.4). Compare the result with the work of compressing
V -nRTInA V~
=
0.5Wdm'
-nRT11lP
1.00 dm'
Because the gas is ideal and the p m e s s is isothermal, n RT i\ equal to either PIV, ur P,V2',,both of which equal 2.03 dn13.bar, and so
Note that I V , ~is~ less than that for the two-stage prucess and that the work for that procesq is less than the work required in Example 5-1 (200 J). (Compare Figures 5.2. 5.3, and 5.4.)
Jusl as the reversible isothermal compression of a gas requires the minimum amount of work to be done on the gas, a reversible isothermal expansion requires the gas to do a tnaximum amount of work in the process. In a reversible expansion. the external pressure is infinitesimally less than the pressure of the gas at each stage. If were any l q e r , the expansion would not occur. The work involved in the reversible isothermal expansion of an ideal gas is also given by Equation 5.4. Because V2 > V, for expansion. we see that wrv < 0 ; the gas has done work on the surroundings, in fact, the maximum possible.
c,
E X A M P L E 5-3
Derive an expression for the reversible isothermal work of an expansion of a van der Waals gas. SO 1U T 10N : The expression for the reversible work i s
FIGURE
5.4
An illusIration of the constant-pressure compression of a gas as described in Example 5-2. The w t ~ required k i\ given by the areas under the two rectangles.
where
192
Chdptcr 5 / The F~rsrLaw ~f Thermodynamics
We substi.titutcthis expression for P into uvmvlu obtain
5 4 . An Adiabatic Proccss Is a Process in Which Nu Energy a, He31 ISTranaftxrrerl
in differential form, ur
frlrni. Equalions 5.9 and 5.10 are statements of the k'irst h w of Thernrudynamics. The First Law of Thermodynamics, which is essentially a statenlent of the law of conservation of energy, also says that even though 6q and 6 w are separately path functions or inexact differentials, their sum is a state function or an exact differerjti;~l. All state functions are exact differentials.
in integrated Note that thic quatinn reduces tn Equ.ation 5.4when a = b = 0.
J 5-3. The First Law of Thermodynamics Says the Energy Is a State Function Because the work involved in a process depends upon how the process is carried out, work is not a state function. Thus, we write Sw = w
(not AUJor uj,
-w,)
(5.5)
It makes no sense ar all to write w,, w , ,up, - ~ u ,or, Aw. The value of TU obtained in Equation 5.5 depends upon the path from state 1 to 2, su work is called apatkfunction. Mathematically, &TI) in Equatiun 5.5 is called an inexactd#erentiul, as opposed to an exat.[ d~flerenriullike d U , which can be integrated in the normal way to obtairi U, - U, (see Mathchapter D). Workand heat are defindd only for prwesses in which energy is transcerrcd between a system and its surroundings. Both work and heat rn path functions. Although a system in ;I givcn state has a certain amount of encgy, it does not pussess work or heat. The difference between energy and wurk and heat can be xummarized by writing
l2
d U = U?-- U , = AU
l
SIU= w
Sq=q
(not w Z- w , )
-
(notq,-q,)
5 4 . An Adiabatic Process Is a Process in Which No Energy as Heat Is Transferred Not only are work and heat not state functions, but we can prove that even reversible work and reversible heat are not state functions by a direct calculation. Consider the three paths, depicted in Figure 5.5, that occur between thc same initial and final statcs, PI,V , , T, and P,,V,, TI.Path A involves a reversible isothermal expansioll of an ideal gas from P,, Vl, TI to P,, V,, T , . Because the energy of an ideal gas depends upon only the temperature (see Equatiun 4.40, for example),
(U is a state function)
(path function)
(path function)
(5.8)
For a process in which energy is transferred both as work and heat, the law of conservation of energy says that the energy of the system obeys the equation
I
v $5 An illus&aiion of three different pathways (A, R + C, and D + E) to take an tdeal g a ~from PI, Vl , TI to P,, V,, T, . In each case, the value of AU IS the same ( A l l 1s a std~t:iunction), b11t the values of q wd w are different ( q and ul are path funct~ons) FIGURE
Chapter 5 / The Firs1 Law of Thernmlynamics
196
5-5, The Temperature uf a Las Decreases in a Reversible M i a h a ~ i cExpansion
We learned in Section 4-2 that 7,= 3 R / 2 for a monatomic ideal gas, so Equation 5.2 1
and
becomes Therefwe,Inr the overall pnxess, dun+. =
6;
+ *liE =
+
jl:l
c~(T)~?T
C.(TIT = o
and
Nnte that AU = 0 for all three processes indicated in Figure 5.5, but that wmyand qm are different for each one.
()
(
)
E X A M P L E 5-5 Calculate the final temperature if argon (assumed to bt. ideal) at an initial ternpraturc of 300 K expands reversibly and adiabatically from a volume of 50.0 L to 200 L.
SOL UT I0N : First solve Equation 5.22 for T,/T,,
Path B in Figure 5.5 represents the reversible adiabatic expansion of an ideal gas from TI,V, to T,, V2.As the figure suggests, T, < TI,which means that the gas cools during a (reversible) adiabatic expansion. We can determine the final temperature T, for this process. For an adiabatic process, y = 0, and so
We divide both sides by T a A n and integrate to obtain
= ,
monatomic ideal gas
Thus, the gas cools in a reversible adiabatic expansioll (V? > V,).
5-5. The Temperature of a Gas Decreases in a Reversible Adiabatic Expansion
Note that the above expression tells us that 6 w = d w is an exacl differential when 6q = 0. Likewise, Sq = dq i s an exact differential if Sw = 0. The work done by the gas (the system) in the expansion is "paid for" by a decrease in the energy of the gas, which amounts tn a decrease in the temperature of h e gas. Because the work invoived in a reversible expansion is maximum, the gas must suffer a maximum drop in temperature in a reversible adiabatic expansion. Recall that for an ideat gas, U depends only upon the temperature and d U = C, (T)dT = n C , ( T ) d T .where C, (T) is h e molar constanlvolume hear capacity. Using the fact that d w = - P d V = -n RTrl V / V fora reversible expansion, the relation dU = d w gives
v,
312
and then let TI= 300 K, V, = 50.0 L, and V? = 200 L to obtain
P
We can express Equation 5.22 in terms of pressure and volume by using n RT to eliminate TI and T,:
Upun taking both sides to the u3 power and rearranging, we obtain
p I
V5/3 I
=p
,,
v5i3
(monatotnic ideal gas)
(5.23)
This equation shows how the pressure and volume are related in a reversible, adiabutic process for an ideal monatomic gas. Compare this result to Boyle's law, which says that
for an isothermal process.
Chapter 5 I The First Law of Thermodynamics
198
Because El = El ( N ,V),we can view d Ej as the change in Ej due to a small change in the volume, d V ,keeping N fixed. Therefore, substituting d E, = (a Ej/aV),d V into Equation 5.26 gives
E X A M P L E 5-6 Derive the analogs uf Equations 5.22 and 5.23
for an ideal diatomic gas. Assume the temperature is such that the vibrational contribution to the heat capacity can he ignored. SO 1UT I 0 N : Assuming that C,,vi, x O+ we have from Equation 4.4 1 that 5 R / 2 . Equation 5.20 for a diatomic ideal gas becomes
5-6. Wurk ~ r l Idlent Have a Simple Muletular Inr~rpreration
F, = This result suggests we can interpret the first term in Equation 5.26 to be the average change in e~lcrgyuf the system caused by a small change in its volume, in other words, the average work. Furthermore, if this change is done reversibly, so lhat the system remains essenr ially in equilibrium at each slage, h e n the p, in Equation 5.26 will be given by Equation 5.25 thoughout the entire process. We can emphasize this by writing
(diatomic ideat gas) Suhflituting T = P V / n R into the ahnw
equation gives
If we compare this result with the macroscopic equation (Equation 5.9)
P,
v"'
=
P,
y7I5
we see that (diatumic ideal
gas)
I and
5-6. Work and Heat Have a Simple Molecular Interpretation Let's go back to Equation 3.1 8 for the average energy of a macroscopic system,
with
T ~ Swe. see that reversible work, S T U results ~ ~ ~ from an infinitehnal change in the allowed energies of a system, without changing the probability disuibution of Its states Reversible heat, on the other hand, results from a change in the probability distribution of the states of a system, without changing the allowed energies. If we compare Equation 5.29 with
we see that
we can identify the pressure uf the gas with
Equation 5.24 represents thF average energy of an equilibrium system that has the variables N , V, and T fixed. If we differentiate Equation 5.24, we obtain
Recall tha~we used this equation without proof in Section 3-5 to show that P V = R T for one mole of an ideal gas.
5-7. The Enthalpy Change Is Equal to the Energy Transicrred as Heat
5-7. The Enthatpy Change Is Equal to the Energy Transferred as Heat in a Constant-Pressure Process Involving Only P - 1' Work For a reversible process in which the only work involved is pressure-vulume work, the first law tells us that
AU=qfw=qIf a
L:
PdV
Let's apply these results to the melting of ice at W C and one atm. For this process, q , = 6.01 k~.niol-I.Using Equation 5.37, we find that
where the overbar on H signifies that AHis a molar quantity. We can also calculate the value of AVusing Equation 5.36 and the fact that the molar volurnc of ice {Vh)i s 0.0196 L.mol ' I and that of water (TI)is 0.01 80 I-.rnuI--':
e pmess is carried out at constant volume, then Vl = V2 and
where the subscript V on q emphasizes that Equation 5.33 applies ro a canstantvolume process. Thus, we see that AU can be measured experimenlally by measuring the energy as heat (by means of a calorimeter) associated with a constant-volume
process (in a ngid closed container). Many processes, particularly chemical reactions, are carried out at constant prcasure ( o F n to the atmosphere). The energy as heat associated with a constant-pressure process, q , , i s not equal to AU. It would be convenient to have a state function analogous tc~U so that we could write an expressLon like that in Equatlon 5.33. To this end, let P he uonbtant in Equation 5.32 so that
Thus, in this case, there i s essentially no difference between AH and AU. Let's lwjk at the vaporization of water at IOWC and one atm. For this prwess, = 30.h L.mol-'. Therefore, q,, = 40.7 k~ .mtli-', V , = 0.01 80 L.rnol--', and
But
where we have used the subscript P on q , to e~nphasizcthat this is a constant-pressure process. This equation suggests that we define a new state function by
At conslant pressure, AH=AU+PAV
(constant pressure)
Equation 5.34 shows that
Thus, this new state function H, called the etlthalpy, plays the same role in a constantpressure process that U play? in a constant-vulume process. The value of AH can be determined experimentally by measuring the energy as heat associated with a constant-pressure process, or conversely, q, can be determined From A H . Because most chemical reactions take place at constant pressure, the enthalpy i s a practical and important thermodynamic function.
Notice that the numerical values of A= and AU are significantly d~fferent(% 8%) in this case ~ecauseATfor this process i s fairly large. We can give a physical interpretation of'rhzse results. Of the 40.7 kJ that are absorbed at constant pressure, 37.6 Icl ( q , = A UJare used to overcome the intermolecutar forces holding the wnter molecules in the liquid state (hydmgen brlnds) and 3.1 kl (40.7 kJ - 37.6 kJ) are used to increase the valume of the system against the at~nospherrcpressure.
E X A M P L E 5-7 Thc value of at 298
4H
is -572
K and one bar for the reaction described by
W.Calculate AU for this reaction as written.
Chapter 5 / The First Law of Thcrrnudynamics
SO L U 1'1 0 N : Because the reaction is carried out at a constant pressure of 1 .MI bar, A H = q, = -572 kJ. To calculate A U , we must first calculate A V . Initially, we have ihree moles of gas at 298 K and 1.00 bar, and so
v=--n K T
(3 mo1)(0.08314
P
=
If the substance is heated at constant pressure, the added energy as heat is q , and the heat capacity is denoted by C,. Because A H = q,, C , is given by
bar.^-' .mol-')(298 K) I .(KI bar
74.3 I,
We expecl hat C, is larger than C , because not only do we increase the ternperdturt. when we add energy as heat in a constant-pressure prwess, but we also do work against atmospheric pressure as the substance expands as it is heated. Calculating the difference between C , and C, for an ideal gas is easy. We start with H = U + P V and replace P Y by n RT to obtain
Afterward, we have two moles uf liquid water, whose vdume is about 36 mL,which i h negligible compared with 74.3 L. Thuu, A V = -74.3 L and AU = AH =
-
PAV
-572 k I + ( l I X I bruj(73.4 I-)
5-8. Heat Capacity Is a Path Function
(&)
= -572
+743
(ideal gas)
H=U+nRT
(5.41)
= --5h5kJ
'The numerical difference &tween A l i md A U in this caw is about 1%.
I
Notice that because U depends only upon the temperature (at constant n) for an ideal gas, H also depends only upon temperature. Thus, w e can differentiate Equation 5.41 with respect to temperature to obtain
Exaniple 5-7 i s a special case of a general result for reactions or processes that invnlve ideal gases, which says that But
dH d T = ($)
where =
( gaseous products
P
= C,
(ideal gas)
and
As Example 5-7 implies, the numerical difference between A H and AU is usually
dT =
small.
(g) Y
= C,
(ideal gas)
so Equation 5.42 becomes C,
5-8. Heat Capacity Is a Path Function Recull thal heat capacity isdefined as the energy us heat required to raise the temperature of a substance by one degree. The heat capacity aiso depends upon the temperature T. Bccnuse the energy required to raise the temperature of a substance by une kelvin depcnds upon the amount uf substance, heat capacity is an extensive quantity. Heat capacity is also a path function; for example, its value depends upon whether we heat
the substance at constant volume or at constant pressure. If the substance is heated at constant volumc, the added e'nergy as heat is q, and the heat capacity is denoted by C , . Bccause A U = q,, C, is given by
-
C, = n H
(ideal gas)
(5.43)
Recall from Chapter 3 that C, is 3 R / 2 for une mole of a monatomic ideal gas and is appruximately 3R for one mole of a nonlinear plyatomic ideal gas at room temperature. Therefore, the difference between F, and C, is significant for gases. For solids and liquids, however, the difference is small.
I
E X A M P L E 5-8 We will prove generally that (Section 8-3)
Chapter 5 / The First Law of Thcrmcdynamics
cp r,
First, use this result to show that = R for an ideal gas and then derive an expression for - C, for or gaas that obeys the equation of state
rp
5-9. Relative Enthalpies Can Be Determined from Heat Capacity Data and tieats of Transitrun
where we have used the fact rhat P = RT/[V - B ( T ) Jin going from the thlrd line to he last line. Notice that this expression is the same as tlm for an ideal gas if R ( T ) is a cunstant.
SOLUTION: Foranideal gas. P V = R T , sn
5-9. Relative Enthalpies Can Be Determined from Heat Capacity Data and Heats of Transition and su
By integrating Equation 5.40, we can calculate the difference in the enthalpy of a substance that does not change phase between two temperalures:
To delermine ( 3 P / a T ) r Cur a gas that u k y s Lhz quation of state, B ( T )P,we first solve for P .
PV = RT
t. If we let T = 0 K. we have
and then differentiate with respect to temperature:
[Notice that we have written the integration variable in Equation 5.45 with a prime, which is standard mathematical notation used to distinguish an integration limit (T in this case) horn the integration variable, T'.] It would appear from Equation 5.34 that if we had heat-capacity data from 0 K to any nther temperature, T, we could calculate H ( T ) relative to H(0). That is not entirely true, however. Quation 5.45 is applicable to a temperature range in which no phase transitions occur. If there is a phasc transition, we must add the enthalpy change for that transition hecause heat is absorbed without a change in T for a phase transition. For example, if T in Equation 5.45 is in the liquid region of a substance and the only phasc c h a n g between 0 K and T is a solid-liquid transitiun, then
Similarly,
and
Therehe, using the equation for
c,
- -
=
r, - C , given in the statement of this exan~ple,
( ) (E) aT , aT
where C;(T)and c ; ( Tstand ) for the heat capacity of the solid and liquid phases, respectively, T,& stands for the mehing temperature, :I( FI is the enthalpy change upon melting he heat of fusion):
..
Figure 5.6 shows the molar heat capacity uf benzene as a function of tertlperature. Notice that the plot of C, versus T is not continuous, but has jump discontinuities at the tempt:raturcs corresponding to phase transilions. The melting point and boiling p i n t of benzene at one atm are 278.7 K and 353.2 K, respectively. As Equation 5.45 implies, the area under the curve in Figurc 5.6 fr,orn 0 K to T 5 278.7 K givcs the mt~lar enthalpy of solid knzene lrelative to H(0j1.li, calculate ttlc molar enthulpy of' liytiid
5-1 0. Enthalpy Changes for Chemical Equations Are Additive Because most chc~nicalreactions take place at constant pressure (open to the atmosphere), the enthalpy change associated with chemical reacrions. A , H , (the subscript r indicates that tlie enthalpy change is I'or a chemical reaction) plays a central role in ~ l i p r mnchemisrry, which is the branch of Ihermodynarnics that concerns thc measurement of the evolutir~nor absorption of energy as heat associated with chemical reactions. For example, the combustion of methane,
F I G U R E 5.6 The constant-pressure molar heat capcity of benzene fron; 0 K to 500 K. The melting point and boiling point of benzene ai one atrn are 278.7 K and 353.2 K. respectivelq
benzene, say, at 3(X) K and unc atm, we take the area under the curve in Figure 5.6 from 0 K to 3(X) K and add the molar enthalpy of fusion, which is 9.95 Id-mol-'. Figure 5.7 shows the molar enthalpy of benzene as a function of temperature. Notice that H ( T ) - H ( 0 ) increases smoothly within a phase and that there i~a jump at a phax: transition.
releases energy as heat and is called an exothermic reaction ( a o = out). Must combustion reactions are highly exothennic. The heat evolved in a cumbustion reaction is called the heat of combustiun. Chemical reactions that absorb energy as heat are called endothermic reactions (endo = in). Exothermic and endothermic reactions are illustrated schematically in Figure 5.8. The enthalpy change for a chemical reaction can be viewed as the total enthalpy of the products tninu~the total enthsllpy ofthe reactants:
For an exolhctmic reaction, HPd i s ~ C S Sthan HKa,,SO A r H < 0. Figure 5.8a represents an exothermic reaction; the enthalpy of thc reactants is greater than the enthalpy of the products, s o 4, = ArH i0, and encrgy as hcat is evolved as the reaction proceeds. For an endothermic reaction, Hpm,is greater than Hmmr,, so A, H > 0,Figure 5.8b represents an endother~nicrcaction; the enthalpy of the reactants is less than the enthalpy of the prcducls, so q , = ArH > 0, and cnergy as heat must be supplied to drive the reaction up the enthalpy "hill." Let's consider several examples of chemical reactions carried out at one bar. For the combustion of one mole of methane to form one mole of CO,(g) and two molcs of
I
Reactants
7
P ~ D ~ Y C ~ S
-
F I G U R E 5.7 The molar enthalpy of ben)tne [relative to H(O)jfrom O K to 500 K.
I
(:I)
(b)
F I G U R E 5.8
An cnthalpy diagram for (a) an exothermic reaction and (b) and endothermic reaction.
Chapter 5 / The First Law ul Thermodynarn~cs
H,O(I), the value of AtH i s -890.36 k.l at 298 K. The negative value of ArH tells us that the reaction gives off energy as heat and i s therefore exothermic. An example of an endothermic reaction i s the water-gas reaction:
5 - 10. Enthalpy Changes for Chemical Equations Arc Additive
As an exart~pleof the application of Hess's Law, consider the use of
and For this reaction, ArH = +I31 kJ at 298 K,so energy as heat must be supplied to drive the reaction from left to right. An important and useful property of ArH for chemical equations is additivity. This
property of A,H follows directly from he fact that the enthalpy is a state function. Jf we add two chemical equations to obtain a third chemical equation, Lhe value of brH for the resulting equation is equal to the sum o f the ArH for the two equations being added together. The additivity of A,H i s best illustrated by example. con side^ the following two chemical equations.
to calculate (he value of ArH for the equation
Wl,(I)
+ Cl,(g)
+
pcl,(s)
In this case, we add Equation 2 to the reverse of Equation 1 to obtain Equation 4:
Thas, from Hess's law, we obtain
If we add these two chemical equations as if they were algebraic equations, we get We now multiply Equation 4 through by 112 to obtain Equation 3:
The additive property of ArH tells us that ArH for Equation 3 is simply and so
In effect, we can imagine Equations 1 and 2 as representing a two-step process with the same initial and final states as Equation 3. The total enthalpy change for the two equations together must, therefore, be the same as if the reaction proceeded in a single step. The additivity propetty of A,H values i s known as Hess's Law. Thus, if h e values of ArH( 1) and A, H ( 2 ) are known, we need not independently determine the experimental value of ArH(3) because its value is equal to the sum A r H ( l ) ArH{2). Now let's consider the following combinaiion of chemical equations.
+
-
1 EXAMPLE
-
5-9 The molar enthalpies of combustion of isnbutanz and n-bulanc are - 287 1 kJ,mol and -2878 kl.rnol-I, respectively at 298K and one atm. Calculate ArH fur the conversion uf one mole of n-butane to one mole uf isobutrlne.
'
S 0 L U TI 0 N: The equaiiuns fur the two cumhustion reactions are
7
H - C , H , ~ (+ ~ ) Ol(gl -+ -1CO,(g) f 5 H,OfI)
and
Bwause Equation 2 is sirnpiy the reverse of Equation I , we conclude from Hess's Law that A t H (reve~se)=
-Ar H (forward)
(5.48)
(1)
I
Chapter 5 1 The First Law of Thermodynamics
If we reverse the second equation and add the result to the first equation, then we
obtain the desired equation
5-1 1. Weals of Reactions Can Be Calculated imm Tabulated Heats of Formation
vaporizatinn [e.g., H20(1) -t H,O(g)J. Table 5.1 lists many of the subscripts you will encounter. The stundard molar enthnlpy of formarion, A,H0, is a particularly useful quantity. This intensive quantity is the standard reaction enthalpy for the formation of one molc of a molecule from its constitue~~t elements. The degree superscript tells us that all reactants and products are in their standard states. The d u e of A,H0 of H,0(1) i s -285.8 W.mol-' at 298.15 K. This quantity irnpIies that the balanced reaction is written as
The heat or this reacliun cannot be measured directly because competing reactions occur.
5-1 1. Heats of Reactions Can Be Calculated from Tabulated Heats of Formation ' I ' l ~ ccr~thalpychange of a chcrnical reaction, A r H ,depends upon the number of moles of the reactants. Recently, the physical chemistry division of the International Union of IZrre and Applied Chemistry (IUPAC) has proposed a systematic procedure for tab~ilatingreaction enthalpies. The s t n d u r d reaction enthalpy of a chemical reaction is denoted by A, H" and refers to the enthalpy change associated with one mole of a
specified reagent when all reactants and products are in their standard states, which for a gas is the equivalent hypothetical ideal gas at a pressure of one bar at the temperature of ititerest. For example, consider the combustion of carbon to form carbon dioxide CO,(g). (The standard state of a solid is the pure crystalline substance at one bar pressure at the temperature of interest.) The balanced reaction can be written in many ways, including
because A , H refers to the heat of formation of one mole of H20(1). (The standard state for a liquid is the normal state of the Equid at one bar at the teniperature of interest.) A valuc of A,H0 for H,O(I) equal to -285.8 kJ.mol-' tells us that one mole of H,0(1) lies 285.8 kJ "downhill" on the enthatpy scale relative to its constituent elements (Figure 5.9h) when the reactants and products are in their standard states. Most compounds cannot be formed directly from their elements. For example, an attempt to make the hydrocarbon acetylene (C,H,) by the direct reaction of carbon with hydrogen
yields not just C,H2 but a complex mixture of various hydrocarbons such as C,H, and C2%, among others. Nevertheless, we can determine the value of A,H" for acetylene by using Hess's Law, together with the available A=H" data on combustion reactions. All three species in Equation 5.5 1 burn in oxygen, and at 298 K we have
and
The quantity ArH ' implies E q u a t i o ~5.49 ~ because only one mole of the (specified) reactant C(s) i s cornbusted. Thc d u e of AIHofor this reaction at 298 K is ATH" = -303.5 k~.rnol-'.The corresponding reaction enthalpy for Equativn 5.50 is
If we multipty Equation 1 by 2, reverse Equation 3, and add the results to Equation 2, we obtain
with
We see that drlI is an e x t e n h e quanlily, whereas ATHoi s an intensive quantity. The ;~rlunntagcof thc tcrrninology is that it removes lhe amhiguity of how the balanced reaulion corresponding to an enthalpy change is written. Certain subscripts are used in place of r to indicate specific types of processes. Frw cxainple, the subscript "c" is used For a combustion reaction and "vap" is used for
5-1 1. T A B L E 5.1 Common subscripts for the enthalpy changes of processes.
Subscript
vap sub fus
trs mix
SOLUTION: The chemical equations for lhe rhree conlbuslion reactions are as
follows:
Rei~tion
Vaporization, evaporation Sublimation Melting, fusion Transition hetween phases in general Mixing of fluids
ads
Adsurp~iun
C
Combustion Formation
f
Heats of Reaction$ Can Be Calculated from Tabulated Heals nf Formation
If we reverse Equation 3, multiply Equtltion 2 by 2, and add the rcsults to Equation 1. we obtain the equation for the formation of CH,(g) frum ils elements.
\
along with
Because Equatiun (4) represents the formation uf one mole of CH,(g) directly irrm its elements, we have A,H0[CH,(g)l = -74.81 kl-mul-' at 298 K.
I
I
F I G U R E 5.9
Standard enthalpy changes involved in the formation uf CO,(g), H,O(I), and C,H,(g) from their elements, based upon the convention that A,H0 = 0 for a pure element in its stable furm at oric bar and at the temperature of interest.
Note that h e s t o i c h i o m e ~ ccoefficients have no uniuj in the IUPAC convention. Because Equation 4 represents the formation of one mole of C,H,(g) from its elements, A,HG[C2H,(g)] = -t226.7 kJ.mol-' at 298 K (Figure 5 . 9 ~ )Thus, . we sce that we can obtain values of A,H' even if the compound cannot be fumed directly from its elettlents.
As suggested by Figure 5.9, we can set up a table of A f H o values for con]puunds by setting the values of A,H0 for the elements equal to zero. Thai is, for each pure element in i t s stable form at one bar at the temperature of interest, w e set A,H0 equal to 7em. Thus, standard enthalpies of formation of compounds are given relative to the dements in their nonnal physical states at one har. Table 5.2 lisls values of A f H Cat 25°C for a number of substances. If you louk at 'Fdble 5.2, you will see that A,Hn[C(diamond)] = + I ,897 kJ-mol-', A, Hc[Br,(g)] = $30.907 k l .mol-I, and A, HU[I,(g)l = +62.438 kJ.rnr11-' . The values of A, H" fur these forms uf the elements are not equal to zero because Cldiamond), Br,(g), and I,(g) are not the normal physical states of these elemeits at 25°C and one bar. The normal physical states of these elements at 25°C and one bar are C(graphite), &,(I). and I,(s).
E X A M P L E 5-11 Use Table 5.2 tu calculate the molar cnthalpy of vaporizatiun A+apH' of brorninc al
25°C. E X A M P L E 5-10 Given that the standard enthalpies of combustion of C(S), H2(g), and CCli,(g) are -393.5 k.l.rnol-I, -285.H kl-mol-', and -890.4 W.rnol-'], respectively, at 298 K, calculate the stmdard enthalpy of fnmaticln uf methane, CH,(g).
I
S 0 L U T I O N : The equation that represents the vaporizatiun of one nlolc of hrtrnilne is
Chapter 5 1 The First Law of Thermodynamics
5.2 Standard tnolar enthalpies of formation, A,H", for various subslances at 25°C and one bar.
5-1 1. Heats uf Reactions Can Be Calculated from Tabulated Heats of Formation
TABLE
Substance
Formula
Therefore,
A. H"/!d.~nol-'
Note that this result is not the value of AVqH'' at its normal boiling point of S8.8'C. The value of Av:pH" fi 558.T is 29.96 kJ.mol-I. ( h r c'cvill learn how to calculate the temperalure varration of A H in the next section.)
Carbon dioxide Carbun monoxide: Cyclohexanc Ethane Ethanol
Ethenc Cilucose Hexme
Hydrazinc
coz(g) co(g)
-393.504 - 1 10.5
c(,H~z(l) C, H, (g) C,HsOH(I) C p , (g) C&H,206(s) c6H14(lj N2H4(1)
We can use Hess's law to understand how enthaplies of formation are used to calculate enthalpy changes. Consider the general chemical equation
where a , b. y , and z are the number of moles of the respective species. We can calculate A r H in two steps, as shown in the following diagram:
II
N*H,g)
mu1 uf A
Products
A , H"
z n ~ o lo f Z
am)
Hydrogen bromide Hydrrlgen cl~loride Hydrogen fluoride Hydrogen lodide Hyrlrugen peroxidc lodine
H20,(I) 1, (g)
Methane
CH, (g)
Mclhanul
CH,OH(I)
-74.81 -239. 1
CH,OH(g)
-201.5
licl(g)
\
HFk)
-aAfH3[A] - 187.8
f Elcments in their
f 62.438
Nitrogen dioxide Di~iirrogcn~etr~oxide
No2(g} N,O,(g)
Octnnc
c8H
(1)
-250.1
PenPane
csH I I(]) C, HE(g) C I ~ H ~ (s) PII
-173.5 -103.8
First, we decompose compounds A and B into their constituent elements (step I ) : and then we combine the elements to form the ci~rnpoundsY and Z (step 2). In thc first step, we have
N20, (1)
P~opanc Sucrrlse
Sulfur dioxide Sulfur trioxide , Tclrachloro~netha~~t.
so,(g) so,(g) UCI,(I)
-22.2~ -296.8 -395.7 -135.44
We have omitted the degree superscripr on the ArH because this value is nnt necessarily referenced to one mole of a particular reagent. The minus signs occur here because the reaction involved is rhe reverse of the formation of the compounds from their elements; we are forming the elements from the compounds. In the second step, we have
ArH(2) = y A,H5[Y]
-+ zA, H , [Z]
The sum of A,If ( I ) and AtH ( 2 ) gives ATH for the general eql~ation:
2 16
Chaptcr 5 / The F~rsiLaw of Thermodynam~cs
Note that the right side of Equation 5.52 is the total enthalpy of thc products minus the total cnthalpy uf the reactants (see Equation 5.47). When using Equation 5.52. you need to specify whether each substance is a gas, liquid, or solid because the value of A,Ho depends upun the physical state of the substance. Using Equation 5.52, wc determine A r H fur the reaction
5-12. The Temperature Dependence OF A, H Is Given in Terms of the Heat Capacities of the Reactants and Products Up to now, we havc calculated reaction enthalpies at 25°C. We will see in this seulion that we can calculate A,H at other temperatures if we have sufficient hcat-capacity data. Consider the general reaction
We can express A, H at a temperature T, in the form
Using the data in Table 5.2, we ubtain where, from Equation 5.45,
Note that A, H0[O,(g)l = 0 because the Af HI' value for any element in its stable state at 298 K and one bar is zero. To determine A r H for
we niuitiply ArH = - 1299.58 kJ.mul-' by
I
etc. Similarly, A r H ( T , )is given by
2 mnl lo obtain ArH = -2599.16 W.
EXAMPLE 5-12 IJw thc A$' dava in Tahle 5.2 to calculate the value ol-A,H' for the comhrlstion clC liquid ethanol, C,H,OH(I), at 25°C:
ctc. Ifwe substitute Equation 5.54into Equation 5.53 and Equation 5.56 init1 Erlu~ation 5.55, and then subtract the resultant A , H ( T l ) from ArH(T,),we obtain
SOLUTION: Reremng to Table 5.2, we find that A , H 3 [ C 0 , ( g ) ]= -393.509 kJ,mol-I; A, H'[H,O(1)] = - 2 ~ 5 . ~ 3 k l ~ m o l - ' ; A,HU[O,(g)J = 0 and A, Hn[C,H,OH(1)J = -277.69 kJ-mu]-' . Application of Equation 5.52 yields
where, as the notation suggests, AC, ( T ) = y C,,, ( T )
+ zC,,,(T)
-uC,,(T)-
bC,,,(T)
(5.5gj
Thus, if we know AtH at TI,say 2SnC,we can clilculnte ATH a1 any other temperature using Equation 5.57. In writing Equation 5.57 we have assumed there are no phabe transitions betwegn TI and T,. Equation 5.57 has a simple physical interpretation given hy Figurc 5.10. 'Lb cnlculate the value of A,H at some temperalure T, given the value of A r H at T I ,we can follow the path 1-2-3 in Figure 5.10. This pathway involves taking the reactants
5-1 2. The Temperature Dependence
219
1
E X A M P L E 5-13
'
The standard molar enthaply of formation, A,H0,of NH,(g) is -46.1 I kl-mol- at 25°C. Using the heat capacity data given below, calculale the standard molar heat of formation of -(g) at 1000 K.
a
+ 7.'
F I G U R E 5.10 An illusvation of Equation 5.57. Along path 1 we take the reactants from 7; to TI. Along path 2 we let the reaction occur at TI.Then along path 3, we bring thc prducts from T,back to T,.Because AH is a state function, we have that AH&) = AH, AH, AH,.
+
+
from teinperaturc T? to T I ,letting the reaction occur at T,, and then taking the products f ~ w nTI back to T,. The mathematical expressions for AH for each step are
L:
iT
< 1500 K
S O L U T I O N : Weuse thteyuation
The relevant chemical equation for the formation of one mole of NH,(g) from its elemerits is
and so
AH, = ArH(Tl)
A H3 =
where 298 K
AC;(T)/J.K-' .rnol-' = ( I ) C;(NH,)
C p(products)dT
= -31.21
+ (30.88 x
-
l r 3 K-')T
(i) Ci(N,) - (:) - 15.895 x
Ci(H,)
lo-* K-')T'
The integral of A C , ( T ) i s
and so AH(I',) = AH, 4-AH,
+ AH,
As a simple application of Equation 5.57, consider
Let's calculate A,
and
The pressure dependence of ArH (which we will study in Chapter 8) is usually much smaller than its temperature dependence.
Problems
Problems 5-1. Suppose that a 10-kg mass of iron at 20'C is dropped from a height of 100 meters. What is the kinetic energy of the mass just before it hits the ground? What i s its speed? What would be the final temperature of the mass il-all its kinetic energy at impact is transformed iniu intcrnal energy? Take the nlolar heat capacity uf iron to be C, = 25.1 J,rnol-I .K-I and ihe gravitstionai awelmlinn constant to be 9.80 m-s-?.
5-2. Consider an ideal gas that occupies 2.50 dm3 at a pressure of 3.00 bar. If the gas is so that the final volume is compressed isuthemally m a constant external pressure, 0.500 dm', calculate the smallest value can have. Calculate the work involved using this value of tX,.
c,,
5-3. A one-mole sample uf CO,(g) occupies 2.00 dm3 at a temperature of 300 K. If the gaq is compressed isothermally at a constant external pressure, Pcx,,so that h e final volume is 0.750 dm3, calculate the smallest value 9,can have, assuming that CO,(g) satisfies the van der Wmls equation uf state under these conditions. Calculate the work involved using this value uT Pcx,.
5-12. One mole uf a monatomic ideal gas initially at a pressure of 2.00 bar and a temperalure uT 273 K is taken to a final pressure of 4.00 h r by the rcvcrsihle palh defincd by P / \' = constant. Calculate lhe values of AU, A H , q , and nl Cor this proce\s. Take F , to be equal to 12.5 1.1nol K - I .
'
5-1 3. I h e isothermal compressibil~tyof a subslance is given by
B = I JP, but fur a liquid, @ is fairly constant over a moderate pressure range. IT p i s constant, show that For an ided gas,
where 1/, is the vulurne at apressure Po. Use this result to show that the reversible isothern~al work of compressing a liquid from a volume V, (at a pressure Po) tu a volume V (at a pressure P) is given by
5-4. Calculate the work invulved when one nlole of an ideal gas is compressed reversibly from 1.00 bar to 5.00 bar at a constant temperature of 300 K.
5-5. Calculate the work involved when one mole of an idzal gas is expanded reversibly from 20.0dm' to 40.0 dm3 at a uctnslant temperature of 300 K. 5-6. Calculate the minimum amount uf work required to compress 5 . 0 0 mules of an ideal gas isothernlally at 300 K from a volume of 100 dm' tu 40.0 dm3. 5-7. Consider an ideal gas thal occupies 2.25 L at 1.33 bar Calculate the work required to compress thc gas isothermally tu a volume of 1.50 L at a constant pressure of 2.00 bar followed by another isothermal cumpression to 0.800L at a constant preshure of 2.50 bar (Figure 5.4). Compare lhe result with the work of compressing the gas isothermally and reversibly from 2.25 L to 0.800L.
5-8. Show that for an isothermal reversible expansion from a molar vulurne 7 ,to a hnal molar the work i5 given by volume
5,
Tor the Redlich-Kwong equation
5-9. Use the result of Problem 5-8 to calcutate the work involved in Ihe isothermal reversible expansion of one mole of CH,{g) frum a vojunie of 1.00 dm' .mol to 5.00 dm"mo1-' at 300 K. (See Table 2.4 f o the ~ values of A and R.) 5-10. Kepeat the calculation
iil
(You need to use the fact that Jln xd.r = x Inx
- x.)
The fact that liquids are incompressible is reflected by p being hnvall, so that @(P P,) << I €or moderate pressures. Show lhat
Calculate the work required to compress one mule of toluene reversibly and isvther~i~ihlly from 10 bar to 100 bar at 20°C. Take the value of @ to be 8.95 x bar-' and the molar volume to be 0.106 rno1.~-' at 20°C.
5-14. In the previous problem, you derived an expression €or the rcversible, isotherlual wurk done when a liquid i s compressed. Given that B i s typically O(lO-') bar-.', show that V / V, 1 Cor pressures up to about 100 bar. This I ' . : :)f course, reflects the fact that liquids are not very compressible. We can exploit this I: & I : ! t)y substituting d V = -8 I'd P from the defining equation of 0 into w = - j Pd V and then ~teatingV as a constant. Shuw that this approximation gives Equation 4 of Pruhlem 5--13. 5-15. Show that
Proble~ri5-9 for a van der Waals gas.
5-1 1. Derivc an expression for the reversible iholhcrmal work of an expansion o f a gas that 0 4 s the Peng-Robinson equation of stale.
-
for a reversible adiabatic expansion of an idcal gas.
Chapter 5 1 The First Law uf Thermodynamics
to water. Calculate Ihe final temperature of the water. The molar heat capacity of coppr is 24.5 J-K..',rnol-l.
5-16. Show that
Cot n rcvcrsible, adiabatic expansion of a monatomic gas that obeys the equation of state P(V hj = RT. Extend this resul~lo the case of a diatomic gas.
-
5-26. A 10.0-kg sample of liquid water is used to cool an engine. Calculate h e hcat removed - from the engine when the temperature of the water is raised from 293 K to 373 K. (in joules) Take C , = 75.2 J.K.' ,mol-' For H,0(1).
5-27. In [hi< problem, we will derive a general relation between C, and C,. Start with V = U(P.T ) and write
5-17. Show that
h r o reversible adiabatic expansion o f an ideal gas.
5-1 8. Show that -
-
P 1 V 1[ ~ I R"C' '-
- P2 -
We could also consider V and T to k the independent variables of U and wrile
-.
-
1
f o r a n adiabatic expansion of an ideal gas. Show that this formula reduces to Equation 5.23 for a n~onatomicpas.
5-19. Calculate the work involvcd when one mule of a monatomic ideal gas at 298 K expands rcvcrsiblg and adiahalicalIy from a prcssure of 10.00 bilr to a pressure of 5.00 bar. 5-20. A quuntity I C N,(g) at 298 K is compressed reversibiy and adiabatically from a volume of 20.0 dm' t o 5.6)dm'. Assuming ideal behavior, calculate the final temperature of the N,(g). h k e C, = 5 R / 2 .
Now take V = V ( P , T) and substilule its expression for d I/ into Equntiun 2 10 obtain
Compare this result with Equation 1 to obtain
and
5-21. A quantity of CH,(g) at 298 K is compressed reversibly and adiabatically rrom 50.0 bar to ?I10 bar. Assuming ideal behavior, calculate the final temperature uf the CH,(g). Take C, = 3 R . 5-22. One nlole of ethane ill 2 5 C and one aim is heated to 1200°C at constant pressure. Assuming ideal behavior, cillculate the values of w , q, AU, and A H given that the molar he^ capacity of ethane is given by
Last, substitute V and Cv to obtain
=
H
- PV
into the left side of Equation 4 and use the definitions uf C,
Show that C, - C, = n R if ( a U / a V ) , = 0,as it is for an ideal gas. over the ahove temperature range. Repeat the calculation for a constant-volume process.
5-28. Following Prublern 5-27, show that
5-23. The value of A, H' at 25'C and onc bar is +290.8 kJ for the reaction
5-29. Starting with H = U
+ P V , show that
Assumirig idcnl behavior, calculale the value of A , U for this reaction.
5-24. Liquid
Interpret this result physically.
5-30. Given that ( a U / a V ) , = 0 for an ideal gas, pmve that (aH/a V)T = 0 for an ideal gas.
5-31. Given that ( a U / a V ) , = 0 for an ideal gas, prove that ( a C , / i l V ) , = 0 for an ideal gas
Chapter 5 1 The First law uf Thcrmudynamics
224 5-32. Show ihat C,
- C, = n R if (a H / a P), = U, as is true lor an i&aI
gas
+
5-33. Differentiate H = U P V with respect to V at constanr temperature to show that (d H / a V ) , = 0 for an ideal gas.
5-34. Given the following data for sodiunl, plot H ( T ) - H ( 0 ) against T Ir>rstdiurn: melling point, 361 K; boiling point, 1 156 K; A,Hn = 2.60 kJ-mul-I; A9apH'= 97.4 Icl-mol I; C,(s) =28.2 ~ ~ r n o l ~ ~ ~ = ~ 32.7 ~ ' ; J.rnolK1.K ~ , ( l ) ';??,(g) = 20.8I.mol l . K I.
225
Problems
5-42. Use the following data to cdculate the value of AvqH" of water at 298 K and compare your answer to the one you obtain from Table 5.2: AyapHYat 373 K = 40.7 !d.rnol-'; C,(I) = 7 5 . 2 3 r n o l - ' . ~ - ' ; F , . ( ~=) 3 3 . 6 J . m o l I - K - ' .
5-43. Use the flnllowing data and the data in Tahlc 5.2 In calculate the standard reaction enthalpy of the wakr-gas reaction at 1273 K. Assume that lhe gases k h a v e ideally under these conditions.
5-35. The A,HCvalucs for the following equaliol~sare
Use these data tu calculate rhe value of ArH For the reaction described by
C;IH,O(g)]/R = 3.652
-+ (I. 156 x
lo-' K
')T + (1.42 x 10-' K - ~ ~ T '
5-36. Given the following data,
1 H,(g) f 1 F2(g) -+ HF(g) H,0(1) H2(g) -k 1 0,(g)
ArH' = --273.3 kJ.rnul-I ArH" = -285.8 k~.rnul-'
-+
5-44. The standard nlolar cnthalpy of furmalion of C 0 2 ( g )at 298 K is -393.5tW kJ.rnol-l. Use the following data to calculate the value uf A, H' a1 IOW K.Assume the gaws behave ideally under thesc conditions.
calculate the value of A,H for the reaction described by
2 F,(g)
+ 2 H:O(l)
+4
HWg)
+ O,(g)
5-37. The standard molar heats of combustion of the isomers m-xylene and p-xylene are - 4553.9 kJ.mol-I and -455h.X kJ.mol-I, respectively. Use these data, together with He\s'sI,ilw. lo calculate the viilue or A, H" for the reaction described by rrr-xylene
-
p-xylme 5-45. The value of the slandard ~ n o l wreaction enthalpy fur
5-38. Given that A,H' = -2826.7 k l for the combustion of 1.W rnol of fructose at 298.15 K, C,HlzO,(s) t h O,(g)
.--* 6 COl(g)
+ h H1O(l)
and the A, H' data in Table 5.2, calcu\ate the value of A, H fur fructose at 298.1 5 K.
is -802.2 kJ.rnol- at 298 K. Using thc heat-capacity dava in Problems 5 4 3 and P I 4 in tldditiun to
5-39. Use ihe A I H L data in Table 5 2 to calculate the value of A r H Lfor thc coti~hustion reactions dcscribed by thc equatio~ls:
+ q O,(g) ---+ CO,(g) + 2 H,O(l) + O,(g) ---.NJg) + 2 H,0(1)
a. CH,OH(I)
h. N2H,(l)
tu derive a gencrnl equation for the value of A, H " at my temprnture between 300 K and 1500 K. Plut ArH" versus T. Assume thal the gaws behave ideally under these conditions.
Compare the hcilt 01-combusliun per gram 01- lhc fuels CH,OH(I) and NIH,(l). 5-40. Using Table 9.2, calculale Ihe heat required
*
to v a p r i ? e
1 .(I0 mol of CCI,(l) a1 298 K.
-
5-41. Using the Al H' data in Table 5.2, calculate the values of A,fI" for rhc following:
+
a. C,H,(g) H,O(l) CIH,OH(I) b. CH,(g) + 4 C l l ( g ) + CCl,(I) +4HCl(gl 111each casc, statc wtlcther thc reaction is endothermic or exothertnic.
5-46. in all the calculalionc thus far, we have assumed thc rc,~ctio~~ lakes place at constant temperature, so dint any energy evulved as heat is absorhed by thc surroundings. Supposc, however, that. Lhe rcaction takes place under adiabatic corldilions, so that all the energy released as heat slays within the system. Jn this casc, the tenqxrature of thc system will inoreaw, and the final ternFraure is called the adiubuficjlum ttrrlperaturc. One relatively simple way to estimate this temperature is to suppuse iec rcaction occurs at lht. initial temperature of the reaceants and then determine to what temprarure the p d u c t s can he r a i d by the quantity A r H 3 .Calcuiatc the adiabatic name temperature if one mole of
Chapter 5 I The First Law of Thermodynamics
CH, (g) is burned in two moles of O,(g) at an initial temperature of 298 K.Uae the results of the previous problem.
porous plug, the work done on the gas is P, V,. The pressure on the other side of the plug is maintained at P,, so if a volume V, eniers the right-side chamber, then the net work is given by
5-47. Explain why the adiabatic Aame temperature defined in the previous problem is also called thc maximum flame tenlpcrature. 5-48. How much energy as heat i s required to raise the temperature of 2.00 moles of O,(g) frum 298 K to 1273 K at 1-00bar? Take
The apparatus is constructed so that the entire process is adiabatic, so q = 0 . Use the First Law of Thermodynamics to show that
5-49. When one mole of an ideal gas i s cotnpressed adiabatically to one-half of its original volume, the temperature of the gas increaws from 273 K to 433 K. Assutning that ?, is i~ldcpendentof lemperaturc, calculate the value of for this gas.
or that A H = O for a Joule-Thornson expansinn Starting with
r,,
5-50. Use the van der Wads equation to calcula~ethe minimum work required to expand one nlole of CO,(g) isothermally fmm a volume of 0.100 dm3to a volume of 100 dm3 at 273 K. Cumpare your result with that which you calculatc assuming ideal behavior.
dH=
("a), -
dP+
(:p
show that
5-51. Show that the work involved in a reversible, adiabatic pressure change of one mule of an ideal gas is given by
whcre 1; is the initial temperature and P, and P, are the initial and final pressures, respectively. 5-52. Ln this prohlem, we wiil discuss a famous experiment called the Joule-Thornson experimmt. In the first half 01- the 19th century, Joule tried tu measure the temperature change when a gas is expanded into a vacuum. The experi~nentalsetup was nut sensitive enough, hrnvever, and he found that there was no temperature change, within the limits of his error. SDOII afterward. Joule and Thomson dcvised a much mure sensitive method for measuring the temperature change upon expansion. In their experiments (see Figurc 5.1 l), a constant applied pressure P, causes a quantity of gas to Row slowly from one chamber to another through a porous plug of silk or cotton. If a volume, V,,of gas is pushed through the
Interpret physically the derivative on the left side of this equation. This quantity is called the Joule-Thom,mn coefiri~nt and is denoted by p,,. In Problem 5-54 you will show that it equals zero for an ideal gas. Nonzero values of ( a T / a P ) , directly reflect intermolecular interactions. Most gases cool upon expansion {a pusitive value of (aT/aP),] and R JouleThomson expansion is used to liquefy gases.
5-53. The Joule-Thomson coefficient (Problem 5-52) depends upon the temperature and pressure, but assuming an average constant value of 0.15 bar-' for N,(g). calculate the drop in temperature if N,(g) undergoes a h p in pressure of 200 bar.
5-54. Show that the Joule-Thumsan cwMicient (Problem 5-52) can k written ns
Show that ( a T / a P), = 0 for a n ideal gas.
5-55. In h i s problem, we will investigate the pressure dependence of the heat capacity, C , Because C, = (aH/aT),, we see that
Initial state n moles at V I + T I
Y
-
p2
t ~ u r o u splug
-
b '
PI-
n moles
Now show that
Final state FIGURE
5.11
A hchernatlc description of the Joule-Thomson experiment.
where fi,, is the Joule-Thornson coefficient Show that C, is independent of pressure fur an ideal gas.
Chapter 5 1 The Fint Law of Thcrrnudynamics
228
5-56. Given that C,(?') for N?(g)can be represented by
for 298 K < T < 1500 K and that {A,, can be represented by
MATHCHAPTEK
THE BINOMIAL DISTRIBUTION AND STIRLING'S APPROXIMATION
for 150 K < T < 5 0 K around one atmosphei-c, estimate the preSSUCE dependence of C,,(7') for N,(g) at 3W K and one atmosphere pressure. 5-57. In Chapter 8, we will derive the lllrrnula
Show that U is i n d e ~ n d e n uivolume t for an ideal gas Show that
fur an isothermal expansion OF a van der Waals gas and hat
for a Redlich-Kwong gits.
5-58. IJse the rigid rotator-hmonic oscillator m d e l and the data in Tablc 4.2 to plot F,(T) fllr CO(g) fro~n300 K to 1OOO K. Compare your result with the expressiun given in Pmble~n5 4 3 . 5-59. Use the rigid rotator-hamuniu oscillator tnodel and the data in Table 4.4 to plol P,(T) for CH,(g) from 300 K to 1000 K. Compare your result with the expression given in Problem 5 4 5 . 5-60. Why dn you think the equatir~nsfor the dcpendcnce of temperature on volume fur a reversible adiabatic process (see Equation 5.22 and Example 5.6) depend upon whether the gas is a muna~omicgas or a plyatomic gas?
E
In the next chapter, we will learn about entropy, a thermodynamic state function that has a molecular interpretatinn of being a measure of the disorder uf a system. In doing so, we will have to put the idea of the disordcr af a system on a quantitative basis. A problem we will encounter is that of determining how many ways we can arnngc N distinguishable objtxts such that there are n , objects in the first group, t i : objects in the second group, and st, on, such that
that is, such that all the objects are accounted for. This problem is actually 'a fairly standard one in statistics. Let's salve the problem of dividing the N distinguishable objects into two groups frst and then generalize our results to any number of groups. First, we calculate the number of permutations of N distinguishable objects, that is, the number of pr~ssible different arrangements ur ways to order N distinguixhable objects. Let's chwxe ouc of the N objects and place it in the first positiun, one of the N - 1 remaining objects and place it in the second postion, and so on until all N objects art: ordered. Clearly, there are N choicer, for the first position, N - I choices for the second position, arid so un until finally there is only une object left for the Nth positiotl. The total numkr of ways of doing this ordering is the product of all the choices:
Next, we calculate the number uf ways of dividing N dis~inguishablcobjects intc~ two groups, one containing N1 objects and the other containing the N - N , = N? remaining objeqs. There are
MathChapterE / T H E B I N O M I A L D I S T R I B U T I O N
MathChapterE / THE B I N O M I A L D l S T E l B U T l O N
ways t o fonn the first group. This product can he written more conveniently as
and
as can seen by noting that
Equation E.3 may be written in a more symmetric form:
- N,)!. You might think that thc tutal number of arrangements is the product ol'the two factors, N ! / ( N - N,)! and N,!,but this product drastically overcuunts the situation kcause the order in which we arrange the N, objects in the first group and the N2 ohjects in the second group is immaterial to the problem stated. All N,! orders of the first gnjup and N 2 ! ortlcrs of the second group correspond to just one division of N distinguishable objects into two groups containing N , and N z objects. Therefore, we divide the product of N ! / ( N - N,)! and N,! by N , ! and N,! tu obtain Tbc number of ways of formingthe sccondgroup is N z != (N
+
E X A M P L E E-1
+
with N, N, f . . . IV, = N. This quantity is called a multinomial cuefficient because it occurs in the multinomial expansion:
where we let W ( N , , N?)dent~tethe result. (Problem E l 2 shows that O! = 1.)
I
-+
where the asterisk on tbc summation signs indicates that only terms with N, N1 = N are included. This symmetric fortn of the binomial expansion suggests thc form of the multinornial expansion given below in Equatiun E.6. Simple nu~llericalexamples verify that Equations E.3 and E.4 are equivalent. The generalization of Equation E.2 to the division of N distinguishable uhjects into r groups, the first containing N,,the second containing N z , and so on, is
I
Use Equation E.2 10 calculate the number of ways of arranging four distinguishable objects into two groups, containing three objects and one object. Verify your result with an explicit cnumcration.
+ + +
where the asterisk indicates that only terms such that N, NZ . . - N, = N arc included. Note how Equation E.6 is a straightforward generalization of Equation E.4.
t If ure let a , h. r , and 11 he the Four di~tinguishableobjetts, the four arrangements are c~br: d, aM : c , ncd : b, and bcd : a . There are no others.
E X A M P L E E-2 Calculate the nunlber of ways of dividing IO distinguishableobjects into three groups conlaining 2, 5, and 3 objects.
I
S O L U T I O N : WeuseEquation E.5:
The combinatorial factor in F4uation E.2 is called a binomial coefficient because the expansion of the binomial (x 3- ')" is givcn by
For example,
If we use Equation E.5 to calculate something like the numbcr of ways of distributing Avogadro's number of particles over their energy states, then we are furced tn dcal with factorials of huge numbers. Even the evaluation of 1 OO! would be a chore, never mind lo2'!, unless we have a good approximation for N !. We shall see that there i s un approximation for N! that actually improves as N gets larger. Such an approximalion is called an asymptotic approximation, that is, an approximation to a function that gets better as the argument of the function increases.
232
Ma~hChaptcrE/ T H E B I N O M I A L D l S T K l H U T l O N
Mathchapter1 1 T H E B I N O M I A L I I I S T R I B U T I O N
233
Because N ! is a product, it is convenient to deal with In N ! because the latter is a sum. The asymptotic expansion to In N! is called Stirling's approxiniation and is given by
The pmof of Stirling's appnjximation is not difficult. Because N! is given by N! = N ( N - I)(# -2)...(2)(1),InN!isgivenby
which is surely a lot elisier to use than calculating N! and then taking its logarithm. Table E.l shows the value of In N ! versus Stirling's approximation for a number of' values uf N. Note that the agreement, which we express in terms of relative ernlr, improves markedly with increasing N.
Figure E.l shows Inx plotted versus x for integer values nf x . According to Equation E.8, the sum of the areas under the rectangles up to N in Figure E.l is In N!. Figure E. I also shows the continuous curve In x plotted on the same graph. Thus, In x is seen to form an envelope to the rectangles, and this envelope bcct~rnesa steadily smoother approximation to the rectangles as x increases. Therefure, we can approximate the area under thesc rectangles by the integral of Inx. The area u~tderInx will pourly approximate the rectangles only in the lxginning. I F N is large enough (we are deriving an asymptotic expansion), this area will make a negligible contribution tu ihe total area. We may write, then,
E X A M P L E E-3 A more refined version of Stirling's approxiii~ation(one we will not have to use in the
next chapter) says that
Use this version of Stirling's approximation tu calculate In N! for N = 10 and conipare the relative error with that in Fdble E.1. SO1 LITION: For N = 10,
1
which is Stirling's approximation to In N !. The lower limit could just as well have been taken as 0 in Equation E.9, because N is large. (Remember that x In x -t 0 as x + 0.) We will use Stirling's approximation frequently in the next few vhaplers.
and using the value of In lo! from Tablc E. I , we see that rclative e m r =
15.104 - 15.096
15.104
= 0.0005
The relative error is signihcantly smaller than that in 'hblc E. I . The relative errors for rile other entries in Table E. I arc cssentially zero for this cxtendd version of Stirling's
approximatio~~.
T A B L E E.l
A numerical comparison of In N ! wilh S~irling'sapproximation.
N
In N !
N ln N - N
Relative error"
5
10
15
20
25
F I G U R E E.l
A plot of In x versus x . The sum of the ;wens under the rectangles up tu N is In N!.
"rebalive ermr = (In N ! - N In N
+ N)/In N !
7
Integrate by parts {letting u = r'-I
Problems E-I. Usc Equation E.3 to write the expansion of (1 + 1)'. Use Equation E.4 to do the same thing.
+
E-2. Use Fquation E.h to write out the expression for (x f y z)'. Compare your result to the one thnt you clbvain by multiplying (x y z) by ( x t y 2 ) .
+
+ +
+
f-3. LJse Fqrluatinn E.6 to write out the expression For ( x y f 2j4. Compare your result to the one that you ohlain by multiplying ( x y 1)' frum Problem 5 2 by itself.
+ +
E-4.How many permutations of the letters a , b, c are there? E-5. The coefficients of the expansion uC (1
+ x)* can be uanged in the follr>wingfum:
Do you see a pattern in going trom one row to the next? The triangular arrangement here is called Pzxal's triangle. E-6. I n how many ways can a committee of three be chosen from nine people?
E-7. Calculate the relative error for N = 50 using the formula for Stirling's approximation given in Example E-3, and compare your result with that given in Table E.1 using Equation E.7. Take In N! tn be 148.47776 (CRC Handbook of Chemistry a d Physics). E-8. Prove that x lnx + 0 as .T + 0. E-9. Prove that the maximum value of W ( N . N , )= N ! / ( N ( H ~ n rTreat : N, as a continuous variable.)
E-10. Pruve that the maximum valuc of W I N , , N,, R, =R, =N/r. 2 -
-
N,)!N, ! is given hy N, = N / 2 .
. . . ,Nr)in Eqmtiun E.5 i s given by N, =
.
E-1 1 Prnve that
E-12. The quanlily n 1 as we have defined i t is dcfined only for positive integer values of n. Consider now the function of x defined hy
and dv = s-'dt) tu get
Now use Equation 2 to show that I'(x) = (x - I)! if x is a positive integer. Although Equation 2 provides us with a general function that is equal to ( 1 1 - I ) ! whcn x takes on integer values, il is defined just as well for non-integer values. For example, show that r(3/2),which in a sense is is equal to 7r1'*/2. Equation I can also be used tu explain why O ! = 1. Let x = 1 in Equation I tu show that r ( l ) , which we can write a< O ! , is cqual to 1. Thc function T(x) defined by Equation 1 is called the gamrnabnction and was introduced by Euler to generalize the idea of a factorial to general values of n . The gamma function arises in many problems in chemistry and physics.
(i)!,
Entropy and the Second Law of Thermodynamics
In this chapter, we will introduce and develop the concept of entropy. We will see thal energy considerations alone are not sufficient to predict in which direction a pruceas or a chemical reaction can occur sponvaneously. We will demonstrate that isolated systc~ns that are not in equilibrium will evt~lvein a direction that increases their disorder, and then we will introduce a thermodynamic state function called entropy thal givcs a quantitative meaure of the disorder of a system. Onc statement uf the Sccand Law of Thermodynamics, which governs the direclion in which systems evolve to their equilibrium states, is that the entropy of an isolated systern always increases as a result uf any spuntaneous (irreversible) prmess. In the second half of (his chapter, we will give a quantitative m o l e c u l ~definition of entropy in terms of partition function. Rudolf Clausius was hurn in Koslin. Prussia (now Koszalin, Poland), on January 2, 1822, and died in Ig88. Although Clausius was initially attracted to history, he eventually received his Ph.D, in mathematical physics from the llniversity of Halle in 1847. Hc held a position for several years at the Un~versityof Zurich but returned to Germany and in 187 1 wtlcd at the University of Bonn,wtlerc he remained for the wst of his life. Clausius is credited with creating the carly fuundations of thermodynamics. In 1850, he published his first great paper on the theory of heat, in which he rejected the then-cumi~tcaloric theory and argued that the energy of a system is a thermodynamic state fundion. In 1865, he published his second lmdnlark papzr, in which he inlnlduced another new thermtdynunic stale function. which he called etltropy, and expressed the Second Law uf Thcr~nodynamicsin terns of thc entropy. Clausius alxo >tudicrl the kinetic thet>l-yr,f gases :~ndmudc i~npjrtantcr>~itributions to it. He wits ct\auvinistic :uld htrongly dcfended Gcr~nanachievcr~lcotsagainst ~ h n he t cunsidercd the iniri~~gclncnts of others. Muft of Clausius' work was done before I870 because of two events in his life. In 1870, he was wuunded while serving in an ambulance corps in the Eranco-PrussinnWar and suffered life-long pain from his injury. Morc tragically, his wife died in childbirth, and he asaunled the respons~bility01- raising six young children.
-
6-1. The Change of Energy Alone Is Not Sufficient to Determine the Direction of a Spontaneous Process For years, scientists wondered why some rcaclions or pmcesseh proceed spontar~eously and others do not. We all know that under the right conditions iron rusts, and that objects d o not spontaneously unrust. We all know that hydrugen and oxygen react explosively to form water but that an input of energy by means of cleutrotysis i s requircd to decompose water into hydrogen and oxygen. At one tilnc scientists believed tho1 :I criteriun for a reaction o r B process tu procccd s p o n ~ a n e o h l ywas Ihitl it shoultl bc exothermic, or evolve energy. This belicf was tnotiv;lted by the [act that the prorlucls of an exothermic reaction lie at a lower energy or enthall~qthan the reactants. After all, balls do roll downhill and opposile chagea do attract each other. Mechanical sy stcms evolve in such a way as to minimize their energy. Now consider the situation in Figure 6.1, however, where uiie bulb contains a gas at some low pressure at which it may be considered to behave idcally, and the othcr hulb is evacuated. When the two bulbs are connected by opening thc strjpcock between them, the gas will expand into thc evacuated hulb until the pressures in
6 2 . Nonequ~libr~urn lsulated Systems Evolvc in a Direction That Increases Their Disorder
-
each of these processes obeys the First Law of Thermodynamics, but using this law, we cannot tell why one direction occurs spontaneously and its reverse does not. Although mechanical systems tend to achieve their state of lowest energy, clearly some other factor is involved that we have not yet discussed.
F I G U R E 6.1
mo bulbs connected by a stopcock. Initially, one bulb contains a colored gas such as bromine and the other one is evacuatd. When the two bulbs are connected by opening the s t o p k , the bromine occupies both bulbs at a uniform pressure as seen by the uniform culor.
the two bulbs are equal, at which time the system will k in equilibrium. Yet a careful determination of the thermal processes of this experiment shows that both AU and AH are essentially zero. Furthermore, the unaided reverse process has never been observed. Gases do not spontaneously occupy only part of a container, leaving the other part a$a vacuum.
Anothcr cxa~npleof a spontaneuus prwess that is not e x o t h e m ~ cis depicted in Figure 6.2, where two pure gases are separated by a stopcock. When the stopcock is opened, the two gases will mix, and both will eventually become evenly distributed between the two bulbs. in which case the system will be in equilibrium. Yet once again, the value of AU or A H for this process is essentiafly zero. Furthermore, the reverse process has never been obsewed. Mixtures of gases do not spontaneously unmix. There are many vpuntrtneous endothermic processes. A simple example of a spontaneous endolhermic reaction is the melting of ice at a temperature a h v e O'C. This spontaneous prucess has a value of AhsHgequal to +6.0 kl.mol-' when the tempcrature is around WC. An espxially interesting endothermic chemical reaction is the reation o f solid barium hydroxide, Ba(OH),(s), with solid ammonium n i h t e , NH,NO,(s):
The energy ahsorbed by mixing stoichiometric amounts of these two reagents in a test tubc can cool the system tu below -20°C. These and numerous other examples indicate that spontaneous processes have a direction that cannot he explained by the First Law of Thermdynarnics. Of course,
F I G U R E 6.2 Two bulbs connected by a stopcock. Initially, each bulb is occupiad by a pure gas, say bromine and nitrogen. When the two bulbs are connected by opening the stopcock, the two gases mix u~liforti~ly, so each bulb contains the same uniform mixture.
6-2. Nonequilibrium Isolated Systems Evolve in a Direction That Increases Their Disorder If we examine the a h v e pmesses from a microscopic or molecular point of view, we see that each one involveh an increase in diwrder or randomness of the qystcm. For example, in Figure 6.1, the gas molecules in the final state are able to move over a volume that is twice as large as in the inilia1 state. In a sense, lwating any gas molecule in the final state is twice as difficult as it1 the initial state. Recall that we found that the number of acccssible translational states increases with the volume of the container, Problem 4-42,A similar argumenl applies tu the mixing ut two gases. Nut only is each gas spread over a larger volume, but they are alw mixed together. Clearly the final (mixed) state is more disordered than the initial (separated) state. The melting of ice at a temperature greater than O"C also involves an increase in disorder. Our molecular picture uf a so11J being an ordered lattice army of its constituent particles and a liquid being a more random arrangement dirzctly implies that the melting of ice involbes an increase in disorder. These examples SUggeFt that not only do systems evolve spontaneously in a direction that lowers their energy but that they also swk to increase their disorder. There i s a competition between the tendency to minimize energy and tc~maximize disorder. If disorder is not a factur, as i s the case fur a simple mechanicai system, thcn energy is the key faclor and the direction of any sponlaneous prwess is that whlch minimizes the energy. If energy is not a factor, however, as 1s the case when mlxing two gases, then disorder is the key factor and the direction of any spontaneous process is that which maximizes the disorder. In genera!. some compromise between decreasing energy and increasing diburder must be met. What we need is to devise some particular property that puts this idea of divnrdcr on a useful, quantitative basis. Like energy, we want this property to be a state functiun k a u s e then it will he a property of thc state of the system, and not of its previous history. Thus, we will rule out heat, although the transfer or energy as heat to a s j stem certainly does increase its dihr~rder.To try to get an idea of what an appropriate function might k,let's consider, for simplicity, the heat transfer a~sociatedwith a reversible, small change in the temperature and volume of an ideal gas. From the First Law (Equation 5.9). we have
Chapter b / Enbopy and h e Second l a w of Thermodynamics
242
6-3. Unlike qXv,Entropy Is a State Function
1
and
whcre Tz is given by (cf. Equation 5.21)
243
E X A M P L E 6-1 Calculate qmvand A S for a reversible expansion of an ideal gas at consrant pressure PI from T,,Vlto T,,V, (path D in Figure 6.3)followed by a reversible coaling of the gas at constant volume V, from P , , T,to P,. TI(path E). SOLUTION: ForpathD(cf.Exarnple5-4),
The poini here is that qm differs for the two paths, A and B t C,indicating that qw+i s riot a state fu~~ction. Now let's evatuate
and so
For path E,S?umv = 0, and so 6qmv,,= dU, = C,( T ) d T
for thcse two paths. For path A from P,,TI, V , to P,, V,,TI, we have, using Eqnation 6.6,
and
For the complete p e s s (paths D + El,
- qmv,,j+ q w v . ~= Pi(V2 - I:)
q r e r . n r ~-
For the reversible adiabatic expansion from PI,V,, T, to P,,V,, T, (path B) followed by a reversible heating at constant volume from P,, V,, T, to P,, V,, TI Cpath C), we have, using Equation 6.7.
and
To calculate AS for path D, we use Fquation 6.1 1 to write
To evaluate the second integral here, wc must knnw how T varies with V Tui thif procew. But this is give11by PI1' = nHT,so
But using Equation 6.8, AS, turns out to be
.
For path E, Aw,, = 0, and using Equation 6.12 for 6qmv gives
and so
The value of A S for the cornplete process (paths D + E) is
Thus, we see, that the AS, (Equation 6.9)is equal to AS,,, the value of AS is independent of the pnth.
(Equation 6.10) and that
A SI , , E = AS,
+ AS,
v
=n ~ l n 2 VI
Chapter 6 / Entropy and the Second Law of Thermodynamics
Notice that this is the very same result we obtained for paths A and B + C, once agaln suggesting lhaf S is a slate function.
I
6-2 We shall prove in Example 8 4 that siinilar to that found fnr an ideal gas, U is a function of only the temperature for a gas that obeys the equation ut' state EXAMPLE
P
But T,,thc temperature at the end of the reversible adiabatic expansion, can he found from
I
Using the fact that d a = ? , ( T ) ~ Tand 6qm, = 0 gives -
-
RT
-
C , ( T ) d T = - P d V = - V7 - db V
Divide through by T and inkgrate from thc ~nitial\tatc to the final state lo get
RT = w v-b
where b is a constant that reflects the size of the mulecules. Calculate y , both the paths A and B + C in Figure 6.3 for one mole of such a gas.
6 4 , The Second Law of Thermodynamics
and A S fur
,
Substituting this result inlo the above cxpreshion for AS,, gives
3OLt~TION: Path A reprewnls an sot hernial expansion, so dlJ, = O because U depends only upon the ternperaillure. Therefore, SyE,, = -Aturn,,
RT = PdV = r - - d V V-b
Therefore, we see that even though qmy,, $ qrCY B t (.,ncverthelefs, -
and
-
AS, = AS,?,
The enwopy change i s given by
For path B, a reversible adiabatic expansion, qw,,fl = 0,so AS, = 0
We will show several times in the following sections that the entropy is related to the disorder of a system, but for now, notice that if we add energy as heat to a system, then its entropy increases because its thermal disorder increases. Furthermore, nt~tice that because dS = Sqzv/T,energy delivered as heat at a lower tetnperature contributes mare to an entropy (disorder) increase than at a hjgher temperature. The lower the temperature, the lower the disorder, so the enegy added as heat has proponinnally mure "order" to convert to "disorder."
For path G, 6111=~,, = U, and SqFov,r = dUr = C , ( T ) d T
and
The molar entropy change is given by
and so
-
AS,,,
=
AS,
+ AS,
=-
& The I . Second Law of Thermodynamics States That the Entropy of an Isolated System Increases as a Result of a Spontaneous Process We all know that energy as heat will flow spontaneousl~ fi-rjm a region of high tenlperature to a region of low temperature. Lel's investigate the role entropy plays in this pmcess. Consider the two-compartment system shown in Figure 6.4, where parts A and El are large one-componen! systems. Both systems are a1 equilibriunl, but they arc not at equilibrium with each other. Let the temperatures of these t w t ~systems he 7, and 1;. The two systems are separated from each other by a rigid, hear-conducting wall so that energy as h>at can R(IW from one system to thc other, but the two-comparttncnt system itself is isolated. When we call a system i.volured, we mean tha~the sysiern is separated from its surro~~ndings by rigid walls that d o not allow malter or energy to pass through them. We may picture the walls as rigid, totally nun-heat cc~nducting,
H. The Second Law of Thermodynamics F I G U R E 6.4 A two-compartment system in which A and B are large, one-cumponent systems.
Each system is at equilibrium, but they are not a1 equilibrium with each other. The two systems are separated from each other by a rigid, heat-conducting walI. The total
two-compartment system itself is isolated.
and impervious to tnntter. Conscqucntly, the system c w d o no work nor can work be done on thc system, nor can it exhange energy as heat with the surroundings. The two-compartlnent system is describcd by the equations
U,
+ U , = constant
we wish to separate the effect due to a change in energy from the effect due to a change in entropy. Because the energy remains constant, the driving force for my spontaneous process in an isolated system must be due to an increase in entropy, which we can expmss mathematically by d S > 0. Because the system is isolated, this increase in entropy must he created within the system itself. Unlike energy, elitropy is not necessarily conserved; it increases whenever a spontaneous process takcs place. in fact, the cntrt~pyo f an isolated system will ct~ntinueto increase until nu mare spontaneous processes occur, in which case the system will be in equilibrium (Figure 6.5). Thus, wc conclude that the entropy of an isolated system is a maximum when the system is in equilibrium. Consequeiitly, d S = 0 at equiribrium. Furthcrmore, not only is dS = O in an isolated systcm at equilibrium, but dS = 0 for any reversible process in nn isula~rd systcm because, by definition, a reversible process is one in which the system remains essen~iallyin equilibrium during the entire process. To summarire our conclusions thus far, then, we write
Vg = constant
VA = constant
247
dS > 0
(spontaneous prwess in an isolated system)
riS = 0
(reversible process in an iqolated system)
(6.17)
s = s* + s, Because V, and V , are fixed. we have for each separate syskni
Because we have considered an isolated system, no energy as heat can fluw in or out of the system. For other types of syslems, however, cnergy ax heat can flow in or out, and it is convenient to view dS in any spontanerjus infinitesimal procers as consisting of two parts. One part of dS is the entropy created by the irreversible process itself, and the other part is the entropy due to the cnergy as heat exchanged between the system and its surroundings. These two contributions account for the entire change in entropy. We will denote the part of dS that i s created by the irreversible process by dS,, because it is produced by the systcm. This quantity is always positive. We will denote the part of d S that is due to the exchange of energy as heat with the sum~undings by dSrxL,because it is due to exchange. This quantity is given by Sq/ T, and it can bc positive, negative, or zero. Note that 69 need not be BqrtV. The quantity Sq will he Sqmv
dU, =JSqTN+J~~irtY = ThdSn (dV,, = 0 )
dUB=&qw-t6wrev=T,dSs (dV,=O) The entropy change of the two-compartment systein is given by
But dU, = -dU, hccause the two-compartment system is ist~lated.so we have
(i &)
d~ =duH
-
(ti. 16) Srner
...
Experimentally. we know that if T, > T,, then dU, < O (energy as heat flows I'mm sysiern B to system A), in which case dS r 0. Similarly, d S > 0 if T, < ?A because d U , > 0 in this case (energy as heat flows from system B ti] system A). We may interprel this result by saying ihat the spontaneous flow of energy as heat from a budy at a higher temperature Za a body at a lower temperature is governed by the condition dS > 0. If T, 5 T,. then the two-ct)mprtrtment system is in equilibrium and
(JS = 0. We can generalize this result by investigating the rule entropy plays in governing the directic~nof any spontaneous process. To be able to focus on the entropy alone, we will cansider an infinitesimal spontaneous change in an isolated system. We chwse an ist11att.d systern because the energy remains constant in an isolated system, and
......................
Spontaneous
.
---
Equilibriunl dS=O
S
F I G U R E 6.5
A schematic plot of entropy versus time for an isolated system. The entropy increases ( d S > 0) until no more spontaneous processes occur, in which cafe the system is in equilihriuln. and d S = 0.
2.18
Chaptcr h / Fntrupy and the Second Law uf Th~rmdynarnics
if the exchange is reversible and 6qInif the exchange i s imversiblc. Thus, we write for any process
Reversible
For a reversible process, Sq = Sqnv,dSpRd= 0, SO
F I G U R E 6.6
A cyclic process in which the system is first is;cllatcd and undergoes an imevcnible process from state 1 to state 2. Then the system is allowed to interact with iis surrt~undingsand i\ brought back to state I by some reversible path. Bcaufe entropy is a state functiun, A S = 0 for a cyclic process.
in agreement with Equation 6.3. For an irreversible or spontaneous process, dSp,, > 0, dSeXEl, = Sqirr/ T. and so
Equations h. 19 and
6.20 can be written
;ts
une equation,
where the equality sign holds for a reversible process and the inequality sign holds for an irreversible p m e s s . Equation 6.22 is one of a number of ways of expressing the Second Law of Thermodynamics and i s called the hequaliw of Clausius. A formal statement of the Secr~ndLaw o f Thermodynamics is as follows: 'Ihcru is a thermodynamic state function of a systcm called the entmpy, S,such [hat lor any change in the thermodynamic siaie of Lhe system,
where the equaliw sign applies if the change is carried out reversibly and thc inequality sign appltcs if the change is carried out irreversibly at any s l a p .
tPr. van US^ Equ~llio~i 6.22 tr) prove quile generally thut the entmpy 01' iun isolated sybrem always increases during a spontaneous (irreversible) process or that A S > 0. Consider a cyclic process (Figure 6.6) in which a system is tirst isulated and undergws an irreversible process from state 1 to slate 2. Now lei the system interact with its surroundings and return it to-statc I by any reversible path. Because S is a state function, A S = 0 for this cyclic prwexs, so accnrding to Equation 6.22,
The inequality applies because the cyclic process is irreversible from I to 2. The first integral here equals zero bxause the system is isolated, i.e., SqIm= 0. The second integral is by definition equal to S, - Sz,LO we have 0 > S, - S?. Because the final state is state 2 and the initial state is state 1 ,
Thus we see that the entropy increases when the isolated system goes from state 1 to state 2 by a general irreversible process. Because the universe itself may be considered tu he an isolated system and al\ naturally occurring processes are irreversible, one statement of the Second Law of Thermodynamics says that the entmpy of thc universe is constantly incre;lsinp. In fact. Clausius summarized the first two laws of thermodynamics by The encrgy of the Universe is constant;
the entrnpy is tending to a maximum.
6-5. The Most Famous Equation of Statistical Thermodynamics Is S = k , In W In this section, we will discuss the molecular inlerpretatiun of entropy more quantitatively than we have up to now. We have shown that entropy is a state function that is related to the disorder of a system. Disorder can be expressed in inurnher oI'ways, hul the way that has turned out to k the most useful is the following. Cmsider an el~semble of A isolated systems, each with energy E , volume V , and number of panicles N. Realize that whatever the valueof E, it must bc an eigenvalue of the Schrodinger equation for the system. As we discussed in Chapter 3. the energy is a [unction ol'N and V , so we can writc E 7 E ( N , V ) (see, fur example, Equations 3.2 and 3.3). Although a l l the systems have the same energy, they may k in diffcrenl rluantum states because of degeneracy. Let the degenewcy associated with the energy 6' be !i2(E), so that we can label the L?(E) degenerate quantum states by j = 1. 2, . . . , Q ( E ) . (The degeneracies
Chapter 6 1 Entropy and the kcond Law of Thermndynnn~ics of systcms that consist of N particles turn out to be enormous; they are numbers of the be the order of P" for cncrgies ]lot too close to the ground-state energy.) Now, let ~f. nu~nbcrof systertls in the ensemble that are in the state j. Because the A systems of (he ensernhte are distinguishable, the number of ways of having a,systems in state 1, 0: systelt~s in state 2, ctc. is given by (Mathchaplet E)
with
If-all A systcms are in one particular statc (a totally 'ordered arrangemetit), say statc 1, then a , = A, a, = o, = . . . = 0 and W = I , which is the smallest value W can have. 111 the other extreme, whcn all the uj are equal (a disordered ardngement), W takes on its largest value (Prt)ble~nE-10). Therefore, W can be taken to be a quantitative Ilwnsurl: ul'the disorder uf a system. We will not set the entropy proportional to W , however, but to in W according to
wIiere k , is the Bolt7mann constant. Note that S = 0 for a completely ordered system (0,= 1, ri, = a, = - . - Oj and achieves a maximum value for a completely disordered cystem ta, = la, = u , = . . .). Equativn 6.24 was formulated by Bnltzinann and is the most f;imr>us equation of stiltistical therniodyr~amics.In fact. this equation is the only inscription on a monument to Boltzmann in the central cemelary in Wenna. It gives us a qualititatirc relation belwecn thc thermodynamic quantity, entropy, and the statistical quantity, I+'. We set S equal to In W rather than W for the following reason. We want S to be such that the total entrupy o f a system that is made up of two parts (say A and B) is
given by
In other wurds, we want S to be an extensive state function. Now if W, is the value of W for system A and W, is the value of W for system B, W,, for the composite system is given by
' h e entrupy of the compbsite system is
b 5 . The Mn~lramnus Equation of Statistical Thcrmdynamicr ts S = k, In W
An alternate form of Equation 6.24 expresses S in terms of the degeneracy St. We can determine this exprcrsion in the following way. Given no other infonnation, there is no reahon to choose one of the 52 degeneratc quantum states over any other: each one should occur in an ensemble with equal prr~bability(this concept is actually one of the postulates of statistical thermodynamics). Consequently, we cxpcct that the enxemhle of iqolatcd sfktcms should cuntain equal nurribers of Fystems it1 each quantum state. Becau~eS is a maximum for an isolated sy'iterrl at eqriilibriu~n,W must illso hc n maximum. The value of W is niax~lniredwhcn all the a, art. equltl (Problerti E-10). Let the total number of systems in the ensemble be A = n Q and let each q = ~ t so , that the set of Q degenerate quantum states is replicated n timcs in the ensemble. (We will never need the value of n ) Uwng Stirling's appruximation (Mathchapter E) In Equation 6.23, we get n
Senrrmhlc = k HIn W = k,[AIn A
rr, 111a,]
1-1
L1
= k,,[nQ In(n0) - x ( n Inn)] = k,[1151 In(n0) - Q ( n I n n ) ]
Thc cntropy of a typical system in the enselnhle is given by SejlsenlMc = AS nQS5yrm,and st"
TYI,I.III
=
where we have dropped the suhxcript, system. Equation 6.25 is an alternate fonn nl Equation 6.24 and relates entropy to disorder. As a concrete example, consider a system of N (distinguishable) spins (or dipoles) that can bt: oriented i r ~one of two pn~sihle directions with equal probability. Then, each spin has a degeneracy of 2 associatcd with it, and the degeneracy of the N spins is 2". The entropy of this system is Nk, In 2. We will use this result when w e discuss the entropy of carbon mont>xide at 0 K in Section 7-8. As another example of the use of h u a t i o n 6.25, Problem 6 2 3 has you show that
for an ideal gas of N particles, where c ( N ) is a function of N and f ( E ) is a function of the energy. Now let's determine A S for an isothermal expansion of one mole of an ideal gas from a volume V, to V,.
Chapter 6 I Entropy and the Second Law of Thermodynamics
h4.
We Must Always Devise a
Reversible Process ro Calculate Entropy Changes
E X A M P L E &4 I n Exutlple 6-2 we stated that U is n function or only the tcmpcrature for a gas that obeys thc equation o f state
Nule t h a ~A S > O kcause V2 > V,. Thus, we see that the enlropy increases in the cxpnnsion of an ideal gas into a vacuum. Because Equation 6.19 tells us to calculatc AS by expanding the gas reversibly and isr,themally from V , to V,, Equation 6.28 holds for the reversible isothermal expansicln. Because S is a state function, hnwever, the value o f A S obtained hum Equnhon 6.28 is the satne a< thc value o f A S for ihe irreversible isothermal expntlsion from V, to V,. How, then, do a ~vzr.cihleand an irreversible isothermal expansion differ? The answer lic:, in the value of A S for the surroundings. (Remember that the condition AS > 0 applies 10 an isolated system. If the system is not isolated, thcn the condition AS 2 0 appliec to 1112 sun1 of the entropy changes in the system and its sur~nundings,in other words. the entire universe.) Let's look at the entropy change of the surroundings, ASww,for h ~ a hreversible and an irreversible isothermal expansion. During the reversible expansion, AU = 0 (the process is isothermal and thc gas is ideal) and the gas absorbs a quantity uf energy ax heat, qrev= -ulrw = n R T In V,/ V , , from its surroundings. The entropy of the surroundings, thcrcforc. decreases according to
whcrc b is a constant that reflects the size of the molecules. Calculale AT when one molc of such a gas at T and V , is allowed to cxpsnd into a vacuum tr, a Inla3 vulurnc of K.
SOLUTION: Westilrtwilh
Becausc U is a function of only the teniperature, and hence is independent of thc volume. dl' = 0 for the expansion. Thcrcfore.
and
The total cntropy change is given by
as it should be hecausc thc cntirc process is carried out reversibly. In thc irreversible expansicln, AU = 0 (the pnlcess is isothermal and the gas is ideal). The value of Prx,is also zero, sn loLrn= O and therefore, gin = 0.No energy as hent is delivered to the system by the surroundings and so
Once again, thc cntropy increases when a gas expands into a vacuum.
Let's look at the mixing of two ideal gases, as depicted in Figure 6.2. Because the two gases are ideal, each acts independently of the other. Thus, we can consider each gas separately to expand from VInltialto Vhnsl.Rlr nitrogen, we have (using Equation 6.28)
'Thus, the total cntropy change is given hy As N*
=n
N~
Rln
VN,
+ llg,
and s o A S =- O as we enpecl for an irreversible process. Did we use qhn= n to calculate 4S,umin this process?We actually did because no work was done by the process. In the general case of an isoihermal process in which 110 work is dune (Stu = Of. the process is one of pure heat transfer attd dU = 69 = d q , where dq is all exact diffcrcntial because U is a state function. Therefore, q is path indcpcndcnt and so w e can use qImto calculate the entrupy is this particular case.
and for brwmine,
=
R In
-11 ' 1
v ~ 2
+
"N:
VN2 ifBl2
255
256
Chapter 6 1 Entropy and the Second Law uf Thermodynam~cs
'I'hc total entropy change is
M .We Must Always Deviw a Reversible Proces to Calculate Entropy Change
There is essentially no work done, su hq,," = dU = C , d T . Therefure,
= -n
Rln ";
hl
v~r:
Rln
-n
v%z+VB,t
nr2
VN2fVBr2
Because V is proportional to n for an ideal gas, we can write the abave equation as
if we take C, to k constant from T, t c ~T,, then 1
(6.3 1)
AS = C , In 2 =1
If we divide both sidts by the total number of moles, nLuld, = 11,;
+ n,,! and introduce
mole fractions and
ynr2=
Now, for the initially hotter piece, TI = T, and T? = (T, + Tc)/ 2 , and so Th + Tc AS,, = C, In -
-
2Th
n,aal
Similarly,
then Equation 6.29 becomes
More generally, A,,,,xSfur the isothermal mixing of N ideal gases is given by
in agreement with Equation 6.26. Equation 6.30 says that amIx3 > 0 because the arguments of the logarithms are less than unity. Thus, Equation 6.30 shows that there is an incrcase in entropy whencvcr ideal gases mix isuthennally. Last. let's consider A S when two equal sized pieces of the same metal at different T, and Tc,are brought into thermal cuntacr and then isolated from their teml~erati~res, s ~ ~ r ~ - t ) u ~ ~Clearly, d i n p s . the two pieces of ~nztalwill come trl the same ti11;11telnprature, T , which can be calculated by
The total change in entropy is given by
We will now pnlve that ( T , + T ~ ) '> 4T,lC,and that A S z 0. Start with
(T,- T ~ ) '= T; - 2ThT + T,'
>0
Add 4ThTc to both sides and obtain
heat lost by hotter piece = heat gained by colder piece
Therefore, the value of the argument of the logarithm in Equation 6.32 is greater than one, so we see that A S > 0 in this irreversible process.
Solving for T gives T = - Th
+ <.
2 We now will cillculate the entropy change for each piece of metal. Remember that we must calculate A S along areversible path, even though the actual prucess is irreversible. As usual, we use Equation 6.19. w
I
E X A M P L E 6-5. The constant-pressure molar heat capacity of O,(g) from 300 K to 1200 K is given by
where T is in kelvins. Calculate the value of constant pressure from 300 K to 1200 K.
AT when one nwle of O,(g) is healed at
Chapter b I Entropy and the Second Law of Thermcdynamics
SOLUTION: As usual, we start with Eguation 6.19
Hut reservoir
Cold r e s e r v o i r
Heat engine
In this cnsc, 6qreV= C , ( T ) ~ so T, F I G U R E 6.7
Using thc givcn expression for ? , ( T I , we have
A highly sche~uaticillustration of a heat engine. Energy as h a t ( q h )is withdrawn horn a high-temperature thermal reservoir at tctnperature T,. The engine does work (a)and delivers an amount of energy as beat (qJ tu the lower-~ernpera~ure reservoir at temperature TL.
and
where bq,,,, is the energy w~thdrawnreversibly as hcat from the high-temperature reservoir at tertlperature T,,and SqrcV,cis the energy discharged reversibly as heat lo the lower-temperature reservoir at temperature T<.Nute that the sign conven~ionfor energy transferred as heat means that 6 q m , i x a positwe quan~ityand that Sq,, ,is a negative quantity. From Equatiun 6.33, we have that the work done by the engine is Note the il~creaseill entropy due tu the increased thermal disorder.
The work done by the engine is a negativr: quantity, so -111 is a positive quantity. We can define the efficiency of the prtxehs by the ratio of the work done by the engine divided by the amount of energy withdrawn as heat from the hot reservoir, or
6-7. Thermodynamics Gives Us Insight into the Conversion of Heat into Work The concepl u i entropy and the Smoad Law of Thermodynamics was first develnpcd by a Prcnch engineer named Sadi Cnrnot in the 1820s in a qtudy of the efficiency of the l~ewlydevelopcd stcam engines altd other types of heat engines. Although primarily oC hlstonual interest to rhernists. the result of Carnot's analysis ih still wclrth knowing. Bnsically, a steam engine works in o cyclic manner; in each cycle, it withdraws energy as heat from some high-temperature thermal reservoir, uses some of this energy to do work, and then discharges the rest of thc energy as heat to a lower-temperature thermal reservoir. A schematic representation of a heat engine i x xhown in Figure 6.7. The ~naximumamoullt of work will be obtained if the cyclic process is carried out rcw-sihly. Ol course, the maximum amount of work cannot be acheived in practice hccause the reversible wth is an idealized process, but the results will give us a measure uf the maximum efficiency that can be expected. Because the process is cyclic and reversible,
-7L
maximum efficiency = -= qw.h
Equation 6.34 says that q
T%,t
qtev.h
+~IP*.C
qrcr,h
= -qIrv ,(Tc/Th), so the efficiency can be written as
maximum efficiency = I
T = -T,-T -rh
(6.35)
Th
Equation 6.35 is really a remarkable result because it is independent of the specific design of the engine or of the working substance. For a heat cnginc working hetween 373 K and 573 K,the maximum possible efficiency i s
200 maximum efficiency = 573 = 35% In practice. the efficiencywould be less due to factors such as friction. Equation 6.35 indicates that a greater efficiency is r~btainedby engines working with a higher value of T, or a lower value uf T.. Note that the efficiency equals zero if T, = IC,which says that no net work can be obtained from an isothermal cyclic process. This conclusion is known as Kelvin's
Clupter 6 / Entropy and the Ser-ontl Law ul Thermrrdynamics
statement of the Second Law. A closed system operating in an isothermal cyclic manner cannot convert heat into wurk without some accompanying change in the surroundings.
6-8. Entropy Can Be Expressed in Terms of a Partition Function We presented the equation S = k , In W in Sec~ion6-5. This equation can be used as the starting point to derive most of the important results of statistical thermodynamics. For example, we can use it to derive an expression for the entropy in terms of the systcm partition function, Q ( N , V , B ) , as we havc for the energy and the pressure:
and
Substitute Equation 6.23 into Equation 6.24 attd then use Stirling's apprnximation for the factorials (Mathchapter E) to get
= k,AlnA
-
kBA- k , x u ,I n q + k , I
= k,AlnA
- k,
&A.
En~ropyCan Bc Exprtustd In Terms of a Partition Funct~rln
But the last term here cancels with the first because
where we have used the facts that AIn A is aconstant and divide Equation h.39 through by A, we obtain
p, = I . If we furthermore
Note that if at1 the p,'s are zero e!=Fpt fur one (which must cqual unity because C, p, = I), the system is C O ~ ~ ordered T C ~ and ~ ~S = ~ 0 . Therefrlre, we see that according to our molecular picture of entropy, S = 0 for a perfectly ordcred syslzrn. Problem 6 3 9 asks you to show that S is a rnaximu~nwhen all the p j ' s are equal, in which case the hystern is maxi~nallydisordered. Tu derive an expression for S in terms nf @ ( N . V, T ) , we substitute
into Equation 6.40 LO obtain
xq J
x u , Inq I
where we have used the fact that a, = A and have subscripted S with "cnsemble" to emphasize that it 1s the entropy of the entire ensemble of A systems. The entrt~pyof a typical system is given by S,,,,cm = Scnm,,c /A. If we use the [act that the probability of finding a system in the jth quantum state 1s given by
We used the fact that pk, = l / T to go from the third line to the last line. Using Equation 6.36 for U gives S in terms of the partitinn function, Q ( N , V jT).
and then substitute o, = Ap, into Equation 6.3H. we obtain
Recall fmm Chapttr 4 that
6-g. The Molecular Formula S = k, In W Is Analugous to the Therrnodynan~icF m u l a dS = 69, jT
Chapter 6 1 Entmpy and the Second Law of Thermodynamics
for a monatomic ideal gas where a l l the atoms are in their ground electronic state. Using Equatiun 6.43, we obtain for the mular entropy of une mole of a monatumic
263
EXAMPLE 6-7 Show that Equatron 6.45 gives Equotion 6.26 h r the mt~larentropy of mixing nitrogcn and bromine as ideal gases.
ideal gas.
SCI I. U TI ON: Eimt we writc Equation 6.45 as S = N k , In
Applying Stirling's approximation to the last term gives
V
+ terms nr)t involving V
The inilia1 stale is given h j Sl = Sl.K>+ %ar,
= r r N 2 R In V
Therefore.
y~
S = - 5R + K n [ ( T ) 21t mk, T 2
-
2'3
=]
+ nEr2A In V + terms not involviog V Br2
wherc we havc wrirtcn N k R = i t R . The tirial state 1s given by
Fgc,
2'
= '2.N2 =n y2
E X A M P L E 6-6 Ust. Equation 6.45 try calculate the molar entropy of argon at 298.2 K and one bar. and
+ '2.kr2
+
Rln(VN1 VR5)
+ nBr2Rln(VN2+ VHr?)+ terms not involving V
Therefore
compare your result wilh thc experimental value o f 154.8 J ~ K - '.rnoi-l. S O L U T I O N : At298.2Knndonebar.
Because V is proportional ton for an ideal gas, wc have (6.022 x 10" mol-])(l hnr) (l).08314L.bar K .mu]-')(298.2K) = 2.429 x 1 02?L-' = 2.431 x 102' rn-?
-
-'
If we divide this result through by nN2+ trRrz, then we obtain Equation 6.26. t
I
and
2.rrnkDT
'!'
Z~(0.03995kg.rnol-' j(1.3806 x
(7 = [- ) (6.022 x 10" mo1-')(6.626 =
J , ~ - ' ) ( 2 9 ~K) .2 x lo-" l , e ) '
I
3i2
6-9. The Molecular Formula S = kg tn W 1s Analogous to the
(3.909 x l d ' m-2)'12
Therefore
Thermodynamic Formula dS = Sq,,/T In this last section, we will show that Equation 6.24, or its equivalent, Equation 6.40, is consistent with our thermodynamic definition of the entropy. As a bonus, we will finally pruve that p = I / k,T. If we differentiate Equation 6.40 with respect to p,. we get
But Thiv va1uc of 3 agree.; exactly wit11 the experimentally determined value
C dp, = 0 because C pi
= 1, so
Chapter 6 / Entropy dr,d the Second Lavv ol Thermodynamics
2 64
Now substitute Equation 6.41 intc~the In pi rerm in Equation 6.46 to obtain
e place in B accompanied by an exchange of energy a% heat GqEv(arbltrarg). 8 ~ a 1 s the composite system is isolated, the First law rtquires that
The lcrln involving In (jdrops u u ~becausc
Nuw use Equation 6.4 to prove that
Therefore, we can say that the definition given by Equalion 6.4 holds for any system.
and so
But we showed in Section 5 4 that X I E , ( N , V)dp>(N,V , B ) i s he energy as heat that a system gains or loses in a reversible process, so Equation 6.47 becomes
6-6. Calculate qm and AS for a reversible cooling of (memole of an ideal gas at a constant volume V, from P , , Vl, T,to P,, V , , T, followed by a revmihle expansion at constanl pressure P2 from P,, V,. T, to P?, Vt, TI (the final state fur all the processes shown in Figure 6.3). Cotnpare your resull for A S with dlose for paths A, I3 C, and D E in Figure 6.3.
+
+
6-7. Derive Equiltion 6.8 without referring to Chapter 5. 6-8. Calculate the value of AS,if one mole of an ideal gas is expanded reversibly and isothermally from 10.0 dm' Ip Zg,U dm3.Explain the sign o f AS.
Equdtion 6.48 shows, furthermore. that Bk, is an integrating factor of 6qm, or pk, = 1 / T , or p = l / k , T . Thus, we have finally proved that /3 = I / k , T . In t h e next chapter, we will discuss t h e experimental determinatinn of the entropies of substances.
Problems
6-10. Calculate the values of qmvand A S along tbc path U + E in Figuw 6.3 for one mole of a gas whose equation uf state ih given in Example 6-2. Cornpre your result with that obtained iri Example 6-2. 6-1 1. Show that ASn+, i s equal to AS, and AS,+, for the equation urctale gikc11In Example 6 - -
2.
6-1. Show that
6-12. Calculale the values of q,, Bnd AS along the path descr~bcd111Problem &6 For one 111ole of a gas whose equation of state is given in Example 6-2. Compare yuur result with that obtained in Example 62.
it Y is a state function.
6-2. Let z = ;(I, y ) and dz = xydx + y 2 d y . Although d z is not an exact differential (why not'!), what combinatiun of dz and x andlor y is an exact differential? 6-3. Use the criterion developed in MathChapter D lo prove that 6qm, in Equation 6.1 is not an cxact differential (see also Problem D-I 1). 6-4. Use the criterion developed ill MathChapter 1) to prove that SqJT cxirct differential.
6-9. Calculate the value of A S if one ~ n o l eof a11 ideal gas is expanded mvcrsibly and isothermall y from 1.00 bar to 0.100 bar. Explain the sign of AS.
in Equation 6.1 is an
6-5. In this problen~,wc will prove that Equalion 6.5 is valid for an arbitrary system 'lij drl this. cunsider an isolated systemmade up of two equilibrium subsystems, A and B, which are in thermal contact with each other; in othcr words, they can exchange energy as heat k l w e e n themselves. Let suh~ystemA bc an ideal gas and let subsystem R be arhilrnry. Suppose now that an infinilehirnal rcvcrsible process occllrs in A accampanicd by ;in exchange of c ~ l c ~ , gi ~y hh u t Gq,,,(idcal). Simultancrlusly, another infirii1cslmal rcvessiblc process takes
6-13. Shuw that
for a constant-pressure process if C, i~independenlof temperature. Calculate the change in entropy of 2.00 moles of H,0(1) ( U p = 75.2 J-K-'.mol-') if it is heated from 10°C to90C.
6-14. Show that
if one mole v f an ideal gas is taken from T, , V, to T,.V,,assumtng tha~?, is independen1 of temperature. Calculate the value of AS' ',f one malie of N,(g) is expanded from 20.0dm" at 273 K to 3Uodm3 at 400 K . Take = 29.4 ~ . ~ - ' . ~ n o l - ' .
c,.
Entropy and the Third Law of Thermodynamics
In the previous chapter, we introduced the concept uf entropy. We showed that entmpy is created or genemted whenever a spontaneous or irreversible process rxcurs in an isolated system. We also showed that the entropy of an isrbl:~tvrlsystem thut is no1 in equilibrium will increase until the system rzaclles equilihriuni, iiom which time the entropy will remain constant. We expressed t h i s condition mathema~icallyby writing d S 2 0 for a process that occurs at constant U and V. Although we calculated the change in entropy for a few proceses, we did not attemp1 'to calculate absc~lutevalues of the entropy of substances. (See Example 6-6 and Problems 6 4 1 through M 3 , --..
however.) In this chapter, we will introduce the Third Law of Thermodynamics, so that we can calculate absolute values of the entropy of substances.
7-1. Entropy Increases with Increasing Temperature Walther Nernst waq horn in Briessen, Prussia (now Wabrzetnu. Poland), on lunc 25, 1864, and died in 1941. Although he aspired to be a poet, his chemistry teacher kindled his in~erest in science. Between t883 and 1887, Nernst studied physics with von Helmholtz, Uoltzmann, and Kohlrausch. HE received his doctorate in physics at the University of Wurzhurg ln 1857. Nernst was Ostwald's assistan1 at he University of Leibzig from 1887 to 1891, after which he went to the University of GGGttingen. where he esrablishad he Kaiser Wilhellil Institute for Physical Chemistry and Electrochemistq in 1894. Upcm moving to the Universi~yof Berlin in 1905. Nernst h g a n his studies of the behilvirx of hubhvnnces ai very low tenlpratures. He proposed one of the early versions of the Third l a w uf Thermudynamics, which says that the physical activities of substances tend to vanish as the temperature approaches &solute rem. The Third Law made it possible to calculate them~odynamicquantities such as equiiibrium collstants from thermal dah. He was awarded the N o h 1 Prize for chenlistry in 1920 -'in rzrrugltition of his a,r>rkin themmhemistry." He was an early automoblile ellthusiavt and herved during World Wilr I a!. a driver. Nernst lost both of his s o ~ in ~ sWorld War 1. His anti-Nu.i stan* io the 1930s led to increasing isolation, so hc retircd to Ilis country home, where he died io 1941.
We stat with the Fist Law of Thennodynamics for a reversible process:
Using the fact that SqW = TdS and 6wlr> = - Pd V , we obtain a combination of the First and Sscond Laws of Therm4ynamics:
We can derive a number of relationships k t w e e n themdynamic quantities using the laws of thendynamics and the fact that atate functions are exact ditlerentials. Example 7-1 derivks the following twu important relationships
I!
6
2 74
Chapter 7 1 Entropy and the Third taw o f Thermodynamics
and
7-2. The Third Law nt Thermodynamics S a p That thc Entropy of a Perfect Crystal Is Zero at 0 K
and substitute Equation 7.1 for dU to obtain
Proceeding in a simitar manner as in E ~ a m p l 7-1 e (Problem 7-1). we obtain I
I
E X A M P L E 7-1 Express U a$ a function of V and T and then use this result and Equalion 7.1 111 derive Equations 7.2 and 7.3. SOLUTION: I i we treat U as a thaptcr H)
function o f V and T , its total derivative is (Math-
and
From Equation 7.7. we get We suhhtilute Equi~liun7.4 into Equation 7.1 and solve for dS to obtain
Thus, if we know C, as a function of T, we can calculate AS. Most processes we will consider occur at constant pressure, so we will usually use Equaticm 7.9tocalculate AS. If we let TI = 0 K i n Equalion 7.9, then we have
Using t l ~ cdefinition t h a ~( a U / D T ) , = Cy,we obtain
If we comparc this equation for d S with the Iota1 derivative of S = S(T. V),
S ( T ) = S(0 K) t
(constant
P)
Equation 7.10 tells us that we can calculale the entrnpy o f a substance if we know S ( 0 K) and C , ( T ) from T = 0 K to the tempenture of interest. (Notice once again that we use a prime on the variahle of integration to distinguish it from an integration limit.)
we see that
Equation 7.2 tells us Row S varies with temperature at constant volume. If we integrate with respect to T (keeping V constant), we obtain
(constant V )
Thus, i f we knt~wC, (T) as a [unction of 2', we can calculate AS. Note that because C , i s always positivc. the~ntropyincreases with increasing temperature. Equatiun 7.5 is restricted to constant volume. To derive a aiinilar equation for uonstanl prcssllre, we start with dH = d ( l l + P V ) = dU f PdV
+VdP
7-2. The Third Law of Thermodynamics Says That the Entropy of a Perfect Crystal Is Zero at 0 K Let's discuss S ( 0 K) first. Around the turn of the century, the' German chemist Walther Nernst, after studying numerous chemical reactions, postulated that A , S -+ 0 as T + 0. Nernst did not make any statement concerning the entropy of any particular substance at 0 K, only that a l l pure crystalline substances have the same entropy at 0 K. We have added thc "purc crystalIinc" condition here to avoid some apparent exceptions to Nernst's postulate that we will resolve later. In 191 1. Planck, who incidentally did agreat deal of research in hennodynamics (including his dwtoral thesis), extended Nernst's postulate by pustulating that the entropy of a pure substance approaches zeru at 0 K. Planck's postulate is consistent with Nernst's but takes it further. There are
2 76
Chapter 7 1 Entropy and the Third Law uf Thermcdynamics
several equivalent statements of what is now called the Thirdhw of Thermodynumicx, but the one we will use is
7-3. A,,S = A,,H/T,,, at a Phase Transition We made a tacit iisxurnptir,n when we wrote Equation 7.14; we assumed that there is no phase transition between O and T. Suppose there is such a transition at T,,, between 0 and T. We call calculate the entropy change upon the phase transition, A,,,S, by using the equatinn as, = % (7.15)
Every substance has a finite positive entropy, but at zero kelvin the entropy may become zero, and does so in the case of a perfectly crystalline substance. Thc 'Third Law of Thermodynamics is unlike the hrst two laws in that it introduces no new state function. The first jaw gives us the energy and the second law gives us the entropy; the third law provides a numerical scale for entnlpy. Althrlugh the Third Lw was formulated hefore the full development ul' the quanrunt theory, it is much more plausible and intuitive if we think of it in terms of ~nolccularquantum states or levels. One of our molecular formulas for the entropy is (Equation 6.24)
where W i s the number of ways the total energy of a system may be distributed over its various energy states. At 0 K, we expect that the system will be in its lowest energy stare. Therefore, W = 3 and S = 0. Anorher way to see this result is to start with Equation 6.40 frlr 3:
TI,,
A phase transition is a good example of a reversible process. A phase transition can be revered by changing the temperature ever so slightly. In the melting of ice, for example, at one atm, the system will be all ice if T is .just slightly less than 273.15 K and all liquid if T is just slightly greatcr than 273.15 K. Furthermore, a phase transition takes place at a fixed temperature, so Equation 7.15 hecomes (recall that A H = q , for a phase transition)
'
I
EXAMPLE 7-2 Calculak the rncdar entropy change upon melting and upon vaporizatiot~ at one atm fur H,O. Use ah5H = 6.01 k.l .rnol-'at 273.15 K and A ~ =~40.7 ~ kH ~-moi-' at 373.15
K.
S 0 L UT I0N : Using Equation 7.1 h, we have
where p, is the probabihry of finding the system in the jth quantum state w ~ t henergy E,. At O K, there is no thermal energy, so wc expect the system to be in rhe ground statc; thu?, p,, = 1 and all the other pj's equal Lero. Therefore, S in Equation 7.12 equals tsro. Even if thc ground state has a deger~eracyof n , say, then each of the n quantum states with cncrgy E, would have a probability of I/n, and S in Equatlon 7.12 would be
S(O K) = -k,
" 1
C ,=]
-
1
and
Note that A,,T is mnuch Imger than A,">S. This make< sense r~lolecularlybecause the difference in disorder between a gas and a liquid phase is much greater than the difference in disorder between aliquid and a solid phase.
In- = k , Inn n
Even if the degeneracy of the ground state were as large as the Avr~ydroconstant, 3 would be equal to only 7.56 x 10-12~ . ~ - ' - r n o',l which is well below a measurable value of 3. Because the Third Law oJThermodynamius asserts that S(0 K) = 0, we can write Equation 7.10 as
To calclllate .I'(T), we integrate C,(P')/'Tup to the first phase transition temperaterm for the phase transition, and then intcgrate C,.(7')/T from ture, add a Atr,H / the first phase transition temperature to the second, and so on. For example, if the substance has no solid-solid phase transit~on.we would have, h r 1' greater thm the boiling point,
2 78
Chapter 7 1 Entropy and the Third Law of Thermwlynamics
where T,"
7-5. Practical Absulute Fntrnpier Can Bc
Determined Calorirnetr~cally
SOLUTION: Wt.useEquation7.18 and get
vaporizatiorl. respectively. 3.39 3.K.' -mu!-' -= 1.13 J - K -rnt)l-' 3
'
- .
7 4 . The Third Law of Thermodynamics Asserts That C, + 0
as T + 0 It hac hccn shown experimentally and thet~reticallythat C i ( T ) 4 T' ax T + 0 for most nonmetallic crystals (C; for metallic crystals goes as QT f bTZas T 4 0, where o and h are constants). Thiq T' temperature dependence is valid from 0 K to abut 15 K and is called the Dr~byeT 3 law, after the Dutch chemist Peter Debye, who first xhuwed thcuretically that C ; ( T ) + T' ax T -. 0 ior nonmetallic sulids.
7-5. Practical Absolute Entropies Can Be Determined Calorimetrically Givcn suitable heat capacity d a ~ aand enthalpies uC transition and transitiot~ternpcratures, we can use Equation 7.17 to calculate entropies baed on the convention of setting S(0 K) = 0. Such entropies are called third-law enlropies, or practical ab\olute entropies. Table 7.1 gives the entropy of Nl(g) at 298.1 5 K. The entnlpy at 10.00 K was determined by using Equatior) 7.18 with = 6.15 1 . K - I .111ol I. AI 35.6l K ,thc solid undergoes a phahe change in crystailine structure with A,,%H= 0.2289 kJ.rnol-I. so = 6.43 ~ . K - ' . m o l I . At 63.15 K, N,(s) rnclts with A,~H= U.71 kJ.mol I . so A,,S = 1 1.2 J . K '.moi-I. - Finally. N1(l) at one atm boils at 77.36 K with A , , , ~ ,= H 5.57 k.l,mol-I, giving AYqS= 72.0 ~ . ~ - ' . r n u l -For ' . thc regions between the plixse transitions, ~ , ( T ) / Tdata wcre integrated ~iumcrically(Problem 7-14). Accorditlg to Equatiun 7.17, the molar entnlpy is give11 by the area under fhc curvc of C7,(7')lT plotted agiiinst the temper;dure. The small correction at the end of Table 7.1 needs explaining. The valucs a f entropies of gases presented in the literature are called standard entropies. which by convention itre corrected for the nonideality of the gas at one bar. We will teurn how
c,,
EXAMPLE 7-3 According to the k b y e theoly, the ltw-lemperaturemolar heat capacity uf nonmetallic
solids grKs as
whcre Towdcpcrld~upon the particular solid, but IS ahjut 10 K to 20 K for must solids, and H, is a consml characteristic of thc solid. The parameter 0, has unit\ of ten~perdhlreand is called the Deby remjwmiure of the sulid. Show that if is givcn by the above expression, the luw-te~tlperature contribution to the molar entropy is given by
r,
7.1 Thc standard mrhr entropy of nitrogen at 298.15 K. TABLE
501. UT I ON: Substitute the given expression for ? , ( T ) into Equatiun 7.14 to get P~OCCSS
-
S/J-K-' .mol-l
35.61 to 63.15 K
2.05 25.79 6.43 ' 23.41
Fu~ion
11.i
63.15 to 77.36 K Vaporizativn 77.36Kto248.15K
1 1.46 7?,0 39.25
O to 10.00 K 10.M to 35.61 K
Transition
I
-
EXAMPLE 7 4 Given t11atthc rnolar heat capacity of solid chlorine is 3.39 J.K-' .mol-' at 14 K and obeys tllc Debyc T' law beluw 14 K. calculate the molar entropy of solid chlorine at
14 K.
1
'
Co~rectionfor nnnidenlity Total
,,'
0.02 191.6
2 80
Chapter 7 1 Enlropy and rhe Third Law of Thefmpdynamics
to make this correction in Section &6. Recall that the standard state of a (real) gas at any temperature is that of the corresponding (hypothetical) ideal gas at one bar. Figure 7.1 shows the molar entropy uf nitrugen plotted against temprature from OK tu 400 K.Note that the molar entropy increases smoothly with tempcraturc between phase transitions and that there are discontinuous jumps at each phase transition. Notc also that the jump at the vaporization transition is much largcr than the jump at the ~lleltingpoint. Figure 7.2 shows a similar plot for ben7ene. Note that knzene dues nut undergo any solid-solid phase transitiuns.
7-6. Practical Absolute Entropies of Cases Can Be Calculated from Partition Functions Recall fmnl Seclion 6-8 that the entropy can he written as (Equation h.43)
where Q ( N , V, T ) is the syste~npartition function
Equation 7.19 is consistent with the Third Law of 'Thermcdynamic<.k t ' s wrile Equation 7.19 for S more explicitly by substituting Fquation 7.20 into it:
We want to study the behavior of this equation as T + 0 . Assume for generality that the first tl states have the same energy E, = E, = . . = En (in ijther words,
.
F I G U R E 7.1 The molar entropy of nitrogen plotted against temperature from 0 K to 400 K.
that the ground state is n-fold degenerate) and that the next m stales have the same energy En, = E,,+2 = . - . = En+m(the first excited state i s ?n-fuld degenerate), and so on. Let's look at the summations in Equations 7.21 as T -+ 0. Writing out Equation 7.20 explicitly gives
If we factor out e-El/kldF, thcn
~ u En+, t - E l > 0 (essentially by definition), so
-
TI K
F I G U R E 7.2 The m o l ~ entropy of benzene plotred against temperalum from 0 K to 503 K.
Therefore, as 7'
-t
0,
2 fl2
Chapter 7 / Entropy and the Th~rrlLaw uf Thcrrnodynamics
In the limit of small T , then, the first terms in each summation in Quation 7.21 durninate, and we have
Thus, as T 4 0. S is prnponional to the logarithm of the degeneracy of the gn~und state (see Equation 7.13). As wc argucd in Section 7-2, even if n were as large as the
Avogadro constant, S itself would be colnpletely negligible. We learned in Chapter 3 (Equation 3.38) that
283
7-6. Prac~ifnlAbsolute Entropies of Cases Can Hc C~lrulatedfrom Partition Functions
Let's use Equations 7.27 and 7.24 to calculate the standard molar entropy of N,(g) K and compare the result with the value in Table 7.1 obtained from heat capacity data. If we substitute Equation 7.24 into Equation 7.27, we obtain at 298.15
The first term reprcscnts the translational contribulion to S,the secoitd represeo~she rotational contribution, the third and fourth represent h e vibratinnal contribution, and the last term represents the electronic contribution to S. The necessary parameters are = 3374 K, and g*,, = I. At 298.15 K and one bar, the various = 2.88 K, factors are
for an ideal gas. Furthermore, we learned in Chapter 4 that for a ( I ) murrarottiic ideal gas (Equation 4.13):
V
-
-
HT - (0.083 14 L.bar.mol-' .K-')(298.15 K)
-
EI,
A
NAP
(6.022
x 1
rnol-')(l bar)
( 2 ) diatoniic idcal gas (Equation 4.39):
(3) linear pulyatotr~icideal gas (Equation 4.57):
Thus, the standard ~nolarentropy S ' at 298.15 K is (4) ~~onlincar ~wlyato~nic ideal gas (Equation 4.60):
(7.26) The various quantities in these equations are defined and discussed in Chapter 4. If we substitute Equalion 7.22 into Equation 7.19, then we obtain
If we use Stirling's approxitnation for In N ! (= N In N
-
N), then (Problem 7-27)
c o m p d with the value of 19 1.6 J . K - I +mol-'given in Table 7.1. The twu values agrce essentially exactly. This type oC agreement is quite cornanon, and in many cases the statistical thermodynamic value is nlore accurate than thc caIorimetric vatuc. Table 7.2 gives standard molar entropies for several substmccs. The accepled literature values are often a combination of statistical thern~odynamicand calorimetric values.
I
EXAMPLE
7-5
Use the equations of this section to calculate the standard molar entropy of carbol~ diuxide at 298.15 K and compare the result with the value in Table 7.2.
I
7-7, The Valucs ul Standard Molar Entropie Uepcnd Upon M o l t ~ u l a Mas5 r anrl Molecular Slructurc
T A B L E 7.2
Standard molar entropies ($") of Various substances at 298.15
K. .
Substance
S"/J.K-] .mol-'
Subhlance
285
hur values of (gVi,,,/T are 3.199 (twice), 6.338, and 1 1.27. L.ah1, R ~ , ,= I . Putting all this togelher, we find that
S / J . K - ' , ~nol-I
which is in excellent agreement with the value in Table 7.2.
I 7-7. The Values of Standard Molar Entropies Depend Upon Molecular Mass and Molecular Structure
SOL U T I0N : Carbo11dioxide i s a sym~nctriclinear molecule with four vibrational degrees uf freedom. We substitute Equation 7.25 into Equation 7.27 to obtain
Let's look at the slandnrd molar enuopy villucs in Table 7.2and try to dt.temline some trends. First, notice that the standard molar entropies of the gaseous substances arc the largest, and the standard molar entropies of the solid suhstanccs are the srnnllest. Thesc values reflect the fact that solids are more ordered than liquids and gaszs. Now consider the standiird molar entropies of the noble gases giver1 in Table 7.3. The increase in standard molar entropy of the noble gases is a ccmsequence of thz~r increasing mass as we move down the pcriodic table. Thus,an incre;rse in nlnss Ira~ls to an increase in thermal disorder (~norc:translational energy levels arc avuil~~hlr) ant1 a greater entropy. We know from quantum theory that the greater the molecular mass, the
7.3 Standard molar entropies (3') for the noble gases, the pa+eous halogens, and the hydrogel1 halides at 298.15 K.
TABLE
Nohle gas S0/J K-I mol-I He(g) Ne(g)
Paralleling the calculatioti for N,(g), we find thar ( 2 r r M l l - , ~ / h ~=) ~2.826 :~ x lu3' rn-, and TIN, = 4.1 17 x 10-'' rn Using the value of (-IhM = 0.561 K from 'lhhle 4.4,we find that T / 2 e n , , = 265 8. Similarly, wc usc Table 1.4 to show that the
-'.
Ark) Wg)
126.2 146.3 174 4H 164 1 169.7
Halogen S'/J.K-' ,mol-'
Hydrogen hallde A - / J . K
F,(g) Cl,(g)
fiF(p) tlCl(p)
202 8
Br2(g)
223 1 245 5
I,(g)
21A1.7
IIBr(g)
HI@)
.mol-'
173 8 186.9 19H7 106.6
Chapfer 7 / Entropy and the Th~rdLnw of Ihcrrnorlynamics
more closely spacd are the energy levels. The same trend can be scen by con~paring the standard rnolnr entropies at 298.15 K of the gaseous halogens and hyrlrogen halides (see Fable 7 . 3 and I;igure 7.3). Gencrnlly speaking, the more atoms of a given type in a molecule, the greater is the capacity of the molecule to take LIPerlergy and thus the greater is its entropy .. (thc gre:iter the number of atoms, the more different ways in which the molecules can vibrate.) This trend is illustraled by thc serics C,H,(g), C,II,(g), and C,H,(g), whose standard molar entnjpies in joules per kelvin per mole at 298.15 K are 201, 220, and 230. respectively. For molecules with the same geometry and numlxr of atoms, the srandard molar entropy increases with increasing molecular mass
I
,.
d
E
? '*
2
/ ,-
HX
4 y r /
200-
. w
x
o.../--fi--
*---D
* ----. <------
Noble gases
*--.------__---
Note that Table 7.2 gives S- = 245.5 J-K-' mol-' for Br,(g) at 2'18.15 K and one bar. But bromine is a liquid at 298.15 K and onc bar, so where does such a value come frorn'! Even though bromine is a liqutd under these conditions. we can u~lculate S"[Br,(g)] acct~rdingtu the scheme outlined~nFigure 7.4. Therefore. we necdthe ~alucs of the molar heat capacity of Br,(l) (75.69J,K-' .mol-I), the molar heal capacity of Br,(g) (36.02J.K-' .~nol-'),the normal boiling point of Br2(l) (332.0 K),and thc molar
7.3 Standard molar entropies (S") for the noble gases, the gasetms halogcns, and the hydrogen lvolidcs at 298.15 K plotted against In M ,where M is the lnoleculm mahs. FIGURf
I
-'
gasenus forms at 298.15 K are 298 J . K .mol-' and 274 J.K-' .rnol-I, respectively. The entrupy of acetone i s higher than the entropy of trimethylene clxide because of the free rotatiotl ul thc rilethyl groups about the carbon-carhnn bonds iri the ilcclone ~noiecule.l'he relatively rigid ring structure of the trimethylene {~xide mcrlccule rzstricrs the tnovemcnt of the ring atoms. This restriction gives rise to a lower molar entn)py becausc the capacilg uf the rigid isomer to take up energy is less that1 that of the more flexible acetone molecule, which has more possibilities for intermolecular motion. For molecules with approximately the same molecular masses, the more compact lhc mr)lecule is, the smaller is its cntropy.
A '
0
7
@
7-?. The Valuer of Standard Molar Enlrupim Depend Upnn Molecular Mas: and Mnlwular Structure
EXAMPLE 7 6 Arrangc [he folluwi~~p molecules
I in
order oi increasing standard n~nlarentropy:
UH,Cl,(g): CHCl,(g); Cti,Cl(g). SO L O T I0 N : The n u m k r of atoms i s tile same in e x h case, but chlurinc has n greater mnsc than hyrlrogen. 'Thus. we predict that
Br,(l)
This ordering is in agreement with the values of the standard nlolar entropies at 2YR.15 K, which arc in unitc of joules per kelvin per mole. 234.6. 270.2. anti 295.7,
-5
.
-.-
332.0 K Path 2
j
i
n"
respectively I
1
An interesting comparison is given by the isomers acetone and trimethylene oxide
(whore molecular structures are shown below), whose standard molar entropies for the
298.15 Rr2(')
Br2(g)
-Brl(g)
K
ER.... 7.4 The sr-hernc used to calrulate S2[Rr,(g)] at 298.15 K.I n h t t h I . Br,{l) is heated to its boiling point, 332.0 K l'hcn Br,(l) is vaporized to Rr,(g) at 332.0 K (Path 21, and finally Br,(g) is cwled from 332.0 K hack to 248.15 K (p,h 3).
Chapter 7 / Entropy and the Third Law of Thermodynamics
290
291
Problems
7-5. Use the data in Problem 7-4 to calculate A S if one mole of ethenc is heated froni 3(K! K to hCI0 K at constant pressurn. Assumc cthcnc bellavcs idcally.
the standard entropy change is given by
For example, using the values of S' given in Table 7.2 for the substances in the reaction described by the chemical equdtion
7-6. We can calculate the difference in the results of Problerns 7-4 and 7-5 in the follox~ing way. First. %howthat because F , - 7, = R for an ideal gaq,
AT,
=
AS, $ Rln
T
-2 TI
Check to cee numerically that your answers to Pmhlemc 7 4 and 7-5 diIfer by R In 2 = nh93R = 5.7hJ~K-'.rnol-'.
This value ot- A,SUreprcsenls the value of ArS for the combustion of o n e mole o f II,(g) or the hrmation of on^- mole of H,0(1). when all the reactants and prrxlucts are in their standard xtatec. The large negative value of A,S" reflects the loss of gaseous reactants to produce a condensed phase, an ordering prwess. We will use tables of standard enthalpies of formation and standard entropies to calculate cquilihrium constants of chemical reactions in Chapter 12.
7.7. The results of Problems 7 4 and 7-5 niust be corinectcd in thc following way. Show that the two processes can be represented by the diagmm
Problems 7-1. Form thc total derivative of If ns a function of Fquaion 7.6 to derive Equations 7.7 and 7.8.
T and P and equate the result to d H in
r,
7-2. The lnolar heat capacity of Il,O(I) has an approximately constant value of = 75.4 J.K-' .mol f r o t ~W~C to 100'C. Calculate AS if two moles of are heated from 10°C to 90'C a1 constant pressure.
'
TO(])
where paths A and B represent the processes in Prr~hlems7-5 and 7 4 , respeclirely. Now, path A is equivalent to the rum af paths R and C. Shrnv that AS, is given by
7-3. The ll~olarheal capacity of butane can he expressed by
range 3(H) K 5 T 5 1500 K. Calculale A S if onc mole of butane is heatcd from 300 K to IW K at constant pressue. over ~tlctclnperature
and that the resull given in k u b l e m 7 6 rollows. 7-4. The molar heat capacity of C,H,(g) can be expressed by
7-8. Use Equations 6.23 and 6.24 tu show that S
-
over the tempxilure range 300 K < T < 1OOO K. Calculate AS if one mole of ethene i s heated fl-om 300 K to 600 K at constant vtllume.
=O
n O K,where every system will h in its
ground state.
7-9. Prove that S '= -k, C p, In p, = 0 when p , = 1 and all the other p, = 0. In vlher words. prove that x Inx + 0 as x + 0.
Ch,~~itrr 7 i
2[lb
Entropy and !he Third Law uf Thrrrrlodynamics
' ~ ~ i o lWhy - ~ ~is. there a dlhcrepancy bctween the c~lculatedvalue and the exprimcntal value? your result with the calculatd value of 160.3 J , K
7-28. Show that Equalions 3.21 and 7.19 are consistent w~thEquations 7.2 a~id7 3
7-29. Substitute Fquation 7.23 into Equation 7.19 and derive the equwion (Prr~hlem6-3 1 )
for one mole of a m o n ~ t o m i cideal gas.
7-30. Use Equation 7.24 and the data in Chapter 4 to calculare the standard ~nolarcntropy of CI,(g) at 298.15 K. Compare your answer with the expl.rin~entalvalue of 223.1 ~ . K - ' . m o l - ' .
and TU5(al-+ 5,) = 61.6 K, T, = 68.1 K, Tvap = 81.6 K, A [ ~ H = 0.836 kl,mol-', A~,
7-25. The ~nolarheat capacities of solid and liquid watcr can be expressed by
7-31. Use Equation 7.24 and the data in Chapter 4 to calculate the standard mular entropy of CO(g) a1 its standard hiling point, 81.6 K. Compare your answer with the experimental value of 155.6J.K-'.rnul-I. Why is there a discrepancy of ahout 5 J - K - ' . mol-' ? 7-32. Use Equation 7.26 and thc data in Chapter 4 to calculate the hlandard mular crltmpy of NH,(g) at 298.15 K. Compare your answer with the experimcntal value of 192.8 J , K '.mol-'. 7-33. Use Equation 7.24 and the &t;l in Chapter 4 w calculate the statldard molar rntropy u l Br,(g) at 298.15 K. Compare your answer with the experimental vnluc of' 245.5 J.K-] .mol". 7-34. The vibrational and rotationai constants for HF(g) within thc hartnunic oscillator-rigid rotator model are 5 = 3959 cm-' and B = 20.56 cm-'. Calculate the standard mdar entropy of HF(g) at 298.15 K. How docs this value compare with that in Table 7.3?
and T,s= 273.15K, A , u 3 g= 6,(W7!d.rnolKL,A,apx(~ = 298.15 K) =43.93 kJ.mol-', (4, = 192 K, the correclion for nonideality = 0.32 1.K-I .mol 'I, and thc vapor pressure 01- H:O at 298.15 K = 23.8 torr. Use these data to calculate the standard molar cnntropy of HIO(g)at 298.15 K. You need ~ h vapor c pressure of water at 298.15 K b e c a ~ ~ that s e is the eyuilihrium pressure of I4,O(g) when it is vaporized at 298.15 K. You musk include the value of AS that results when you compress the H,O(g) from 23.8 torr to its s t a n d ~ dvalue ofonc bar. Yuurancwer should come out to he 185.6 J-K-'.mot-', which dnes not agree exactly with the value in Table 7.2. There is a residual c n w p y associated with ice, which a detailed analysis of the slructure of ice gives as ASrchdud, = R ln(3/2) = 3.4 J .K-' .rnol-I, which is in g o d agrcelnent wich - SeKp. 7-26, l l s e the data in hohlern 7-25 and the enlpirical expwsalon
7-35. Calculate the standard molar entropy uf H,(g) and D,(g) at 298.15 K glven that the h m d length of buth diatomic molecules is 74.16 pin and the vibrational tenlperaturcs of II,(g) and D,(g) are 6332 K and 4480 K, respectively. Calculate the standard molar cntnqy of HD(g)at298.15K(RL, = 7 4 . 1 3 p m a n d 8 b , , =5496K). 7-36. Calculate the standard molar enuopy of HCN(g) at 1000 K givcn that I = I .I(B l 0 x lo-* kg.rn2, i , = 2096.70 cm-'. G2 = 713.41cm-I, and ir, = 33 1 1.47 cm-'. Kecall that HCN(g) is a linear triatomic molecule and therefore the bending mode, I>,, is doubly dcgenewk. 7-37. Given that i, = 1321.3 cm;], ir, = 750.8 cm-I, C3 = 1620.3 cm-I, = 7.9971 cm--I, B = 0.4339 cm-I, nnd c = 0.4103 cnl-I, calculate the standard molar enmpy of NO,(p) at 298.15 K. (Note that NO2(@is a bent trintomic muleculc.) How does your value comparc with that in Table 7.2? 7-38. 111Problem 748,you are asked to calculalt: thc value rlf ArS at 298. I5 K ubl~lpthe k ~ t u In Table 7.2 Tor the reaction described by
to plot the standard molar c$ropy of watcr from 0
7-27.Stlow for an iirltdl gas thal
K lo 5(X) K. Use the data in Table 4.2 to calculate the standard molar entropy of each of h e reagenls in this reaction [see Example 7-5 for the calculation of the stilndard mular entropy of COl(g)]. Then use these results to calculate the standard entropy change for the above reaction. How does yuur answer compare with what you obtained in Pmblem 7-48?
Chapter 7 1 Entropy and the Third Law of Thermodynamics
-
7-39. CaIculate the value of ArS' for the reacliun r l e s ~ r i b dby
4
H21@ + O,(g)
H,Olg)
at 500 K using h e data in Tahlec 4.2 and 4.4.
7-411. 111 cnch cave helow, predicl which molecule of the pair has the greater n~olarentropy undcr tllc same ctlriditions ( a ~ s u m egaseous species).
7-45. Table 7.2 e k e s S'[CH,OH(I)] = 126.8 J - I - ' .md-' at 298. I 5 K. Given that T = 337.7 K. A v w H ( ~ ,=) 36.5k3,mr,l-', C,ICH,OH(I)] = 81.12J.K -rnol-', and L J P C,[CH,OH(g)] = 43.8 J . ~ - ' . r n u l I , calculare the value oC S [CH,OH(g)] at 298.15 K and colnpxe your answer with the experin~en~al value of 239.8 J - K - '.rnol-l.
'
7-46, Giventhc followingdata, ThL= 373.I 5 K, = 4n.65 kJ-mol-I, C,.[H,O(I)I = 75.3 J.K-' .rnol-I and^,[^,^(^)] = 33.8 I.K-' -mu{-'. show thatthevalucs nf S'iH1O(l)l end S'[H,O(g)] in Table 7.2 are consistent. 7-47. Use the data in Table 7.2 to calculate the value of ArS" for the folluwing rcactions at 2SC and onc bar.
a. U s , graphite)
+
2 O,(gj C2H2(g) lT,(g)
b. CH,(g) C.
+ 0,(g) -+ CO,(g)
+
---t
CO,(g)
+ 2 H,0(1)
C,H,(g)
7-48. Use the data in Table 7.2 to calculate the value of A,S" €or the following reactions at 25°C and one bar. 7-41. In each ~ 1 1 % below, predict which molecule of the pair has the greater m d a r entropy undcr the same condilions (assume gaseous species).
a. CU(g) + 2 H,(g)
b.
c. 2 cow
7-42. Arrange the followil~greactions accordir~g10 increasing values of A,S" (do not consult ;illy rererenues).
7-43. Arrange the follox~ingreactions according to increasing value< or ArS" (do not consult any references).
7-44. In Problem 7 6 0 , you ,& askcd to predict which ~nolecule,CO(g) or CO,(g), has the greater ~nolarentropy. Usc thc data in Tables 4.2 and 4.4 to calculate the standard molar cnlropy nf CO(g) and CO,(g) at 298.15 K. Does this calculatiun confirm your intuition? Which degree of freedom makes the dominant contribution to the molar entropy of CO? 01-C0,?
-+ CH,OH(I)
C(s, graphite) C TOCt) + CO(g)
+ O*(g)--c 2 CO, (g)
+ H,(g)
CHAPTER
8
Helmholtz and Gibbs Energies
The criterion that dS > 0 for a spontaneous process applies 011ly tu an isolated system. Consequently,in the variuus p m e s s e s we discussed in Ckapter 6, we had to consider the entropy change of both the system md its surroundings tr) determine the sign of AS,, and establish whether a prmess is spontaneous or not. Although of great fundamental and theoretical importance, the criterion that d S > 0 for a spontaneous process in an isolated system is too restrictive for practical applications. In this chapter, we will introduce two new state functions that can be used to determine the direction uf a spontaneous process in systems that are not isolated.
Hermann von Hetmholtz waf born in Potsdam, Germany, on August 31, 1821, and d~edin 1894. Although he wanted to study physlcs, his family could not afford to send him to the University, so he studied medicine in Berlin h c a u x he could obtain state financial aid. He was, however, required tu repay his stipend by service a$ n surgeon in the army tr>r cight years. He was later appointed professur at the U~livcrsityof KBnigskrg, and he also held positions at Bonn, Hcidelberg, and Berlin. In 1885, in recognition of his position as Ihe foremost scientist In Gcrmany, he was appointed president 01- the newly fuunded Physiccl-Technical Institute in Berlin, an institution dzvoled tn purely scientific research. Hel~nholtzwas one vf the greatest scientists of the 1Yth century, making imponant discoveries in physiology, optics, acvustics, electromagnetic theory, and thermodynamics. His work in physiology showed that physiological phenon~enaarc based upon the laws 01- physics and not on some vague "vit:J force." In therrnodynanlics, he derived the equation now known ah the Gibbs-Helmholtz equalion, which we will discuss in this chapter. I3elmhultx wrts always generous with his s l u d c ~ ~and t s other scien~ists,but unfortunately he was a barely intelligible lecturer, even to the likes of Planck, who was astudent in several of his c1as.w~.Helritholtz's great influence in German science was wcogllized by Lhe Kaiser, who bestowed him with Lhe titlc "wri:'
8-1. The Sign of the Helmholtz Energy Change Determines the Direction of a Spontaneous Process in a System at Constant Volume and Temperature Let's consider a system with its volume and temperature held constant. The criter~onthat dS 2 0 for a spontaneous process does not apply tu a ,> , I ) ,tt constant temperature and volume because the system.is not isolated; a systcrn must he in thermal contact with ilthermal reservoir to be at constant temperature. If the criteriun d S 2 O does riot apply, then what I S the criterion for a spntaneuus prnrc* ,llh,ir we can usc fur a syq1t.m at constant temperature and volutne? Let's start with tht: expression of the F m t L a w oT Thermodynamics, Equalion 5.9,
.,
Because Su) = Pex,dV and dV = 0 (constant volume), then 6 w = 0. If we suhsti~ute Equation 6.3, dS 2 Sq/ T , and 6 w = 0 into Equation 8.1, we obtain
&I.
The equality holds for a reversible process and the inequality for an irreversible prmess. Note that if the system is isolated, then dU = O and we have d S > O as in Chapter 6. We can write Equation 8.2 as
If T and V are held constant, we can write this expression as
d(U - 7's) 5 0
(constant T and V )
Equation 8.3 prompts us to define a new thermodynamic state function by
so Equatian 8.3 hecomes
dA
0
(8.5)
(constant T and V )
The quantity A is calted the Helmholtz energy. In a system held at cotlstant T and V ,
the fIelmhr,ltz energy will decrease until all the possible spontaneous prtxesses have occurred, at which time h e system will be in equilibrium and A will he a minimum. At equilibrium, d A = 0 (gee Figure 8.1). Note that Equation 8.5 is the analog of the criterion that d S 3 O for a spontaneous prmess td vccur in an isolated system (cf. Figures 6.5and 8.1). For an isothermal change from one state to another, Equation 8.4 gives
Directinn nt R S l ~ ~ t l t n r l r ~ uf'rmt~~j5 !. in a Systcm at C(mstant i:olun~e 2nd Ic~riipr,~turc
where the equality holds for a reversible change and the inequality holds lor an irwversihle. spontaneous change. A process for which AA > 0 cannot t&e place xpontaneously in a system at constant T and V. Consequenrly, something. such as work, must be done on the systcm to effect the change. Notice that if A U < 0 and A S > 0 in Equation 8.6. then both the energy change and the cntn~pychange contribute to A A being negative. But if they differ in sign, some sorl of compro~nisemust be reached and the value uf AA is a quantitative measure uf whcthcr a process is spontaneous or nut, The Helmholtz energy represents this compromise between the tendency of a system to decrease ils energy and to increase ils enlropy. Because A S is multiplied by T. we see that the sign of rlU is rnore importan1 at low tempcratures but the sign of A S is rntll-e important at high temperatures. We call apply the criterion that AA < O for an irre\,crsible (spontaneous) pnxess in a system at constant 7' and V lo the mixing uf two ideal gases, which we discussed in Scction 1 7 4 . For that process, A U = 0 and ~3 =,!R In y, -- j2H In y2. Therefore, for the rnixirlg of Lwo ided gases at constant T and Y. AX = RT (y, In ,! J? In?;), which is n negative quantity because p, and q are Iess than one. Thus, we see once again that the isolhermal mixing of two ideal gases is n spontarleoua
+
pnxes';. In addit~ontu scrving as our criterion for spontaneity in a systcm at collstant temperature and volume, the Helmholtz energy has an important physical interpretation. Let's start with Equation 8.6
for a spontdncous (imevcrsible) process, so that AA < 0. In this process, the initial and final states are well-defined equilibrium states. and there is no fundamental reason we have to follow all irreversible path to get from onc state to the other. In fact, we can gain some considerable insight into the process if we look at any reversible path connecting these two states. For a reversible path we call replace A S by qrm/ T, giving
Using Equation 8.5, we see that
But according to the first law, A U
Spontaneous
AA = I
- qm> is
U , ~ ~
equal ta wmv,so we have
(isuthem~al,revcr~iblc)
(8.9)
Eqn~ltbrium dA = 0 - --
t
f F I G U R E 8.1 The Hclmhdtz energy. A , of a system will decrease during any spontaneous processes t h t c~ccurat const;lnl T and V and will achieve its minimum value at equilibriu~n.
If AA i0, thc process will occur spontaneously and turw rcpresents the work that can be done by the system if this change is canied out reversibly. This quantity is the maximum work that could be obtained. If any irreversible prvcess such as friction occurs, then thc quantity of work that can be obtained will be Iess than u ! ~If ~ AA. > 0, the process will not occur spontaneously and 11)- represents the work that must be done on the system tu produce the change in a reversible manner. If there is any irreversibility in the prmess, the quantity of work requircd will be even greater than I I J ~.~ ,
3 04
8-2. Direction uf d Sponbnmus Process for a Syrtem d t Constaml Pressure brlci Tcrnp~taturc
8-2. The Gibbs Energy Determines the Direction of a Spontaneous Process for a System at Constant Pressure and Temperature Most reactions occur at constant pressure rather than at constant volume because they are open to the atmosphere. Let's see what the criterion of spontaneity is for a syswnl at constant temperature and pressure. Once again, we start with Equation 8.1, but now we substitute d S > Sq/ T and 6 w = - PdV to obtain dU 5 l'dS
- PdV
or dU
-
TdS - PdV
0
Because both T and P are constant, we can write this expressinn as d(Il
-
TS + PP') 5 0
The equality holds for a reversible process, whereas the inequality holds for an irreversible (spontaneous) process. If AH < 0 and A S > 0 in Equation 8.15, both terms in Equation 8.15 contribute to A G being negative. But if A H and AS have the same sign, then AG = A H - TAS represents the ct>mprornisebetween the tcndency ul' a system to decrease its enthalpy and tu increase its entropy in a constat~tT wd P process. Because of the factor of T multiplying A S in Eyuatiun 8.15, the A H term can dominate at low temperatures, whereas the T A S term can dominate at high fcmperatures. Of course if AH > 0 and A S -;- 0,then AG > 0 at ail relnperatures and the proccss is never spontaneous. An example of a reaction favored by its valiic oT hut dishvorcd by its valr~e ArS is
(constant T and P)
The value of A,!{ for this reaction at 298.15 K and one bar is -176.2 kJ, whereas the corresponding value of ArS is -0.285 ~ J , K - ' ,giving n,G = ArH - TArS = -91.21 kJ at 298.15 K. Therefore, this reaction prmecds spontaneously at 298.15 K and one bar. A process for which h e sign of AG changes with a small change in temperature is the vaporization of a liquid at its normal boiling point. We represent this process by
We now define a new lhermodynamjc state functir~nby
so Equation 8.5 h o m e s dG 5 0
The analog of Equation 8.7 is
(constant T and P )
(8.12)
Note that Equation 8.11 is the analog of Equation 8.4. The quantity C; is called the Gibbs energy. In a system at constant T and P, the Gibbs energy will decrcace as the result of any spontaneous pmesses until the system reaches equilibrium, where dG = 0.A plot of G versus time for a system at constant T and P would be similar to the plot of A bersns lime tor a system at constant T and V (Figure 8.1). Thus,we see that the Gibbs energy, G, is the analog of the Helmoltx energy. A , for a process that takes place at constant temperature and prchsure Equation 8.1 1 can also he written as
L
The expression for the molar Gibbs energy of vaporiation, Ay3pG, for this process is
The molar enthalpy of vaporization of water at one atm near 100"C, AbdpH, is equal to 40.65 kl-mol-' and buap3 = 108.9 J.K. '.rnol-'. Thus, we can write A+4,,G as
+
where H = U P V is the enthdpy. Note thal the enrhalpy plays the same role in a constant T and P process that the energy U plays in a constant T and V process (uf. Equation 8.4). Note also that G can be written as = 40.65 kJ.mol
thus relating the. Gibbs energy and the Helmholtz energy in the same manner that H and U are related.
' - 40.65 kl.mol-' = 0
The fact that avavG = 0 means that liquid and vapor water are in equilibrium with each other at one atm and 373.15 K. The molar Gibbs energy ut'liquid water is equal to the molar Gibbs energy of water vapor at 373.15 K and one atm. The transfer of one ~riolc
!{-.I, Directinn uf a Spurilanmus Prncfis lur a Syrlptn nr Constant Prc5s~rrcontl r~rnp~rarurc
of liquid watcr to water iapjr under these conditions is a reversible process, and so A,,,pC; = 0. Now let's consider n tcmprature less than the normal hoiling point. say 363.15 K. At thls temperature, ALhpG= + I . 10 kJ.nivl-I. The positive cign means that the formation 01' one molc of water vapor at one ahn from one mole of liquid water at one atrn and 363.15 K is not a spontaneous proccss. Ifthe temperature is above the normal boiling point. however, say 383.15 K, then ArapG= - 1.08 kJ.mol-I. The negative sign means that the fwmation of one mole of w;iter vapor from one mole of liquid watcr at one atm and 383.15 K is a spontaneous process.
If bG < 0, the process will occur spontitneuuslp, and wnom, i s the work exclurive of P - V work that can be dune by the system it' the change is carried out reversibly. This i s the maxinlu~nwork that can be obtained tmm Ihe process. If any ~rreversibility occurs in the process, the quantity of work obtained will be less than the maxiniurtl. If A C > 0, the precess will not uccur spontaneously and u:"~,,,is the minimum work, exclusive of P-V work, that must he done on the system to make the process uccur. For example, it i s known experimentally chat A G for the formation of one mole of H,0(1) at 298.15 K and one bar from H,(g)and O,(g) at 298.15 K and ane bar is -237.1 kJ.rnol-I. Thus, a maximum of 237.1 kJrnol-' of useful work (that is, work exclusive of P - V work) can be obtained from the spontaneous reaction
E X A M P L E 8-1
Thc n~olarenthalpyof fusion o f ~ c c a273.15 t K anrlnne arm is A,~H= 6.01 kl .rnol". and then~olnrcntropyuf fusionunder lhe fanlcconditiuns ic Alu5S= 22.0 J - K - I ,rnol-l. Show that A,,,G= 0 at 273.15 K rind one atm, that A,$G < 0 when the tclnperature i s greatcr than 273. I5 K, ant1 thal A ~ >~ 0 when ~ E lhe temperature is lesu than 273.15 K. S O L U T I 0 N : Assulning that A , , . R and A,u,3 do not 273. I5 K,we can write
Crmverseiy, it would require at least 237.1 kJ.mol" of energy t o drive the (nnl3spr)ntaneous) rcaction
vary appreciiibiy around
I
E X A M P L E a2 The value 01- AE for the decornpnsition of one mole u i H,O(I) to H,(g) and O,(g) at mlc her and 298.15 K is +237.1 kJ.rnol-l. Calculate the minimum voltage required tu decnnlposc one mole of H,C)(I) to If,(g) and O,(g) at one hilr and 298.15 K by
If r = 273.15 K,(hen A,,~Z = 0, indicaringthat iceandliquid watcr are inequilibrium with cach other ilr 273.15 K and onc atm. If T -z 273.15 K, then A,*E > 0, indicating that ice will not sp)ntimeously melt under these conditions. If T > 273.15 K, then A,,,
e~txtrolysis. SOL U T I 0 N : Electrolysis represen& the non- P-V n.ork required to carry out the deco~nposition,su we write
I
I
'l'he valuc c~fA G can he related to the maximum work that can be obtained from a spontaneou~process carried out at constant T and P. To show this, we start by diflercritiating
C;
=U
- TS + PV
dC = r l U and substitute d I l = T d S
+
-
You lnipht r e ~ r ~ e r ~from ~ h c rphysics thnt clcctrical work is given hy cllage x vultagc. The chargc involved in e)ectndyzi~rgo ~ mole e r>f H,n(l) can be determined from tllc chemical equalion of the reacliun
to get
1dS- SdT
6 1 0 , ~for ~
+ P d V -t V d P
dU to gel
+
The oxidation state of hydrogcn goes from 4-1 to 0 and that of oxygen gces from -2 to 0.Thus two electrons are Iransferred per YO(1) molecule, or two times the Avogadro constan1 of electrons per mule. The total c h a r ~ eof twu moles of electrons is
dG = - 6 d T t V d P - k S ~ o ~P~d V
+
Because ihe reversible pressure-volume (P- V) work is - Pd V , the quantity SIU,, Pd V is the reversible work other than F- V work (such as electrical work). Therefore, we can write dE; as
T h e minimum voltage, &, required to decompose onc Inole is giwn by -
AG E= 1.929 x Id c
whcre ~ U J ~ , represents ~ , , the total work exclusive of P-V work. For areversible process taking place at constant T and P, dG = B T U ~ ~ , ~ or ,, A(; = tunrm,,
(reversible, constant T and
P)
(8.16)
lo3 J-mo[-' - = 1.23volts 1.929 x 10' C
- 237. I x
-
where we have used rhc fact that one jrwle is o coulomb times a volt ( I l = ICV).
1
1
8-3. Maxwell Relations Provide Several Useful Thermodynamic Formulas
A number of the tbermodynan~icfunctions we have definedcannot be measured directly. Consequently, we need to be able to express these quantities in tcrnms nf others that can be experimentally determined. To do so, we start with the definitions o f A and G, Equations 8.4 and H. II . Differentiate Equation 8.4 to obtain
For a reversible process, dU = T d S
- Pd V , so
By comparing Equation 8. I7 with the formal total derivative of A = A ( V , T ) ,
we see that
( )
= -P
and
($),
F I G U R E 8.2 The molar entropy uf ethane at 4I)0 K plotted against density ( p = I /TI.The value of 4 l O K is 246.45 ~.mol-'.~-'.
s ' at~
Equation 8.20 allows us to determine the entmpy oC a substance as a function 01volume or density (recall that p = 1/ V ) from P-V-Tdata. I f we let V, in Equation 8.20 be very large, where the gas is sure to behave ideally, then Equation 8.20 bccorries
= -S
Now if we use the fact that the croas derivalives of A are equal (MathChapter D), Figure 8.2 plots the molar entropy of ethane at 400 K vcrsus density. (Problem 8-3 involves calculating the molar cnuopy as a function of density using the van der Wadls equation and the Redlich-Kwong equation.) We can alsu use Equatiun 8.10 10 deri~ean equatlc~nw e dc~ivedearlier 111 Scction 6-3 by another methnd. For un ideal gaq, (d P / a T ) , = t i R/ V , so
we find that
Equation 8.19, which is obtained by equating the second cross partial derivatives of A , is called a M u ~ c l relution. l There are many useful Maxweli relations involving various thermodynamic quantitics. Equation 8.19 is particularly uscful because i t allows us to deter~ninchow the entropy c ~ f asubstance changes with volume if we know its eqwition of stiite. Integrating Fquation 8.19 at constant T, w e have
''
):(;
dV
E X A M P L E 8-3 Calculatc AS Tor an isothern~alexpar~sionfrurn V , to V2 fur equiltion of state
(constant T )
Y
we have integrated ( a S / 3 V ) , ; in other words, T is held constant in Ihe derivative, so T must he We have applied the conditirln of constant T t c ~Equation 8.20 because
held constant when we inlegrate.
I
SOL U T I O I\;:.Wc use Equation 8.20 to obtain
I o gac
that
obey:.
t11u
Chapter 8 / I Ielmholu anrl Gihhq Energies
Note that we derived thisequation in Example 6 2 , but we had to be told that dU = 0 in an isothermal prtwess for a gas obeying the aFmve equatiotl o f ftatc. Wu did not need [hi:, l~~for~llatio~~ to derive our result hcrc.
We have previously stated that the energy of an ideal gas dcpcnrls only upon tempcraturc. This stntement is not gcncrally tl-uc for real gaxes. Suppose we want to know how thc cncrgy uf a g~ changes with volume at constant temwralure. Unfortunately, this quantity cannot bc measured directly. We can use Equatiun 8.19, however, to derive n practical cquation for (i3U/d V),; in other words, we can derive an equation thal tells us how thc cnergy OF a substance varics with its volume at ccunstmt temperature in tern~sof readily measurable quantities. Wc differentiate Equation 8.4 with respect to V at constant teinpcrature to obtain
,
Subqtituting Equalion 8.18a for {a A / a V ) and Equation 8.19 for ( a S / a V ) , gives
0
200
400
P f har F I G U R E 8.3 The molar energy U glutted against pressure for ethane 14.55 kJ. ~nnl-I.
I
at 403
K. The value of
EXAMPLE 8 4 In Example k 2 , we ftated we would prove later that the energy of n gas that obeys thc equalion t>C stale
is independent olthe vulume. Use Equatiun 8.22 to prove this
Equatic~n8.22 gives (ilU / 3 V ) , in terms of P-V-T data. Equations like Equation 8.22 that relate thermodynamic functions to functions of P, V, and T are sometimes called
SOI.U'I ION: Fur
P ( V - tr)
= R7-,
tl~ertnodyna~nic equations of state. We cat1 integrate Equation 8.22 with respect to V to determine U relative to the ideal gas value, and SO
where V'"S a large volumc, whcre the gas is sure to behave ideally. This equalion along with the P-V-T data gives us U as a function of pressure. Figure 8.3 shows U plnttcd againsl pressure for ethane at 400 K. Problem &4 involves calcultlting as a fu~lctionof volume fur the van der Waals equation and the Redlich-Kwong equation. We can also use F4uation 8.22 to show that the energy of an ideal gas is independent of thc volu~ncat constant iemperature. For an ideal gas, (3 P / a T ) , = n R / V , so
which proves that the energy uf an ideal gas depends only upon teinperature.
1
vdi s equ:~lro
E X A M P L E 8-5
Evaluate
[a~/av), for one molc of n Rcdlic11-Kwong gas
SOLUT tON: Recall that the Redlich-Kwong equatio~~ (Equation 2.7) ic
I
Chdpter 8 1 Hel~nhnltzand Cibbs Energie
&I The . Enthalpy of an Ideal Car Is Indupnden~of Pressurc
and
We derived the cquatiun The experimental value of C, is 24.43 J.K-] .rnul-'. Note that C, compared with
in Problem 3-27. Using Equation 8.22 for ( 3 U / 3 V ,,)
we obtain
Foranidealgas(aP/aT), = n X / V a n d ( a V / a T ) , = n R / P , a n d s o C , -C, = n R , in agreement with Equation 3.43. An alternative equation frx U p - C y that is more convenient than Equation 8.23 [or solids and liquids is (Problem 8-1 1 )
Each of ~e paniai derivatives hcre can be expressed in terms of familiar tabulated physical quantities. The isothermal compressibility of a substance is defined as
and the coefficient of thermal expansion is defined as
- c, is sll>all
z,(vr C,,)and i s alsu much sn~allerfor solids than for
gases,
a% you
might expcct.
I
I
8-4. The Enthalpy of an ldeal Gas Is Independent of Pressure Equation 8. i8r4 can be used directly tu give the voiurne dependence (IC the Hclmholtt energy. By integrating at constant temperature, we have
For the case of an ideal gas, we have
Notice that this result i s -T times l?qualiun 8.21 for AS. This result must be so hecausc AU = IIfor an ideal gas at constant T, so AA = - T AS. If we differentiate Fxpation 8.1 1, F = U - . T S F V , and subsritutc c l I J = 7-dS - P d V , we get
+
Using these definitions, Equation 8.24 becomes dG = -SdT
+VdP
(8.30)
By comparing Equation 8.30 with
EXAMPLE 8-6 The cwfficient of thermal expansion. a,of copper at 298 K is 5.00 x K-I , and its isothermal compressibility, K , is 7.85 x I()-' atm-I. Given that the density of copper is 8.92 g . ~ r n -at~298 K, calculate the value of C, - C, h r copper.
w e see that
I
SO I. U T I 0 N : Fur cupper, the molar volume, F, is given by
Note that Equation 8 . 3 1 ~says that G rlecreascs with increasing tempcralure (kS 5 0) and that Equation 8.31h says that G increases with increasing pressure (because V > 0).
cause
Chapter 8 1 Hclrnhultz and Gibbs Enelgies
1f we now take cross derivatives of G as we did for A in the previous section, we find that
31 5
W. The Enrhalpy of an Ideal Gas 15 Independen1of Pressure
EXAMPLE 8-7 Use the virial cxpansion in the pressure
to derive n viriaI cxpansion for A S fur a reversible isothermal change i n p s < u r c .
This Maxwell relation gives us an equation we can use to cdculate the pressure dcpcndence uf S. We it~tegmteEquation 8.32 with T constant to get
A s = - ( )
dl'
(constant T )
SOL U T I 0 N: Solve the abote equation for V :
and write
f
Eq~~atinn 8.37 can he w e d to nhtain the molar entropy as a functio~lof pressure by integrating (il V / i f T ) , data frnm some low pressure, where the gas i s sure to behave ideally, tu some arbitrary pressure. Figure 8.4 shows the molar entropy of ethane at 400 K obtained in this way plotted against pressure.
Substirute this result into Equation 8.83 : ~ n dintegrule h m P, to ?'f
tu obtain
We can alho use Equations 8.31 tu shuw that the enthalpy uf an ideal pas is independent of the pressure, just as its energy is independent of the volume. First. we differentiate Equation 8.13 with respect to P at constant T to obtain
Now use Equation 8.31b for (aG/a P ) , and Equation 8.32 for ( a S / B P ) , to obtain
0
200
400
600
800
P l bar F I G U R E 8.4
The molar ccntropy of ethane at 400 K plotted against pressure. The value of 246.45 ~.rnol.-' .K1.
S' at 400 K is
Note that Equation 8.34 is the analog of Fquation 8.22. Equation 8.34 is also called n therrnodyna~nicequation of state. It allows us to calculate the pressure dependence of H from P - V - T data (Such data for ethane at 400 K are shown in Figure 8.5). Fnr an ideal gas, ( a V / B T ) , = n R / P ,so ( i ) H / a P ) , = 0.
E X A M P L E 8-8 Evaluate ( a H / aP), for a gas whose equation of state is
For an ideal gas, (3 V / a T ), = n R / P , so muation 8.33 gives us
S O L U T I O N : Wehave
This results is not really a new one for us because if we let KT/ V , , we obtain Equation 8.21.
tr
P, = n RI'/V,
and P, =
Chapter 8 / Helmholtz and G~bbsEncrg16
so Equation 8.34 gives us
Notc that
%5. The Var~ousThermodynamic Fun~tionsHavc Ndlural Independt.n~Varidt>l@
takes un a simple form, ill the sense that the coefficients ol' dS and d V are simple thermodynamic functions. Consequenlly, we say that the nalural variables of U are S and V ,and we have
This concept of natural variables i s particularly clear if we consider V atid T instc;irl oF S and V to be the independent variables of U , in which case we g t (cf. Equation 8.22)
( ( a H / a ~= ) , 0 when B ( T ) = 0.
I
1 Certainly U can be considered to be a function of V and T, but its total derivative is not as simple as if it were considered to be a function of 5' and V (cf. Equation 8.36). Equation 8.35 also givcs us that a criterion for a spontaneous process is that dU < O for a system at constant S and V. We can write Equation 8.35 in ternla c ~ dS l rather than rill to get
which suggests that the natural variables of S are U and V . Furthermore, the critcriun for a spontaneous process is that dS 2 0 at constant U and V (Equation 8.2 for an isolated system). Equation 8.39 givcs us P I bar F I G U R E 8.5
The molar enthalpy of ethane a1 400 K plotted againsl pressure. Thc value of 17.867 kl.rnnl-'.
3'' at 400 K is The total derivative nf the enthalpy is given by (Equation 7.6)
8-5. The Various Thermodynamic Functions Have Natural Independent Variables We may seem to he deriving n Lot of equations in this chapter, but they can be organized neatly by recognizing that the energy, enthalpy, entropy, Helmholtz energy, and Gibbs cnergy depend upon t~aturalsets of variables. For exampie, Equation 7.1 summari~es the First and Second Laws of Thcrmodynan~icsby
which suggests that the natural variables of H are Sand P. The criterion of spc>nraneily involving H is that d H < 0 at constant S and P. The total derivative of the Helmholtz energy is d A = -SdT
-
PdV
,
from which we obtain
dU
= TdS
-
PdV
(8.35)
Notc that when S and V are considered to he the independent variables o f U . then the total derivative of U ,
(;),=-S
and
(g),
=
-.
18.42)
Equation 8.42, plus the spontaneity criterion that d A < 0 at constant T and V, suggebt that T and V are the natural variables of A. The hlaxwell relations obtained fmm Equation 8.43 are useful because the variables held constant art. more cxperime~~tally
Chapter 8 / Helmhul~z~ n Gihbs d Encrgies
con~rollabiethan are S a n d V ,as in Equations 8.37, or U and V, as in Equations 8.40. The Maxwell relation h m buations 8.43 is
which allows us to c~lciilaletbc vrlli~rnedependence of S in tenns of P-V-T daia (see Figure 8.2). Last, let's consider the Gibbs energy, whose toval derivative is dG = -SdT t V d P
A-6. Thc Standard Statc fur a GASat Any Tcrnperalure Is the Hyputhetical Ideal Gas a t Orle H J ~
If we add d ( P V ) and subtract d ( T S ) from Equation 8.48, or subtract d ( T S ) from Quation 8.49, or add d ( P V) tn Equation 8.50, we pet
The other equations of this section r011ow by comparing the total derivative of each function in trl ms of its natural variables to the above equations for d U , dFJ. d A , and d G . Tahlc 8.1 su1nmari7xx wrne of thc principal equations w e have derived in this and
previrlus ciiaptcrs.
(8.45)
Equation 8.45, plus the spontaneity criterion dG < 0 for a system a1 constant T and P. tell us that the natural variables of G are T and P. Equation 8.45 gives us
T A B L E 8.1 Thc four principal thcrmdynamic encrgies, their ditfcrcntinl expwsrions, and the corresponding Maxwell relations.
Thermodynamic enerev
Differential
Corrcsponding
expression
Maxwell relations
'l'lic Maxwell ~.eli~tion we obtain fruin Equations 8.46 is
d A = -SdT
which we can use to calculate the pressure dependence of S in terms of P-V-T data (F~gure8.4). This section i s mean1 lo provide both a sutnlnary of many of the equations we have derived so far and a way to bring some order to them. You do not need to memorize these equations because they can all be obuincd from Equation 8.35: d U = 7-(IS -
PJV
(8.48)
which is nothing m o r r than the First and Second Laws of Thermodynamics expressed as one equation. If we add d ( P V ) to both sides of this equation, we obtain
If wc subtract J ( T S ) from both sides of Equation 8.48, we have I
d ( U - TS)= TdS - PdV
- T d S - SdT
- PdV
(g)T(Elv =
8-6. The Standard State for a Gas at Any Temperature Is the Hypothetical Ideal Gas at One Bar One of the most important applications of Equation 8.33 involves the correction for nonideality that we make to obtain the standud molar entropies of gases. The standard molar entropies of gases tabulated in the literature are expressed in tcrms nf a hypothetical ideal gas at one bar and at thc same tetnperature. This correction i s usually s m d l and is obtained in the following two-step procedure (Figure 8.6). We first take our real gas from its pressure, nf one bar to some very low pressure P ' ~where , it is sure to behave ideally. We use Equation 8.33 to do this, giving
-
s[pid)- -$(l bar) = -
n-7. The GibkHelmholtz Equation 13escrlbt.s the Temperature Dependence of thc Lihbs Encrgy
Real gas = ideal gas
we have
r e r y low pressure ( P ' ~ J
f
\
E q n . 2 2 . 5 2 using
Eqn. 2 2 . 5 3 u s i n g ideal gas equation of s t a t e
cqunlir)n of state
Substituting this result into Equation 8.54gives
S.(at 1 bar) = ?(a I bar) +
,
Real gas bar
Hypothetical
_7 ideal gas Correct~un l o r non-ideality
I bar
F I G U R E 8.6 Tile scllcrne to bring the expen'menralentropies of gases to the sttandard s i a k <>I-a(hypathetical) ideal gas at the same temperature.
f!dT %
x (1 bar)
+
,,
.
(8.56)
where we have neglected P ' with ~ respec1 to one bar. The second term on the right sidc of Equation 8.56 represents the correctiun that we add to 5 to get S ' . We can use Equation 8.56 to calculate the nonideality comctio~ltu the entrupy of N,(g) at 298.15 K that we used in Table 7.1. The experimental value of ri B , , / d T for N,(g) at 298.15 K and one bar is 0.192 cm3.mot-' -K'I. Therefore, the correction for nonideality is given by corrcctinn for nonjdeality = (0.192 cm3.mol-'. K-I)( I bar)
The superscript "id" on P emphasizes that this value is for conditions for which the ga\ behaves ideally. The quantity ( a F / i ) ~can ) , be determined from the equation of state uf the actual gas. We now calculate the change in entropy as we increase the pressure back to one har, hul rrrking rhr g a . ~to hc idr,rzl. We me Equation 8.52 for thia prtxess, hul will) ( i l V / i ) ~ )= , K I P , giving
= 0.192cm3.bar.mol~' .K-'
which is what was used in Table 7.1 :The correc~ionin this case i s rather mall, but that is not always so. If second virial coefiicent data are nut available, then an approximate equation of state van be used (Problems 8-20 through 8-22). The superscript o of S'(1 bar) emphasizes that this is the standard molar entropy of the gas. We add Equations 8.52 and 8.53 to get
S m ( a 1t hu)- S(at I bar) =
(8.54)
In Erluation 8.54, 3 is the molar entropy we calculate from heat-capacity d a ~ aand heats of transitions (Section 7-3). and S" is the molar entropy of the corresponding hypothetical ideal gas at one bar. Equation 8.54 tells us thal we can calculatc the necessary c o m c ~ i u nto obtain the standard entropy if we kntw the equation of state. Because the pressures involved are around one bar. we can use the virial expansicln using only he second virial cwfficient. 2.22, IJsing Eqi~;~tic~n I
8-7. The Gihbs-Helmholtz Equation Describes the Tcrnperature Dependence of the Cibbs Energy Both of Equations 8.31 are useful because they tell us how [he Gibhs energy varies with pressure and with temperature. Let's look at Equation 8.31b first. We can use Equation H.3 1 b to calculate the pressure dependence of the Gihb~energy:
aG =
L:
VdP
For one mole of an ideal gas, we have
(constant 1.)
(8.57)
Chapter 8 1 Helmholtz ~ n Cd i b b ~Energies
We could have obtained the xmrle result hy using
G ( T , P) = G'(T)
+ HI' In( P / 1 bar)
The Git)t>s-tlelmholtz Equatiun Describes the Temper;lrure Depcnder~ceOIthe G~bbsEncrgv
Now differentiate partially with respect to T keeping P fixed:
For an ixolhennal change in an idcal gas, AH= 0 and AS is given by Equation 8.2 1. It is custo~naryto ICI PI = 1 bar (exactly) in Fquation 8.58 and to write it in the fonn -
8-7
(8.59)
wlicrc F (T) is called the standard molar Ciibbs energy. The standard molar Gibbs cllergq in this case is the Gibbs energy of one mole of the idea! gas at a pressure uf nrle bar. Note that G ' ( 1 ) depends upon only the temperature. Equation 8.59 gives the Gibbs energy c~fa n ideal gax relative to the standard Gibbs energy. Accr~dingto tiquation 8.59, E(T, P) - G o ( T )tncreaws logarithmically with P, which we have see11 is cntirely an entropic effect for an ideal gas (because H is independent of P for an ideal gas). We will see In Chapter 12 that Equation 8.59 plays a central role in
These last two terms cancel because o f the rdation (aS/d l'), = U , ( T ) / T (Equation 7.7),so we have
Equation 8.60 is called the Gibbs-Helmholiz eqruttion. This equation can be directly applied to any process, in which case it hecomes
cl~en~icnl equilrbr~ainbolving gas-phase reactions.
E X A M P L E 8-9 Solids and l~quidsare fdirly incompressible, so V in Equation 8.57 may be taken to be constan1 lo a good apprvritllntion in this case. Derive an expression fur G(T,P) analugous to Equation 8.59 for a solid or a liquid. S O L LIT IO N : We inlegrate
Equalion 8.57 at constant T to get
We let P, = I bar and E(I',= I bar. T ) = G ' ( T ) to get
This equalion is simply another form of the Gibbs-Helmholrz equaLion. We will use Equatioris 8.60 and 8.61 a number of times in the following chapters. Fur example. Equation 8.61 is used in Chapter 12 t o derive an equation for the temperature dependcnce o f an equilibrium constant. We can dctcrrnine the Gibbs energy as a function uf temperature directly from equations wc dcrived in Chapters 5 and 7. In Chapter 5, we learned how to calculate the e~tthalpyof a substance as a function of temperature in terms of its heat capacity and its varlous heats of transition. For example, if there exists only one solid phase, so that there are no solid-solid phase transitions between T = 0 K and its melting point, then (Equation 5.46)
where P must be expressed ill b m . In this ca@,e,Z ( T , P ) increases lineilrly wlth P,but k c a u l e the volumc of a condcnscd phaseis much <mailerthan that of a giu, the slope of C ( T .P) versuq P ,or ( ~ G /PB ) ] = V , is very small Consequently, at ordinary pressures G ( T ,P ) if almost Independent of pressure and is apprr)xirnatcly equal to I C ( T ) for a condensed phaw.
Equation 8.3 la determines the temperature dependence uf the Gibbs energy. We can derive a useful equation for the temperature dependence of G by starting with Fqualion 8.31r1 (Problcm K24), but an easier way is to start with G = H - TS and divide by Z'to obtain
for a temperature above the boiling point, where Figure 5.7 shows HIT) - H(0) versus T for benzene. We calculate H(T)relative to H ( 0 ) because it is not possible to caIculate an absolute enthalpy; H ( 0 ) i s essentially our zero of energy.
Chapter 8 1 Helmhultz and LiWs Energies
I n Chapter 7, we learned to calculate absolute entropies according tu Equation 7.17).
8-8. Fugacity Is a Measure of the Nonideal~tyof a tias
325
also shows that there is a discontinuity in the slope at each phaw transition. (Ben/.ene melts at 278.7 K and boils at 353.2 K ai me atm.) We can understand why there is a discontinuity ill the slopc of G ( T ) versus T at each phase transition by looki~lgat Equation 8 . 3 1 ~
Because entropy i s an intrinsically pnsitive quantity, the slope of C (T)versus T i s negative. Furthermore, because S(gas) > S(1iquid) > S(solid),the slopes within each single phase regionincrease in going from solid to liquid togas, s o the slope,{i)G/ilT) is discontinuous in passing from one phaae to another. Thc values of H " ( T ) - H2(O), S 3 ( T ) , and G'(T 1 - H " ( 0 ) arc tabulated for a variety of substances. We will use these values to calculate equilibrium constanls it1 Chapter 12.
,,
Figurc 7.2 shuws T ( T ) versus T for benxne. We van use Equations 8.62 and 8.63 to because calculate E ( T )-
n(0)
Figure 8.7 shows E ( T )- H(O) versus T for hcnzene. There are several features of Figure 8.7 to appreciate. First note that G ( T )- H ( 0 ) decreases with increasing T. Furthermore, E ( T )- H(0) i s a continuous function of temperalure, even at a phase transition. To see that this is so, consider the equation (Equation 7.16)
&8. Fugacity Is a Measure of the Nonideality of a Gas
I n the previous section, we showed that the molar Cibbs encrgy ofan ideal gab is givctl by Because A,,G = AmH - TbAvSS,wc see that L I , ~ $=G0, indicating that the two phases are in equilibrium with each other. Two phases in equilibrium with each other have the same value of G, so G ( 1 ) is continuous at a phase trtnsitiun. Figure 8.7
-
G ( T , P) = G 3 ( T )4- R T l n
Y , P
(8.W
The pressurc P" is equal to one bar and C"(T) is called the standard molar Gihhs energy. Recall that this equation is derived by starting with
and then integrating, using the ideal gas expression, R T I P , fur 7. Let's now generiilizc Equation 8.64 to the case of a real gas. We could start with the virial expansion,
TI K I
F I G U R E 8.7
A plot uf C ( T )- H(0) versus T for benzene. Note that G ( T )- H{o) is cuntinuous but its derivative (the s l o of ~ the curve) is discontinuous at the phase transitions.
and substitute this intu Equation 8.65 to obtain a virial expansion for the molar Ejihhs energy,
Chnlser A I Hclrnl~olr~ 2nd Gibbs Enrrgies
where we are intepnting from some low pressure, say Pid,where the gas is sure to behave ideally, to somc arbitrary pressure P. The result of the integration is
Now according to Equation 8.64, G(T,PId) = GV(T)+ R T 1n p L d / P 3where , GJ(T) is the tnolar Gibbs energy of an ideal gas at a pressure of P" = I bar. Therefure, Equation 8.66 can be writtcn as
Equa~irln8.67 is the gencralizatiot~o f Equalion 8.64 to any real gas. Although r gas, depending upon the vali~esof B,,(T), Equation 8.h7 is exact, it diirers f r ~ each B,,!T), and so011.It turns out to he much morc convenient, particularly for calculations involving chemical equilibria, as we will see in Chapter 12, t c ~maintain the form of Equation 8.64 by defining a new thermodynamic function, f (P, T). calledfugnciv, by the eqi~ation
8-8. rugnrily Is a Mcasure of the Nn~jide~litv uf a L a 5
suggested by Equation 8.69, because vtherwise f ( P , T ) would not reduce to P ahen B z p ( T )= B,,{T) = 0. This choice of standard state not only allows all gasef to be brought to a single common state, but also leads to a procedure to calculate f (P, T ) at any pressure and temwraturc. To do so, consider the schcme i n Figure 8.8, which depicts the dilferencc in molar Gibbs etiergy between a real gas at P and T and an ideal gas a1 P and T. We can calculate this difference by starting with he real gas at P and T and thcn calculating the change in Gibbs energy when he pressure is reduced to essentially zero (step 2), where the gas is certain to behave ideally. Then w e calculate the change in Gibbs etiergy as we compress the gas back to pressure P, but taking the gtas to hehave ideally (step 3). The sum of steps 2 and 3, then, will bc the difference in Gibbs cncrgy of an ideal gas at P and T and the real gas at P and T (step 1). In an equation, we have
Substituting Equations 8.64 and 8.68 into Equation 8.711, we hwe
But the standard state of the real gas has been chosen yuch that The nonideality is huried in f (P, T). Because all gases behave ideally as P -+ 0, f~ig;~t-ity must have the property that
= P" = I bar, so
We now use Equation 8.65 to calculate the change in the Gihhs energy along steps 2 md 3:
sn that Equatinn 8.68 reduces to Equation 8.64. Equations X.67 atid 8.68 are equivalent if
11 might sccm at this point that we are just going in circles, but by incorporating the nonirleality of a gas thruugh i t s fugacity, we can preserve the thermudynamic equations we have derived for ideal gases and write those corresponding to a real gas by simply rcplauing PIP' by f/SC. All we need at this stage is a suaightfurward way to determine the fugacity of a gas at any pressure and temperature. Before looking into this, however, we must discuss the choice of the standard state in Equation 8.68. Being a type of energy, the Gibbs energy must always be taken relative to some chosen
s t ~ i ~ d a rstate. d Nc~lethat the standard p o l a r Gibbs energy G - ( T )is taken to be the same quantity in Equations 8.64 and 8.hS. The standard state in Equation 8.64 is the ideal gas at one bar. so this must be the standard state in Equation 8.68 as well. Thus, the standard state of the real gas in Equation 8.68 is taken to be the corresponding ideal gas at une bar: ill other words, the standard state of the real gas is one bar after it has been adjusted t c ~idcal lxhavior. In an equation, we have that f " = Po. Note that this choice is also
Real gas
A El
Ideal gas (T.P)
(T+P)
Real gas (T.P+O)
-
Ideal gas (T,P-+O)
F I G U R E 11.8
An ilIustraiion of the scheme used to relate the fugacitj of a gas to its standard state, which is a (hypothetical) ideal gas at P = I bar and the temperature T of interest.
Chapter 8 1 tielmholtz and Libbs Energies
328 The sutn of
AZ,and AC;, gives anolher expression for AE,
Equaling this expression for AZ, to
AG,in Equation 8.71, we have
0
200
400
600 P /bar
SO0
1000
F I G U R E 8.9 A plot of ( 2 - I ) / P versus P for CO(g) at 2M) K. The area under this curve from P = 0 to P gives In y at the pressure P.
Given either P-V-T data or the equation of state of the real gas, Equation 8.72 allows 11s t ~ ci~l~uliltt: l the ratio of the fugacity t c ~the pressure of a gas at ilny pressure and temperature. Note that if the gas behaves idcally under the conditions of interest (in od~crwurds, if 7 = qd in Equation 8.72). then In f./ P = 0, or 1 =; P. Therefore, the extent of the deviation of j/P Cmm unity is a direct indication of the extent uf the deviation of the gas from ideal behavior. The ratio f / P is called the ,fugaciv
uutficirtlt, y ,
For an ideal gas, y = 1 . By introducing the compressibility factor, Z = P I / / H T , Equation 8.72 can be written as
F I G U R E 8.10
A plot of y = j / P against P for CO(g) at 200 K. Thew values ot f / P were obtained frurn a numerical integration of (Z - 1)/P shown in Figure 8.9.
Even though thc luwer limit here is P = 0, the integrand i s finite (Problem 8-27). Furthcrmoir, (Z - I ) / P = 0 for an ideal gas (Problem 8-27), and hence In y = 0 and f = P. Figt~re8.9 shows ( Z - 1 )/Pplotted ag;unsl P for CO(g) at 200 K. According lo Equation 8.74, the area under this curve from 0 to P ih equal to In y at the pressure P. Figure 8.10 shows the rzsuldnp values oF y = f / P plottcd againat the pressure for UO(g) at 200 K. We can also calculate th; fi~gacilyif w e know the cquatiun ul' stale of the gas.
1
E X A M P L E 8-10
Derive an expression fur the fugacity of a gas that obcys tbe equation o l
where b is a wnstant.
ht;lte
Chapter A I Helmhultz and Gihbs Energier
SO 1U TI O N : We solve for 7 and substitute into Equation 8.72 to get
Problems 8-33 through 8-38 derive expressions for In y Tor the van der Waalf equation and the Rcdlich-Kwong equation.
We can write Equation 8.74 in a form that shows that the f i ~ g a c i t ycoefficient is a function ( I € the reduced pressure and the reduced temperature. If we change the integration variable to P, = PIP,, where PL is the ciitical pressure of the gas, then Equation 8.74 takes the form
Now recall from Chapter 2 that, to a good approximation for must gases, the compressibility factvr Z i s a universal function of P, and T, (see Figure 2.10). 'rilerefore, the righ~side of Equation 8.75, and s o In y itself, is also a universal function of P, and T,. Figure 8.11 shows the experimental values of y for many gases as a family of curves ol'cunstnnt T, plotted against P,.
E X A M P L E &I1 Use F i ~ u r e8.1 1 and Table 2.5 to estimate the fugacity of nitrogen a1 ti23
K and FIGURE
S O L U T I O N : We find from Table 2.5 that Tc = 126.2 K and q = 33.6 atm for N,(g). Therefore 7, = 4.94 at 623 K and P, = 29.8 at 1 0 0 am. Reading from the curve!. ir! Figurc 8.1 1. we find that y w 1.7. At 1030 ahn and 623 K, the fugncity of nitrogen is 1700 atm.
8.11
The fugacity coefticientsof gases plotted against the Mduoed pressure. P/PL,Tor variuus values of the reduced remprature, T / Tc. I
and 82.4 J.K-'.tnol-', r e s p e ~ ~ v e l yCompare . your results with those you ohtained in Problem 8-1. Are any of your physciaI interpretatio~~s differem'!
Problems 8-1. The molar enihalpy of vapurization of benfxne at its normal builing point (80.09"C) iu 30.72 kl.rnul '. ,4ssu&ing that AmpH and A - ~ Sstay constant at their values at 80.09C, calculate the value of AvapG a1 7S.O'C. 80.09"C,and 85.UC. Interpret these I-esultcphysically. -
8-2. R d o Problem 8-1 without assuming that Avap H and dvwSdo tlot vary with temperature. Take the molar heal capacities of liquid and gaseous benwne to be 136.3 J-K-' -mul-'
8-3. Substitute ( a P / a T ) , from the van der WaaIs equation into Equation 8.19 and integrate from TIdto V to obtain
---
.Y(T> v)- F ( T=) R 1" V - b
-
-
V
-h
Chapter 8 I Hclrnholtr and C i l h Energies -ad
NOW
let V' = R T / P l d rpld= P = one bar, and V
>> b tu ubtain
Given that u = 327.7 pm, s l k , = 95.2 K, and h = 1.58 for N,(g), calculate the value of AUfur a pressure increase from 1.00 bar to 10.0bar at 300 K.
8-8. Deternine C, Gwen that
5''
= 246.35 J.rnol-'
.K ' for ethane at 400 K, show rhat
-
P(V- h ) = R T .
'
8-10. Use Equatiun 8.22 tu \how that
Show that (aC,/a
fur the Redlich-Kwong equation for ethane at 400 K.Calculatc 3 as a fi~nctivnof p = 1 / F and compare yc)ur results with the experin~cntalresults shown in Figure 8.2.
V), = 0 for an ideal gas and a van der Waals gas, and that
for a Redlich-Kwong gas.
8-1I. In this prohlem you will derive the equation (Equation 8.24)
8-4. Use the van der Waals equation to derive -
for a gas that obeys the equation uf s t i l k
8-9. The cwficient of thermal expansion af water at 25°C is 2.572 x K I, ,and its isulherma1 compressibility is 4.525 x 10 bar-'. Calculate the value of C', - C, fur one n ~ v l e or water at 25'C. The density of water at 25'C i s 0.997I15 g . n l ~ - ] .
Calculate 5 as a functiun of p = l/V fur ethane at 400 K and compare your results with ihe experimental results shown in Figure 8.2. Show that
-
C,
-id
U ( T . 1')- U ( T ) = -2
V
Use this result along with the van dcr Waals equaticln to calculate the value rlf U as a function of 7 for ethane at 4M)K, given that = 14.55 kJ.~nol"'.To rlo this, s p ~ c i f y7 (from 0.0700 ~ - m o l - to ] 7.00 L.rno1-I, see Figure 8.2), calculate both U(V)and and p l o ~U(V)versus ~ ( 7Compare ). your result with the exprri~ncntald i ~ t in ; ~Figure 8.3. Use the Redlich-Kwvng cqun~ionto derive
vid
~(v),
Repcat the above calculation for ethane a1 4(M)K.
8-5. Show that ( B U / d V),. = 0 for a gas that obeys an quation of state of the form Pf ( V )=
To start, consider V to be a function of T and f and wrile out d V . Now divide thrclugh by dT at constant volume (dY = 0) and then substitute the expression for ( a P / d l ' ) , that you obtain into Equation 8.23 trl get the above expression. 8-12. The quantlty ( W / d V ) , has units o f pressure atid is called the rr~ten~ul pre.rsure, wtirttl i s a measure of the intermolecular forces within thc M y of a substance. It i s q u a i to zcro For an ideal gas, is nonzero but relatively small for dense gases. and i s relalively large tor liquids, particularly those whose molecular interactions are strong. Use the frhlowing tlutn tocalculate the internal pressure ofethane as a functiun uf preswt: at ?NO K. Compare yhlur values with I t ~ evalucs you obtdn frum the van der Waals equation and the Redl~ch-Kwor~g equation.
RT. Give two examples of such equatiuns (11state that appear in the text.
8-6. Shnw that
8-7. IJse the result of the previvus problem tu show Lhnt 8-13. Show that
-
Use Equation 2.41 for the square-well potential tu show that
Chapter 8 1 Helmholtz and tiibbs Energies
USCEquation 2.41 for the square-well potential to obtain
8-19. What a l t the natural variables of the entropy?
8-20. Experimentally determined cnwopies me commwly adjufted for nonideality by using iln equation of state called the (niudiiicd) Berthclot equation: Given t h a ~o = 327.7 pm, e / k , = 95.2 K, and h = 1.58 for N,(.g), calculate the value of ( a H / aP), at 300 K.Evaluate AH = H(P= 10.0 bar) - X ( P= I .O bar). Compare your rcsult with X.724 J.rnu1-'. the value nf Z ( T )- H(o) for nitrogen a1 30() K.
8-14. Show that the enthnlpy is a function of only the temperature for a gas that obeys the cqu;ttiun orstotc P(V -- h T ) = R T , whcre b is aconstant. 8-15. Use your results Tor the van der Waals equation and the Redlich-Kwong equation in R o b l e ~ n8 4 to calculate H(T,V ) as a function of volumc for ethane at 400 K. In each casc, use the equation Ti = + PV. Compare your results with the experimental data shvwn in Figure 8.5.
Show that t h i ~cquatior~leads to the corrcftinn -
S ( a t one bar) = S(at one bar)
27 RT"
+ -+(I
32 %T
bar)
This result n c c d ~only the critical data for the substance. Use this equalion dung with the critical data in Table 2.5 to calculale the nonideality correction for N,(g) at 29H.15 K . Cntr~pareyour result with the value used in Table 7.1.
8-21. Use thc resuIt of Problem 8-20 along with the critical data in Table 2.5 to determine the
0-16. Use Equatio118.34 tu
nonideality correction for CO(g) at its normal boiling point, 8 1 .h K. Compare your ~,crult with tI~cvalue used in Problem 7-24. 8-22. 11%the result of Problern 8-20 along with the critical data in Table 2.5 to determine the nonideality correction for Cl,(g) at its normal boiling point, 239 K. Conlpare your r r d t with lhe value used in Problem 7-16.
Use a virid expansion in P to show that
8-23. Derive the equation
U Use the square-wcll second virial coefficient (Equation 2.41) and the p m m e t e r s given in Problc~n8-13 to calculate thc value of (aC,,/a P), for N,(g) at O'C. Now calculate F p at I(Wl iktln and 0. C, using C: = 5 K / 2 . 8-17. Show that thc molar anthalpy uf a substance at pressure P relative to its value at one bar is given hy -
H ( T . P ) = H(T,P = I bar)
I.
T2
which is a Gibbs-Helmholtz equation for A .
8-24. We can derive thc Gibhs-Helmholtz equation directly frum Equation 8.3 la in the fullowing way. Start with {nG/aT), = - S and suhstitue Fur S from G = H - TS to obtain
+ Now
Calculate the value of H ( T , P) - = ( T , P = I bar) at WC and 100 bar for mercury given that thc molar vvlumc of mercury varies with temperature according to
where is the Cclsius temperature. Assume that v(0) does nM vary with pressure over this
-
rnnge and exprcss your answcr in unit.: ni kJ.mol-I. 8-1 8. Show thal
dH=
[
V-T
(K)J -
dP+C,dT
W h a ~docs t h i ~etlualitln tell you a h u t thc natural variables of H?
8-25. Use the following data for henrene to plot Z ( T )- v(0) vekus T. Iln this case we will ignore the (usually snlall) comctiuns due to nonideality vT the gas phase.]
Chapter 8 / llelmholtr and Grbbs Energies
336
Compare your result with the result you obtained in Problen~8-30. 8-26. Usz the following data for propene to plot Z(T) - E(0)versus T. [In this case we will ignore the (usually small) comctions due to nonideality of the gas phase.]
8-32. Use the following data for N,(g) at O'C to plot the fugwity cuetlicient as a function of pressure.
8-33. It might appiar that we can't use Equation 8.72 to determine the fugacity of a va11 dcr Waals gas because the van der Waals equation is a cubic equation in so we can't solve it analytically for V to carry out the integration in Equation 8-72. We can get around this
v,
problem, however, by integrating Equation 8.72 by parts. First show that
8-27. Use a virial expansion for Z to prove (a) that the integrand in Equation 8.74 i s finite as P + 0, and (b) that (2 - I ) / P = 0 for an ideal gas.
where P ' ~-+ 0, 'i7" -+ m, and P'V" + K T . Substitute P from the van der Waals quation into the first term and the integral on the right side of the above equation and integrate to obtain
KI'V a
8-28. Derive a virinl expansion in Lhc pressure for In y . 8-29. The compressibili~yfactor for ethane at 600 K can Ix fit tu the expression
for 0 5 P/bar 5 Wl.Use this expression to determine lhc fugacity coefficient of ethane as a function of pressure at MH) K. I
8-30. Use Figure 8.1 1 and the data in 'lhblc 2.5 to estimate the fugacity of ethane at 360 K and IMKl atm.
8-31. Use thc following data for ethane at 3M) K to plol thc fugaci~ycocfficient against pressurc.
RTIny==----V-b Now use the fact that
+
V
V-6
a
RT-RTIn7-=-KTIn, V -b V
P
P
and that P ' v ' ~ = RT lo hhuw Lhirl
This equatlon gives the fugacity coefficient of a van der Waals gas as a function of V .Yuu can use the van der Waals equation itfelf to calculate P frum 7,so the above equation, in conjunction with the van der Waals equation, gives In y as a function of pressure.
8-34. Use the fi11a1equation in Prtlblern 8-33 along w ~ t hthe van der Wads equntmn to plot In y against pressure for CO(g)a! 200 K. Compare your result w1111Figurc X.10.
Chapter l( / Iielrnhol17and Gibhs Etiergies 8-35. Shnu- thal the expression for In y for the van der Waals cquation (Problem %33) can be writlrn i n the reduced fwm
Problems
Show [hat
because the entropy of each cumpartmen1 can change only n?a result of a change in energy. Nuw show thal
Use this equntioti nlotig with the van der Waals equation in reduced form (Equation 2.19) tn p l o ~y again$( P, lor TR = 1.(lo and 2 . M and compare your results with Figure 8.1 1.
8-36. Use the method outlined in k n h l e m 8-33 tu show that
Use this rcsult to discuss the direction nt the Ilow of energy as heat from one temperature to anothcr. 8-42. Modify the argument in hobien~&4 1 to the case in which the two compartments m separated by a nonrigid, insulating wall. Derive the rcsult
for the Kedlich-Kwong equation. You need to use the standard integral
Use this result lo discucs the direction of a volume change under an isothermal pressure difference. - - - -
8-43. In this problem, we will derive virinl expansions for U . H, S, A , and c. Substitute 8-37. Show that In y for the Redlicb-Kwnng equation (see Problem 8-36) can be written in the reduced form
Z = l+B,,Pf
R,,P'+...
into Equation 8.65 and integrate from a small pressure, Pd,to P to obtain
Z ( T ,P ) - C(T,P'"
= RT Tnl
, + P P
RTB3P +2
~2
RTB,,P
+
Now use Equation 8.64 (realiae that P = P id in Equation 8.64) Lu get 8-38. 1Jxe lhe expression fur In y in reduced furni given in Problem 8-37 along with the RedlichKwong equation in reduced form (Example 2-7) to plot In y versus P, for T, = 1 -00 and 2.01) and compare ycmr rt.sulls with those ynu obtained in Problem 8-35 for the van der Wxals cquation. 8-39. Cumpare In y frlr the van der Waals equation (Problem 8-33) wit11 the values o f In y Fur ctlianc at h(Kl K (Prnhlcm 8-29).
at P' = 1 bar. Nuw use Equatior. 8.3 la to get
-S(7'. 1') - S ' ( T )
= - R l r ~P -
at PP'= 1 har. Now use 5 = H
-
d(RTH?,)
p-
dT
) -1-d ( R T B J p p2 2
dT
+
,, ,
(2)
TT trl get
8-40. Compare In y for the Kedlich-Kwong cquatim (Problem 8-36) with the values of 111y for ethanc st 603 K (Problem 8-29).
8-41. We can use the equation ( a S / a U ) , = I / T to illustrate the consequence of thc fact that entropy always increases durirlg at1 irreversible diabatiu prmess. Consider a twocompartrllent system enclowd by rigid adiabatic walls. and let the two compartments he scpmatcrl by LI rigid heal-conducting wall. We assume that each compartment is at equilibrium but that they are not in cquilihrium with each other. Becausc no work can k done by this twu-compartment system (rigid walls) and no energy ar; heat can be exchanged with the surroundings (adiabatic walls),
U =U,
+ 11, = constant
Now use the k t that
C , = (az/a~), to get
We can obtain expansions for and by using the equation andG=;i$~~=~+~~T,Shwthilt
H = V + PV = + R T Z
Ch.~pter U / blelrnholtz and G~hbsEnerg16
and
at
and is a direct measure of the expecled temperature change when a gas i s expanded through a throttle. H1ecan use one of the equations derived in thi* chapter to obtain a convenlcrlt working equation fw JL,,. Shuw hat
P' = 1 bar.
8-44. In this problem, we will derive the equation
Use this result to show that pIT= 0 for an ideal gas
8-49. Use the virial equation of stare of the form
v'
where is a w r y large ilr~olur)volumc, where the gas is sure to hehave ideally. Start with d H = TdS + Vd P 10 derive
and use one of the Maxwell relations for (aS/a V ) , tu obtain
to show that
It so happens lhat B,, is negative and dB,, /dT is pusitivc for T' < 3.5 (see Figure 2.15) so that p,, is p s i t i v e for low temperatures. Therefore. lhe gas will cwi upon expfllsiun under these conditions. (Sce Pmblem 8-48.) 8-50. S ~ O that W h
P,l
Now integrate by parts from an ikal-gas l~rnitto an arbitrary limit io obtairl tile desired equation.
CP
for a gas that obeys ihe equation of state
8-45. Ilsinp the result of kublern 8 4 , show that H i~ independent of volume fur an ideal gas. What about a gaq whcse equation of state is P(V- b ) = RT? Does V depend upon volume for this equation of state? Account fur any differeilce. 8-46. Using the resull ol~Problem8 4 4 . show ihnt
RTb !I-If'=--V-b
-
= --
P(T- b ) = R T . (SeeProblem 8 4 8 ) .
8-51. The second virial coefficient for a square-well potential is (Equation 2.4 1)
Show that 20
V
(or the van der Wxals equmion.
8-47. Using the result of Problem 8-44, show that
where b, = 2 n p 3 ~ * / 3Given . the following square-well parameters, calculate , I , , at 0' C and compare your values with the given experimenlill rdues. Takc C, = 5 R / 2 fur Ar and 7R/2for N, and CO,.
for the Kedlich-Kwung quatiun. 8-40. We introduced rhe Joule-Thornson effect and the Joule-Thomson coefficient in ProhIcms 5-52 through 5-54,The Joule-Thomhon coefficient is defined by
8-52. The temperature at which the Joule-Thomhon coefficient changcs sign is called the Joule-Thumsnn inversion trmperatuw, T .'The low-pressure Joule-Thonlsun irlvcrsion tcinperalurc for tk square-well potential is obtuined by setling p , , = I3 in Ruhlern 8-5 I . This procedure leads to an equation for k , T j ~in terms of >."hat cann<>tbe solved snalytically. Solve the equation numerically to calculate T for the three gases given in the previouh
Chaptcr 8 / Helmholtz and Gibbs Energies problcm. Thc cxpcrilnental values are 794 rccpectivcly.
K,621 K, and 1500 K for Ar, N,, and CO,,
8-53. Ure the rltlta in Prr~hlem8-51 lo estimate the temperature drop when each of the gases l l n d e l ~ iin ~ eexp:~~i\ic>n ~ for I(H) a m 10 nnt. alm.
8-54. When a rubber band is slretched, it exerls a reslrlring force, f ,which i s a function of its length I . and its tempzriiture T . The work irivolved is given b y
Why is there no negative sign in front of thc integral, a% there is in Eqllation 5.2 for F - V wt~rk?Given ttrat the volun~echange upon ctretching a rubber band i s negligible, show that
Problems
8-55. Derive an expression for AS ror the reversible, isothermal change of one mole of a gas that obcys van der Waals equation. Use your resl~ltto calculate A S for the isothcr~nal compression of one mole of ethane from 10.0d1n' to I.Mldm3 nt 400 K. Catnpare your result to what yuu would get using the irleal-gas cquation.
8-56. Derivc an expre$sio~lfor A S Ibr the reversible, isothermal change of [me mole of a gas that obeys the Redlich-Kwung cquation (Equation 2.7). UP^ your result to colculale A S for the isotbennal co~npressionof one mole of ethane from 10.0dm' tu I Wdm' at 400 K. Cumparc your result with the result you a v u l d get ucing the ideal-gas equation. In thr lu.~trighr problems, we explnra the con.fequenr.Psof IAP far# f / ~ a~qullibn'um t zlote~ ure srable; tl~u? is, if a s y t e m at equilibrium Bperturhdin some way, it will s~mntanrously return f o its equilibrium state. The .smbility of cquilibvim stares buds lo fl namhtrof jien~rul inequalities c a k d stability ~wnditions.Exmq)les uf stubility rondifinns [ha! wc derivt, in the,fi)clllnwingpm)b/ems are (a P/a V), -; 0, U , > 0, C , > C , , and ( a S j d T ), > 0.
rlnd that
Uhi11~ [he definition A = U
and derive
-
8-57. Equi~ron8.2 says that dU 5 0 lor any spontaneous prwesc that occurs *t constant S and V. Thus, the energy u'i1l always decrease as a recult of a sponlaneous prwess and will attain a n~inilnurnvalue at equilibriutn. Consider now o two-canip:~rtrnentsy $ternlike that shown in the ligure below.
TS, show that Equation 2 becomes
the Maxwell relalion
Suhslitute E!.quation 5 into Equation 3 to obtain the analog elf Equatirm 8.22
For many elastic system$, the observed temperature-dependence of the force is linear, We define en idenl rubber hand hy
f = T4M.j
(ideal rubber hand)
(6)
Stuow hut ( R U / aI , ) , = Ofor an ideal r u b k r hand.Comparethisresult with ( a U / a V ) , = 0
fur an idenl gas.
Now Ict's considcr what happens when we stretch a rubber band quickly (and, hence, ndiahnticnlly). In this case, dV = d111= fdL. Use the fact that U depends upon only the ternperature for an ideal r u b k r hand to show that dC' = The quantity
(a.U/a?j,
("a), -
dT
=
fdL
The entirc system i s cnclosed by a rigid, adiabatic wall (so that it is isulated). Thc wall thal sepiuateq the system into two cumpanlnents is tutatly restrictive (rigid, nonheat-conducting. and nunpermeable). The two identical cornpartmen& are at equilibrium with U = U I S . 1') at constant N . Now let's partition the volume unequally hetween the two compartments (by moving the wall slightly), so that one cumpartment has a volume V AV and the oihcr has a volume V - A V . Argue that
+
u(s,V + A V ) + U ( S , V - A V ) Expand both U ( S , V f A V ) and U ( S . V A V to obtain
- AV)
2 U ( S . v)
(1)
in a Taylor series (Mathchapter C) in
is a heat capacity, so Equation 7 become<
Argue now that i f a rubber band is suddenly stretched, then its temperature will rise. VeriCy this result by holding a ruhhzr hand against your upper lip and strelching it quickly.
Equation 2 is one of several conditions for an equilibrium statc to be slahlc with respect to small changes in its extensive variables.
Chapter 8 /
Helmhollz .lnd Gihhs Energies
Starting with d U = T d S - P d V , show that Equation 2 implies that the so-called adiabatic compressibility, K,, must obey the inequality
Put this a11 logcther to show that
or that K~ must be a positivequantity.
Thus, we see once again (see Prublern 8-57) thut
8-58. In the prcvious problem, we showed that one of the stability conditions of an isoiated system i s that ( i l ' ~ / a ~ ~>) 0. , We did this by repartitioning the volume between the two cvmpartrnents. Now consider the effect of repartioning the entropy between the lwo compartments keeping their volume^ equal. Show that
Now use the equation dU = T d S
-
PdV to show that this inequality implies that
Nulice that we determine the sign of the second dcrivative of H with respect to the exlenslvc variable, S, by using the repartiuning that we used in Problem 8-57 irml the sign uf he second derivative with respect to the intensive variable, P, by first determining (3H/11P j s , differentiating it, and then elating its reciprrcal to (a'Ujav2),, whose slgn w e already knew.
8-60. Of the twu natural variables of the Helrnholu. energy (at consrant N),one is an exlznsibc variable, V , and the other is an intensive variable, T. Using the repartioning approach shown in hublem 8-57, show that or that C, > 0.
8-59. The enrhalpy, H, (at constant N ) has natural variables Sand P, one an extensive variable ( S ) and the other an intensive variable ( P ) .To determine the sign of ( i 3 2 ~ / a ~ 2 we ) , , use the repartioning approach that we described in Problem 8-57. Show that
and that
Now show that the ist~hcrrnalcvmpressibiliry, K,, i s always positive, or that Now show that this inequality implies that
or that C, > 0. The pressure is an intensive variable, so we are unable to repartion it. Tu determine the sign of we first determine ( 3H / a P ) , . Show that
The Helmhult~encrgy is also a function ul- cine it~tcnsivenilural vwiablc, 7'. S l h ~ that (see problem 8-59)
(a'~/a~~),,
Thus, we see thal the cntropy increases monotunically with temperalure at constant volun~e. N o w differentiate once again with respecl lo P keeping S constant.
8-61. The two natural variables of the Gihbs energy Cat constant T and P. First show tht
n: ) are both intcnsivc v;jrirlhles.
I
Now show that the reciprocal of (a l / / a P ) , is given by
Thus, we see that theentropy increases monotunically withtc~nperalureatconstantprrhhurc.
Chapter A I Helmhottz and Gibh Energies
8-64.-Equation 8.45 gives 3 for a monatomic ideal gas in terms of T and V . Use the fact that U = 3 R T / 2 to show rhat'~quation8.45 can be xritten as
Now show that
id 7.Show that where u is a constant independent of mn Equmions 3 and 4 of Problem 8 4 3 . Show also that
8-62. Use the resulk u T the previous problem and Equation R.27 to provc that C, > U,. 8-63. In Pi-oblcnls 8-57 and K-58, we rcpartioncd the volume of each cumparlment wilhoul chunging the cntropy of ench compartment, or we repartion4 the entropy without changing the volume. Let's now look at the case in which hoth V and 5 ' change. Show that in this CilSe.
U ( S + h S , V $ A V ) + U ( S - A S , V-AV) >2U(S,V)
(1)
1.l1.c ' l j y l o ~ . ~ u p a ~ ~of s i oa nfunction of two variables is
f(.~t.Ax.~-+Ay) = f ( ~ . yt. )
+ cuhic terms in Ax, Ay, and their products Using this result, show that Equalion I hecomes
Equatiun 2 is valid for any values of A S and A V . Let A V = 0 and AS = 0 in turn to show that
Now let AS = to obtain
(a2u/asav)
azu/asz),
and
AV=l
I
It lunlq out that Equations 3 and 4 are the malhematical conditions that the extremum given by dU = O is actually a n~inirnumrather than a maximum or a siuldle point.
given by Equation 5 salisfic~
Phase Equilibria
Josiah Willard Gibbs was burn in New Haven, Cunnecticut, on February I I, 1839, and died there in 190.7. He received his Ph.D, in engineering from Yale Univeraiiy in 1863, the second doctorare in science and the first in engineering awarded in the Unitcd States. He staycd on at Yale, for yeal-s without salary, and rclnaincd there for the rest of his life. In 1878, Gihbs published a lung, uriginal treatise on therlnodynamics titled "On the F~ptilihriurnof Ht.~en)geneuusSuhstanccs" ill the Tmnsnrrions rfihr Coirner.tir,uiAcademy r,SSc'ie,rtas. In addition to introducing the concept of chemicnl potential, Gibbs introduced what is now called Ihe Gibbs phae rulc, which relates the number ol~components(C) and the nulnbcr o f phases ( I ' ) in a cyclem t r l lhe nurnher of degrees of freedom ( f - ,the numhel- o f wrinhles such as ttmpsrarure and pressul-e that can bc varied il~depelldenrly)by the equation I.' = C' + 2 - F. Bctwccn its austcre writing style and lhe ohhcurity of Ihe journal In which it waa published, huwcver, this importan1 work was not as widely appreciated as it deservd. Fortunately. Gibhs sent copies to a number of prornlnent European scirnticts. Maxwcll s ~ varl ~ ddcr Waals inrmcdiately apprccintod the sig~ifilnce(IF ihe wcrrk and made it knciwn irl Europu. Eventually, Gihbs rcctivud thc wcogtlition that was his rlue, and Yale finally otl'ered him u sul:lr~edv)sit~on in IXHO. Gibbh wah an urlashulr~i~~g, n~vdestperson. living in New Haven in hi:. l'a~nilyhome his enlire lilt..
The relation between all he phases of a substance at various temperatures and pressures can he concisely represented by a phase diagram. In this chapter, we will study the infomation presented by phase diagrams and the the~modynamicconsequeticcs ol'this information. In particular, we will analyze the temperature and pressure dependence of a substance in term? of its Gibbs energy, particularly using the fact that a phase with the lower Gibbs energy will always k (he more stable one. Many thermodynamic systems nf interest consist uf two rlr more phases in equilihrium with each other. Fur examplc, both the solid and liquid phases of a s u b ~ I ; ~ ~ i r e are in equilibrium with each other at its melting point. Thus, an iu~alysisof s u c l ~;I system as a function of temperature and pressun: gives the pressu~-depcndenw~ ) f the melting point. One of the many unusual properties of water i s that the melting point of ice decreases with increasing pressure. We will see in this chapter that thih property i s a direct consequence of the fact that water expands upon frcel.ir~g,or that the mtllar volume of liquid water is less than that of icc. We will also derive an expression that allows us to calculate the vapor pressurt of a liquid as a function of temperature from a knowledge of its enthalp? . ; . ,b.Jhilation. These ~ t s u l r acurl all be undcrsttx)d using a quantity called the chcm~calpotential, which is one ol' the most useful functions of chemical thermodynamics. U'e will see that chcmical putential is analogous to electric potentid. Just as clev~riccurrcnt AOM;S from a regi011 of high electric potcntial to a region of low clectric potential, matter flows from a region o f high chcrnical potential to a region oT low chelnical potcntial. In the la51 sectioti of the chapter, wc will derive a statistical thermdynamic expression I I ) ~thc chemical potential and shc~whow to calculate it in terms of molecular or speurruscopic quantities.
349
9-1. .4 P l i a w niagram 5ulnrnnri7w I ~ SarlidP I iquicl-Lab Bchavior of a Substance
9-1. A Phase ILIiagrarn Summarizes the Solid-Liquid-Gas Behavior of a Subst;~ncc. b r u might I-ccilll from gerleral clicmirlry thal we can su~nrrlarizetl~csolid-liquidg:~slbchavic~~ot a
Thc lint'; tll:~r aepardte the thrcc I-cgionc indicale pressures and tetnperaturcs at which two ph:ucs can coexict at equilibrium. For exalTlple, a1 all poirits along the line 11i;lt wparates the solid and gas regions (lirle CF), benzene exists ;IS a sulid and a gar in crl~rilihriumwilh e;~chother. Tliix line is called the solid-gas coexistcncc c i ~ t v c .Ar .;uch, it spcciiics the viper p1,essulr of solid benzene as H l'unctiun of lerrlperaturc. Sirtiilurly. the line that separates the liquid and gas rcgions (lint. FD) g i w s ihr vapor Imssurtt of liquid ~ T I ~ C I ~5I C*C fur~ctionof tcnipcraturc, and the line that separates tht. solirf a n d liqi~iriregions (lir~cFE) givcs thc melting point of benter~ens a function 01' ~ ) ~ - f i sNutice ~ ~ e . that t11c lhree lines it) the phase diagram intersect at one poinl (point F). at whicll solid, liquid, atid gaseous benzene cncxist at equilibrium. This ~)nitilis called the 11-iplepoint, which occurs at 278.7 K (5.S0C)and 36.1 torr for hetirene.
F I G U R E 9.1 ptiasc diagran~of berlz,ene, {a) drsplayed nc w l - w r T rlicplay conde~lse~ t11c vertical axis.
Tllc
P against T , and (b) log P versus 7.. The log
P
I
E X A M P L E 9-1
Expll-imentidly. he v:lpur plessurc of llclutd k n z c l ~ eis given hy
and Ihc vapnr przscnre
rlf solid l~enzeneis givcn by
Calc~ll:itethe presvrire nntl the ternpcl aturc ar rhc tri~ilc~ ~ r ) ~IICr i hcrwerle t
S (3 1 I.: 1 I C).\: Solid, liquid, ; I I I ~ B S C O U~S C ~ I L C I I CatCthc ~X Iriplz X ~poinl. ~ ~ Therrhre. nt thc triplc point, these I U , ~rqualionc lrlr the vaprjr pl-essurr must give tl~csame valuc. S c ~ t i n g[he two expressions ahuve f o ~111 P cqunl t ( each ~ ott~erpbes
or 7 = 278.7 K,ur5.5 C . ~I~huprcssurci~t t t ~ c~riplr1)ciinl i5girt.n h!: or P = 36. I torr.
In(P/tr,rr) = 3.5K
Within a r inglc-phase regioti, borh the pressure and tlic tcmpcruturc nlukt hc spzcilicd. and we .say that tlicrc arc two degrees of freednrri within a single-ph:~.;c region 01 a pure suhstancc. Along any uf the coexistence curves. either the presstire or the tempel-;rturc alonc is sufficient to specify n point otl ihr curve, so we say tli;rt thcrc is one degree 01- frredom. The triplc point is ;I fixcd point, so there are r ~ orlegl-re:, ol freedon1 there. If we think of P and T ;!a dcgrccs of freed0111of the systrrri. [hen the nuinhcr o f dcgrees of freedom, f , at any poit~tin a phase diagram of a pure sub5tancc is givcn by ,f = 3 11, where p is the nurtlher of 1)h:tses thi11coexiqt at cquilihrium :tt lhat point I f we start o n tlic pressure axis at 760 torr (2.88 o r i ihe vertical axis in Figurc q. I h ) and move horizontnHy to the right in the phase diagram of hen7cne. wc c:ln scc how benzene behaves with iticreasirig terriperature at a collstant prcssrlrc of 760 torr (one atmosphere). For terrlperaturzs h r l ( ~ w278.7 K, hcnzcnc exists as a solid. At 278.7 K (5.5-C). w e reach ~ h c.solid-liquid cocxistcncc curvc, and benzene ttlel~sa1 this poiril. 'l'his point is called thc t~ormnlnrplting ~joinr.(The meltinp poini a1 a prerhurz of' one bar is called thc sirind(rrd rnt.lring point.) Then (or lrrnprraturer between 27X.7 K and 353.2 K (80. I'C:), benzene exists as a liqi~id.A1 the liquid-gas voexistcncc curvc (353.2 K ) , ben7cnc boils and thcn cxists as n gas at IeIiiperalures highcr than 353.2 K. Noie t h ~ lif we w c r ~to
3.5 l
Chaplet
i Phase tquilit>rid
solid-liquid ctxxistence curve as the melting point as a funclion of pressure. Figure 9.2 shuws the melting point of benzene plotted apninst prcssure up to 10000 atm. The dope of this curve is 0.0293 'C-atm-' around 760 torr, which shows that the melting puint is fairly insensi~iveto pressul-e. The ~r~elting point of bzrrzene increases hy about one degree it) going honi a pressure of I atnl lo 34 atm. By L ' O I I ~ ~ A SFigure ~, 9.3 is a plot of the boiling point oT benzene as a function uT pressure; it shuws thal the borling point depends strongly upon prcssure. For examplt, the ~ a ~ r r n aimosphmc al pressure at 211 elevation of 1000() tect (3 100 mctcrs) is 500 torr, so acvt~rdingto Figure 9.3, h e n ~ e n ehoilu at ti7 C at this clcvation. (Recall thai the hoil~ngpoint is defined to he that temperature at which lhe vapor pressure eqilals lhe atmo~phericpres.si1re.j The boiling point at exactly one ntm is called the nonnal boiling poirlt. Ttle builit~gpoint at cxactly anc har is callcd the starldi~l'dboilirig point.
9-1. A Phase Dingram Summ,lri/es the
Sul~d-i~qurd-c;,~s Bcha~~vr ui a
Substance
E X A M P L E 9-2 The vapor pressti]-eof henzenr call bc exprcsscd hy thc c~r~prrical IorltjuIa
Use this formula tv show that benzcnc boilc ilt h7 C u t ~ c the l ~ alrrjohpher~cprehturr I, 500 torr.
SOl.llTl(-jh : Ren~eneh i i s when its rapur pressure pwswre. Therefore P = 5 0 0 torr, so we have
1s
uquul to thc ilt~rjos~lhcric
ur T = 340.2 K. ur 67.1'C.
-
V l ,.--
0
0
5000
FIGURE
9.2
A plch 01- the melling pulnt vl benzene as a fu~lctionof prcssurc. Noticc that I
ooO('
P l n1r11
rhc niclring palin! increases slwwly with prc~surc,(Notc that thc scalcs o f thc hori~ontulaxec in Figures 9.2 ant1 0.3 are very different.)
Example 9-1 shows that ihe pmssure at the triple point of bcnzcnc is 36.1 tori-. Note from Figure 9.1 lhat if the pressure i s less than 36.1 ton; h c ~ ~ z c ndws c not n ~ u t t as we increase the lemperature, hut rather .~rthlirtit,s;hat ia, it paascs dircctly frurn thc s l h d pha.ce to the gascuus phitsc. If thc pressure at the triple poitit happens to he gre;ktel, than one atm for a substance, it will suhlirrie rather than ~nelt;it onc iatltl. A I ~ O I C ~ suhs~anvewith this property is carbon dioxide, whose solid phasc i:, callerl dry ice hccause it doesn't liqucfy at ailnospheric prcssurc. Figure 9.1 shou)s the phase diilg~,ib~~~ fur carbon dioxide. T h e triple point pressure of CO, is 5.1 1 atrn, and so we see that CO, sublimes at one atnl. The nonnal sublimation temperatul-r 01 C:O. ix 195 K (-7X. c).Figure 9.5 chows the phase diagraln for water. Witcr haa the unusual 1,rol)crry that ils melting point decreases with incrrasing pressure (Figurc 9.6). T h i h hrhavior is reflected in the phase diagram 01' willer hy the .slope 01 the solid-liqi~idvoe-ristcncc curve. Althougti it is diflicult lo see in h e phase diagram h c v a ~ ~(he w slope ofthc srdi~lliquid utexister1c.e curve is so large, i~ does point upward 11, the lclt (1ia.s a negative slope). Nunierivally, the s l o v ol' the curve around onc arm is - I 3 0 atln. K I . Wc will
"1
Solid
1
?h
Liquid F I G U R E 9.3 A plot of thc hnil~ngprririt or benzene as a luncliun of pressure. Notice that
30 0
I0I)O
f l torr
2000
thu boiling po~ntdcpcnds ~tmnglyon rhe pressure, (Note that the scales of the horizontal axe!, in Figurcc 9.2 and 9 3 are
rliffrrent.l
point
Gas
1.511
251)
TI K
F I C; LJ B E 9.4 'l'he phase diagrart~of cnrhorl dloxtdc Note that the triple point prehsurr uf cnrbot~dinxidc i~ grunter rhan onc a1r11 Consequently, MI-bundioxide sublin~ssat ntrnospheric pressure.
equilihriuni with each other along the liquid-vapor coexistcncc curve (such densities are called r~rthnharird e t ~ . ~ i r i we ~ s ) see , h a t these densities apl7roach each olher and becolnc equal at the critical point !Figure 9.7). The liquid phase and vayxjr phase simply merge into n singlc fluid phase. Si~nilarIy,the molar enttlalpy nf vnpori7ation decryases along this curve. F i g ~ ~ 9.8 r e shows expcrirr~cntnlva1uc.s 01. thc molar rnlhalpy of vnporiz:~tin~~ ot benzene plulled against tertlperature. Noticc that the value of Alnl'Hdecreases with increasing temperature and drops to zero at the critical temperature (2x9 'C for henzerje). The data ill Figure 9.8 reflcct thc fact thaI the difference hetwee11a liquid and its vapor
I I C; U H E 9.5 The phase diagran~o f w3tt.r. (a) dixpla~eda!, P ngninst 7'. and (b) log 1' versus 7'.'The log P !el sus I' display condenses the horizontal axis. Allhnugh 11 15 difticult to di~cernt)ccausc 01- thc .\cult.r d the i~gurc,h e rr~ellrrlgptint of water decreases with increasing presstire. ('rilicnl
point
F I G U R E 9.7
A plut of the urttluharic dcnsitics of the liquid and vapor ph;~+ef ~ ~ ~ ~ C T I ill L Ccquilibriu~n IIC along the liquid-vapor coexistence curve. Nutice Ihat the der~fitie~ of thc liquid and vapor phares approach orlc another and become etlual at [he critical point (289-C).
FIGURF
9.6
A plot of thc nleltinp point of water versu\ plmrure The 111ellingpcyin1 01-wuter drcreaws nit11
incrcaqinp prescurr-.
sec i n Scction 9 -3 that thc rcason the melting puirlt uf water decrease5 with i ~ l ~ r t ' i t ~ i r l g prrssurr is that ~ h t molar . v~durnrot. ice is greater than that of water under the same condil1on.s. Ani~monyand hisrnulh are ~ w other o such substar~cesthat expand upon I I P ~ I ~\lost ~ E . suhqtanceq. hou-ever, contriic't upon freezing. In each ot 1:lpurcq 9. I (be117cne),4.4 (carbon dioxide), and 9.5 (water). the liquidIS cocxistcncc culu-e endlahmptly at the critical point. (Recall thal w e discuhaed the ut.itical beliavior of gases in Sectiurl 2-3.) As thc critical point is iipproitcl~cdalong tilt liquid-gas coexislenre curve, the difference between the liquid phase arid gaseous 11l1;1sebecomes increasingly leas distinct urltil the difference disappears entirely at the point. f:nr cxarnple. if we plot the densities of thc liquid ant1 vapor phases in
F I G U R E 9.8
Bxperimenlal vuluec of- the molar enthalpy of vaporization of hen~cneplotted agairi~t temperature. The value of A ~ decreases ~ ~ H~ i t increasing h temperature a ~ drops ~ d to zelo at the critical lelnperature. 289'C.
decreilaes 3s the crilical pc~inti s appruached. Because the two p h a x s hccolne less and 12bs distinct as the critical point is approached and then merge into nnc phasc at the critical polnt, 4,,,pS= S(gas) - S(lic1uid) bccu~ncazero at thc critical point. Theretire A L , , p l = l 7'AvlySalso becomes zero there. Above the critical poitlt, there is no distinction betrveei~a liquid and a gas. and a gas cannot be liquef ed no rrlatter how great the pressure. A nice lecturc demonstration illustrate:, the idea of the critical temperature. First, ti11 ;I pliiss tube with a liquid such as sulfur hexafluoride. ('The critical temperature of sulrur hexaflut)ridc is 45.5"C, which is a cu~nenicnttcnipcraturc to achicvc.) Aftcr evacuating all the air so that the tube cuntairis only pure sulfur hexaflitoride. sea1 off lhe luhz. Rcl(lw 45.5-C. the tube will cotitain two layers, the liquid phase and the gas phase scpiiratcd by a mer~iccus.Now. as thc tube and its contents are warmed, thc mcniscus becornes less dislirict and just as the critical teIrlperature is reitchtcl, the ~ncniaci~s disappears elltirely and the tube becortles transparent [SF6(g) is colorless]. Whcn thc tuhc and its contents are cooled. the liquid phase and thc ~ncniscussuddenly appciu- at thc critic:~ltcmpcrilture. A lluid v e v near its critical point constat~tlychangcs trow a liquid to a vapor st;~lr.,r,:ti~singHuctuatir,~~'; in Ihc dcnsity froln cme r c p i o ~to~anothcr. Thcsc Huctuatinns scatth.1. light wry ~ t m l ~ g(hrll~lt.x.h;~I ly liar ;I 1111ely dial)ursed fug) ;inti lhc syhtelll ;iyptars rnil k!. This cR-tct i.c known as c-rifrt,r~lo p r i l t w r,rlr'e. These tluctuatir,nc are rlif hcult to study experirrien~allyhecauhe gravity cau.se'; lhe density Huclua~ionsto hr diqlrrted. To cnercome the effcct ofgrav~ty.a tcnlti of scientists. engincen, and technicians rie.signctl an experiment to Ineahurc thc laser light scattered by xcnun at its criticul point o n board thc C o l u ~ l ~ bhpace ia shuttle. After several prelirnirlary experiments, they were able to measure the de~ailsOC lhe fluctualiona tu withln micn~kelvinsof the critical temperature of xznorl (289.71 K) 011 the March 1996 flight c)t Columbia. No othcr microgravity cxprimcnt has logged as many hours as this one. and the results will provide us with a detailed understanding 01- thc 11q1tid-vaporphase ~ransilionand the 11quid-vapor i~~terface. of ii criliciil poir~t,a gi15 can bc ttansformetl ir~tcin liquid Rccai~xeof (he existc~~ce u i ~ l ~ o ever u t passing thrrugh a two-phaac stutc. Simply +,tart In the gas region of thc phase d~agramanti go intc~thc liquid region by tl-avcli~~g out i~rnunrlthc critical point. The gas passcs gl-adu:llly iind cor~tinuouslyit~tothc liquid state without a two-phase rcgiun appuriing and without any apparent uortdcnsation. You might wundcr if the solid-liquid coexistencc curve ends nhruptlq its the liquidgas ctlexistcncc cun7cdocs. ?'he very riilture of a critical point requires that we pass from one phase tc~the otlie~in a gradui~l,cot~tinunlmnlmcr. Rcc;lusc thc gas and liquid phases are but11 fluid ptliises, the diltzrznct. hclweetl tht.1111s purely one OF degree ralher lhon actual ctructurc. On the other t~arld,a liquid phase aitd a solid phi~se,or I W O diflerent solid ph;lsts for that matter. are qualilativrly different beci~use(hey have intrinsically dif'celtnt structulrs. It is nut possible 10 pass f r o ~ none phase to ~ l chher ~ e in a gradual, continual manncr. A critical thererol-e. cannot exisi [or such phases. and the coexistence curve separating these phases musl continue indefinitely or intersect the coexisterlce curves of olher phases. In tact. many substances exhibit n variety of solid phases at high pressul-es, anti Figurc 9 9 shows the high-prcssure phase diagram of
Liquid
FIGURE
9.9
The phasc diagratn tor \v:iter at high pressures \ho\ving c n alablc phnhck of i i u
water, showinp various distirict solid phirse*. Icc ( I ) i.c the "nr,t-ln;~l"1r.e th;ll occur-s at one attn, and the clthcr ices are different cryst;llline I'urms uf solid Hln that dl-c st.ll>lc at vcry high pressul-es. Nok. fur example, that icc (VII) ia stable at temperatures \veil above O' C, and evcn a b w e I OWC', hut i t is for111edor11y at high prchsurcs.
9-2. The Cibbs Energy of a Substance Has a Close Con~lcction to Its Phase ~ i a ~ r a r n Recall Figurc 8.7 where the molar Gi bbs enetgy of bcnzenc i.; plolled again51 ICIIIpernture. As thc hgurc shows, the molar Gibbs energy is n contirluous fi~nctioi~ 01tclnperalure, but there i s n discontinu~tyi n ~hc.slope t , f c ( ] ' ) hersus I' ; ~ t c;lc.t~ph;15~. transition. Figure Y.l Oa IF a rnagnificatior~of ;I plot of ~ ( 7 'vel-\us ) T i l l thc I - C ~ I O I I arourid the melting point of benzene (279 K). 'rile dashed exter~sionsrcprcacnt t IIC G~hh.; energy of the .supercooled liquid and the (hypothetical) .superlleatcd solid. Picluvc. tnu\ ing along the curve of G ( T )versus T in Figul-e g , 10s with increasing ternpzrittulc. A l m g Lhr solid-phase bmnch, G ( T )dccreaqcs with a slopr (ilcla?.),, = -3' . Whc.11 we rcauh thr melting point. we switch to thc lirluid branch ht.c.a~tsethe (iibt)s enerpy of the liquid phase i:, lnwer than that ofthc solid phi~he.The slol~r01 [he licluirl t)~,itncl~ i\ steeper thi~tlthat r,E the solid branch hecause {3c/i)?') ,, = -3' and s I ;, s '. Thel-e fore, the rnol:lr Cilhhs energy of the licluid pliasc must hc luwcr than Illat ol'thz solirl phase at higher tcnipcraturzs. The dashed extetision of the solid branch rcprcwnts the (hypothetical) auperheated solid. and evet~if it wcrc to occur, it W ~ I L I I Lhe I ~tnsl;ihlt relative to the liq;id and would convert to liquid. The daslicd lir~csrcprcscnt what are called metastable states. Figure 9.10b xhows the transition from liyuirl to gas at Ihc normal boiling point (353 K) of ben7.cnc. The hoil~ngpoint occurs wherl the liquid alvi gas branches of the E ( T )versus T cur\.es intersect. The slope 01' the gas branch is 5
9.1U 11 plot uf G ( T )vcrqus T for h e n ~ r n ein the region around(a) its melting point (27'3 K) and (h) its hrlili~lgpoillt (353 K). flGURE
F I G U R E 9.11
A plot U C ~ ( Pagainct ) t' shuuring he gn?. l~quid,anrl valid hrancllcs at a tcrnpx;ltur,t ncar -I . thc triple print. (a) A "llormal" suhslance ( V ' < V ) 1s depicted at a tclnpcrature ahrlrc thc triple-p)lrlt telnprature, where we EEC a pas-liql~id-solidprogressioll with increasing pressure. (b) A suhsvar~celike water (V' > ') i s depicted at a ternpraturc lower than !he tr~plcpoint temperature, where we see a gas-did-liquid prngres~ion.
s',
strrpcr than tt~atof the liquid branch because S S, and su the molar Gibbs encrgy of the pas must he Iower than that of (he liquid at higher tetnperaturcs. We can see frorn the equi~tinnC; = H - T S why the solid phase is Cavored at IOU ternpcmrul-cs whereas the gnscuus phase is filvored at high telnycraturcs. At low tt.iirperatures, thc T.7 lerm is srnall compared with H: thus. a solid phnsc i.c favored at Iorv teiriperaturcs hecausr it has thc Iclwesl enthnlpy c ~ the f three phases. At high ltnrrlpt.raturcs. on the uther hand. I I is xrnall co~nparedwith ttie 7 S term, xo we see thnt thc gas p h ~ s ewith its relatively large cntrrjpy is favored at high temperatures. Thc lir111idphase, which is intcrlncriiate in both cnergy and disordcr to the sulid atid gaseous ~~liaxcr. i x favored at it~tcrmediatrtmlperarul-es. It is also instructive lo look at thc molar G i b b ~cnergy as a futlction of pressure at a lixed tcmpcraturr. Recall that P), = V , so that (he slope of G vcrsus F is always
(ac/a
positive. Pos nlost substances, Y >> 1; I : , V ', SO thc slope or the gas branch is much greater than thiit ot a liquid hranch, which in turn is greater than that of a solid hranch. Figure 9.1 l a skctchcs a plot of G ( P )agaitist P showing (he gas. liquid, and solid hraliches at a tempcratiu-ejuut greater than the triple-point t e m v n l u r e . As we incruse the pressure. we move along ihe gas branch nT C ( P )until we hit the liquid branch, at wl~iclipoint the gas contlenses ro a liquid. As we continue to increase the pressure, we rci~cllthc solid bri~nch,which ncccsaarily lies lower than that of the liquid branch. The path we havc lust f(~llowedin Figurc 9.1 l a corrcspc)nds to moving up along a verlical line (hat lies just to the right of the trilrlc poinl in the phase diagram of a "normal" suhS ~ ~ I ~ I like C L ' k n z c n e . For a mbstance such as watrr, however. > V I at least for mod?.rate prcssul-cs. so a plot of G(P) against P looks like thal given it1 Figurc 9.1 I h. 'I'raci r y nlor~gthc curve f-or E(P)for increasing pressures in Figure 9.11b corresponds to n ~ o v i ~up~ ag vertical lit~e.justto thc let-t ofthc triple poinl in thephnsc diagram 01waler. Figure 51, i 2 phows thc bclinvior uf P) Venus P at a nulnbcr of temperatures for ; Inul-ma1 s u h s ( a ~ ~ such c e as h e n ~ e n e Part . (a) shows vcrsus P for a tenlperature
v'
c(
c(P)
l e s ~than the triple-point 1errlper;iturc in Figure 9.1. In this case, we gu dirccllp from the gax phase to thc solid phase as we incrcase the pressurc. The lnolar Gibbs cnclgy of the liquid phase at thcsc temperalures lies higher than that of eithcr the solid ur gas phase and does not cntcr the piclure. ['all (h) shows the trlolar Gihhc c n e g y situation at the triplc-pint temperature. At tlic triple puint, the curvcx lor Iht: Ciibbs c n c ~ y i c s01 each of tllc three phases intersect, and f-or a "normal" substance like benzcnc, thr liihhs energy of thc d i d phase lies lowcr than that of the liquld phase for psessulrs above the trilllc-point pressurc. Part ( c ) shows the Gihhs energies at a tcmpelature slighr ly lcss than the critical temvralure. Noticu that the shjpes of thc fa5 and liquid brnt~chca art. alrriost the s:11ne at the point of intcrscction. ?'he reas011 for this similarit!. is that the slopes of thc curves, (ifc/il~),. arc equal to the rl~ularvolumes of the ~ w phascs. o which are approaching each other as tlic critical point is nppruached. Part (d) shows the - Eiibbs energies at a tempratuse grcatcr than (he critical ternpcraturz. In this case, G ( P ) varies smoothly with pressure. There i s nu discontinuity in thc slope in this case hccause only a single Ruid phase is involved.
9-3. The Chemical Potentials of a Pure Substance in Two Phases
in Equilibrium Are Equal C'onsider a systcm consisting of two phases OF a pure subsvancc in equilibrium with each othcn For example, we might liuuc waler vapor in equilibrium with liquid watcr. The Gihhs energy of this systcm is given by U = G I + Gg. where G" and G' are the Gibhs encrgies of the liquid phaqe and the gas phase, rcspectivtly. Now, suppose that
9-3. The Chemzra~Pote~~t~als ut Purr. Sub>l,~r~r,r III l w n
: L.iquid ;
-:,-
Liquid
Phares rn Equilil)r~urb~ Arr
t
36 1
11l1al
The partla1 derivatives In Equatiutl 9.2 are central quantities in the trcatmenl oi cquil~hria.They art: called ckem~rnlputmt~ul.\and are denotcd hy 1 1 ~dnd IL':
-
G
Sol~d
Sulid
G
In tcrms of chen~icalpotentials, the^^, Equation '1.2 reads i
I
-
) d g
(corlstant
r and F')
(9-4)
If the two phases are in equilibrium with cach other, the11dU = 0. and because dtry# 0. we find that p L p ' . Thus, we find that if two phases of a singlc suhsvance are i11
----Solid
Sulid
-
FIGURE-9.12
h plot of G o ' ) aguins~P a1 a nunlhcr of ternperstures for a "normal" substance llkc benzene. III ( n l the ttmpcmturc is icss than the triple-point temperature; in (b) thc tenlperature is equal to the tr~l>lr-poii~t 1enlperatui-c;in (CJ the ternFrature is a little l e u than thc critical tclnpzruture; and ill (d) the
=(
te~nperalureis grcotcr than the cr~llcaltc!~~pc~-;$tu~-t..
equilihrii~rnwith each othcr, then the chemical pc>tcntialso T t h a ~substance in the lwo phases are equal. If the two phaaes are not in cquil~hriurrlwith cach utlirr. sporiktneous tran4c.1of matter from one phwe the other will uccur in the direction such that riC; < O. If pg > F 1 ,the term in parentheses in Equation 9.4 is positive, so rlnF must he negative in order that dC i0. In c~therwords, matter will transf r rrorn the vapor phase lo (he liquid phase, or from the phase with highcr chemical polential to thc phncc wilh lower chemical potential. If, on the other hand, p+ p l , ,hen drrQvill bc positive. nlz:~r~i~lg that matter will transfer from the liquid phase to the vapor phase. Once agarn, (he transfer occurs from thc phase with higher chemical potcntial to the phase with lower chemical polzr~tial.Notice that chemical putetitinl is analogt,us lo eleclric potcntial Jusl as electric current flows from a higher electric potcntial 11)a 1owe1 elcctric prltcnt~~l. matter "tlowh" from a highcr chemical potential to a lower chenlical potcntial (hcc Problerri 9-1 9). Although we have defined chcmical po~entiiilquite gel~crallyin Eqi1atio119.3, i t rakcs on a simple, F~niiliarform for a pure substance. Bccause G , I ~ k eU , H . and S, is an cxtenqive therniody~iamicfunctic~n,C; is proponional to the si1.e of a system, or I; cc n.We can express this proportionality by !he equa~ionG = tip( T . I ) ) . Note that this equatiorl is consistent with the def nition of p ( T , P) hecause
d t ~lnoles are transfemed from the liquid ptlnsc: to the vapor phase, whilc T aiid P are kept constant. The inliriitesimnl changc in Ciibbs energy for lhis process is
Pi-
Rut On1= -dn' fur t l ~ ct~ansfcrol'drr niolea tlum ~ h liquid z phase tu thc rapor phase, so Equatic~n'I.1 becomes
Therefore, for a single, pure substance, f i is the same quarltity as thc molar C;il,hs energy and pcf, P } is an intensive quantity like temperature aitd pressure. We can use the fact that the chemical potentials ot a single substancc in two p h a w in cqullihriunl are equal to dcrive an expression for the variation clf eqililibriurn preasurc with temperature fur ;my two phascs of a giver! pure substancc. I.et Lhr two phases be a and f l , s o th>~t p" (1'. P) = "I
(T,P)
(equilibrium hetweeri ptlases)
(9.61
0 I The
N o w take
I ~ I ?total
Chcm~cdlI'uterili;il? r)i a Pule S L I ~ S ~ I 111 I I LIWOV Phases 111 Equif~hr~urn Arr
Fqi~al
Wc can take rhe reciprocal n f this result to obtain
derivatiXmeof both sides nf Ecjuiltion 9.6
d7' dP
= 0.0292 K atm
'
Thus, wc see thal l t ~ ert~eltingpoint o l ' k n ~ e r l iticrcascs e hy 0.0292 Kper atmosphere of prevsure i~r~)urlcl otie atm. Tf A,",Z and were independetlt of prcssurc. we c t ~ ~ l d me this reqult lo predict that the rnelting point of hcnzene at 1000 atln is 29.2 K higher [hall it is at one atrn, r ~ 307.9 r K. The experimental value ir 306.4 K, so our asvumprion of constant h,,,pand A,>V is fairly satiqfilctory. Neve:ertbelesr. Equation 9. I 0 i \ strictly valid only ovcr a lirnilell mtlge hccausc A,",H and Atu,i7do vary with I' and P . Figure 9.2 rhows the cxperimerital melting point uf benzene vcnus prescure lip to IOO M atm. You can see from thc figure Ihat the slopc i q not quile constanl.
atu
Rccnusc 11 is sin~plythe molar Gihhs encrgy for a single substLtnce,we h a w in analogy will) i5qu;ttions 8.3 1
I
- -
I
E X A M P L E 9-3
Determine [he vnluc of ~ l I ' / Pd lr)r icc at its nornial mellir~gpoint. Ihe rnnlar e r ~ l t ~ a l p ~ of fusion or ice a1 273.15 K anti one atm I F W10 J mol-I, and A,(,<\'under the sarrle conditions is -- 1 .h3 cln'. tnol ' (recall ih:it unlike lnosr suhstancez warer ~xpandsupcu~ freezing, SO lhat A,,E,I'= TI - F' c 0.) Lstiinate the nrelting point of ice at ICHH) a~rll.
where I 'and 3 are the rriolar volumc and rhc molar entropy, respecli~rely.We suhslitute this I-esultinlw Eyualiutl 9.7 lo obtairt
SC) I
I! I I Cj I\::
Wc use the reciprocal of Eqrlation 9.1 0:
Solving for d PlrlT g i w s
rlP -. dl-
- -xR-T -
At2
~ ~ A,~V7
Eqtlnticln 9.9 applies to t ~ , phases o ill cquilibriurn with each other, h(> wc may use the 7' and write lict that ~~~~3 =
Lquation 9.10 1s called thc L'/rrp~ymmrquulton, and relates the slope of (he twu-phase boundary line in ;i ptlilsc t l i ~ g r n ~with n the valurc of A , ~ and H A,,T for a transition hctr\,ccn thesc two phares. Let's use Equation Y. I0 to calculate the slope of h e snltd-liquid coexistence c u v~e lor bcn7enc around o r ~ atm c (Eigurt 9.1 ). Thc nlolar cnthalpy of fusion c ~ fbenzene at its ntlrttlal nlel~ingprlint (278.7 K) ic Y .95 kJ mnl-' , and ~ , , v u n d e rthc same conditions I \ 10.3 cni' mol ' 'Thuq. rQ'/dT at thc normal rneltlng point of bcn7erte is
Acvurrl~r~g that d T / d f is conslar\tup tn 1 WO arm. we find thal AT -. -7.5 1 K. r)r 11lat the melting prlir~to f icc at IOOO a m I S 2hj.b K. The enpt.rimet\tnlvalue is 263.7 K. The discrepancy arkec Irom our assumplion that thc values of A,~Vand A,,Z are indqwrldctlt of pressure F~gurc'3.6chows the exprinlcntnl melting p i n t elf icc vcrsus preswrc up to 2000 ntm.
I
I
Notice that the melting point of ice decreases with increasing pressure, so that the slope of thc solid-liquid equilibrium curve in the pressure-temperature phase diagrarr~ of water ha5 anegalive slope. Equation 9.10 shows that this rlope is a direct result o I the fact that A,V is negative for this case. Equaticln 9. I0 can be used t o estimale the molar volume oT a liquid at its hoilirig pint.
EXAMPLE 9-4 'I'hc. valm prt.$su~,c of hrn/cnc is found 10ohcy lllc c~npiricaicquatiun
t r o r r ~298.2 K 10its normal hoilir~gpoinl 153.24 K Given [hut the molar cnthalpy ot vapori~ationat 353.24 K is 30.8 kl.~nol' iind that Ihe rr~olarvolurne of liquid benzene a1 353.14 K lc 9h.O cid-mr)l-'. use the above equation tu lete ermine (he nioln~~olurtle ul' the vapor ;it its cquilibriurn prcssurr a1 353.24 K and cnmparc this d u c with the
idtal-gas value. -
5 0 I LJ T I U I\;: We
htalt
9-4. The Clnusius-Clapeyron Equation Gives the Vapor Pressure of a Substance As a Function of Temperature When we ~rsect t q i ~ i ~ t i o9.10 n to ci~lculatcthc var-ialrrin (li' the r~jeltir~g points of ice (Example 9-3) and benzerie, we assu~lledthat A , ~ Iand I Atr,Vdo no1 vary appreciably with prehsure. Allh~,ughthis approxittiatiorl is fairly satisfactcl~yfor qolid-liquid and solid-solid transitiorrs uvcr a srnall A T , it is 1101 jitlisfii~tor'yfor liq~iid-ga.;~ n r sultJi gas transitions hccausc thc molar volume of a gas varies st1'onply with prcssurc. It thc klnperdture is nut too near thc critical point, however, Eqilatiurl 9.10 can bc cast into a very useful form for condcnscd phase-gas phase ~ransi~ions. Let's apply Equ;ition 9.10 to a liquid-vapor equlllbriu~rl.111this ciisc, crc Ii;ivc
with Equation 3.10, which w e solve lor A(,>,,l/
C:mg lhc above cn~pi~~lcnl v;ipor pressure dquatiun at 7 = 353.24 K.
Equation 9.1 1 gives the slope of thc liquirl-vapor equilibri~ltrjline in the phase diagr,um of the suhstance. As lung as we are not too near the crilicai pnint, 7' >> 7'. so that wc can neglevl V ' uonlpared with in thc ~ l ~ n o m i n a t o r Erl~lation9. I I . Punhcrmtirc. if thc vapor pl-e.csureis nut too high (oncc agaln, if w e are no1 loo close to the a-itic;~l point), wecan asaulnc the vapor i s ideal and replace 'li"hy R T / P. s o thi~tEclu;~tit>rl 'j. I I
vY
hccon~es 1 dP
P
'l'he molar volume u r the vapor
~h
Ttw corresponding value Iro~nthe ideal gas equntlon is
uhich is slightly larger t h n ~the~ aclual value. I
In I' - ri--
dl'
-
AbdI>H -
i?r2
This equation, which was Rrst derived by Clauxiils in I XSO, is Lrluwr~as the C1r;u~itriClap~yrnneqrdation. Remember that wc havc neglected the rr~olarvolume of the liquid ctlmparcd with the molar volurr~eof thc gas and that we assurr~t'clrt~evapor can hc treated as an rdcal gas. Never~t~eless. Equatio~i9.12 has the advanrage of t,ei11g ~norr: corlvenient to usc than Equalion 9.10. As ~niphtbe expcctctl. however, Eq~ralio~! 9.10 is Inure nccuritc rhan Eqi~i~tion 9.12. The real a ~ d v i ~ ~ ~uft aEquation ge 9.12 is that it cikn be reiidily in~cgratcd.If wc assume A ~ does ~ nu1 ~ vary H with tcmpcrature over the integratiori limits of T , Equat~on9.12 becomes
Equation 9.13 cikrj be used to c;ilcularc rhc vapor pressure at some tetnpcraturc givctl rhc molar cnthalpy ofraporization and the vapor p ~ a s u r t :11 : snrrie other temperature. Fur ertnlnple, the 11ormal boiling point of benzene is 353.2 K and A%,,, H = 30.8 k l -lrlol I . Assuming A14,,,H dues not vary with temperalur't', let's onlculaic t l ~ cr,;ipor prcqxure ol
C - h ~ f ) l')~ r! Fhasc Equi1ihri.l
366
hcnzcne ai 371.2 K. We suhslitute PI = 760 tom, TI = 353.2 K.and T, = 373.2 K into Equation '1.13 to obtain
P
In= 760
(
30800 1,mnl-'
'
H.314 I - K .lrlol.
19.8 K
'
or P = 1330 turr: The experilrlental value is 1360 tom.
E X A M P L E 9-5 -Ihe vapor przshtrru watet.;tt 363 2 K i q 529 inrr Use Equation 9 13 10 dererminr the average value of A,,,p11 o f waler hctwecn 363.3 K and 373.2 K.
1 0 0 0 K 1 7' F I G U R E 9.13
A plot o f the logarilhm u T the v a p r preshurr r d liquid h e n m n e again\! the rrc~prr)calkelvin temperatrll-euvel a temperature mnge of 3 13 K to 353 K .
5 0 1 I I T I O N : W e 11ltethe tact t h ~ tthe nornlal b;oilirlg point of waler is 373.2 K (I' = 760 ton.) nr~du-rite --
where k is an ~r~tcgration constarlt. Equat10119.15 cxplrsscs flit- iarrntlon 01 kapor pressure obrr a larger temperature range than tquation 9.14. 'I'hus. a plul of Iri I' against I / T will nut hc cxactl!, I~ncar,ill aglcelncnt u ~ t hthc exper~mentnldata fur most liquids and solids ovcr an cxtcndcd tcmpcraturc range. For cxamplc. thc r:lpur pressure of solid ammonia in torr 1s found to ohcj thc cquatlon Ttlc valuc of A,,,pH[rir wntcr at its norrnal boiling p o ~ n i~ t M.65 kJ-mul-I. I
1
If wc intcgrale Equation Y. I2 indefinite1 rather than k t w e e n definite limits. u'e obtain (assuming A + x r His constant)
from 1 4 6 K t o IYSK
I
I
l:rl~r;~tion 9.14 says that a plot 01- the log~rithmof thc vapor presst~rcagninht - the rcciproual of thc kclvin telnperature should be a stmight line wirlr a slope 01- - A v z p H/ I ( . Figure 9.13 ahow:, s u c l ~n plot for herlzef~cover h e temperature range 3 13 K to 353 K. The slope r,l the line givcs A+dpH= 32.3 kl-tnol I. This value represcnts an average vnluc of Aydp%over thc given lernperature interval. Thc value of ~ % at~the ~ normal f l boiling point (353 K ) ia 30.8 k J , n ~ o lI . We car1 recogni~ethat AvopHvaries with temperature by writing AvaVHin the form
-
A , , ~ = = A t- RT
+
+ ...
\r>hert:A. B. C. . . . are constants. If this equation is substituted into Equatiun 9.12, lhen inlegratio11gwes
E X A M P L E 9-6 Use the CI:trlsius-C'lapcynm etpiatinn ; ~ n dEquation 9.16 to thnlpy nf uuhliln;~tionrir :llnrntmi;~from 146 K to 19.5 K , SOL CI T I O N : According to Equation 9.12
dete~nlinethe molar erl-
9-5. Chemical Poient~al Can UP Lualuaterl I-ro~r~ .r P,lrt~lirrn I unclll)n
'I'he Clausius-E:lapcyron
equatlon
can bc used to show that the slopt. of the solid-
501. U T l O N : T h e denvalives u i h l h expl-essiuns ,lt
the triple poillt are
ga5 coexistence cun7enlusr he grcater than
the slope of the liquid-gas coexistetlce curvc ucar the triple point. wherc thcse two curves mccr. According tu Equation 9.12, the slope ol' the solid-gas cun7eis given hy and
rl P' A,,,,iT = p'dl' R I.2
d PI dl'
-=
and the slope of the liquid-gas vurvc is given b j
so the ratio uT the a l u p
I!,
(2.303P,)
K (1330T )3.52 ton.. K ' =
4 3 1/3-52 = 1 22
I At the lnple point, P' and P I , the vapor p r e s s u u of the sulid and liquid. reqx.ctivcly, are equal, bn the ratlo of the slopes from Equal~rms9.17 and 9.18 is
at lhe triple point. Because et~thaIpyis a statc function, Ihe enthalpy change in going directly from the solid phase to the gas phasc is the same as tirst going from the solid phasc to the liquid phase and Ihcn going from the liquid phase to the gas phase. In an equation. we have
9-5. Chemical Potential Can Be Evaluated From a Parlition Function In this section, we will derivc a convcnicnt fortnula for the chemical potential in tern15 of a partition function. Recall that the currespunding formulas for the energy i111d entropy me {see Equations 3.21 and 6.43)
and whcrc the three A ~ must S all be evaluated at the lame tc~nperature.I f wc substitule Equation 9.20 into Equatior~0.19. wc see that
Using the fact that the fIelmholtz energy A i s cqual to I J
- TS. Equi~tinns9.21
and
9.22 give Thua, we see that the slope of the solid-gas curve is greater than t h a ~or the liquid-gas culvc at the triple point.
I
EXAMPLE 9-7 'l'lh: vapor pressulcs of cchid and liquid an\tnorhia near the triplc prlird art: given by
k t ' s nuw include
N in our discussiun of natural variahlcs, and writc
I
The last tertn i11 Equation 9.24 i s expre~sedin 1ernl.s ol' N , the number c ~ f ' ~ n o l c c ~111~ l c x the system. It i s mirlre conventiunal t c ~express this quantity in tcrms of n. thc numher of moles in the system. We can do thi.; hy noting that Calculntc the ratio o f tbc clrlpes of the sulld-gas curve and the liquid-gas ctlrve at the rriplc pninl.
hecausc 11 and N dilTrr by a constant factor of the Avogadro conscant. Therefore, we may uJrite Equation 11.24 in the for111
We'll now show thai (;I A / a n ) , , ,is just nnothcr way of wiling lhe chemical potential, 11. If we add d ( I 1V ) both sides of Equation 9.25 and use the equation G = A l1V , we get
+
rlC; = d A
+ d ( P I : ) = - S d l + VrlP + (
)
Recall now thal y ( V , T) cx V for an ideal gas, and so we can write Equation 9.28 as
wherr y (L'. 7')/ V is a functic~nor terriperature only. tyuatiun 9.2Y also gives us an equation Tor G because G = np. We can make Equation 9 29 look exact1 llke Equatiun X.59 iC we substitute L,T/P I'or \'/IV:
dri 7 C
But it-wc compare this resull to thc total deli%-ativeof G = I ; ( J , P . I ! ) ,
If we cotrlpare this cquntion wilh
ive see [hat
I'tlus, H C call u\e either I; or A to dttcrrnine fi as lung as !be keep thc natural variahler
of each olle tixed when we takc the partial derivative with respcct to tt. Uk can now suhslitutc Equatiori 9.23 inlo Equation 9.26 to ohtait~
We have gonr irom the secclrid lerm to the third tcrln by mul~ipyingk , and n by the Avogadro coriscant. Equat~ol~ 9.27 lakes on a lirirly qirnple form fur dn ideal gas. If we whqtitutc the ~rle~~l-ga., cxprcsrion
we see that
Once again. rccall that q / V is a function of T unly lor an ideal gas. To calculate '(T). we must remeniher that P is expressed relalive t~:,the standard state prcssurz P ', which is equal to one bar or 10' Pa. We elnphasize this conwrition by writing Eqr~ation9.3 1 as
If we compare Equntic~ng . 3 3 with kquation 9.30, u-e see that
The arguttlent of thc logarithm in Equatioll 9.34 is unitless, ax it must be. Equntiun 9.34 gives us a tnolccular formula lo calculate p C { T ) ,or G " ( T ) .For example, for Arrg) at 298.15 K: whcre we have used Stirling's approximation tor In N ! . If we subs~itutethis rcsult into
Erlr~atiotl9.27, w e c~btaitl
k 1' - RT H4-.P
-
NAP
'
(8.314 J 1no1 K ')(298.15 K ) -- -. (6.022 x I d 3 mol ')(1.00 x 10' 13a)
The panilion function .yn(v.T ) lor a diatomic molecule is
and Notice that this expression is the same ua Equation 4.39 except for the factor of e h u / L ~ T e " ~ '= k oelJoiR7 T in Equation 4.39, which account5 fur the ground-state cnelgy being taken to be - [ I , , . The gnlund-state energy associated wilh ril'(v.T ) ~ 1 ~ by ~ 1 Equation 9.37 is ~e1.0.1 . ~ 1 'use ~ Equation 9.36 along with Equation 9.37 tu calcublc p - Eo for HI{& at 298.15 K in the harnlonic oscillator-rigid rotator app~~onimallo!l. with HRI1= 9.25 K and 0>,,2 3266 K (Table 4.2).Therefore.
V This resull is in excellent agrecnicnt with thc cxpcrirt~cntalvalue of -3'3.97 kJ,mol-I. Being essenlially an cnergy. thc valuc of thc chcmica1 potential rtlust be based upon some choice of' ;I zcro of cllcrgy Thc cht~nicalpr)tentiul wc haveiust calcula~ed is I ~ i ~ >upon r d the gnluntl state of [he ; I ~ O I huit~g ~I x r o . For di;iio~l~iu rr~t)lecules,we haue choberi grourid-srale energy (vihratin~~al and clectro~~ic) to he -11,. as illus~rated in Figure 4.2. In tabulaling valucs of $(TI. ~t 15 customary to take the groutid-state CIlergy I)[ thr ~nr>leculc rather than the separated atorns as iri Fig~~rt: 4.2 to he the zuro of energy. To .see how this definition uf the zcro of energy changes the h)rm af thc partition fuoction, write
1"'
(1n)(0.12791;g.rnol ' ) ( 1.3X06 x 10-'' .l.K')(298.15 K) (6,022 x 10'' rnol ')(h.h2hx 1 (I-'" - s ) -
RT --
NAP'
- (8.314~.mol~'.~-')(298.15K1 (6.022 x 10'hmui-')(lO' Pa)
and
I f wc factor out .e
'l,''rl1,
we have
where wc havc writtet~q " ( ~7'). lo emphasize thal he grt~und-statecnegy of the molecule is takcn to be zero. Subs~ituling(his result into Equation 9.34 gives
Thc lileraturc value, which include\ nnhiirmonic and nunrlg~d rot:ktur et'Kects, 1.; -52.94 kJ.mol I. We will use balues of p ( T ) - E l , when we d~scusschcmici~l etluilibr~ain Chnptc~12.
Problems 9-1. Sketch the phakc diagra~nfor oxygen uh1n.g the follotving data: lriplc pr>lllt.54 3 K ;111il 1.14 torr: CI-iticnl puinl. 154.6 K and 37 828 torr: nnrrr~alrnelritlg point, -2 1 H 1'C: ;ud nu]-ma1boiling point. - 182.9 C, Drle~oxygen nlctt u~idcrurl applied pl'ess111-ear n;itt.ldoes?
('
whew E,, = N,?, and P: = 1 bar = 10' pa.
9-2. Sketch the ptalc diagm~nfor I, g~vcllrhe Ir~llu\vingdata: triple polllt. I I 3 :LIILI 0 I 2 atm; critjcal pink. 5 1?C and 1 16 atrn: nt)r~nalrneltj~~g point. I 14 C; nvrr~lnlbuilrng point. 184°C; and density uf liquid > clen~ityol'solid.
1
9-6. The s l o p of tllc melting curve uf melhanc is given by
Freezing liquid
dP (1 T
= (0.08440 bar, K
' ")T'"~
from [he triplc point to arbitrary tetnpcratures. Using the filct that thc telnperature and pressure a1 lhe triple point nrr 90.68 K and 0 1 I74 har, cnlculattr the rn'lling prewuic of tr~cthar~c nt 300 K.
9-7. The vapor prcssure uf methanol along thc entire lirluid-vapor exprerhc.rl vcry acc~lratelyby the crrlpirical cquntiorl
CklCS1FtCtlCP
cuwc c:lrl hr
F I G U R E 9.14
A dut~sily-temperaturepl~ascdiagram c~fbenzene where I. = T I T L ,and Tt = 512.60 point of mcthnnul i u 337 67 K.
9-3. 1;igllre 9.1-1 shuws n dc~~sity-lemperalure phase diagram for benrene. Using the follow~ng dala for the triple point and the critical point, interpret this phase diagram. Why i s the triple point irldicnted hg a line in this type r l C phase diagram?
Triple poit~t Critical poi111 Nurmal frccziijg pcyinl Norm;il boiling point
I'IK
Phm
278.681) 56 1.75 278.h8 353,240
0.04785 48.7575 1 .O 1325 1.01325
/)/rnul.L" Vapur Liquid '
0.002074 3.90
1 1.4766 3.90
0.035687
10.4075
9-4. The vnpor pressures of solid and liquid ch101-incwc giver1 by
K. Use lhis formula to show that Ihe norrr~nlI)oilin~
a I~quidis the temperature at which the v a p r prescure 1% 9-8. Thc standard huiling point exactly one bar, l!se the cmpil-ical formn1:r given in thc previous pl-ohle~nto
whcre x = T / T r , and Tc = 561.75 K. Use lhis formula to show that the norrnal hrlilirlg poi111of be117enei \ 353.24 K. Usc the nhove exprehsiorl to calculate the ct:indnrd h r > ~ l i ~ ~ c p u i ~ of ~ t ben7ene.
9-10. Plot thc follow~ngdata for the drn\it~cscrf liquld and g:iseuuc rltiane in cqutlihrium with each uthcr as n function of tempruturc, nt~ddetermine the critical telnpxiturc of eth;~lle.
uhcre P is tllc abnolule temperaturc. C:alculate the ternperaturc and pressure at thc triple point uf c h l o r l ~ ~ e 9-5. The pressurc along the melting curve frnm the lriple-point tcmprature to an arbitrarq tempclaturc can be tit empirically by ~ h Sirnun c equation, which i s
and u are constant< w h ~ f eralues depend upon the fuhstance. Given that Ptp= O.(W7X5 bar. TP= 278.hH K , n = 4237, and a = 2.3 for hznzene, plot P against 1' and compilrc your. rcsuIl wilh that givcn in Figure 9.2. wherr
(i
9-11. Use the (lam 111 the preceding prc~blcirito plot ( ( I ' + p"/2 apiiinsl T -, T . wit11 1: = 305.4 K. I he res~iltirtgstraight line is rill enlpirical law called lhe luw rifrrcrllinen~.dium~ters. IT this curve is plotted on the same ligure a
9-19. I n this p ~ o h l c ! ~we ~ . will clernunstratc tlliit cntropy always increases wher~there is a tna! t i - i ; ~ l flow- fro111a rqir)n r l t ' hiphcr concentration In one of lower c n ~ l c e r ~ ~ r a ~ (Comparc ion. X A I and 8 4 1 . 1 Cunsidcr a two-conipar~mentsystc111erl~lnvedby rigid, ~virliP~luhlc~ns iinpel-~uent~lc, odiuharic u.all5, and Ict thc two compartments be separated by arigid, insuI;itil~g.but pcrrrreahle wall. 1Vc assulrlc that the lwo cn~npnrtlncrltaare in equilibriurll but Ih:it thcy arc Ilnt i r ~eqr~ilihilu~n wit11 c : ~ l lothcr. Show that
where P is iobars - and 1. is in kelvins. Given that T,\,< = 85.45 ~ n n A,
'.
9-24. l!sr the rapnr p1,c~sui-edata given in Rr>blem 9-7 and the dcnsity data give11in P r o h l ~ m9I ? to calcr~latcAval,fifor rneth:inr,l frnm the lriple point ( 175.6 K) to the critic:il ~lninl ( 5 I?,(? K j, l'lcjt yolir reuilt. -
9-25. 1 J e the recult of the previous prohlem lo plot Asdp.Tor 111ctlla1101ft-on) the triple pnii~trt> the criliciil pclinl. 11,
4
11.
-- constant
Nn\t show that h)i- I h i ~<],utc~r~.
(IS =
P + -dV 7'7'
dlT --
-
g -rln
9-26. Uqc thc vapol PI-essured a ~ alrjr ~irethar~nl givcu in Proble~n9-7 tu plot In P aguin+t I i T . Ucir~gyour calculations from Problem 9-24. uker whal tclnpxiture range du you Ihir~kItle Cli~u.;iu\-Cl;ipt.yrrm equation will hc v;~lid?
7
9+27. Tllc lrlolnr c ~ ~ t h a l puTvaprjri,a~inn y of watcr i~*O.h5 k l . ~ n o l' at its normal hoillrlg ~!uirlt. Use the E'lauslus-Clspyron equation t o calculate the v n p r prcqsurc of watcr ;it 1 I 0 T. The cxptri~ncntalv:ilrle i u 1075 lrlrr.
S ) \ ~ C I I I:w ~ this rcsult I ~ ~ I C C I I Ythe ' I direction of a(icotherm;~l)1natcriaI How under r,lltfr~icalpotrtlti:lI tliflcrencr.
[(H-t l l i q :1
9-20. I)vtcrtnlrl~.lhe villue o f (17 / d P lljr ualer a1 i t \ normal boiling point c ~ f373.15 K givcn t l i ; ~ t t l ~ clrlchar tlnthalpy of vnporifalioi~is 40.65 kl.ruol-I, arid the de~~sitics ol- Lhc licluid ' and O.(>OlO f L I . rcspeclively. Estirnn~c[he hniling point nf ant1 baprlr are 0.9584 g,rr~I.V.al?r baric clcr~\l~re.:of liquid o l ~ dgaseous ethyl acetatc art. 0 826 g-inL- arid 1) 00319 g . ~ ~ t ~re\prctively. -l. a1 its rlorrrlnl hoillng point (77.1 1 ('1. The rate of change uf v;lpor prt'shure wit11 trII1pcrature i < 9.0 tc~rrK at thc norrnal hoilinp pnlnt. Eslimate thc rnolnr erilh:111>y uf rapori/airt>n of ethyl acetate at its normal boiling po1111.
9-28. The raprlr presnurc of benzaldehyde is 100 turr at 154 C and its normal hrliling point is 17Y('. Estimate its molar enthalpy 01- rapr!rrzatiorl. 't'l~ccxprl-ininnliil v;ilkle i c 42.50 kJ mol-I 9-29. Use rhe followirlg data lo ektirtlntc thc nortl~alboiling point nnrl the 111olilrtl\tll:~I~)y 111 vaporiz:~tiu~i of lead.
9-30. The vapor
preahurt
of solid iodi~lcis given hy
9-22, The v;lpc71 prcqsurt. uC rne~-c~iry from 4W'C lo t 300'C can k expreu\cd by
Jhc dencily of thc vaprlr at its normal b n ~ l i r ~pninl g is 3.82 E . ~ - and l [hat of tllc liquid is 12.7 g -nlL- I . Estirii;ite thc ~ r ~ u lrnthalpy ar 01- vaporizatiot\ 01- mercury at it< rlnrmal huilinp 1wi111.
-
9-23. The p r v w r e s nt the solid-liquid cwxister~ce boundary of prupune are givcn by [he empiric;~le q u a r i u ~ ~
Use this etluation tu calculate the nortl~nlsuhlin~ntiontemperature and the rrlolar et~tl~illpy of sublitnation of i:(sl nt 25 C. The experlrnrr~talvaluc of Ah,,,
9-31. Fit the folldwing vapor precsure data of icc to all cquation uf the form
r:~riir~s r , in cquilibriurrt wittt a vapur at a PI-cssure 1'. and a flat curlace 01- the same liquid in cquillhri~lnlw i h a vapor at a prcssurc {,. Show that the change in Gibbs energy for the isothcrnml Iransier o f dn moles of the liquld I m n ~the Hat surfice tu the d ~ o p l cis t
'I'his changc in Gihhs energy is due to t l ~ cchnllgc in surface energy of the droplct (the cl~nllgcin surface rncrgy r,f the largc. flat surlncc is ncgligihle). Shn% that
whel-e y ic tl~usurtilce tencion of the liquid and (1A is the charlgt. i n Ihe s11rf:ice arco of a dl-uplrt. Assumlng the rlruplet is spllcricnl, sl~owtllnt
d,= ~
4n r 'dr --
v'
--
9.15 A plot of reduced prewire. Pk, vercuc r e d ~ l c ~vdo l u ~ t ~I e'.,.for tllc vnrj d r r Waals cquatirrr~ar a reduced lelnpr1nturc, T , ,or 0.85, FIGURE
r l A = Brrrdr
9-47. The isolhcrmal comprcssihility, K , , ic dcfir~cdhp
P
hl- = I:,
2 y ~ I rH1'
Brcauw the ripht sidc is positive, we vee that Ihc u p o r pressurc of a droplet is greater than that of a ~ ~ l a l l surl:~ce. nr What if r + x:J
9-46. Figurv V.1 S $ l ~ c ~4 +rcri~lced + ~,re~surc. P,. plottcd agnir~slrt.rluced volunlc, i',. lrlr the a1 a reducer! trmpclntule. &. of 0.85. The srl-ci~lledvan dcr W u l + van der W:j,tls erluatic~r~ Irwp appalrnt in the fpurc wilt occur fur any reduced tclnlxraturt. less than ullity and is n cnn\equence of thc silriplitied limn of the v;1n der Wanls cqualinn. 11 turn? out that arb ot 9tatc ( m e lhar can he written as a hlaclaurin expansion in thc rcduccd arlalylic etl~~;itinl> drr~\itg.l;T, ) will giw loops for subcritical tcnlpcratuws (T, c- I). ?'he vorrcct behur,ior its the. prcqsurc is increased is riven by the pattr a h d k in Figure 9.15. The honri-rtmt;it regioli bdt. rjo~given hy the rat1 dcr tVs11c equation. represents thc cotldencalion of thc gas to a IiqilirI n! a fixed preswI,e. VV"V' can tlraw Ihe hwrizontni lirlc lcnIled a !if line) at tllc corrrcl posilian by rccogniring [hat the chenlical potcntials oithz liquid and the vapor musl hr equal at the poir~tsh arid f. Using this rcquircmcnt. Maxnrll showed that the hori~ontal IIIIC rcprcsrr~firigcondellsotiot~shuultl he drawn such that the a r m of the loops above and below the line must be equa1. Tu prove MII.YHYII'Sf q w ~ I - ~ r CotISrrucrIofl e(~ mltv.integrate (lip,,';3P), : :V by parts alol~pIlle puth hcdel and use the fact [that ,I' ([he value of at puirit f ) = ~ ~ V t ~h ael u e l l i a( j ~p o i ~ h~ )t to obtain
Bccnust. ( 2 f /a V )7 = Oat thecrilicnl poillt, h, rli\trges thcrc. A question that has perier:lred a great deal of cxprriment:il and rhet)rer~calresearch 1% Itre question of the rnanner in w h ~ d ) K~ divc~fics nh 7' a p p r < ~ a c l ~ Tc.~D t w s i r divcrgc a< ln(7' - 7;) or ~ r t i a p sas ( I ' - -' wtlcrc y is arlme ct.ttrtut t,.rpo!ic.rlt? An tally thrrlry o f thc bc11uvirlr of thrrn~odyrjanlic furlcliur~\ such as K ) tery near the crtt~calpoinr was propwed hy van dcr Waals. w h o prediclrd that u , divuges as (l' - T')-'. To cer how van dcrW;ial> arl-ived at this pretlicrion. wc cr)r~\idrrthe (doublc) Taylor enpa~~siotl u i the pressure P(V. 1 . 1 :lhout 7;, and 1:
Why are thrre no terlrls it)
Now show that
and Ihal
e,
w l ~ c ~ c i $ Itie prersr~i-eccurrcspnr~din,vtn Ihe tie line. ll~terprctthis rehull
(V -
y)or (I' y)'.?W ~ i t cth~sTaylor series a$ -
how let
1.' = vt to obtain
'Thus, ihe van der W x a l ~thcory prcdictc that the cnlical cxponeur h e n shown expcri~nentilllythat
:iccuru~ee*pe~-ir~icnt,il n ~ r ; ~ > u r c t ~ l c . n t as s u1f ~+ , '/;,suggest t h : ~uT d i ~ ~ r g i~ e little b niom
clrongly than (P'- Ti,) I. I n particular, it is found thi~tK, 4 (I' - 7')- whcre y = 1.24. Thus. the thewry of van dcr Wnals. nlthough qualipativrly correct, i~ not quu~titatively corrcct. 9-48. We can use the ideas of thc prcvious problem to predict how thc diffcrerlcc in the dc~~sitles (pl and p G )of the crex~stingliquid and vapor states (urthubarir det~sinps)behave its T + Tc. Substitute
111
Itus v;lsc is 112. It
11~15
where 0 = 0.324. Thus, as in the previuu\ prohlcm, alrhouph qunlrtatively corrcct. the V:III der Waals theory is not quantitatively correcl.
9-49. T h e folluwing data give the temperature, ~ h t vapor . preuurc, ntld he density uf lhr coexisting vapor p h s c OF butane. Use lhc van der Wwls cquatiwn and the Kedlich-Kwim:! equation to calculntc the vapor prcssure and compare your rcsult with the cxperimcl~lal values given below.
Into the hlaxwell equal-urea c.un$truction (Prr>blern 9 4 6 ) lo gel
v',
For P < PL.Equation 1 g i ~ e sloops and s o has three rclotc, TL,a ~ i d7'fur I' = I;,. We can obtain u hrct approximation to these rrmts by assuming t h a l v , x 7')in Equatitln 2 and ucrlting
:(v'+
9-50. The Fullowillg data give the tempmture, the rapor pressure, and the density of the coexisting vapor phase of benzene. Use thc van der Wil5 cquation and the Kedlich-Kwong equation to calculate the vapor pressure and compare your rc.\1111 with the cxperirnc~~t:~l s arid 2.18 with TL= 5h1.75 K and Y = 48.7575 bar valucs given M o w . Use E q u a t i o ~ ~2.17 trl calculate the vall der Waals para~neterhartd the Kedlich-Kwong pmrrjuters.
To l h ~ approximalion, c the lhree roots tu Llualion 1 are ohpained from
Show tllnt the three rtmts are
VI = ,Ly-
v:
=
v,
h o w show that
and that this c q u a t i o ~is~ uqulvalcnt to
(d ) i z C
l
,
-
T)]!~
9-51. In Problems 8-57 to -3, wc considered only cases with conslanl n . Use the approach of Prohleln 8-57 to repartition n between two in~tiallyidentical comparlnle~lthto show that ( a ' ~ , ' i t ~ ' ) ,I > 0 and tl~nt( d p / a nj , . , > 0.
CHAPTFR
10
Solutions I: Liquid-Liquid Solutions
In this anri thc next chapter, w e will apply our rherrnudyliatt~icprirlciplcs to sr~lulinr~.;. This chaptcl- fbcu';cs on sulutic~nsthal consist oC Iutj volatile liquitls, such as nlcnholwater solutions. LVc will first discu<sparlial rrlr~larquatjtitirs, which probide thc Inort cotlvenic~ltsrt c ~ thcrmorlynamrc f kariahles to describr .soluticlns. l h i s diccussion will lead to rllc Gibhs-ntrlicm cquallt)l>,which gives 11sa relaliori betwrer~the chaoge i r ~ the prowrties nC utle con~ponet~t of n soluticln in rcrms ul- the change in the propertics of (he olher corrjpurient. '1 he si ttlplest 111odel of a solution is an icIcal solution, in which both co~nponenlsobey Kaoult'h law o$er the entire cnmpnuplin f i x st~lutionsbehave alrrlost ideally. triost solutions are not ideal. Just as nonidcal pi~xc.; can be described in tcrms of fu'ugacity. nonideal snluiinns can he described in tesfns of n q u a d ty called activity. Activity rnust k calculated with rcspect to ;I qpecific stanrhrd stale. and in Section 1 ( k X we ilitroduce two com~nonly-uscdstandart1 states: a wlvent. or Raoult"; Iilw standard state, and a solute, or Henry's law statidard state.
10-1. Partial Molar Quantities Are Important Thcrniodynamic Propcrt ips of Solut inns lvrl Hildebrantl 11 ill Citrtldcu, h'l. on Nr)vcrnhei- I h. I H X I , : I I I ~dicd in 1083, t lc 1x11 hiz, Ph.l) in <,l~cin~.;rry Irom the Ilni~rrsityof FJenns)-lvar~in 111 1406. After spc~~dirlp u ~cccir~.d !car ~ I Ihe L Ilniktr~ityor Bcrlin will1 Yer-11st.he rrrurned to thc lirliver\ily uf Rnusylvnrria au ;a11 i r i < I r i ~ ~ I111 c ~ 1' , 4 13. tic jiliile(1 [he I>cpnrtlnctitof Cllctnistry a1 rile I 'niversity of Calili~i-nia :jt 1lt.1kclcq, where he xlayetl for thc rcrrlairlder of his lik. Althuugh hc ott~ciallyelired in 1952. 111.rcrr~;llrit.dprt>fi<siot~u[ly octivr unlil his death, puhtishi~~g h ~ last s paper in 1981. Hildehrand ~i~;ictr siy~ihcnritcontnhritiul~sto thc 6cIdh or liquids and nunelectrolyte snluLirln\. He rctnincd ;i Iorip interest ill dcvlationc Iroin ide;~lqolutionc 1Kaoult's law1 and the tl~coryur regular sulutior~cHis hooky. Y hr Smluhilif~r!J'~T~~:,llzplrrtrv)Ivt~s and Ilr~gubrS o l u r i o ~ published ~~. wiih Rrlhert Scott. ncrc ?taritlartl rel'rrer~ccsit1 tl~cficld. Hildcb~lndwar 3 Fanicd. uxcellent teacher of p~'t~uraI C~CIIIIC i ~IU I[ -eyr k c l c ~Hi5 perieral chemisrry text. I'rit~r>iplear!f Ch~lt~isrrx, influcucetl ~ l t l ~ c . r>c.l~oolz, trr pl:~ceglc,Itcr c l ~ ~ p h ar~l ri ~kp~iric~plcs;III(I lcsr on thc ~rlcrllorualionuf cpccific 111:1rrrial ill tl~cttxhing :ri1e1:11 chctlli~try.Hildct~ronrl\%,a\a great luvcr Itit. ourduors nrtd cq'c'c~allyei~juyctlqkiil~gar~ilcamping. Ile managed thc U.S. Olympic Ski 'lcnlrl irr 1436, was I'residc~~t nT the Sierra Clrib fion~1437 to 1940, and wrote ;I book on catnpir~gwith ltis daugh1t.r 1 ruli+r.C-btt~p ( , ' ( I / C , I - ~ I I Kor. HoL~, ro r ~ i ~ ~r l ~< - , ~ i h l ihiktv-,~, / ~ roitIprrs, I ) I O I ~ I I I ~ I ~ I TC( (, PI I~~T( ,I ~ ~ S I , T , / l f l l l l l ~ l - ! ti;<,:,,<, .
(llll/ f i , ~ / l l ~ l - l 7 1 ~ 1 1 .
Up lo thi\ poit~t,uc havc riiscusscrl thc t t~crrnt~dyr~iuiiis r>f o ~ I yonc-ct>lnponc.r~~ syhtclnq. LVc will now diccuss lht. tIic.r~nr,dynamicxo1'rr1i1ltic.o111p011eril systerrls, althtli~yh. for si~npl~uily, w e will discusx only syslerns oC lwu crllnpoIirrlts. Most of tllc coilsystcms. L.ct's cepls and r ~ s u l l sw e will develop are npplicnhle to multicompo~~cr~r c~)risidera solution cor~sistit~g of H~ 1110les of cornponcnt I and 11: molcs of cumponcnt 2. The (iihhs energy nl' thiv solution is a functiorl of 7-and P and the two ttlule numhcrs n , and 1 1 , . We ernphasi~ethis dependencr of G ori ~hesevariables by ~ . r i t i n g G = G ( T . P,n , , n ? ) .Thc total derivative uf- G is givrn by
10- I . P , j r l ~ .?rllll;lr ~l Q L I ~ I ~ I I I Arc I ~ + lnlporta~itT h ~ ~ r n i ( > d v r ~I(' ~r (r ~t ~p~r ~ r l ~0 1~ S \ ~~Iul~otl~
IC the cutt~positic>nof t l ~ csolution 1.; I~xed,s o Lhat rill, is the sarne us Bquiitir)n 8.30. ant1 wc have
=
rl11, = 0, the11 L q ~ ~ i ~ t i10. u nI
At coilstant I' o t ~ fP . ruc hakc
Now, imagine that
w e increase the size of thc system unifi)rmly by Inearls of :i s c : ~ ! ~ parameter i, such that On, = I I , d l and dn? = r~,u'h. Notc that as we ~ ~ 1 i - fro111 y 0 11) I the nurnher of moles of componcnta I and 2 \ a n t s from O to 11, and 0 to 1 1 ,, I-especli\el). Bccausc G dependsex~ensivelyo n r t , andr;?, wc musl h;~vethul OG = Grlh. Tlicrc.icjrc. the total Gibbs cncrgy varies h ~ Om to some final value G as 1 I < va~-ied.Ir~truducit~y rlL into Eqilation 10.5 gives
.
.4s irl tlic prcvious chapte~;the parlial derivatives of C; wilh respect to lllolc numbers
are ci~lfedcl~emicalpotenti;ils, or partial rnolilr Gibba energies. The standard notatiun lor chrrrliv~lpotential is p , so wc can write Equatiol~10.1 a< n , . and tr2 are final valuca (antl so do no1 depzr~dupon h ) antl / I , and p , arc iintcnaive variahles (arid so do not dcpcnrl upon lhe size pamfnetcr >.). wc can wi-ltc. the abovc cquation as
Bccause G, where
We will see that thc chcmical potenlial ot- each componenl I r l llje solution plays a central role in dctcnnining the thern~odynaniicpmperlies of the solution. Other extensive thcrmt~dyna~nic variables havc associaled panial molar valucs, al1t11,ugtionly the partial molar Gibbs energy i.s given a special symhol and rlaIne. For cxsmplr. (aS/a,i, j, ",-, is called the partial lnolar entropy and is ticnoted by S,, and
7.
i i t , is callcd the partial molar vnlun~canrl is deno~zdby Gcncrally. f. if Y Y ( T . 1', , I , , n , ) is some ttx~e~lhive thcr~norlyni~rr~i property, thcn ilx ;ic
-
7.
is a Inrasure of how Y changes when 11, is Physically, (he putial ~nulasquantity changed while keeping T. P , and the other mole numbers fixed. l?i~tial mol:~r quanti~iesare in~ensivetherintxiynamic quantities. In fact, for a ptirc systcm. thc chemical potential is jusl (lie Gihhs energy per mole. We can usc the il)tcutc: cwrnplc. w e ail1 cot~sidcrit hitrrrr;~sol~rrior~. tlxit is, onc cornl>osctlr,t two dii-fercn~liquids. 'Ttic Gihhs energ! o f a hinarj solution (Equatlon 10.2) ix
or. up011 itltegratioi~.
Note that G = /LIL for a one-c~rnponcntsystcm, which shows once again thal ,r i:, Ihe Gibbs energy per mole for a pure system, or 111ore generally, that the partial lntllar qi~ilntityof ;I p~rrc~ u h s l ; ~ ~ i.s l c ri l h lt~ol;~r q~lanlityor any extensivc thcrmc~dyni~miv V:I~IIC.
I'artial I I I O I ~ I - qui~111ities have ii pii~-lic~~l;r~-iy I I I C C pl~yhic;~l I I ~ ~ L ~ ~ ~ > I -111C [~C; II - ~I I~ILII ~ I I I volunle, for which the equivalerlt equation to Equation I0.h would br:
Now,whet) I-p1,opanol and water fire mixed, the final v o l ~ ~ of ~ n[he e solutiotl is rlot c q u : ~ ] IO the sum of thc volumes of piwe I -propand n11d water. We can urc Ilrluntion 10.7 lo unlculatc thc final volume of a solurion of a n y compositiorl if wc ~ I I O I V rhc p;lrll;ll ~uolarvolulncs uf I -l>ropan~>l and water iit that cornposiliorl. kigurc 10.1 .;how.c 111c. pi~rtialmc-!la1 v u l ~ ~ m c01.c I -p~-oparlol; I I I ~cvatcr a.; a lunclior~of the rrlolc I'r;iclirlii ol I -propant>l in I-PI-opanol/watcrsoli~lionsat 20 U. Wt:cirn uhc Ihls I I ~ L I Ilo - r .e\ljrll;itc the final volu~ue'of solution when I(X) rt~Lof I-plnl>;rnol is rr~ixedr\ith 100 tnl. ol wakr at 20-C. 'The dcnsititc I' 1-propano1 artd watcr at 20. C' are 0.803 g.n-11. and O.LlY8 g.rriL-]. respcctir.cl> . .,ing thrse densiticq. we ste (ha1 100 mL ciicli of I propano1 and water corresponds to a niole fraction of I -prr,piulol of 0.194. Rct'crri~~g
10-2. The Gil,I)s-Duhpni Equation Rclatcs tlw Change in the Chemical Putcnt~al
301
Furthermore, by using the fact that cross second partial derivatives are equal, we gel
and
If
we
subsiitute these two rckults lnlo
10.1 The partial molar volumes uT I -propanu1 and water in a 1 -propanol/water solutivn a1 20LC plr~l~rrl against the mule fraction of I-propanol in the <elution. FIGURE
-
= 72 rnL,mol- and we see that this corresponds to roughly l/MdIrr = 18 ~ i i ~ . n i u l -Thus. '. the final volume of the solution is
to f-irurr 10.1,
-
which is an exlension of Equation 8.30 I O rnulticoniponent xysterns.
E X A M P L E 10-1
Derive an eyuatir)t\ for thc temperature dependence of p,(7-.P ) in analogy with t l ~ c Gibbs -Hellnholtz equa~inn(Equalion 8.W).
S OF L' 1 I (IN: The Eiihbs-Hcl~uhullz eq~ralinr!i s (Fquation 8.60)
rompared wilh il total vcllurne of 200 m1,before mixing. Problems 10-8 through 10- 12 irwc>lr.cthe deterrrlination of partial moIar voluines from solution data. Now differenhate nith ~ ~ s p eto c t11, ; ~ n dinterchanpc t t ~ corder o f differentiation oli Ihc
10-2. The Cibhs-Uuhrm Equation Relates the Change in the Chemical Potential of One Component of d Solutiun to the Change in thc Chernical Potential of the Other
left side to get
Most of our thermodynamic fi~rmulasfor single-component systems (pure substance>) have a~~alogous formu1;ts in terms of partial molar quantities. For example, if we start with G = H - TS attd diverentiate with reqpect to n, keeping T, P. and n,+, fixed, ~e obtain
where
g, is the partial molar el~thalpyo f component j . 1
We will now derivc one of the tnost useful cquatinns inv~dvingpanial molar quantities. First we differentiate Equation 10.6
an,
and subtract Equation 10.5 to get
rh;lpter 10
1
Solut~onsI: L~quid-L~qu~d Solutions
10
(
AI Equ~l~hriurn, thc
If w e riividc h t h sides by n , -tn,, we have t ,I
,
+/
=0
(constiint T attd P )
(10.11)
and n, are mole fractions. Either of tcjuatlons 10.10 or 10.1 1 1s called the Gibhb-T)~dhml ~qumttutt.The Gibbh-Duhem equation tclls us that ~f we know thc of one component as a fi~nctionnf colnposition, wc c;in dctcrminc chclnlci~l~lotent~al the c~tlicr.For example. suppose we were to know that whcrc
A,
over rhc wholc rangc of x2 (0 to 1 ). A superscript * is thc IUPAC notation for a property of a pure substance, so in (his equation, p; = p , ( x , = I ) is the chemical potential OC pure component 2. We call diffcrentiate 9, with respect to x? anti suhstirute into Equtltlon 10.1 1 to gct
But
= - d x , (because .il t I;, = I), so
a h e ~ eO 5 x , 5 I because 0 :- x1 5 1. Now ~nlegrateboth sides frotn n, = 1 (pure cutnponent I ) to arbitrary x , ti) gel
where 11; = p ,( x , = 1). We will see later in this chaptcr that this result says that if one compment uf a binary solution obcys Raoult's law over the complete cuncentration rangc, the other component dues also.
I
I
E X A M P L E 10-2 Derive a Gihh\-Tluhrm rype of equation
for rhc volu~ricof a hinary holr~liun.
S (71 (1.1 I C)N : We start with Equation 10.7. which is Ihc analog of Equation 10.6 -
V ( T , P , n , . isi)
= n l Vl
+ n, V?
and differentiate (at uollst.?l$ T ntid P) to oblalri dV =
r,,r~-ci +Yrln, +
rr2tfK
+Fdnl
Si1blr;ic.lIhe analog o f Equation 111.5
+
d 1' = y d t ~ , F d n 2
(cvnstilnt
T and P )
vJap
Chemrcal Pott.r~t~al # i f L A ~ Component I
n,dy
f n,dV, = 0
H.15
thc 5.1rr1c. l',>lurIn Larh Phasc
(conslant T and P j
This equation cays that if &c k ~ ~ othe w change i n lhe partial ~ ~ l o l wlurne ar uf one component of n hinary syhtcrn over a r a ~ ~ g e co~npo~~licjll, wc can determine Ihe change in the partial molar volulnc of the other cr,mponerit orcr thc same rangc.
10-3. At Equilibrium, the Chemical Potential of Each Component Has the Same Value in Each Phase in Which the Component Appears Consider a binary solution of two liquids that is In equilibriutn with its vapur phase, which contains both components. Examples are a solution of I-propanol and water or a solution of hcnzenc and loluene, each in cquilihriurll with its vapor. We wish to generalize our treatlncnt in the previous chaprcr, in which we treated a pure liquid in equilibrium with its vapor phase, and develop thc criterion for equilibrium in a 1~in;iry solutioti. The Gibhs energy of the solution and its vapor is
Let nyn, n;" and nYP,niq be the mole nutnbers of each component 111 each phase. Fur generality, Ict J dent~leeither componcnt 1 or 2, su n , denotes thc number id' ~nolesof cornpoilent J. Now suppose that dn, moles of component j are transferred ftonl thc solution 10 he vapor at constant T and P, so that dn? = +(in, and drt;'" = -dn, 'The accompanying change m the Gibbs energy 1s
If the transfer from the solution to the vapor uccurs spontancously. then d l ; c. 0. Furthermore, d n y > 0. so must be less than In order that dG < 0. Thrrefore, molecules of component J move spontaneously from the phase of higher chern~cal potential (solution) to that of lower chemical putentral (vapor). S i n ~ i l ly, a ~1Fp;"" 2 I L ~ ' " . then moleuules of component j move spo~itaneou~ly [rot11 the vapor phaw to the solut~onphdse (d11:"~ < 0). At equilibrium, ahcre d G = 0. we have that
Equation 10.12 holds for each componcnt. Although we have discussed a solution in equilibrium with its vapor phase, our choice of phases was arbitrary, so Equation 10.12 is valid for the etluilibriurn between any two phases in w h ~ c hcomponent .j occurc.
393
Cl~.~pter 10 / Snlutions I. Lialr~id-L~qu~d 5nlutinn~
'The irnporlaru rcqult here is that Equation 10.12 says that the chemical potential uf each colnponetlt in the liquid solution phase can be ~neasuredby the chemical potential of thal uoinponcnt in the vapor phase. If the pressure of thc vapor phase is low et~oupli that w e can consider it to bc ideal, then Equation 10.12 becomes
~vhcrethe sta~ldardstate is taken to be lion 10.13 hecornes
q"= 1 bar. For piire
FIGURE
&t~crcthe superscript * represents pure tlirluid) component j. Thus, for example, @ : ( I ) is the chemical potential and q"is the vapor pressure of pure j. If we subtract ~ ~ u a t i oI U. n I4 from Equation 10.13. \ve obtain
Equalion 10 15 1s a ccl~tralcquatlorl i n thr study of binary sc~lutiotls.Nnte thal 1~:1" + 11: rlb q -+ q*. Strictly spe,tki~lg,wc should lice I'ibgacitic'i (Section 8-8) irlstead of pressures i n Equation 10. IS. but usually the magnitudes of vapor pressures are such that prrssures ;Ire quile adequate. For example, thevapor pressure of water at 293.1 5 K i~ 17.4 torr. or 0.0232 bar.
10-4. The Comporlcnts of an Ideal Solution Obey Raoult's Law Cor All Conccntrations
10.2
h n~oleculardepiction of an ideal sulution. The two t y p s of molecules are distrihuled through nu^ the sulutiun in a random
component j , Equa-
manner.
j rnc~lcculcs.Rccausc the j 1nolecules o t ~the surfisr are (he molecules that can c\cnpr into the vapor phitsc, the palqial pressure is just .r, q*. According tu RanuIt's law (Equation 10.16) and Equalion 10. IS, the chcmical uC utmponent j in the solution is give11 hy poten~ii~l
7
Equation 10.17 also serves to d e h e an Ideal solutir,n ~t ~t 15 valid for all values of A, (0 5 .Y, 5 1) Fu~qhermore,we sho*ed in Seci~on1 0 . 2 that if one cu~nponentobeys Equation 10.17 from .r, = 0 to x, = I. then $0does the other The total vapor precsure over an deal qolution is givcn hq
Therefore, a plot of
Sot,, against x2 or.^,) will be s straight line as shown in Figurr 10.3.
A I'ew solutions have the property that the partial vapor prcssurt: of each cornpatient is given by the sirrlple equation
E q u ~ ~ ~ i10. o nI (1 is called Kuorilt '.I Iuw, ant1 a solution thal obeys Rar,ull's law over the elitire colr~l~ositic>n rarlgc is said lu be an ideal anluriolr. The ~nolecular~>icturt.behind an ideal hinary solution is that the twc~lypes of ~nulccr~les are ranrlomly distributed thrtlughout the solution. Such a disu-ibulion will occul- it ( I ) the ~llnlcculesm roughly the same size and shape. and (2) the inlermoleculitr fi~rcesin the pnre liquids I and 2 and in a ~nixturt.of I and 2 are all similar. iVc cxpect ideal-solu~ion6eh;jvior only when (he molecules of the two conlponents are sitnilar. Fnr example, benzene and toluenc, o-xylenc and p-xylenc. hexane and Iirptar~c,and hrorrrocthane and iodocrhane [urn] esserltinlly ideal soludo~ls.Figure 10.2 depicts an ideal solution, in which thc two types r ~ molecules f are randomly dihtributed. '1-l~e~imlefriic~ion1,reflects the Craction uf [he solution surface that is occupied by
F I G U R E 10.3
A plot of PtVIaI againsl x ~ Lor a~ soiutivn , uf ~ benzene ~ and tnluctie at 40.C. T h i s plol shows that a benzeneltnluc~~r solution is essentially ideal.
I
E X A M P L E 10-3 I -1)n)pilllol;~nd2-prol>arir)lCorn1 ehsenlially all idleal colutioil al all cuncc~~tr,llicln:, nt 25 C. Leiling thc ~LIbs~liptk I und 2 denote 1-pmpanul and 2-propanol, re\ptlcliveiy, ant1 gikc11that P i = 20.9 torr nod Pi =: 35.1 turr at 25 C. ci~lculalethc total vapur pcssure and tt~ccnrrlpusilirln oi the rapor phave at - 3 , = 0.75.
I
t:,,.,,,( A ? 1 0 . 7 5 ) = A , P ; + x 2 P ; = (0.25)(20.9tom)
+ (0.75)(45.2 tom)
= 39.1 torr
dentltc thc lilvle fraullor, ot cach co~!rponenlin lhc vapor phase. Then, by Llalton's law ut partial preswres. I ut
);
I
'-
' 1
-
-
, =.-=P,
-
p
~
39.1 lorr
P7" -=-~
--
., .
(0.75)(45 1 torr)
1,
- 2
+
, I ' = (0.?5>[20.9torr) . - 0.13
p,c>l,,~ pa~>~.l~
~ p ~ L~> ~ , 8d~
~
= 0.87
39.1 tom
Note (hiit y, y, I . Alho rmtu thal the vapur vr)latllc component.
ik
FIGURE
10.4
A pressure-conipr,sition diagram for a I -pmpwol/2-propanol solulir)n, uhich for-nn un essenlially iclcill solution a1 25'C. This figure can he calculated using the approach in Example 10-3. Ttre upper curve (called thc liquid c ~ ~ r v represcrlts e) ~ e r \ u cr , , thc lnolc fractio~iof 2-pl-c~panrd111thc liquid phace, and thc lower curve (callcd ~huvnprv LurYc) rcpIcserlts I:oldl vcrsus sl, the mclle lractiun of 2-prclpanol 111 ~huvapul p h a w Tlic rrrcl poinls marked fiy x repwscnt the values o f xl and !1 from Exarr~plc1I) 3.
of liquid and vapor phase in the fullowing way. The mule fractiot~sin the liquid vapor phases arc
rlchcr than the solution in 111e more
Proh1c.n) 10-1 5 has y o u expand Example 10-3 by calculating I:L>lol as- a t unction of A ? ([he molc flaclion of 2-prupannl in the liquid phase) and as a funclinn of y2 (the mule friiction nf 2-propanol in the vapor phirse), nrtd then plotting $t,lJi against x. and v2. The rest1111ng plol, which IS tart i l t ~ h tpuirl~ ' q,..I , iri Figure 10.4 :111dInwcr [Ire ~ ~ r t ~ + s u r c . At t t ~ cpoinl PC,,r(,, t t ~ cprrrsure cwcceds rhz rdpor p ~ s s u r cof the solu~ir>n,so the region above the liquid curve cunsiats ol'orle (liquid) phase. As (he pressi~reis lowered. we reach (he point A, where liquid starts to vaporize. Alnng [he line AD, [he system d in equilibrium with cach other. At the point B, all the liquid consista c~fliquid n ~ vapor ha5 ~ a p c ~ r i ~ atid e d . the regio? below the vapor curve consists c~fone (vapur) phase. Lct's conhider the poinl C in tl~cliquid-vapor region. Puint C: lies or1 a line cnnnectir~gthe composition of liquid (x, = 0.75) and vapor (y2 = 0.87) phases that we calculnted in Example 10-3. Such a litie is called a rir line. The overall vompusition uf the two-pl~asc(liquid-vapor) system is ,r',. We can determine the relative amounts
A', -
=
ni - 11;
n',
+
H!
n'
and
11
y 2 = -n;dP --
yr
+ Hyl'
i~~iil
>:,;,
H:
-
where nYnPand n1 are the total numhcr uf moles in thc vapor arid liquid phases, rcspcctively. The overill1 mole fraction at .x4, is given by thc total nurrtber of mtdes (11 component 2 divided by the tola1 nil~riberof moles
Using a material halance of the number of rnolcs of conlponent 2. wc have
Thls equatlon represents what is called the Irvpr nil? because I I ' " * ( V ~ - x u ) = ,,I( t a balance or each value ut "n" lime^ the distiance frorit each curve to the point C in Figure 10.4. Note that rrl = 0 when I,<= x, (vapor curve] ~ n d that n'" = 0 when x(, = x? (liquid curve).
x,) can be intcrpreled as
398
Chapter 1U / Sulutinns 1: Liquid-Liquid Sulutionr
r
E X A M P L E 10-4
Calculate thr rclatire nmuunts ofliquid and vapor phase< at an overall compwsition of 0.80 for Ihe ~,;tlucsin Exa~nplcI 0 3. S O L U T I O N : I n t h ~ ~ c a sA;,r .; .0.81).
x, = 0.75, and!,
= 0.87 (see Example 10-3).
SO
Accrvdinp to Example 10-3, the Inole fraction of 2-propanc~lin the vapor phase cquilibriuni with a I-l1ropanul/2-prop;inol solution is grcater than the mole fraction of 2-prupanol in the solution. We can display the con~positionelf the solution and vapor phases at varic~ustemperatures by a diagratn called a ternperatuw-composition ciirr~rorn.To construct such a diagram, we choose some total ambient pressl~resuch as 700 torr 311d wrirc ill
F I G U R E 10.5
A te~npcrature-compositiondiagram uf a I-prupanull2-propandfolu~iot~, wh~chis c~scntiaHy an ideal solrltiun. The boiling point oI- I-pnjpannl is Y7.2'C and thnt of 2-propanol is 82.3.C.
because the total pressure i s taken (arbitrarily) tn be 760 torr. We saw abuvc that = 0.59 at 90 C, so w e have th;it
x,
which ia I;~hcllctlhy poinl h in Figure 10.5.
\Vc then chi)osc some tcmperdturt. I,ctween the boiling points of the lwo corrlporlents and solve the above rquaticm f u r x , , the cornpositon of the solution thal will give a totdl pressi~rcor 760 lurr. A plot of tertlperature against x1 shows thc boiling temperature (at = 760 lorr) nl-a solution as a Cunction of its uornpasiticln (.r,). Such a curve, Iirhelcri the solution curve, is shown in Figure 10.5. For example, at I = 9WC, P; (the vapor presqure of I-propanol) = 575 torr and P; (the vapor pressure of 2-propanol) : :1027 torr. I'hcrefi)re.
- 1
P; - 760 torr ------=I' - P
1027 torr - 76U tom = 0.54 1027 torr - 575 tclrr
E X A M P L E 1U-5 Thc vapo~pressul-es (in tom) o f I-propannl and 2-prrlpur~ola h n iutlctio~~ uft11cCelsius ternper:ilu~-e.I , itre giben by [he c~~~piric:ll Ir>~-lnuI:~s
and
Use thesc fnrtnula~tocnlcu1atexI and values given I n Figure 10.5.
at 9.1.OC, and cumpare your rr\l~ltswith ihe
Tlie pcl~ntcorrcspottding 10t = 40 C and x, = 0.59 is labcled by poit~trr In F~gurc10.5. \lie can al\o calcr~latethe cnrrespundi~~g ccmtposition of the vapor phase as a function of
telnpratulc l'tte ~ n o l efrxtion of componerlt I in the vapor phase is given by Dalton's I~IW ),
r
=-.L-- X1 p;
760 torr
7M torr
or
P,' = 647 turr. Similarly. P; = 1150 lorr Tl~erefnre. XI
=
P; - 760 tom - 1 '
-
-
- 760 torr -. I1.77 l150lorr-6371orr
1 150 tom
I
10-5. Most
5ulutlons ,\re
Not 1dc.d
where G ; and G;are tlic Gibbs cnergies ot the pure components. LJsing Equalion 10.17 for an ideal solution gives
and
A
G ' =~H I P ; " '
ill agreement w ~ r h[he valurc st~ownin Flgulz 10.5.
'l'hc tcnlperiirure-composition diagram can k used t iIluslrate the process of fractional distillation, in which a vapor is condensed and lheli re-evaporated Inany time< (Figure 10.6). I f we wcrc 10 start wirh a I-propanul/Z-propa~lolsolutiun that has n mole Craction of 0.59 in 1-propanol (point a in Figure 10.5). the mule fractir~nof l pmpar~olin the vapor will be 0.45 (poitlt b). If this vapor iscondensed (point c) and thcn re-cvaporatcd, then the mole fraction rlf 1-PI-opanulin thc vapor phase wrll be abuut 0.30 (point d). As this pnlcess i s continued, the vapor becomes increasingly richer in ?-p~upanol,eventually resuIting In pure 2-pn~paaol.A fractional distillation column diCl"ers from anordinary dislillation colu~nnin that the former is packed with glass heads. which provide a large surface area for the repeated condensation-evapora1ion process. propenies upon forming an ideal We can calculate the change irr thenlindyl~iim~u solution from its pure components. Let's lakc the Gibhs etlcrgy 3 s an cxi~mple.We clelinc the Gibbb energy nf mixing by --
I;;(T. P , n , ) - G i ( T , P . n,)
- ,l,)4; - IJ+;
= R T ( n , 111.ul -k n l In x,)
I
A,,,,% c; = G'I" ( T . P , n , . I [ , )
+
(10.20)
(10.11)
This quantity is always negative because x , and x, are less than one. In other uords. iin ideal solution will always f u n spclntaneuusly fro111 ~ t separale s components. '1.11~ entropy uf mixing of an ideal solutior~is giver1 by
Note that this rcsult for an ideal suli~tionis the same as Equation 6.26 fur lhe mix~ngol' ideal gases. This sin~ilarityis due to the fact that in both cases the molecule:, in rhe final sotutinn are randomly mixed. Neveflhzless. you should realize that an ideal solution and a mixture of i d u l gascs differ markcdly in the interactions involved. Althouglr thc molecu1c.s do not interact in a lnixlurc of ideal gikses, thry interact strongly i n an ideal solution. LII an ideal ~olution,the interaction:, in thc mixture and those in thu ~ ~ I I ' liquids are esser~tiailyidentical. 'I'he volu~ncchange u p n mixing of an ideal xoli~tiunis given by
and the enthalpy of mixing
IS
C
(see Eq~~atirjns lo,?1 and 10.22)
( I 0.?4) A rnrx H ' L11mA xG ~ ~ s T A ~ , , ~ s ' ~ = o
Therefore, there is nu volunle change upon ~nixing,nor i5 there any energy l i c , ~ ~ absorbed or evolved when an ideal solutjo~lis formed froin 11spure cornpunelits. Both Equations 10.23 and 10.24 result from the facts that the moleculex are roughly the same size and s h ~ p e(hcnce Aml1vtd= 0) and that the t ~ r i o u sinteraction energies are the cnmc (hencc A , n , 5 ~= ' d0). Equaticlna 10.23 and 10 21 are ~ndecdobserved to tK: true cxperi~iientallyfor idcal ~ o l ~ ~ t i uFor n s .most solutions, ~ O W C V C T ,A,,, H and A,,,,T I' clo not equal zero
Pure l i q u i d to cul)ection H a c k
.
Cr)lu~rlnp a t k e d w i l h glass rings
10-5. Most Solutions Are Not ldcal F ~ G U K E 10.6
A si~nplt.fraclior~aldiclillatiun column. Because r e ~ a t e duondensat~onand n.-evaporation wcur along the entire cululnn, lhe VRptlr bcconleh pwgresivcly ncher in the more vulahlc vulupmlent ac 11 mo\,es up i t ~ cuulurnn.
Ideal solutruns yrt: not very common Figures 10.7 and 10 8 show xfapor pressure diagram5 for carbon d~sulfidelditncthoxymcthdt~e [(CH,O),CH,] solutionh and tnchlornmethanelacctone solutions, rt.spectively. Thr behawor in Figure 10.7 5hows so-called positilc deviatic~nsfrom Rauult'x law hecnusc the llart~alrapor prcssures
F I G U R E 10.9
F I G U R E 10.7 T t ~ cv:twr , ortssure diacram ul-n carbon disultideldimethoxymethane solution at 25'C. Thi\ system chows pocitivc deviations from ideal, or Kauult's law, hehaviur. --
.
'I he vapor prescure diagrarn ot nlcol~ol/wdtrrsolutiuns i i s a f ~ ~ n c t l rol l n the riunjhcr o f carboll atvrns in the alcohols, showing increasing deviation from idcal behavior. The dashed line correspund~to ~netha~lul. Ihe dotted line 10 et11at1nI.and the dashed-dotted line la I-prrlpar~ol.
Figurc 10.9 sh(~wsplols of tllethanol. ethanol, :~ntlI-prc)panol vapor pressures i r ~ alcohollwlitcr 5olutions. Nole Iha~the positive dcviation from ideal hchaviur increases with thc size of the hyrlrocarbot~part of thc alcohol. Thih hehavior occurs because the water-hydrocarbon {repulsive) inieractions bccomc incrc:~singlvprevaler~las the s i ~ e of the hydl-rxnrhon ch;un Inureithes. There are sutne inipcl~Vantfcatul-es lo liotice i n Figures 10.7 and 10.17. I.ct's hlcus on cotltponerlt 1. Thc vapor pressure of uornponenr 1 approacl~csits Raoult's lam ~ a l ~ r r as x l approaches I. In an equation. we have (hat
F I G U R E 10.8 Thc vapor pressol-e(Iiagmln of a ~richlorun!ethanelacetonesolution a1 25°C. This system shows ~~cpalivt? deviakionq from ideal, or Raoult's Iaw, behavior.
of carbon di.sitlfide and dimcthoxyrncthane are grcatzr tha~lpredicted on the hahis of- R a o ~ ~ l tlit%. ' s Physically, positive devialions occur because carhon disulfided i r n e t h o ~ ~ ~ r ~ e tintertlctions h:~nr arc more rcpulsivc than either carbon disulfidexarbon
disulfide or dimcthoxytnctlia~ie4i1nethoxy1n~thane interactions. Ncgalive dcvia~iotls. or1 the other hand, like those .shown in Pigure 10.8 for a trichl(m>methane/acetone solution. are rluc to stronger unlike-njolccule ~ r ~ t e r ~ ~ c than t i o t ~likc-molecule s interactions. Problcm 10-36 asks you to stlow that if one component of a binary solution exhibits 1)ositivc deviations Crom ideal behavior. the11 the other component must do likewise.
10.25 trom Figi~res10.7 and 10.8. it is generally truc. Altho~rghu e deduced Equaticl~~ Physically. Lhis behavior lllay L x attributed tu the fact t h a ~there are so few compunei~t-2 molecules that 1r1os.t component- I ~nolcculevsecc~nlyother componetlt-l molect~lcs,co that the snlu~innbehaves ideally. Ranult'c law behavior is not observed forcomponcnt 1 as x , 4 0 in Figures 10.7 and 10.8, howcver. Althwgh not easily seen in Figures 10.7 and 10.8, thc vapor pressure of component I as x , -+0 is lincal. in x,, hut the slope is not equal to P; as in Equation 10.25. We emphasize this behavior by writing
,
,
In the special case of an ideal stllution. k,, = P;, but ordinarily k , S; P,' Equation 10.2h ls called Ilrtin~'sInti8, and k,, i'i called the Hrrrm's Itrw uonsrnnr c ~ t comporlent 1. As x, -. 0, the cunjpurient-1 ~noleculerare cntnplctclq si~rrt)unded by camponent-2 nlolecules, and thc value of k , , retierls the ~r~terrr~olecular i~~tcmctions hetueen the two co~nponcnts .4s u , + I , on the olher hand. the compnncnt- l
,
,
10-6. 1 hr Gil1t~5nuhem Equatiun
~noleculesarc completely surruundcd by curnponcnl- l molecules, and Pl 1s whal reflccts the interr~~olecular interactions ir~the pure liquid. Although wc havc focusscd
Kcldlt.s Ihr V,qilnr Pressure ut thc T ~ v oC o r i y r b ~ i ~ ~ i l ~
(Recall that thc argument orthe logarithm is actually this fr)mof I L , ( T , P) allows uq lo write
q / f i.where P - is unc bar.) Now
our disvucsion o n component I in Figures 10.7 and 10.8, the same situtation holds for cclnqx)ncnt 2. I!quat~ons 10.25 and 10.26can hc wrilten as and
Th~ls,in a vapor pressure diagram of a solutiutl of two volatilc liquids, thc vapor pressure of each colnponent approact~esHaoult's law as the mole fraction of that component approaches one and Henry's law as the ttlule fraction approaches zcm.
E X A M P L E 10-6 The v u p r preksure (in torrl oFcr)rnpo~~ent I over a bi~larysolution
B u l d x , = - - d x , (kuause .T, -tx, =
is givu11t q
Dcrenn~rie[he Lapor prehsure (P;)and ~ h t He~lry's . law constant (k,, , ) uf pul-e cornpunent I. 513L U I ION: In the limit a h 1,-. I . Thereture.
sa P; = 1x0 torr. As A , r',' k c a u c e r2 4 I ac I
that .r-, + I , the exponenlial factor
-, I hetauhe
1,
I), so Eq~~ation 10.28 hecnnica
which is another t'ornl of the Gihbs-nuhem equatio~~. If componetlt I obcys Raoull'h law as x , -+ I , then P, x , P ; and (8In P , / a x , ) , , , = I/x,#sothe left sideof Equation 10.29 becomes unity. Thus. we have the ccrnditic~n
+0
We now integrate this expression indefinitely to gel
-. 0. on I
Substitute these t w o expressions into the Gihhs-Uutlem equation to get
thc othcr hand, the exponential filctor apprvaches 4 0. Thu.c. we have
In P, = Inx,
+ constant
as
.rl
4
1 or .v: + 0
,
o ~ k,, ~ d = 807 torr
1 Wc will now show thar thc Henry's law hehavior of component 2 as .r, + O i \ a therrnorlyna~nicconsequence ot- the Raoult's law behavior of colnpunetlt 1 as r , 4 1. Tu prove this uunnec~~on. we will slart with rht: Gibbs-Duhetn equation (Eqi~;itioll10. I I )
+
I-,dp, ~ , d & = 0
(oonstnt~t1' and P)
Now. assuming that the vapor phase may be trcatetl as an idcal gas, both chemical potentials can be expressed as
Thus, we scc rhat ~ fcomponent ' 1 obey\ Raoull's law ;is x, + 1, then conlponcltt 2 must obey Henry's law as r, + 0. Pn>blem 10-32 has you prove the convcrae: 1 1 cumponent 7, oheys tlenry's law as x, + 0. then cofnponcnt 1 must obey K;in~~lt'.c law asx, + 1 .
10-6. The Cibbs-Duhem Equation Relates the Vapor Pressures o i the Two Components of a Volatile Binary Solution The ft>llowirlgexalnple shows that if rt7cknuw Lhe vapor pressure curve of one of tf~c compunenls over the entire cornpositiot~range, we can calculate the vapur pressure of the othcr coinprmerlt.
10-7. The Ccmtr,~l Thcrrr~c~dyriarriir Quarllily for Nnr~idpdlSnlutinnr Ir the hrtibilv
41 1
The exponential factor here accounts f o r the nonideality c ~ the f system. The chemical potential of component I i n this rase is given by
In Seclion 8-8. we introduced the idca o f fugncity to preqerve the I I ~ oI f lhe therniodynamic equations we had derived for idcill gascs. We will follow a similar procedure for solulions, uving an ideal solution as our standard. Tu carry over ihe l o r n of Equation 10.32to nonideal solutions. wc dcfinc a quantity called the nrririp by the equation 1 ~ : ' ' = 11: t RT Inn,
1 l C C f R E 10.13
( n l A tcmperaturc-compuc111~>ti di:~gram lllr thc system illustrated In I'igure 10.12. [hI A ~cnb~wrature-cnrrlpsition diagran~f o r n water/pher~oIsgslenr.
( 10.35)
/L; is the chemical potential, r)r the molar Gihbs energy. of the pure liquid. Equaliori 10.35 is the generalization of Equatiun 10.32 to nonideal s~)lulions.The firs! ol' Equations 10.27 says that $ = x,c'. as x - c 1 . If wc suhstitutc this resull inlo Equation 10.3 I. we obtain
where
Fig111-c10. I 3rr look<"upside down" compared with the orie in Figure 10.12. hut nole that tl~cre~r~periiturc decrz;~sesas you go up it1 Figure 10.12. whereas thry decrease as you po down in Figi~re10.13. F~gure10,13h shows a coexistence curve fix a waterlphcnt~l system.
If we compare this equaticln with Equation 10.35. which is valid at all concentrations. we can define the activity of cnrnponcnt j by
10-7. The Central Thermodynamic Quantity for Nonideal Solutions Is the Activity Thr chetnicnl potenlial uf cumpnnent j in a liquid sthution is giver1 by (Equation 10.15)
if wc ;lsxulme. 3s usual, that the vapor prcqsures involved are low ericlugh that the vapors can bc concirlexd to bchavtr ideally (othcnvise, we replace the partial pressures by partial fugacilies). An idcal solutiorl is one in which = xj 5' for all concentrations, so that Equat~rm10.31 bccuines /L;'" =
+ R R 7'11 x ,
(ideal solution)
(10.32)
Equalion 10.31 is st^; valid for a ntlnide;~lsolution, but the relation between ?/ q' and composition i s more cott~plicatedthan sitnply = x, q*.For example, we saw in Example 10-7 that parlial vapor prcssurt: data arc rlften fit by an expression likc
' = -P'
(ideal vapor)
X; + I. Iri other words. the ac~ivityof a pure liquid is urlity (at s11chthat 0 , + .Y, a total prcs>ure o f one bar and a1 the lenlper:tli~rr of irilerrs~),For at1 ideal sulution. y = x, q* for all conccntr;jtions. and s c ~thc activily of componeni i in ari ideal snlu tinn i x given by = rj. Irl a t~onidealsolution. (1, still is cqual to T i ? ' . hut this ratio is 110 longer equal 10 xi, al~tloughu, 4x, as ,r1 -+ 1. According to Equations 10.33 find 10.36, the activity of component 1 can hc reprcscntcd empirically by
Nule thal (I, + 1 21sn, + 1 ( x 2 -+ 0). The ratio r > , / . r , can bc uscd as a rne;Isurr uT the deviation of t+c solution frclrn ideality. This ratio'is called h e ortivih r.rwfirier!l of cornpollent j and is dcnotcd by y,:
Chapter 10 1
Solut~onsI. L I ~ L I I C - L I ~SLuI lI (~ ~ l ~ ~ r > $
104.
Activit~cshlust
lie
Calculated wrth Kt.srirrl
1,
Stanclard 5tato
= x , . 'l'h~l\. WL- see orlce again that il one c t > ~ ~ ~ l > oobeys r ~ c nRaoult'b l liiw ovcr 11ie entire colllyusit~onrange, the other cornponetlt will also.
II y = 1 1o1.all collc.cntr,~tlt)ns. the solutiorl is iclcal. If y,f 1. thc ~cllutiunis not ideal. For cxample. the partial vapor pressures of chlurohenzene it1 equilibrium with a chlorobenzene/l-nitropropanc aoluticln a1 75' U are listcd below:
01- ( I ,
E X A M P L E 10-8 Show that if
Accordrng 10 Lhese data, thc vapor prcssure ol pure chlorohcnzenc at 7.5 'C is 1 19 ton, so the activities and a c t n ~ t yc~)eficientsare as follows:
I
thcn U,
= x,cUXi
.
5 0 L U T I O N : We hrst differentiale In r r ,
Figurc 10.14 shr~wsthe aclivity coefficient of chlorohenzene in I-nitropropane at 7YC plotted against the mule fraction of chlorohcnzenc. Autivily i s really just another way cif expressing chenlical potential because the two quantities are directly related to each othcr thl-ough F, = p; RT In a,. Therefore, just a.s the chemical pote~ltialof one cornponcnt OF a b i ~ l nsolution ~ i\ rt.lik1t.d to the chrnl~c;~l pu~er~tiitl of thc othcr component by way of the Ciihhs-Duhr~nequi~tion,thc ;ictivi~ics;Ire reli~trdto each other by
+
For example, if u, = x , ovcr rhc entlrz c o r n p o s i ~ i orange, ~ ~ mean~ngthat cott~ponent1 obeys Raoult's law over the entlrc colnpos~lit~n range. thcn
Irilegrate fro111.r2 = I tu arbitrary .u2 and ilse the fact (hat ri:
+
1 as x,
--+
I
tc? get
dlna,
=
wlth
rcspect lu .T, :
dx 1- 2ull
-
.rk)drl
Xl
t
and suhctitutc into Equa~ior)10.38 to obtain
Nuw change the integration variable frwn .rl to A::
and integrate frorn xz = I (where
til = 1 )
to ,~rhitraryI-,'
10-8. Activities Must Be Calculated with Kespuct tu S t a n d a r d Statcs
F I G U R E 10.14
I 0 0.0
1 0 A~t~~llrljbc~l~~llr
The activity coefficient uT chlrlrrlhen~tneill I-rlitropropane at 75'C plotted against the inulc fraction o f chloroben~ene.
In nnc sense, Illere arc two types of binary solutions. tho.;t. in which t t ~ ctwo compr>Ilerlt> are miscible in a1Lprc)portions and those in which they art. nol. Only in the latter case arc the designations '+sc~lvent"and -'solule" ur~ambigut~us. As we will see In (his section. the differen1nature of these two types c~t'solutionsleadh us to defirlc differen1 .sl;~rlclnrd StiitCS.
/\Ilhougt~we have no1 said so explicilly, we have tacitly assutned both ut~mponents of the sululiorls we have considered thus far exist as pure liquids at the temperatures or ~ h rsnlutions. : We tlavc det~nedthe activity of each cotrlpollent by (Equation 10.3h)
h a t (1, - + x, HC A- c, I and 4 = I whcn P, = (;,A n activity rlcl~nedby Equlitlon 10.39 is s;~irlto be basrd upon a solverlt, or Raoult's law standard state. Becnusc 01. the relation ( F ! ~ ~ a t i o10.35) r~ / I , = EL; -I KT In a , , thc chemical potential o f cotnpnrlent j is alst~h a x d lrpon a srdvenl. 01,Kaoult's law, standard state. You need to 1eali7c ~ h n tactivities or chemicnl pole-n~ials are ~nclininglessunless it is clcar just what ha< her11u ~ r das thc qhndard statc. If' the two liquids are rniccihle irl all propc>~-lions. there is no clistinctiun bc~wccnsol\rrlt ; i r ~ d
Bt.c~;~i~se cotnponent j is spill-inglj soluble, llic use the second of Pquaiions 10.27, wt~iuhsays that P, + x , k , , ns x, 4 0. where li,, is the H e r q ' s law conslant of cnlnponcl>tj . If we substitute the li~nitirlgvalue of . ~ , k , , ,irllo ~ Equation 10.40 for P , , tcc obtain
,
or kl,,j = I<*. ?'he slandard state in this case requires (hat k,, = P,*. This st;~nrlard uate may nut exist in practice, so it is called a hypothetical standard state. Neverlheless. tlic rlefiniticm of activity invnlvirlg Hcnry's law for dilutu corrlpnrlrllts giu-tl Equation 10.43 i s n a t i ~ r and ~ l useful. The n~~rnerical kalur o f a n a c l i ~ i l yoran activity coefticrcnt depend x y mcthanc xt~lutic~ns at 35.2 I', and ~licscrintanrc plottcd i n Fipurc 10.15. Notir'c I h ~ hotli l curvcq approach Raoult'q law a< their ct>rrespontl~nglnrdc tr;~ctionsapprt~achunity. Thc di~shctllincs in the tigure rcprcxcnt Ilie 1inc;ir rcgiol~s;IS tlic currcsponding rnulc fractiotlk apl>rr~ncli7err). The s1npc.s of thcsc lincs givc tlic iicn~y'slaw constant folcnch component. Thc values come clut to hc k,, = I 1 30 torr s l ~ d k,, ,,,,,,rIl, = 1500 101-r. Wc call usc tlicsc valucs iind tlic valucs of thc vipcjl. prcssurc< of thc purc compuncnts to calculate activities and activity cocfticients based upon each standard statc. For example. Table 10.1 gives PC',,= 407.0 turr and Pd,rr,e,,t = 277.8 torr at rrs, = 0.6R27.
,
Thcrcforc.
,\, = 1 Pr, 407.0 torr = 0.701 1
(I(
P:,,
+ I, as x,
0, as can he seen by cumparing Equations 10.41 and 10.42. bhuatiot; 10.41 hecornrs equivnlcnt tcl Equation 10.35 if wc define u, by c,
P<,$> Itnrr -
f',,,,<, -l,/~orr
o.1100u
n.wo
587.7
0.0489 0.1070 U. I040 0.2710 0 1470 0.453h 0 4Wh 0.5393 0.6071 0.6827 0.7377
54.5
558.3
1119.3
524.1
159.5
501).4 45 1.2 412 7
0.7950
and vhorlse the standard state such that
514.5 turr
Vapor prehcure data uT cart)ur~ dis~ilfidelrli~nctho~ymethane sulritiui~r:it 35.2 C.
-
s o that a ,
- -
T A B L E 10.1
>c<. --. ...-.
We tie fine the i~ctikityoT currlpnncnt j by
>,
0.8.1.35 0 9108 0.9554 I .OOIH)
?34.8 277 h 324.8 340.2 357.2 38 1.9 407.1) 424.3 442.3 458.1 4H I .K 501.0 514.5
37H.(1 7hO.X
342.2
313.3 277.8 250 1 217.4 184 9 124.2 65.1 O.(X)O
10-8
Actlvltic5 Must Hc CI~I~uldleCI w i ~ hR ~ q p ~tor tStandard St,~tcs
and
FlGlJRE
10.15
Vapor pressures of cmbon disullidc
dimcthoxyr~~ethnne over their sulutiuns at 35.2"C. The d i l l hlrilighl lines represent itleal behavior, and the dashed Ilnes represent rhe Henry's law hchavinr lor each crlnlponenl as the correspon~lingmole Iractions approach 7cro
where the superscript (H) simply e r n p h a s ~ ~(hat e s tlleae vaI tics arc hasetl upon a Henry 'b law, or solute, standard statc. Figure 10.1ha shows the Raoult's law, or solven~-based. a c t i v i ~ i ~and s , Figlire 10. l bh shows the Henry's law. or solutc-hiixed. aclivilies p l o ~ ~ e d ngaitlsr the tnole fraction ofc;uhun diculfide. M1c will scc in rhe ncxi ctlapter that i~ solute, or H c n ~ 's y law, slandi~rdstate is particularly appropriik filr ii substarlcc that doc< n o t exist as a liquid ai one bar and a t tlic temperature ul' the bulution under siudy. The activity cr~rfficjenlsbased upon thc Raoull's la^ stnrtdard sstatc (which i c the usual standard state fur miscible liquids) arc pluttcd in Figure 10.17. Nolicc thiit yr,l + 1 as A, --t I and that i t goea 10 2.2 as x, -- t 0. Ruth of these li~rlitingvalucr may he deducei horn the definition o f y ( ~ ~ u a t i c10.37) il
<
-
NOW + as A,+ I , and so yj 41 as.5 1 . At the c~lherlimit, howevcr, P, -. ~ , k , ,as- ~x, 40, so we see that y, + as x, -+ 0. The value of k t , for CS:(l) ia 1 1 30 torr, sn y,, + k,,,,, / P:?> = ( 1 130 1orr/514.5 tori-) = 2.2, in iigreerrlent w it11 FigurclO.17. Thc activity coefficient of dinlcthnxymethane approaches 2.5 as ,i(,,,,,cil, 40 (.rL5, -+ I), i l l agreement with ydlmcrh kH,r,,mc,h/ Pimelh= (1500 torr/587.7 lorr) = 2.5.
k,,-,/T
-
with
w h e ~ thc c superscript ( R ) sirnyly cmphasizes that thcsc values are baked upon a Raoult's law, or solvent, standard state. .Siniilarly.
,,, , 407.0 rnrr acs2 = -= -= 0.3hO k , ,CSZ 1 130 torr
F I G U R E 10.16 (:I) The Kaoult's law irclir,iliec ui carbon disulfide ant1 dime~hr~xyr~>cthanc in carbon rliculfitlz/dimeth~~xy~nctl~ane solutions at 35.2C plnIted against the nlole fraction of-carbondisulfidr. ( 6 )'I'he Henry'\ law activities for the same sysrem.
10-'1.
Can Luristruc 1 .3 Molecular Mudel
Non-ideal Solut1un5
If we divide GE by the total number of moles n , G i h h errerg! nfnirirag,
cE:
I
and show 10.17
we ahkin the m d a r r:rrc.cs
I
E X A M P L E 10-9
Derlvc a forinula For
FIGURE
+ n,,
that
ckCOI
a binary sr,lulion In which the vapor pnlswrcs can he
cr is syn~nlctricahnu1 the 11111: -1, = 2 ; = 1/2
SC) L U 1 I(:) h': According to
Thc Kaouh's law activity culllicicnts 01- carbon tlisulfide (qolid line) and dirncthoxymcthane (dashed line) plo~rcdap~inct.r,,, for carbon disnlfideldrrncthoxy~nethane solutiur~sat 35.2'C.
the above t.xp~.cssionfor PI 2nd P).
Substitute these expws~ionsinto Equation 10.47 to obtain
+
We can calculale the Gihbs encrgy of mixing of binary solutions in terms of the ;rcrjviry cocRicients. Recall frorr~Equation 10.20 that Al),,x t i
= t l ] / ~+ ; ~tI2,1yn - n , , i ;
= r u x , x ~ irx,x:
- r12/l;
Rut, acconling to Equalions 10.35 nrtd 10.37.
which
II we divide Ami,G/K?' hy the - total n u n ~ b c rul' moles, C;~hhrojerxj nJ ~nixir?y.A,,,,xG.
11,
+ n,. we o b l a i ~the~ ntolrrr
ih
hyrrln~etricin I , and r , , and therefore, nhour thc lit~cc,
-- r,
= 1!2.
Many solr~tio~is can be desctl'hcd by the equ;~tion::in Exarnplt: I(&Y, and such solutior~s are callcd regulmr aulurio~s.Problems 10-37 through 1 0 - 4 5 i11volve rcgular solutiotls. e tn calculale U' . Wc can use yrS7 and ydlwllkthat we calculated tilr F i g ~ ~ r10.17 which is shown in Figure 10.18. Note that (he plot of G' vcrsus xCs is nut symmetric about x,.,. = l j 2 . This asymmetry ilnplics that 0 in the elnpiri:al vapor pressure formula ~isedirl Examplc 10-7, and that carbon disullide atid dimetboxyrnetha~ledo not form a regular solutio~lat 35.2'C.
+
'l'hc first ~ w tcrms o it1 Equation 10.44represen1 the Gibbs energy nfn~ixingof an ideal stduticln. To focux or1 thc elTect of nonideality, we definc an pxces.r tiihhs e n c v ? of 1 1 1 i . t - i j 1 ,GI-.: ~.
.
c;" 1% sce
fro111Equation 10.44 that
A . I; mil
-
A,,%G'~
( 10.46)
10-9. NIP Can Construct
d
M o l ~ c u l a rModcl of Non-ideal Solulions
I n this section, wc shall in~mducen simplc molecular rriudel for a non-ideal solution that qualitatively displays many uf the experi~nentalproperties of t~on-idealsululions. We shall assrlmc that the molecules uf components 1 and 2 al-c dislributed randomly
10-9. iVp Call CDIIS~~UCI d MOICLUI~I~ Model ut No111d1edl Scduti#~ri'l
We can locus on ihe nr)n-idciility ol'tllt' sc~lut~on hy introduuirlg rhc qt~;otIit):
Far an ideal solution. E , , = ( r , , 4-~--. , ) / 2 .and 50 vr) = 0. Thc ~nagriitudcof zu. ~hell,ib n measure of the deviation of' ~ h solutic~n e frclrn ideal behavior. Eqilation 10.48 1ahz.t on a fairly silnplc f ( ~ r mif w e clilliinate el, in l'ilvor of 111 using Eyrlatitm 10.49
F I G U R E 10.18 The molar exccss Gibhs energy of mixing of carbun d~sulfideldiructhoxy~~~etI~a~~c hc~lutionsat 35.2 C plottrd againft the mule lraction of carboll disulfide.
th~-i)~~ghout the solutiol], so that the entropy will bc the same as that of an ideitl solution. Therzt'ore, the dit'ference hetween the Gibbs ctlergy of our model sr>lutiolland that of an irlr,iil si~lutiuttwill he due to an energy tcl-m (actually a pnlenlial encrgy term). We exlwss rile potcntiiil energy t ~ the f solution in the form
where IV,, i $ the tiumhcr OK neighboring pair5 of molrcules of type i and j and where is tie interaction energy c~fan i and , j rnoleculc when they are nexl to each uther. Because we have assunled that the nlolecules are ~nndoinlydistributed, w e can dcrive qirnple exprz~aionsfor [he At,, . Lct's cottsidcr ,V,,. the rll~lrlhcrc~fneighhonng 1-1 pairs, first. We car] intcrprel the mole fraclion A , as thc probability that any neighbor of ;b ~nolcculei s a cr)mponerlt I mc~lecule.On dlc avcrage, thc tolal number of type I ncigIih(lrs of any given molecule, then, is givetl by x , , where 1 is the coordination riu~l~ber o f molecules around a central ~nolrcule.Typically z is around 6 to 10, hut we shall not reqt~irea precise vnluc. There arc N , corrlyrjnent 1 ~noleculesin ihe solutiun, so tlicrcfore the rlu1nhc.r of 1-1 ncighhoring pairs is (N1)(zx1)/2. where the l'xtor of 7, is inserted to avoid courlting each 1-1 pair twice. Similarly, we have h',> = :A? N 2 / 2 . Wr use the same value of z because wc assutile that molecular sizes are ktbuut thc s;inlz. The llurnhcr 01- 1-2 ncighhoring pairs is given by zx:Nl, or zx, N?. (he two expressions hzittg zyu~valent.Tlic total interaction energy in the solulian, then, 1s givcn hy
The last term here is equal t o zero when lr! = O and so reprcstrlts the non-irleiil hch;lv~tl~of the solution. Therefore, we can express the Gibhs energy uf the soluticm in the f o ~ m
We can express Fqualion 10.51 1n tertna o t nurrtber'; of iiioles rather thatl nurnbc.~. nf moleci~lesby dlv~dingN , arid N? by the Avogadru curlstant (:V,) to ohtaln
E,!
,VI 7
[ , ' = --
,,
i
is the chemical potentisl ol all idcal ~rll~rtinn [I!rlua
The term (;IC;,,r,,,,';$nl),,,, tion 10.17)
and the derivative in the second tenn in Equation 10.53 is equal to -
an,
,
I
-
II ,II,
01,+n2)'
>tT,;.i-,
22_c?; +
Using lhe fact that 1 ,= N , / ( N , + N,) and x 2 = N2/(lV,-t N,) gives
11, = p ;
7ulN a,'
+ RT Inn, + 2 ' - '. 2
Chapter 10 / Solutions I . ~iquid-LirlrridSolut~un~
where ii = ,-N,,w/2. If wc compalr Equation 10.54 to Eqwations 10.35 and 10.36, then 34% see thal
I
E X A M P L E 10-10 UECEqltatio~)~ 10.5.5 and 10.5lito deriw forrnul:~.,for A m , % Gand
1 cFlilr u h~nary
~olutinn.
11. we l i d diffcrcnliated Equarinn 10.52 with respect to n l instead oftr,, then we would t ~ a v cfound t l ~ t
Substitute these exprzscionc inlo Eyuot~or)10.45 to obtain
10.55 and 10.56 are tflc empirical expressions for I', anti Pz that we havc F.q~li~rin~~< u,c(l .;e\er:~ltinies in rarlier sections. For example. Exa~r~ple 10-8 shows that Equalions 10.55 and 10.56 ;ire relatcd by the Gihbs-Ilr~hetncqi~ation. Eqti;~~inr~ 10.55 is plotted in Figure 10.19 fur M / R T = -f I , U, and - 1. The value o f u = O gives ideal behavior. For positive values of u , wc have yusilive deviations I'I.OI~ itleal bchavior. FI-om E(1rlatiot1 10.49. pohitive values o f rc rncnn that 1-1 and 1 -2irltcl-actions ;Ire liiorc l';ivtjmble Iharl 1-2 interaclio~~s. '!'he increasing numbcl- of I-? inleractions h a t result when the solution is fnr~nedfrom its components causes rr~nlcculesto ehcape i n t o the vapor phase, ~ h u sgiking a positive devi;itir>nfrom ideal bch;~vior.For negative valucs of 11, the 1-2 intzractinnc are more fwt~rablethan the 1 - 1 and 2-2 inleractions. In this case, the mnolcculea mix rvcll. thus PI-nducingnegative dcvii~tiol~s f-roni idcal (Raoult's law) bchavior. -l.hc followil~gExample shows thal we can use Equations 10.55 and 10.56 to derive all expression for the mol;u Gibhs energy of mixing.
RUI
ux,xf
+ u;r?x: = u ~ , . ~ ~I .(. ~x, I , = u x , r 2
SO
A I,,II
c=R7'(,r,In~, t ~ , I n x , ) + u ~ , r ~
(10.57)
Similarly. Equation 10.47 givcs -F
G
- UX,.T?
(10.581
Prohlein 1 0 - 6 I has you show that
and
The entropy of mixing i q the .same as for an ideal soli~tionbecause we have assunled that the rnolecuies arc distribilted rantlomly thruughuut the soltilion. Urilikr at1 ideal salulion. AmlXH # 0 for a non-ideal solution. Equation 10.57 can k wrillen as
10.19 ,I plot uf P , / I',' rcrslh .r, given hy Equation 10.55 for u / K l = 1, 0.and - I . The valr~e u = 0 i c l d s an idcnl solution. Povitrvc val~~cs of u / H T yicld positive deviations fro111ideal (Knuolt's law) bchavior. and rregative salucs of ii:RT yield negalivc deviations from idcal FlGlJRE
hrha5,ior.
+
Figilre 10.20 shows plots of A , l , , x ~ for / r ~rcvcral values rlt R T / u . Note thal the xlopes of all the curves equal zero at the midpoint, r , = .w? = 1/2. Thr c u r t r tor K l ' / i i = 0 VJi~ qpecial in the \enst thal curves for values uf R T J u greater t l ~ 0.50 r ~ are concave upwards for all values of r , , whereas curves for valucs of R T / u Icss than 0.50 me voncale dowtlward at r , = I/?. In mathematical term$, ~ ' ( A , , ~ G 13.r; / I ~ )1~
F I G U R E 10.20
Plots of A,,,G/rr lr)r R T / u R T / l r = 0.40 (trip curve).
=
O.(a (hutturn curve).
RT,'u
posirive (a t t ~ i n i ~ n u m at) n, = w, = 1/2 for thc curves that lic betow the curve with R T . , u = 0.50.whereas a'(~~,G/cr)/ilx:is negative la maximum) at x, = .r, = 1/2 filr ..me> t h ~ li i t a h n t it TIK replan ahere h 2 ( l ..-> C!u):/h.l-' is negatlle r:prcwnr\ ail ,.,:>~~bl:rcgi,)n I Pr~>bl?rnl M 2 i and 15 31n1ilu IU dx l t ~ > p 01. > t k d:r \VuL ~ y i u r i ~ i,r l n ht R d l ~ ~ . h - f ; u o requatiirn g w k n T < I [ (Figurn 2 S). md 111 rh15 CLW ir,rrc.,p)nd 10 rt~iorlsIn u hiih thc. M-o liquid, ut not miu-ible. lhz critical I-alue RT . t = O.50 <.orr<jpul~d~ to a solr~rlnncriric.af tempznrure. 7 . lihere rllz rno lrquidrj .~rc~ ~ u > ~ , i~ hI lall I t . pmponion. ~t ternptratureb above T = 0.50l1,'H ;111d~rnrnihc~hlr at tzniptrature:. hclnn T = r i / R . Lrt'scr~n~it1c.r thc C L I T Vw~ ~ t hR l /I( = 0.10 ~ IFigure I 10.20.The 1 u . k ) ~~~iriirlla~,cprcsent two immiccihlz solutions in equil ihrium wirh each urhcr The oornpositiot~sof these Using Equation 10.62. two solution\ are given by the values of x , at each mirli~uun~. we
FIGURE
= 0.50 (middl: curvc). and
have
10.21
-
+
I
+
EXAMPLE I s 1 1 C x F ~ u i l ~ i o10.63 n w ~~alculatz thc compjsltiun oi r h ~rucj 1rr~irl1,cihI~. .olur~on, In equilibrium a ith each [*[herat a temperature given by R T / u = 0.41. SOL IJ 7 I ( 1 N , W e uukc l l ~ cNuu'tun-K;~pl~<~>~l rnr~hncl~ h n ruc i~ltnduccdIn M;~thC'l~.lpter A. The functron {(XI oT Equation A . I ic
Equation A. 1 kcorrles
a s Lhe conditior~for the extrzn-ia O K AKc,,,G/u. Firvt 110tt:that .T, = 1 /2 solves Equalion 10.63 for any value of R T ! u , which accouiltq for the fact that all the curves in F~gurt.10.20 have eithcr a lnaxirnun) or a minitnum at .r, = 1/2. F3g plolting (XT/'la)llnx, - In(] - A-,)I ( 1 - 2 x , ) agiinsl .u, for various values ot' h ' l ' / ~ iyuu , can see that only .r, = 1/2 salislies Equation 10.63 h>r K T / u 1 0 . 5 0 . nhzreas two 0111er routs occur fur R T / u i0.50. The two rt~otsgive thc caml>o.silion ol- the two miscible solutions in c q u h i h ~ u r nw i ~ heach other. For (he case in which R T / u = 0.40, the two vill~~es of x, are 0.145 and 0.855. Figurc 110.21 shows the mule Iruclion of cornpoilent 1 in each o f the two imrriiscible solutions as a fil~ictiollof temperature ( H l ' / ~ i Note ). thiit Figure 10.2 1 i s sitt~ilrtrto Figure 10.13.
+
-
A lempraturc-colnposition diagram lur a hinary systcrn for which A,,,,lG!~r ( R T / u ) ( . Y ,In x, x2 Inx,) x,x2 (Equation 10.62). The curve gives the cnnlposltions o f the twu irnmisc~hlesnlut~o~ls as a function uf temperature. There is or~lyotle I>omugeneuuc phace i n the rcgion above the curve and there art. I w n inlrnisciblc solutiuns in equilthr~urrlwith each other in the regitjn heluu, thu curve.
with K I ' / I I
-
0.40. For one of the soIuticjl~s,we htart w i ~ hA , , = O.lo() u ~ l dgct
I
I b r thc other culution. wc st;jrr xith w,, n
,
= 0.9011 and get A
,
,
f'k)
10-4. Apply Euier'\ theorem
10
11 = I:( S . V , 1 7 ) . Do ?nu rccogllizc tllc resulting cqu:ltion'!
1U-5. Apply Eulcr's theorem to A = A ( / ' , I'. ~ r jVu . you rectlgnii.e Ihe rev~lringe q u a i ~ ( ~ n "
10-6. Apply Etller's theure~nto V = V ( 7 ' . P. n , . 1 1 , ) to cleriw Equalion 10.7. 10-7. The properties of many \olutionc are given au a iunc~ionof the mass Ixrccr\t of tlh: corllponclltq. If we let the Inass percent o f curnponen~-2he ti.. then tlcrixe ;I rzl;ilion het~seen,4, :inrl the n ~ u l efraclions. 1,and I ? .
Wc lrlust cmphasixc once again that (lie rcsults of this section are n result of the ai~llplt.t;l~lclomly-dislributrd nod el (fiat wc used. Although the model gives a nunlbcr of-qualitative res~ilts.i~ m u s t k borne in mind that it is just a n :ipproximale m o d e l that n c introduced to g ~ v erume rr~olccularor physical insight into the rlaturr of non-idei liquid binary solutior~s. blc will continue our rliscrlssic>rlof solutions in (tie next chapter, where we focus on hcllut ions ill which thc t w o colnpclnents iue not soluble in all proportiorls. In particular, Re u,ill rlircusk solutions of solid< in liquids, where the t e n n s solutr a n d solvent are ITI~~II>~II~CUI.
10-8. I'he CRL' Ho~rdl~ook of Uic7rni~lr?.(IIZI! Ptfvrrt.~givec Lhe d e r ~ s i t i eelf ~ tr\arly aqucouq sulurio~~s nq n function of the mass percentage of solute. Uure denntc (he dtnsity by p :~nrl the rn;r\s percenhge of cnrr~po~~t.r~t-2 hy A ,.the llrl~actbvokgivcs p = p ( A , ) (in g . lnL I ) . Show that thc quantity V = Or, M I 4 n , M , ~ / ~ ( A is, Ithe volume of the suluiicm crlntairlinp N(nv show that n , rrioles of cnnlponerit I and r t z lrlolcs o f con~pollcl~t-2.
and
I
I -
Problems
Show that
10-1. In ~ h ctext, wc aenr 11,onlEquation 10.5 to 10 h using a physical argumenl involving varying tflc cize of the ky~tcmwhilc keepir~g7' and P fixed. We could ; I I F ~have uscd a m;~!lic~l~ntical prucesf called EuIer's thrortm. Before we can learn about Euler's theorerr\, wc musl lirst delirlc a Ilr~mo~e.er~cnrr.r~rnction. A function j ( t , . z,. . . . . i , %is) said t be II(IIJ~O~CIICDIIS 11
in ;lgreeinenl with Ey11:ltion 10 7 10-9. Thc dcrlsity (ill g.mol I ) of a I-propanol-water solution at 2NC as il function oC ,I2. ~ h c mass percentage or I-pmpnnol, can he cxprcsscd n?
Argue that ex~cnsivct h c ~lnt~rlyrlnmicquantities are hurnogencr~u
.
10-2. Lulei-'c thcol-en1 says that if j'(;, 2:.
. . . . z , ~ is ) homogeneous, then
where
Prove Eulcr'u theorcm hy differenlint~~lg Ihe equation ill Prilblcin 10-1 with respect 10 1. urjd then sctting I. = 1. Apply Eulcr's Ihrcrteln r c ~(: = Gin,. l a , , 7'. P) to derive F4uarion Il1.h. (Hlnr: Uecausc 7 arid P ;ire inEnsive variablec, thcy re simply irrevelant variables in thif case.) 1U-3. IJ+eCr~ler'sthcoren~(Prohlern 1 0 2)
pro\:ethat
Use thir exprewion trj plol those i r ~Figure 10.I.
v, ,,and -
p,panol
versus A, and corrlpart. your vnluch wltli
10-10. Given thc density nf a bi~lnrysolutiol~as a function of the nlolc fraction uf c o m p o n e ~ 2~ t [p = o(.t>)I,show that the volume of the ~olutioncontaining rr, n ~ u l e so f c u n ~ p o n e ~I ~; IlI I ~
lnolcs uf comprlnent 2 i s givcr~hy 1' = ( r z , MI rliusc of coln1xmcnt j . Now shuw thal
11,
+ t ~ ~ M , ) j p < ~rvherr , ) , Mi
1\
tlh:
ruolm
10-16. Calculate thc relativc anlounts of llquid and vapur phacec a1 an overall cornpusition of U.5U for one of the pair of values, s: = 0.38 and s, = 0.57, that you obtained 111 Proljlcnl I(bI 4
10-17. In this prohlclrl, we w11l derlve arlalylic expreh+iorlxlrlr Ihe pres~~~rr-c.r~mpusi(lun cr~i-1r.h in Figure 10.4. The liquid (upper) curve is ,just
which is a straight linc, nc sccn i n Figurc I 0 4. Solvc the cyuaurjrl
Show that
for x, in terrrlq o i y; and cuhslilule intrl Equalion ( 1 I tr, ohl;~in
'
10-11. Thr dcrlc~ty( i r l g . t11ol ) of ;I I-propanollwuter solution (11 20 mde 1r;lction of 1-pr~,p;rnul. call he expre5hed as
r us a functiu~~ uf x2, the Plot t h ~ srcsull versus y, and show thal it gives ttlc vapor (1uwt.r) currr In Flgurr 10.1. 10-18. Prow t11at J: tally.
;.
.T:
i f P,* z P; and that
j1 .r
.r, i f P-: c P ; . Irrtrrpret t t ~ i rraull + phyw
where
a,,-- 0.49X23
lJae this expresclon to calculntc the values
10-19. Telrachlurumerhane and trichlomr~hylenrfornl ;In e ~ s ~ n ~ i ;idc,lI ~ I l y !,olu~iotl , ~ t40 C' .II all co~~ce~ltrations. Give11that the vapor prcssure of tetrachloror~~ctt~a~~t. and t n c h l ~ ~ r t ~ c h y lene a1 lo' C are ? I 4 lorr and 138 lurr. respectidy, plot the preshurc-unlnposiliunJi;ly;l~n for tl~i.;systcnl (scc l'rohlcrn 11L17).
u , = -0 171b3
I',,, atld ,
-
VI-l,lnp31,ula$ a filllction of
1,
10-20. The vapor prcssurcsot tctrachlornn~ctha~lc (1 alld lrictllon~.lhy Icl~c121 h c l ~ c c76.8 r~ and 87.2'C cw he expresced empirically by the formulas
r
accurtling 10 the equation in Prwblc~n1W10 10-12. Use the d;11;1 in Ihe TRI- I I ~ r r l l t ) o o kof C-/;~mir.tryc111r1 Physir.s to curvc tit the density ul i~ walcr/glyccml sulution tu a lillh-r~rderpolyllomial i n Ihe ~nolefraclion r l l gl ycernl. arid then detcrlnine the parliol rnnlnr volumcs of water and plycernl a + a tur~cliot~ of mols
and
Irucllorl. PI01 you1 1s,ult 10-111. Just h'fnr? Exairlplc I[)--?. we shuwrtl Ihar 11 orlc cunlpr>nrnl ut il I)I~,II).solulirln ot)cyc Knuult's I;IW r,iCer [he er~tirccolupusitiu~lrange, thc other crlmpjnerlt does also. 5 1, then 1 1 , = 1 1 ; i R T l r l 1, for Nou chou ttxit if / 1 , = 11 $ R T l r l .r: tor .r! ,<-,,, 5 II c .rl x: I - r l n,,v,). Notice Ih;lt lor thc l,ariEe owr mhich J L , u b c p the \implc iorrrl givutl. /.I, obeys a .i~lnilarlysin~plrli,rn~.I t wc lct A-,~,,,,,, = 0. we rlhlain @ I .- 111 R T 111 -I I (0 5
+
.\
5 I j.
10.14. ('olltirluc the c;llc~~l;itiun\in Exalr~plc10- 3 to ohtail] ); as a function 01- .rl hg r wyinz .r. l r r l t r i 0 to 1. Plot yourreyllt. 1 U-15. Llse !our ~-esl~lls frrlnl Prrlhlcrll 1t1- 14 to construct the pressure-cumposilii)ri dingran1 i n Figure It) 4.
where t i s the Celsiuc temperature. Ascurninp Ihul leirachlurc>nielhaneanrl ~richlr,ruetliylrn' li)mi an ideal culution a1 all compobilirln\ hclwcen 76.8 C' ;lnri 87.1 I.'. ~~11~1il;ite the values of A , anrl r, a1 82.0 C (at ;in ambient pressure uf 760 torr).
10-21. Use the data ill Prr>blcln10 20 to collstnlct thc c1ltlr.c ~ C ~ I ~ ~ C I ~ ~ ~ ~ L I I ~ C - Cd1a~r.ilt11 ~~III~>OS~~~~II 01-a te~rachlo~nme~hanrltrichlororeth~lene suIuIiun. 10-22. Ttic ruprlr pres+ui-eso l hen~eneand ir)luene between 80 C: ant1 I 10 the Kelvin terngratrire are gih-en by the empirical forrnlllas
r a\
a functlr,ii
1,l-
Chapter 10 1 Sululinns 1: Lirluid Liquid Solr~tinns
and
70-30. Suppow the vapor pressurc of thc two conlponcnts of a binary solution arc givcl~by
In[ Fc:51,'torr) =
4514.6 K -
1-
+ 18.347
Assrlmirrg that kn/,enc 311dttlluenc form an ideal ~ ~ I u t i nUSE n . tllcse fonrlulas tu conqtruct :I t e ~ n ~ ~ ~ - ; i ! r ~ r t . - c ~diagram ~ t ~ ~ p ~ot ~ sthis i t icyktem ~ ~ ~ ~ at an at~~hienl pr,cssure nr 7hO turr.
10-23. C'onrtrucc tllc re~nperaturc-oompo
V,
-
4
E', for
it"
irlenl snlrltir>r!, w h e r ~7:is the 1no1:ir volumc of pure
10-25. Thc volume of ti~ixing01-~niqcibleliqutds is defined as Ihe wlurnt or tflc solution minus tllc volume of the individual pure components. Show that
nt cr,nslanc P n~tdT , ~vhcreI ': is thc inular wlurne of pure cotnponent i. S h o ~that A,",,v = 0 fur 311 ldcnl srllution (see Prohlc~u11C24).
p
1 - -rl
p&&.";4a; f
and
10-31. Tllc clnpirical cxprcssion for thc val)nr prcsqurc that wc uscd ill Bxn~nplcsI0 h ond 1(&7, for ex:t~npie,
i~ sr)[Iitt~tllefcalled the ~ W u r ~ u I efqutrfiorr. .% IJse Equation I O.?Y to prow that thcrc call I)c f Ht'lrry'h Id\, no lrnear tern1 in the expunenlial !actor in PI,li)rolheru.ict. P, wrll r ~ o tatrcl). as .t: 0.
-
10-32. 111 llrc text, wc nent I a\ x , 4 I I r l ~ h l r pmhlerri. We will prme Ihe converse: the Ratlu11's law behavior nt coinpotletit 1 ac x , -. I is a dircct consequence of the 1Ienry's law behavior nf co~r~ponent ? as .\: -. 11. Show that the cherr~icalpr~ter~tinl of compr>rlenl-2 as .r, + 0 ic
10-26. Supposc the v;lpcir pressures uF the two cnnlpbnents of a binary solulirrti are ghen hy Diffcrcntiatc p 2 n,ith respect to x : allti subctitute rhe result into tltc Cibbs U~tliclncqr1;itioll to t ~ h ~ a i n
dl
d p , = R7'(;~c.un th:~t P; = 75 0 ton. R I I ~I',' -- 160 torr, calculate the total \apc,r prersuw and the c o n ~ p i > + i t01~ rI ~he~ vapor l pIi;~w:I[ .rl = 0.4~1.
10-27. Plut J , verxux x, fur the hyctctu described io the previous problem Why does the CUrVE lie h c h w the htrarght line cotitlccting [he origin with the p~irlta , = 1. XI = I ? nescribc n c.ys~ctt~ fol- which ihc curve v+nuld lie ahrjvc the diagonal line.
10-28. UFCthe exprcs%inns lor P, ant1 P: given in Problem 10 26 to c(!~rstructa pressurecnmpositio~ldiagra~n. 1U-29. The vnpor prehcurc (in turn) of the two cornponenu in a hinxrcry solution wc given hy
nnd
C
1ntegr:lrc t h i ~exllres
U, t,
;--
I
nnri
+0
w e the fact that
L I , (A, = 1 )
-
11;
which is the Rat~ult'slau crprcsaiuti inr chetnicnl potetlI~nl
10-33. I11 Example 10 7, wc saw that if
then
Shnw that this wsult fnllows dtrcctly frnrrl Equatinri IO.2Y.
10-34. Suppose we express the vapor press~~res of the components of a binary solution by
to
arrd
Use the Gihbs-Ddlcln equation or Equfit~on10.29 to prrlvr that n must equal 8,
10-35. Use Equation 10.29 lo show [hat if one component of n binary solution obeys Raclult'f law fur all c o ~ ~ c c r ~ ~ r a t it11e11 o o s ,the tlther colrlponent also obeys Ruoult's law for all cuncen~raiionr. 1U-36. Use F q u a t i o ~10.29 ~ tv show that if one component of a binary solution h:~b posilive deviations from Rarlult's Inw, then the other cnmpunent must also. The J o l l u In!: ~ ~ n11 / e prob1~111.i dt1~1(1p the ~ d r!t~ (1 nI Y - ~ U I I I~olution. T
10-37. If the vaporprcssurcs of the tuu cuniporlcnts in a hinary solution art given by
10-41. Plot A ~ , ~ ? ; / In U Problem 10-37 versus .r, for K I / u = O.hO.0.50.0.45,0.40, ant1 0.35 Note that svrne ot ttic curves have region\ ~ ' h c r c2'~,,,,~G,'il rf c 0 Thcsc rcgions corrr<pond to regirlnc in which thc two Iiquirls ;Ire no1 tn~scible.Shou that RTIla = 0.50 i\ ,j critical talue. in [he scrisr: that unst;ihle regloll> occur only n-hen X T / I I :. O._iCI. (See ~ h c prcvious problem.)
"'
10-42. Plot both PI,'P; = x , uw'i:" and P , I ' P"L - x,rv'; . rllr R T/rr = 0.hO. 0.50,0.4.i.0.40. and 0.35.h t e [hut the loops occur f& values ot R T / u < 0.50.
10-43. Plot both I', / E ; = a , P ' * ; ! ~ ' and P. > /P:. = * . p " ' / " '
for X ? ' i u = 0.40 Thu luops ~ntlicate regions in which the two liqurds arc not luiscible, ah rxplairltd in Problem 1&39. Ilmu a horizontal line connecting the left-side and ~ h r~ght-side c intcrseclinrls of thc tm-o cur\r\ This line, which connects statec in which the rapor prrslurt. (or cl~cmic,~l polri~tlal)ul c;itI~ cuml)nnent is the same in thc two sulutir>ns01- rlltfcrcrlt colnpclsi~iun,cclncy)o~~ds to OII? 01 the hori,untalll~lcs in Figure 10.12. Nnu.sct PI/f'; = r l r " ' f ' x ' cqual to I : ! I;- - r.r"'i and solve Tor R T / u in ternls of 1,.Plnt RT,iu agdinct A , and U ~ ~ R I i~ I Icoc*i\leilce m r \ c likc the one in Figure Ill. I?.
"'
10-44. The inrllar enthalplcs of mixing o i sc)lutions o f tcrrachlr)rorricthanc ( I ) ;md c y c l t j t ~ c ~ u ~ j ~ . (2) at 25'C are lihted bcluw.
A svlulion that satisfies thebe equatluns is caller1 a wgrrlur ~olutii~n. The statistical therlnodynamic mudel of-binary sol~rtiunspwwntcd i ~ Seorirrn r 10-9 shuws Lhat u is proportional to 2t ,I --- c l l - E:,. where t , ,is thc intrracliur~energy bctween ~nolcculesof compnnc~ltsi urld j . Nolr ~hitr1) =; (1 if sll = ( e l , + ~.,.. ) , / 2which , meanc that energelitally, n~oleculcs01ctrlnpc~nentsI arlri 2 "llkc" the opposite ~ ~ ~ r ) l c c uah l ewell s au their OWII. -
10-38. Prtrr,c that A ,,,,, G, '1 3 , and 3 point .rl = X? = 1/2.
3 in Ihe prcblou', prohle~narc sgmmelnc about the
10-39. Pint P I / P; = xlel"~:" versus s,for Kl'!,r = O.M),0.50,0.45,0.40, and 0.35. Note that snnle of the curvec have rcgions where the h p c is negative. The fullowing pfl?bicn~has you show that this behaxcinr occurs a h e n HT/sr -: 0 . 5 0 . 'These rcglons are similar to the loop\ r)P thu van der W u l s equatiot~or the Rztllrch-Kwong equation when 'I' c TL (Figure 2.8). iind in t h ~ case s con-e\prlnd to r.cgio11sin which thc two liquidh a r t not ~niscihlt..Tllc cr~ticnl v;jliir RT/rr = 0 . 5 U vurrcspondc to n soiutioo CI-ir~cal tcrr~per;iturz.0.50u/'K. 10--10. Ll~ficrc.t~~i:lrc t J , = t r ~ ; r U ' l ' 1 '- '' ~ v i ~I lL l' ~ C ) L ' Cd~) r, 11) I)I.OV~' that i', Ilil!. a I ~ I I X ~ I I ~ ~ I I ~ I or ,I nli~li~nun, al rh' pni111s.ti = f [ 1 - 22 ) I " Shrrw th,~lKI'jtr 5 0.50 !'or either a iilaxirnum tw a minirnum tv occur. no thc positions t ~ thesc f extl-ernes whcn K T j u = 0.35 vr>necpontl to rhc plot you obtainctl in the prcvioub prohler~~'!
I
Plot A ~ , against ~ H x,x, according tu Proble~ul W-37, Drj tt.trachlo~-oinetl~ane urld cyctu hexane form a regular solution? 10-45. 'The inular er~thalpicsof mixing of sulutir~rlsof tetrahydrtlfurat~arid tricl~lurr>nie~t~ar~c ut
25 C arc licted below.
-0 464 -1.374 -2.1 18 -2.308 -2.383 - I HXX - 1.4h5 -0.801 Do tctriihydiofuran and trichloro~nethaneforrn a rcgular sulu~ion"
434 10-50. Some vapor pressure data lor cthanoVwatcr solutions nt 25 C are listed beluw.
10-46. Uurivc the equation
hy \tiirlinp with Equation 10.1 1. LJse this equatttlon to obtain the wmc rcqult as in Exam~ I C10-8.
10-47. 'I'he vapor prewure dala for carhon disulfide in
Trrblt.10.1 can he curve t
~ by t
Usirlg the re\ull\ ol Exatr~y>lc10.7, show that the vapor pressure of dimcthorymethanc is givt.11by Plot ttic?e data to dctcrn~il~e the I Ienry'~I:iw crlnctanr Ttlr elhanr~ltri watt.r or~rlfor 65:ttcr cttianrd 31 25 ('.
Vow p l u ~P, vtrsu?
.i,
a d coinpare thc rcault wit11 the data i n Tahlu 10.1. Do carbon
tliculfidc and dinletlwxyrnerhanr furln n rcgulnr solutioo at 35.2'r:) Plot tltc plot ~yinnieri-ic;11>c>iit n vcrticnl linc :I! r l = 1/2?
10-51. Usin? the data in Prohfem 1&50. plot [he activity c t d h c i c r ~ ~(hnscd ? upnll Knor~tt-s Ian) ol hrlltl t.ltianrd and wntcr against tllc lr~olefraction uf ethanol.
ELag:iinst s,.Ic
I0-48. A lnixtult. uf tricl~lnrcl~ncthilne ;mri ;icelonr N ilh xdcc, = 0.71 3 has a lotal vapor prcssure = 0.8IF; (il\eri of 220.5 trvr at ?X 2 (:, a r ~ dt t ~ ctnolc fraction of i~cetunein (tie vnpor is [hat thc vnynjrprcsstire uf pure trichlororr~ett~atlc nt 28.2':C is 221.8 l r r , cnlculatc the acctih-ity and the aclivily cWfticicnt (based upon a Rar)ult'~law standard state) of Irichlororncthanc in Ilie ntixture. ~ \ ~ < L Ithc [I~ yap01 c b ~ h i ~ itieally ~es
10-53. Surrw 5apor pressure dntn for a 2-prup;1not/benzenesolulitm
a!
25' C are
10-49. Cotisidcr o binal y solution form hich the vapor pressure (in torr)of one 01-thccornpone~lts
(bay corllporieut I l i + give11 cmpiricall!: h!:
Plrlt the activi~tecand the activity cocfficicl~tsof 2-propanol and benzene rzlatiw rr, a Kaoult's law ~t;~ndarrl (lare rercuc the rriule 1-ractiunt1[2-propn[1ol. r;~lculutc111c activity nlld the a h ity coellicien~of cornponcnt 1 ahen .r, = 0.25 hased on ;I rrdveril nllrl a aolutc sta~ldnldstmr.
10-54. Lluing the (lala it\ Fmhle~ri10-53. plot ?;'/KT vcrsus x ? .,~F.,,,~,l. ~~
10-55. .k~cts\fhermm~yntlrrr~c rlurrn/iticr are defined rclativc ~uthe values the rluantitics would have 11- the pure colrlponenls formed an ideal solu~ionat thc samc givcn terriprrature and pt'cssure For exi~~nple. wc haw [hiit [Equaticln 10 47)
at 1,= rl = I!?.
IEt h i ~result consistcnl wllh the graphs you ubtaincrl in P~,i>hlcln10 --ll'!
10-59. Use the data in T;ihle 10. I to plot Figures 10.15 through 10.18 10-b0. Use Equi~l~r,rllO.h2 to show t h ; ~lhc ~ slopec 01- all thu curveh in Flgure 10.2(1u c cqudl ru Lero when x , = u, = I/?.
Show Ihal
-
10-62. 111this problem. we will prove that ( a ' ~ , , , , ~ G j i l must x : ) bc greater ~ h a n,ern i l l a stal>lr. region of a binary solution. First choose - ~ o ~ point n c sy in Figure 10.20 and -.draw a strilighl line at x: tangen1 lu the curve of An,,%G against .TI. Now wgue (hat Aml,GI I I U S ~lic nhuvc thc tangent line fur the wglon around tu he spahlc, or that
Show that thc cquation for the tangent line is
-
Now expand A,t,,,Gin a Taylor series (hlathChapter C) n h x t thc point sj' and show that
for thc binmy ssolutivn to be stable. for d regular soluliun (see Roblcrr~l lL37) 10-57. Example Ifl 7 cxpressek I he vapor prcasure:, rd the two components of a hirlary avlution as
Shun lhal these
CXPICSSIOI~!,
are equiv;rlent to
~ s i r l gthcse expressions for 1t1c ac~irity ~ w t h c i c ~ ~Jtrive t \ . an exprebsio~lfur fiLI n tcrtlls ul'n and p , Show thai your s x p r c s s i o ~reduce, ~ lo that for kttor a regular \ol~~liun.
10-58. Pr0r.c thal Ihe maxima or n l ~ r ~ ~ufr ~A,,,,,C la dchr~edin Frr~hlcm10 -37 occur at x- = 1 !2 for :iny value 01- K1':u. NOW pnlre that
"
>0
a' A , , , , ~ E -i1.r;
- (I
for R T / u
;-
0.50
f u r R T I I I = 0.511
A-,
=
Solutions II: Sol id-Liquid Solutions
Peter Drbye (Icft) w;tu horn i n M:rilctrivht. the Nether1;tnds. on March 24. IHX4 and died i l l 1966. Dehye u a \ rli-igir~ally lruir~eda$ an electrical etlgincer but turncd his attelltion to physic<. 1-eceivi11;his Ph.I). from the University of Munich in 1908. Altcr t~oldirlgpositions tn Su ir/.erlnr~d.111ch'cthcrlntids,and Gertnany, hc lnovcd to tl~cU~~iversity of Berlin in the early 1930s. Although he Iurl been assured that he would be able retnin his Dutch citizcnsllip, Dcbyc fnut~dthat hc would k uuriablc to corltinuc hic work in Bcrli11unless he became a German citi~rn.Hc rclused and lelt Germany in 1939 li)r Cr)rnclI U I ~ ~ V Cwhrre ~ E Ihe~ remained ~, for the rekt r,l hir liCr, hecrjrnii~gan American citi~enin 1946 Debye was nwardcd thc Noh1 Prizc I ' Ic~ hcl~~i\irg i n 1036 "Tor his crlnlrihulionc t o our knt)wlrdgc 01- nlolcvuli~rstructure thru11g11 hi.; in~cstig;~tiru~s un dipole nunl lent^ ant1 on the diffraction of X rays arid eleclror~si r ~gafef." Erich Hiickel (risht) uaq horrl in (iiittir~gcn,(icrrrlany, 011 August 19. 1896 and died in 1980. Hr rrceivetl hiz. Ph.D. i n physicc irom Ihe tjni~erxityof Gtiltirlgerl iri 1421. Hc latcr workcd with Petel- llehye in Zurich. nnd together they developed a theory for the lherrnrldynamic pmpc1,tics of snlutiot~~ ot strong clcctrolytcs that ic now kllown as the Dehye-t Iiickel theory. Huckel also d'ielupld Hiickel inwlecul;ir orbitill theory, which we learned i r Chapter ~ I0 applie~ to cnnjug.?tcd and arotllntic n~olcculcs.Hiickcl was appointed professor of theoretical physics ;iI the GTnivrrsitx uf Marburg in 1W7. where he remained until his retircmcnt.
In the prcviou\ chapter. we studied biniiry solutions, such as ethar~nl/watc~solulioris. in which the two cumponents were ~nisciblein all proportions. In such so1ut1on.s~ ei~her component call bE treated as a solvent. In this chapter, w r will sludy solutions in which one of the componcnts is prewnl at much stn;jllcr concentralions thari the othcr. LO that lhe lerms "solutc" and "solverjt" are ~near~ingfr~l. We will introduce a solutc qtu~dard state bared upon IIenry's law such that thc activily of the solute becotncs cqlral 10 its concentralion as its concentration goes to zero. In the fir<[few seclinrls, we will study solutions of nt>tielectrolytcs,ant1 then soluliuns of electrolytes. lJnlike for solutiorls of no~~electrolyles, we will be iil>lc tu presenl exact expressions fnl- the activi~iesand activity cucfticients in dilute solutions of electrolytrs. I n Sections 11-3 and I 1 -4, H e will discuss the colligative properties of solutions, such as osmotic pressure, as wcll its the depression of the freezing point and elevation of the boiling point of a solvtrlt by the adlditic~nof solute.
11-1. We Use a Raoult's I..aw Standard State for thu Solvent and a Henry's Law Standard State for the Snlute for Solutions of Solids Dissolved in Liquids 111 Scction 10-8, we considered sulutiona in which one of the compuncntq is only sparingly soluble irl the other. Tn cases such as ihese, we use thc tcrlns solurr fur the sparingly soluble component and sol~~errt for he cutrlporlerlt in cxccss. We cus~oninrily denote xolvent quatltities by a subscrip1 1 itnd solute quantities by a qubscrip~2. The activities w e defined for the solvent and st~luteart: such that 0 ,+ x , as x , + 1 and + x2 as +* 0. Recall that a , is dcfined with respect to n Raoult's law s t a n d ~ r d state (Equation 10.39)
a, =
11 ' -
p:
(Rauult's law standard state)
C h a p f ~ r1 1 ; Sulut~onrII: S n l i c l - L i c l t ~ r tSollrt~unr l
11 -1
-
p2
Rauult's a ~ ~I rlenry's l Standard Sldle
Ld\*is
We define the solute activity in ternis of molalit hy requirit~gthat
nt~dthat a , ic dcfiried with respcct to a Henry's law standard slate (Equation 10.43)
(L., - -
.
(1 1.2)
(Heriry's law standard slate)
kH1
whcrc the subscript .r enlphasi~esthat m 2 , and kl,,r ilrz bascd on a mole fraclion scale (P: = k , , xx,). Even if the soll~tcdoes nut h ~ v ae measurable vaporpressurc. defining the acl~vilyby Equation I 1.2 I Snevertheless crmvcnient because the ratio is still meaningful; evcn though 1': and k,,? may be exveedir~glysmall, the ratio P,/ k , is tini~e. Although w r havc defirlell the activities of the srdvcnt and solute iri terms of ~llole f~ractiuristhe use of mole fractions to exprehs the conce~~tration of a snlule in a dilute solutiur~is not numerically convcnicnt. A rtinre canvcniei~tunit is molrilifi (rn), which is delir~edas thc number ot.mt>lcsof s o h ~ t zper 1000 granis of solvent. In ill1 equation, we have
,
where the subscript In ert~phasizcsthal a?,"is bnscd on a molality scalc. \jie can exprcs5 Henry's law in terms of the molality rather than thc mole fraction by P: = k , , ) , , I I I . where once again [tie subscript ni eniphasizcs that k , ,,,, is basud 011 il ~ ~ ~ o l : ~sc;llc. l i t y In terms of k,, ",, the ac~ivityof the aulute i s defincd by
Another common concentration unit is trlolurity (r.), which ih the numbcr o l - n ~ r ~ l z s of solute per 1 OH) n1L of sdution. 111 an equation, n,
=
(11 Kl
.--
lO(l0 inL snlutic~n whciz 1 1 , i~ the number uf rnt~lesof solu~e(subscript 2). Note t h a ~the units of molality are ~nol.kg-I.We say that a solution containing 2.00 n~oleso f NaCl in 1.00 kg of water is 2.00 molal, or thal it is a 2.00 rtlul. kg-' NaCl(aq) solution. The relation between the mole fraction of snlutz (x,) and molality ( n t ) is
Note that lnolarily has units of rnu1.L I . Wc say that n solution containing 2.00 n1(>1e> of NaCl in 1.00 liter of solution i s n 1 . 0 h n o l i i 1solution, ~ or that it is u 2.W rriol-I. ' NaCl(nq) solution. We define the solute activity it) tenlls uf moli~rityby I-cquiring that a,,, + c.
wherz M I is the molar mass (g-r r i t > l - ' ) of thc solverll. The tern] 1000 g.kg I / M , is the number of moles (Tfbolvent ( n I ) in 1000 g of solvent and m , by definition, is lhe numbcr of moles ~ )solutc f in 1000 g of solvent. In tIiz case of water, I O(K) ~riol.kg-'/ M, is equal 10 55.506 mof .kg-', Erluatiorl 1 1.4 bccu~nes A-
nl -I -
55.506 mul-kg-'
+ In
Note h a t w, and nt are directly proportiot~alto each other if which is the case for dilute solutions.
1
tn
<< 55.50h mol.l;g
',
as
(,
( 1 1.9)
+0
where the subscript u etnphasires that u2, is hased nn a nioI;~~-ily scale. Wc can express Henry's taw in tern~sof the molarity rather than the n ~ o l efraction ot solute by P , = k,,,(c, where nrice again rhc subscript r. emphasizes thi~tk,, , is h:~scJo n a rllolarjty scale. 111temis 01- k , , the activity of the solutc is delined by ( ,
Curlvesting firom niolarity to molality is easy ifwe know the density of thc soluiirul. which is available for masly solutions it) handbat~ks.For ex;irnple, the density (>I' a 2.450 mol .L aqueous sucrose solutinn nt 20r is 1.3103 g .rn~.-I. 'Thus, thcl-e art' 838.6 g of sucrose. in 1000 1nL uf solutiun. wl~ichhas a total mass of 13 10.3 g . Of ihcse 13 10.3 g. 838.h g are due to sucrose. so 13 10.3 - 838.6 g = 471.7 g i ~ dilz c to water. 'Thc ~nolalitythe11i s given h!:
'
I
E X A M P L E 11-1
C'alculatc the mole fr:~ctionof-a 0.200 mu1 .kg-' (:I. ,H,,Ol, ( a q ) solution. . . 5 0 I U 1. 1 O N : The sulutiul~contarrls 0.200 mules ot wcrosr per 1000.0kg of water.
The iuolc fraciion of
sucre; is
I
E X A M P L E 11-2
The density
[ill
f - m ~ , - of l ) nrl nqueou~+ucroEI: c01t111011L A I ~hc cxpresbrri
ah
r h ~ p r e r1 1 I Snlutluns II: 50l1d-Llquid Sulutiuns
442
C,,I 1(ul;~te , thc mularitp of a 2.00-n10Ii11aqueous sucrose colulion.
Table 1 I . I sutnmarizcs the eqilations for the activities we have delined for the various conccntrution scales. In each casc, the acLivily coefficient y is dcfilied by dividing the activity by the appropriate concentration. Thus, lor example, ym = a , , " / ~ n Prob. lem 11-12 asks you to derive a ~ I a t i o nhctween the v a r i c ~ i ~solute s activity cncfticienls in 'Table 11 - 1 .
S (.) L U T I U N : A 2.IX)-molal ;Iqlleous sucrose svlution conpiins 2.W n~olzs(b84.bg) of sucrosc pcr 1000 p of H!O, or 2.(M nlolcs of sucrose in 1684.6 p o l solution. The dcrlcily nT 111c~ ~ ~ l u t is i ngivcri n by
11-2. Thc Activity of a Nonvolatile Solute Can Be Ohtaincd imrn I ~ F : Vapor Prcssurc of the Solvent T h e cqu:itions frlr the solute nctiviticx in Table 1 1. I are npplicnhlc. t o non\ol;~lilr as well a< volatile suluies. ?'he vijpcjr pl-cssure of ;I r ~ u n ~ o l a t i wlr~tc lc is so low, hourrer. tllat tllcsc cquaiic~nsare not practical tu use. Forlunately. thc Gibhq- Duhern q u a tion provides us with a way t o dctcrniine lhe activity of a nonvolatile s o l u ~ eli;clrrl a measurement o f the aclivity of the solvent. We will illustrate this prnccdurc using 2tr1
I hcrcfore. the n~olaritpof the solution is
t'rohlct~~1 1-5 asks you to dcrivc n gcncral rctatiol~between c ilntl m.
I
11.1 A sumrnary of thc equation.; fur the activities uccd for the varirlr~scr)ncentra~ionscales for dilute solutiuns. TABLE
I
E X A M P L E 11-3
Inw standard state
Solvent-Raoult's
.
-
11
( i ~ v c t l~~ cclct~sity1 0 ) ot tilt s01~ti1i11ill fi.n~L-I,dcrivc n gcllei-al rel;~lic>nhetween x, a r ~ dr,.
5 0 I U I 1 (.) '4: Cunhirler exactly a one liter sample uf the solu~iorl.111this casc, c = I : , , the nuniher- of n ~ o l eo f~ {rjll~tein flie one-liter sample T h e 1rinw of Ihc solution is yirrn by
Sr~lutc tlcllr-y's law standard stale
--
-. .
Molc tr~ctionscale mast
\(I
(IT the soluliun per liter = (IDCIO ml:~,-l)p
- ,
the rn;ls< o f the s o l ~ e n i\t In;].;\ of the colve~ltper liter
2=
nlass of the soiution - maw 01-[he solule
= (1000 n1L.L ' ) p - rM: wl~err.JW, is the
rr~olar~ r l i ~ \f he - ~ ~ 1 t > 1 - ' ) o i [he rulute. Therefom, n , ,thc llurnber of
inoleu rjl sulvcnt, i.;
(I2v,
= - p,--
Lrn
-;
YZ,,,
1(21.
l7:,"
4
3< 111 +
rn
0
P, . + k,, ",m a< m + O
(Henry's law)
Mrliarity scale n,
+ I-
ac
P2 3 k l l , ,c
r, + O RS
c +0
(Henry's law)
Chapter 1 1 / Solut~uns11: S n l ~ dLiqu~d Solutions
iIqueous solution of sucrose. According to a Raoult's law standard stare. the activiiy of' the water is giveti by P , / P , * .Now let's consider a diltlre solutiotl, irl which case a, : :x , . We now want to relate a , tu thc ~nolality of the solute, t n . For a dili~tesolution. III < < 55.506 mol-kg - I , LO we can neglcct m compared with 55.506 mol-kg-] in (he dcllolninatnr of Equi~~ion 1 1.5 and write
T A B L E 11.2
The wprlr pressurc of water (PI) in cquilihriur~lwith an arlueutls stlcrore ar)lullori 81 25°C as a fu~~ction uf molality (n~). Add~tiollaldata are the achviiy o f thc water r n , ) . the umotic crlctficic~~t ($), and the acti>ily coeffic~cnt( Y , , )~of the sucl-ace.
Therefore, for small concentrations.
where we have used (he tact that In(1 - .r,) 2 -x, for small values u f s , . Table 11.2 and Figure I 1.1 give experittientnl data tor the vapor pressure c~fwatcr iri equilibrium wilh an aqucui~ssucrose solution a1 25-C: as a funclion of molality and ttlolt. fratlion. rcspcctivcly. The eqi~ilihriumvapor pressure of pure water at 25-C is 23.756 torr, so u, = P , / P ; = P1/23.756 is gircn in thc third cululnn of Table 11.2. Equalion 1 1.12 rclotes ( 1 , to the rnolality w ( ( ~only r a d i l u ~ esolution. For cxa~nplc, Ltble 1 1.2 .chows that a, = 0.93276 at 3.00 molitl, whcl-eas Equalion 1 1 . I 2 gives It1 ( I , = -0.054048, 01. n , = 0.9471. To accuurlt lor this discrepancy, w e now define a qi~ari~ily 4, called thc nsntotir>roqfiricnt. by
Nut? that # = I if the sulution hehaves as arl ideal dilule solution. Thus, the deviation 01- 4 horn unity i s a 111easureof the ~lonidealilyol' Ihe scllution.
EXAMPLE 1 1 4 Using the d:bra 111Table 1 1 .?, calculale thc valr~eo f $ at l .On r r i o l kg-' S O L U T I(>N : We simply use Lquation 1 1.13 arid find that 'wager
" F I G U R E 11.1
Thc vapor ple\wre of wilter in cquillbriun~with dn aqucuus sucruse cnlution at 25 'C plotted again\( the ~rlolefractiorl nt water. Nore that Rao~llt'slaw (thc straight line i n thc figure) holds frrml I = 1.oU to about 0.97, hul [hilt devintiolls occur at lower valucs o f r s,,kr
in agreenicnt wlth the entry in Tahle 1 1.2.
11 -3. Culligativc I'ruperties Art. Solution Properlies
t~:,obtain
z:
where we have uscd the fact that 77: = A,,,,HIhr rllc port solvcnt. It' we intzgr=~tr: Equation 1 1.17 from pure solvent, where a , = 1, T = T,;,,to a solutiun with arh~lrarj values of a , and TtIII,me obtairl
F ~ C ~ U R 11.3 E
The lugarithlu of the aclivlty c u e f l i c ~ e ~(111 ~ ty,,,>) of sucrose in an aqueous sucruse aolution ill 25-C plotter1 against the molulity ( r r t ) .
11-3. Culligative Properties Are Solution Properties That Depend Only Upon the Number Density of Solute Particles A 11umbernf solution propelties, called c'uiiigriiz~,tp , q ~ r r t i e depe~ld, ~, at least in dilute soli~tion,upon only the number of scllute particles. and ntjl upon their kind. Colligative
properties include he lowering of the viipor prcssure of it solvent by (he addition of a soluie, the elevation of the boilill? point of a solu~ionby a nonvola~ilesolute. the dcpr-cssionof the freezing pomt of a solurion b y a atllute, and osrnotic pressure. We will discuss orlly frzrzing-point depression and osmotic pressure. A t the freezing point of solu~iorl,solid solve~ltis in equilrhriurn wilh 1t1e solvent in solution. 'The thcnnodyt~amiczorldition ol. this equilibrium 1 7 ~ t ~ t
\t71icrcas usual the subscript 1 clerjotcs sr,lvt.rjt iinri T,", is the freezing prjirlt of the solution. We use Equation 10.35 tor 1 1 , t o obtain
We have wntten p\ tor
~ 1 ;sllnply
Equation 11.18 can be uscd t o detcrrnine thc aclivity of the solvent in a solution {f'roblem 11-20). You may have calculated freezing-point depressions in gcncral chcmi\lry using [he formula AT,,,,= K,m
( 1 1.19)
where Ki i s D constant, called thefre~zitrg-/joitlrrlepressian coilstant, whose valuc ~ l c pends upon (he solvenl. We can derive Equation 1 1.19 from Equation 1 1.18 by making ;I few approxi~natinnsappropriate to dilute solutions. If the solution is sufficiently dilute. then Inn, = In x , = In( I - .r2) % -x,, and i f ' we assume that A,\,,His in depend en^ of ietnpcralure over the telnperiture range (?it,,,Ti,),wc obtain
- 7 ,:; <: 0, Because x, and A,"\Hare positive quan~ities,we see inimediatcly (ha1 nr that q.u> < Ti,. Thus, we find that the addition of a solute will l o ~ ) cthe r frcc~ing point of a salution. We can express xl in terms c ~ tnoliilily f by using Equnticln i I ..I.
to cofnpaie ~t with pi. Solving (or In (1,. we get
Nuw differentialt: with resFect to tcniperature and use the G~hbs-Helmholtz equaliotl (Example lo-)},
for small valuer of nt (dilute solution). Furtherrl~orc,because T,:, - TllpI \ uhually o111y a few degrees (dilute solution once again), we can repldce T,,,- In the denornlnatu~of Equatiun 1 1.20 by T i - to a good approximation to get fin,~lly(Problem 11--23)
Chapter 1 1 / Solutions 11: Sol~rl-1.1quidSulut~uris
1 1 4 . Osmotic Pressure Can Be Used to Determine the Molecular Masses of Polymers
where
We can ci~lculatcthc value ol' K, for watcr.
-
Equation 1 1.22 tells us that the freetit~gpoint of a Il.20-rnolal solution of sucrose in w ; ~ t ci< ~ -- ( I .Hh K , k g . m o l ')(0.20~riol,kg' ) -0.37 K.
+
E X A M P L E 11-6 Ttic trcczing-point tlepression c o n s ~ i ~rnl l -~wntcr is 1.86 K.kg,n~ulI. Calculate the wlue a i K , for cyclohexane. n11oscficezing poin~i s 279.6 K and nlotar enthalp!: of t u ~ i mis~2-68 k.1-tnol-'.
'1 hu\, thc ircc/.rng point of a 0.20-molnl solrltion 01- hexanc i n cyc.lollc.rnt~eis 4.1 K Irnwr than thc trecaillg p i n 1 01- purc cyclohcxane, or T,us= 275.5 K . 1
Figure 1 1.4 illustrates the developnlc~lto f osmotic prcssurc. In the init~ali~gh.Such a ttlerrlbrane is called a srtnipt,rmeoblt>mta~nhrt~rrt,. (Many biologic;~lcclls are qurrtlunded by membranes srn~ipermeableto water.) The levels of the two liquids in Pigurc 1 1.4 arc initially the sanie. hut water will pass through thc scmipcrmcahle membrane unlll the chemical potentials of the water on the two ~ i d c aof thc mcmhrane are equal. -1'his process results in the situatinri shown ill the cquilihrii~illcliIlr, where the ~ w liquid o levels arc rln longcr ctlual Thc hydrostatic prcsburt. ht.iid lhat is h u i l ~up is called orrnorir pr-t'ssurr. Because the water is free to pass thrnt~ghtlic semipcrrneahle membrane. thr cherrlici11 poleri~ialof the water must bc the satnc on the two aides of thc memhrarie at equilibriunl. In other words. thc chclnical potential c~fihe pure waler at a pressure I' must equal the chemical potential of the water in the suluticln at a pressure P n and an activity a,.In an equaliori,
Diluted sululiun
I
I
Hydrustotlc prcssurc ticad =
We can derivc an cxprecsiun for thc hc~iling-pintelevation of a soluticln containing a not~volntilcsolute. ?'tie ailalng nf Equation 1 1.22 i'; (Problem 11-25) 7
-,here thc boilirig-pclint rle~wrionmri.rtunt is given by I n i t i a l ctatc F I G U R E 11.4
The value of K , for wntcr is only 0.512 K.kg.mol-', so the bailing point elevatiun i s for aqueous sol~tticms.
;I F ; I ~ ~ C~Tt ~ i lcnCct ll
.
Equilibriurrl s t a t e
Passage of water through a ripid, semipt.rmeable membrane separaling pure water from an aqueous sucrose solution. The water p a s w through thc mcritbrmc until the chenlical potential of the watcr in the aqueous sucrose solution equals that of the pure water The chemical potential of watrr in the sucruse soluliun increaqes as the hydroftatic prescurc above the solution increases.
1 1-5. Solut~unsui tler
where r i , = PI/ P;. We car1 rewrite Equation 1 1.26. as
lrnlytes Are
Nunicle,~lat
Kelat~vclvI uw ( orl+entratrrms
where u is thc molarity. n,/ V , of lhe solution. Equi~liori1 1.31 is called the van't I I l j f l equation for osmolic pressurc. Using h i s equatiun. we calcul;jtc the osmotic preacurz uf a 0.100-~nolnraqueous solution of sucrose at 20 C 10 he
n
The lirst two terms in Equariori 11.27 are the differe~lccin the chernical potential of the pure solvent at lwo different pressures. Equation 9.8
= (0.100 111r~l.~-')(0.08206 I:a~rri.K
I
.rnul -')(293.2K)
= 2.40 atm
Thus, we see that ostnotic pressure is a largc ef'fict. Because ot'lhis, ohmutic przs~~ll-c can be used to detern~jnemolecular masses of solutcs, parliculiuly sulutes with Iiirgc molecular masses such as polymers and proteins.
v'
where is the molar volume of the pure solvent, tells us how the cherrlical potential varies with pressure. We can use Equaliun 9.8 to evaluate p;(T. P n) - p ; ( T , P ) by integrating huth sidcs from P to P n to gct
+
+
E X A M P L E 11-7
II i s lllund that 2.20 g or a ccnain pc~lgmerdl$sulved in enough !vatel- to rllalc 3011 1111. of solulinr~has an osmotic pressure 01'7.45 torr at 20'C. nelerrninc the mulerulnr mass of the prdynie~.
If wc substitute Equation I 1.2X inlo Equalion 1 1.27, we obtiiin
Assuming
7'does nut vary with applied prcssurc, wc can write qua ti or^ 11.29 as
Further~norc,if rhe solution is dilutc, then ( 2 , r= x , = I - x z , with x, small. Therefore, we can write In u , as In(] - x , ) = - x 2 , su that Equation 1 1.30 becomes
Funtie~,more,becausc .r, ih small, n 2
<< rr ,
anrl
S O L U T I (.) '4. Tilt lnulariry 01- [he solutiun i\ giver^ Ily
Thcrcfore, there ore 4.07 x lo-.' rtjolcs of polymer IK\T liter U I S ~ I U I ~01 CIII, (0.300)(4.07 x 10 .') = 1.22 x 10 rnulc~per 300 m l . ursolution. Thuc, wc f nd th;il I -22 x I0 ' mules cn~.re~ponJs to 2 2 0 &, or t h ; ~rht. 1no1ccul;lr IlIilsc is I X.(H)O.
'
If a pressu1.c in excess of 26 aim i s applicd to sea\vaIer at 15 C:. the chemical potential of the water in the seawater w ~ l lexceed that of pure wiiicr. C:on~~quc.r~tl). pure water can he obtaincd from henwatcr by using a rigid semiper~r~zahlc ~nc.rnhl-,~~~r. and an applied pressure in excess ot the osmotic prcssurt. of 2h ;ilnl. 'l-hib ppl'occs~i % krlown as r ~ r r r . so~mosi,?. ~, Reverse osnicrcis units are corn~ner~.iallpavailable ant1 ;II'C used to ohtam I're.sh watcr from halt watcr uslng a variety of serriipern~cablcm z r ~ ~ b r a l l c ~ , the triost cummon of which is cellulusc :jcelale.
Substitute this into the above equation lu gel
11-5.
whcrc wc havc rcplaccd a,"** by thc total volume t,t the solution. V (dilute solution). Thc above equation is usually writtcn as
When s~)llitl~n ctilt)ride diswlvcs in water, the solution co~itainss o d ~ u mions ant1 chloride ions and essentially no undiqsuciated sodiurn chloride. The ioris intel-acl with cach other thl-uugh a coulombic p o ~ e ~ ~ t iwhich a l . varies as I l r . We should corrlpnre this inleractio~lwith (he one hctween neutral solule moleculca (noneleutrulyte~jsuch as sucrose, where the inleraction vi~ricsas something like I / r " . Thus. lhz interactior~ between ions in s o l ~ ~ t i oisneffective over a ~ r ~ u cgreatcr h d~stancethan the itjteraction
Snlutiol-ts of Electrolytcs Are Nonideal at Kelativcly Low Concentrations
11-5, Solut~un$rlf Elcctrufytcs Are Nuriidcal a t Kcbtively Luw Ci~n~cntratiori,
brtweeti neutral solute particles. so solutions of clectrol ytcs deviate frortl ideal behavior no re strongly and at lower concer~tratio~~s than do solutions of nunelectrolytes. Pip u1.e 1 1.5 shows 111y,_ for sucrosc, sodium chloride, and calcium ctiloride plotted versus rnolntity. Note that CnCll(aq) nppcars to bchuvc more nt~nideallythat) NaC'lraq), which in t u 11 ~ hehaves nrorc. nonirleal1y than sucl-ose. Thc charge of +2 on t l ~ ccalciuin ion leailh to ;i slrorlger coulr~rr~bic ir~tctaclionnrld hcncc a stronger devialion frum idei~li~y thar~for NaCI. At 0.100 m o l , k g I, thc activity cr,efficienl of cucrose is 0.998, whereas thal t~f-CaCl,(aq)IS 0.5 1 X and that of NnCl(aq) is 0.778. Belore vie discuss the dctennination ot activity coet'ficier~tsfor electrolytes, we rrlusl first introduce nutalion rleeded to describe the thermodynamic p r o p t i e s of solutions of electrolytes. Consider the g n e r a l salt CV,Av, which dissucintes inta 17, cations and v ailions per forniu!n unit as in
w11c1-c1 1 , :, $- I. := 0 by clcc~ronegatiri~y For example, I!, = I and v - = 2 fur i';1C1, iln~l1 1 , = 2 and 1 1 = 1 tor Na,SD,. Thtrcfore, CaC1, is called u 1-2 electrolyte anti Na,SO, i a called a 2-1 electrolyte. We write the chemical potential of the salt in tcrlns of the cllctnical potentials of its constituent ions according lo
and
Thc sul>crscript zcl-os licrc I-cpresent the chohen standilrd qtatr. which w e can lravz unspecilled it1 this point but is usually taken to he the solute or Heruy's law standard state. If we suhslilute Equations 1 1.34 into Equatior) 1 1.32 and equate the srsult to Equation 1 1.33, we oblain
+ v-lr'
=LP+~: where we have used the rclatiun We can rewrite the above equation as
in analogy with Equation 1 1.32.
For many of the formulas that occur in the thermodynamics of sulutions oI'elcctrolytcs. it is convenient t c define ~ a quan~ityn,. called ihr mcori imtri(: actit~i@, h!:
+
where I! = I]+ 11- . Nole thal o, is raised to the same power as the sun1 of the exponents in the last term in Equation I 1.3h. For example, we write
and
Even thuugh we cannot delemine activitiec of singlr ions, we can still drji~aesit~yle-ion activity crx0icients by m
+
= m.+y-
and
(I-
= rn- y-
where t n , and n are the molalities of the ittdividual iotls, which are given by m t = v t In and m - = v -m. If vie substitute these expressions for u , and rr ir11o Equation 1: 1.36, wc get I
,
u,=uy=(m+m F I G U R E 11.5 The logarithm r d I he aclivity cwfficient (In y,,) of ~rlueul~s solutiuns of- sucrose, sodium chl(>ride,and calcium chloride ptoned against ~nolality( r t t ) at 25-C. Note that the electrolyte ~ o l l ~ t i odcvratc rl~
coi~ccnlrations.
fro111ideality (Ill y,m = 0)riluch rI1orc stmngly than does sticrose at small
,
I
"
)(yl-y
(11.37)
In analogy with the definition of the mean tunic uutivitj a, in Equatton 11 3 6 , we define a Inearl ionic rnolnlity rn, by
1 1 -5 Solutions arl kl~c.lrnlytfiAre Nonidedl at Relat~vclyLow Cunccntratluns
and
;I t ~ t e u ~ iur~ic l ut.livtl~uorJi~>iir'irirt
yt hy
Again. r~oiicethat the s u m of the exponent.s on both sides of Equations 1 1.38and 1 1.39 are thc samc. Given these delinitions, w e can now write Equation 1 1.37 as
E X A M P L E 11-8
Writc out Equatiorl I 1 40 c.rplicitly lor CilC1,. SC) L 1J TIC) N : In this tn)n
fit. hee
case, v+ = 1 and
that m , = III and rlr _
v-
= 2. Furlher~nore,accordillg to thc equa-
Mean ionic activity coefficients call bc determined experimenlally by the samc methtds used for the activity coefficients of nonelcctrt~lytcs.Wc will illustrale their determination froln the measurement of the vapor pressure of the solvent as wz did fix an aqueuus sucrorje solution i n Section 1 1-1. 111anslogy wilh Equatiun 1 1.13, we defiriz an osmotic coefficient for aqueous electrolyte solutions by
Notice that this equation differs from Equation 11.13 by thc inclusion of a fautc~rol' 1, here. Equation 1 1.41 reduces to Equation 1 1.13 for t~onelectrolytesr>lutionshecauw 11 = I in that casc. Problem 11-34 asks you to show that with this factor of 11, C$ -+ 1 as m + 0 for solutions of electrolytes or 11une1ectrolyte.s.Scarling with Equation 1 1.31 and the Gi hba-Uuhem equation, you can dcrivc ihe analog of Equalion 1 1.15 straighlfrjrwardl y:
= 2 m . Thur.
Table 11.4 gives the vapor pressure c ~ an f aqueous aulution of NaC1 as a functior~ of tnulality. Also included in the table are aclivilies of the water (calculated from Tllc rclntlons bctwcen Tahle I 1.3.
(I,.
m , i111r1 y, Inr uther ty~wsof elcctrolytc~are givcl~in T A B L E 11.4
The u p u r prersurt: ( P ), act~vityof thc water (aw),osrnoric coefficient (4). H ~ U and log~trlthn~ of the mean ionic activily coefficienl ( I n y,) ot thc NaCl in an aqumo"9 9olutlon of NaC1 at 25-C as a function or nithalily ( m ) . TABLE
11.3
lhe relations between the aclivi~y01- a
r}~/n~ol -kg-'
P,, ,,/turr
0.OW
23.76
0.200
23.60 23.44 23.29 23.13 22.97 22.H
0.400 0.h00 0.800 1 .oOO 1.300 I ,son 2 200 2.(100
ZIISO (aq)
nl=n+cl-
-
1
1
,
= o ; = r n ; y ; = ! ~ ~ ~ , ) ( )~y~: t=tlr2y:
7-1
Na,tC(CN),(aq)
u! = c i i c r -
-- 02 = rn:
y: =
irrt , ) ' ( I T S ) y :
= (3trz)'(tti11'~= 27rti4y:
3.000 3.40() 3.8'00 4.4W 5.000
22.30
21.96 21.59 21.22 20.83
20.43 14.81 19.17
1l
4
1.WW 0.034 0 9x68
I .On00 0.9745 0.13205
O.YX(12
0.9227
0 Y71h
0.9285 0.9353
0.9669 0.9532 o.4384 0.9242 0.91189
0.8432 0.8769 O.Xh(M)
U.8339 0.80h8
0.451)2
0.972 1 0.9944
1.0 196 l.I)449
111
y=
0 00[)0
-0.3079 -0.3b85 -0.3477 -0.4143 -0.4234 -0.4207 -0.4 1 66 -0.3972 -0.3709
-0.33'36
1.0723 1 ,1015 1.1357
, , 0.304t~ -0.2666
1.191
-0.1389
- 0.2053
Chaprcr 1 I ! Snl~~lionr I I : Solid-Lirlu~dSolutions
n , = PI/ P ; ) . orrnollc cuefficicnts (calculated from Equation 11.41), and mean ionic ,icl~vitycr,efficlrnt~(caiculated from Equation 1 1.42). For sucrose in Sectlor1 11-2, n e curve fit 4 to a polynorntill ill In and then used that polynomial to calculate the value of y,,. As we will see in Sectiun 11-6. the osmotic coemcient of elzctrulytes is better described by an expression of the form (a ~xdynoniialin m "')
difference lies in Equation 11.21 for x:. For a strong electrolyte that dissociates into v+ cations a ~ v~. danions per formula unit, thc mole fraction of solute particles is ~ i v c nby
Xz
=
vln
1000 g . kg-'
"-
+ urn
vriiM, 1001)g. kg-'
( I 1.U)
MI
Note that the right side here contains a factor of v . If this expression for xz is cairied through in derivations of the formulas for the colligative effects, we obtain The osmo~iccoefficjcnr dikh for sodium chloride given in Tihle 1 1.4 can bc
9=I
-
(0,3920 kgl:'.mr,l-"2)nll!2 t- (0.7780 k g tnol
fit
hy
ATvpp= v Kbrn
')nl
- (0.8374 kg"2-~nt~l-"')~n'"+ (0.5326 kg'.niol-Z)m2 -
(0.1 671 kg"' -mnl .'!')1n5''
AT,? = v K , m
and
n = VMRT
+ (0.0206 kg3. mol-')m' 0 5 tn 5 5 . 0 mol-kg
'
( 1 1.43)
This cxprcssion lor 4 along with Equation 11.42 were used to calculate Ihe values of It) y , given in 'Ihble 1 1.4.
I
E X A M P L E 11-10 A 0.050-n~olalaqueous solution of K,Fc(CN), has a freezit~gpoint of -0.3C;'U, liow ln,any ions arc formcd pcr fortnula unit of K,I;c(CN),'? SOLUTION : We can solve Equluatin~l11.45 fur I. to obtain
E X A M P L E 11-9 Vc:clify the elily tor
111
vl at 1.00nrrllal 11) 'lhhle 11.4.
S O L I!T I n N : We tirst write (r~eglectiogthe units in the coefficients of the powers of I T S 111 Lqurttiu~i 1 1 11)
and add t h l ~ rcsult
to @
In y , =
-
1 to obtai~~
5m':z -(t1.3920)(3rra1:2)+ (0.7780)(2m)- (0.8374)
3
1 (0.5326)-
3nt'
2
-
7,n5!7
(0.1673)-
5
+
4rn3 (0.0206)3
I
Thus. at I.oU 1nol:11,In y,
: :
-0.4234, ur yr = 0.655.
Thc fclnnulas w e derived in Section 11-3 for the colligative properties of solutions uf' r~nr~electrolytcs take on a slightly different form for solutiuns of electrolytes. Thc
Thus, the dissolution process of K,Fe(CNI, can he written as
11-6. The Debye-Hiickel Theory Gives an Exact Expression for In y, for Very Dilute Solutions In the previous sectiorl, we expressed the osrrlotic coefficient for solutiuns of elect~,ulytes irl the fortn 4 = 1 rirnl!' $ hnl t- . , . rather than as a simple polyrlomial in rn as we did for sucrose in Scction 11-2. Thc reason we did so is that in 1925. Peter Debye and Erich Huckel showed theoretically that at low concentratio~ls.the logarithm of the activity coefficient of ion j is given by
+
Chapter I 1 / Solutlun5 I I Snltd-L~rlu~dSolutiuns
and thal the logarithtn of the rnean ionic aclivity coefficient i s given by (see Problema 1 1-50 through 1 1-58)
where q , = z + t b and y = :-c are the ciiargcs on the cations a~ldanions, (unirtess) relativc ~ r m i t t i v i t yc ~ the f sulvent, and K is given by
5 is thc
whew .I is thc number uf ionic species and IVJ 1' is the nutnbcr density ot specics , j . If w e ct~rlvertN , / V to molarity, F ~ u a t i o n1 1.50 hecn111e.s
It is customary to define a y~iantilyI , called thc
ior~ica~rength, by
I I 6. The Debyc-Huckel Theory
Equatioll 11.49 is called the Debye-1 Iiicket limiting law hecause it is thc cxar.1 form that In y , takes on for all electrolyte solutions for sufficiently low cuncentratioi~:,. just what is mcant by "sufficiently low conuenlrations" dcpends Itpun ~ t l esystcnl. Nuic that In yL goes as K in Equation 11.39, thal K goes as 4''' in Equariotl 11.53, and that goes as c':' in Equation 11.52. Cunsequently, In yf varjes as r.'!'. This cl,' dcpendrnce is lypical for electrolyte solutions, su when we curve tit 4 in Secrion 11-3. wc tit it to a pulyno~~iial in cl:' (or ml") instcad of-(,(01- m ) . Most c ~ fthe experimental data for In y, arc given in tenris of moliility ralher tllaii ~nolarity.In Figure 11.6,we plot In y, versus m ' j 2 for a number of 1-1 clectn)lytzs. Note that dl the curves mergc into a xingle straight line a1 small cnnccntralic~ns,i11 accord with Lhe limiting law nature of kquatinn I I .49. At srt~allcuncenti-ati0n.s whrrc thc limiting Ianr is valid. the molality and molunty scales differ by only i1 tnultiplicative crmstant, so a linear pl(>tin cl!' is also l i ~ ~ cin a r,nl:' (Problem \ I - 5 ) . The quantity K in Equation 1 1.50is a cerl~ralyr~a~ltity in the Uebyc-Iluck~-1theory and has the following physical in~erpretation.Consider an inn with charge q, situated at the origin of a spherical coordinate syswn. According to Dehye atid Hijckcl (see a l ~ oPmble~n11-51), the net chargc in a spherical shell uf radius I- and thicknehs dr surrounding Itiis central ion is
If we inregrate 111;sexpression fro1110 to ,x.we ob~ain whcrl: r , is the molarity uf Lhe jth ionic s l ~ c i e sin , which case (Pmblcm 11-46)
This result simply say5 lhat the total charge surrounding an ion of charge q, is equal and of the opposite sign t c ~q r . h other it expresses thc eleclrooeutraiik): r)f llle E X A M P L E 11-11
First show rhat unillcss. as it
K huc
units
rll-
111
' ;~ndthcil shuw that
In y,
111
Equatior~1 1.-i9 is
be.
5 ( )I. I J T 10N : \Vc star1 % ~ t hLlualitl~~ I I ..iO Thc u ~ l i t c o q, f are C. s,, a r c ~ ' - \ ' , I : : I. m-', = kg.rl12.s-:-~--'. T are K,and N , / V are rK3.Therefow, thc unilc of A - ;ire
A,! :ue J - K - I
Using Equation 1 1.19 for h y*. F I G U R E 11.6 Values of In y, versus rn':' fur aqueous alkali halide +olutiuns at 2 5 C . Notc that even though the four curves are different, they all merge into one. the nrhye-Hiickel limiting law
(Equation 11.49) a1 small concentrilticlnc.
461
C:hsr~!cr 1 1 / Snlutiuns II: Colirl-Lrt(~~irlSolutiur~s
s~hutiun.Equation I 1 53,which IS plotted in Figure 11.7, xhows that there is a diffuw shell of net charge of opposilt: sign surround~tlgany given ion in solution. We say that Equation 1 1.54 describes an iunir utmosphere about the central ion. Furthermore, the In~utlrnilmin the curve in Figure 11.7 occurs at r = K so we say that K - ' , which Evaniple 11-1 I s11rlu.s tins units of rn, is ;I rneasure of the ihlch~casof the ionic atmosphere. For a 1-1 electnllyte in aqueous solution at 25'C, a handy formula for K is (Prohletn 11-51)
',
where r. is tbc molarity of'ttle solulion. The thickness of the iotlic atlnosphcre in a 0.01 0 tnoli~rsolution is approximately 3000 pm, or nh(~irtI0 times the s i ~ of e a typical ion. For an aqueous soiution at 25'C. Equation 1 1.49 hwomcs (Problem I 1-59)
According to Equalin11 11.52. [ related t c ~the ct~nuerltr;ition,but the relation itself rlcprrjds upon the type of clectmlyle. For exatnplc, for a I - 1 electrolyte, z+ = 1, : = - 1 , r _ = c , and r. = 1.. so I = (,. For a 1-2 electrolyte such as CaCII,, z+ = 2. : = -- I . r-+ = r.. and(, = 2r,so 4 = f (4r $ 2 c ) = 3r. Generally. I, is equal to sorrje i~u~ucrical tictor limes r.. whcrc the vitiue of the nun~ericalfactor depends upon the type of salt. Thercfure. Equation 1 1.56 says lhat a plot of In y, versus r'" should be a f straight line and thal the slope of the line should depend upon the t y p c ~ electrulyte. The slope will he - 1.173 for a 1-1 electrolyte and -(1.173)(2)(3':") = -4.06 for
a 1-2 elect~olyte.Figure I 1 8 s h o a~ plot ~ of In y, bersus rl,' lor NaCl(aq) and CaCl,(aq). Notice that the plots are indeed linear fr)r smaH concenirations and Ihar dcviatioris from llnear behavior occur at higher concentrations [ ~ l x' ~0 05 mol L-' ob c = 0.003 ~nol.l,-' Cor CaCl,(aq) and r'" 0.15 mol-id-' or c = 0.02 n ~ o l . ~ , for -' NaCl(aql1. The slopes of the two linear portions are in the ratin of 4.06 to 1.17
*
11-7. The Mean Spherical Approximation Is an Extension of thc Debye-Hiickel Theory to Higher Concentrations ?'he Dehye-tliickc.1 theory a?r~111es (hilt the ions arc simply point io11s ( 7 ~ 1 . 0riitlii) and that they inler8ct with a purely coulonlbic potential ( U ( r ) = ~+:-r'/4xt,f,rl. I n addition, the solvent i s considcrcd ;I co~~tinuous tnediuln with a u~liforn~ rebtivc permittivity F , (78.54 Il)r water at 25°C).Although the assumptions of point ions ;111d a continuulr! solvent may seem crude. they are quiie sntishv~orywhen the iau arc far :ippart from each othcr on the average, as they art. in \cry dilute solutions. C:onsequcntly, the Llebyc-liiickel expresqion for In y, given hy Equation 1 1.49 is exact in the limit of srnali corlccntr~tions.Thcre is no correspoitding theory for solutions of noncleclrolytes because, being ~ieulralspecics, nonelectrolylt. molecule^ do not intcract with each olher to any signilicarlt cxtent until they approach cacl~othcr ~rrlalivelyclosely. whcrc the solvent can hardly he assurned t o be a co~ltinuousmediuln. Figure 1 1.R e~nphasi~es that the Debye-Iliickrl theory iq a lirrliting law. It should not he considered a quantitative theory with which to calculare activity coet licierits exccpt at very Iow concentratiuns. Nevertheless. the Debye-Hiickcl theory has played ari i~lviiluahlerole as ;i strict limiting law that a!l electsolyte srdu~ionsobey. 111 addition. any theory that attempt^ io describe solutions at higher concentrations must reducc t o Equation 11.49 for s m a l l uoncentraticms. Many attempts have heen tnade tu construct
F I G U R E 11.7
F I G U R E 11.8
A plclr of the nel charge in a spherical shell of radius F and thickness dr s~lrroulldinga central i r r r ~rjf charge q:.Thi* plot illuctratcs the ionic atmusphere that su~rvuncl~ coch ion ill solutiun
A plot oCthc logarithm of the mean ionic activity coefficient (In y,) for NaCl(aq) and CaClltaq) at 25°C versus Note thal both curves approach the Debye-Hiickel limiting law (the ctrnight lines) as the moiarity prlcri to zero.
Chapter 1 1 / Sulut~uns11- Solirl L~quidSolutions
Most of these theories that have h c c ~dcvclopcd ~ rcquirc nu~tlericalsolutiot~s10 lairly complicated equations, but one is notable in that it prrlvidcl analytic exprecqiclns lor the variuu:, thennodynamic propediec r)l- elec!rolyle solutions. The nun~t.o f this theory, the mean spherical approxitr~ation(MSA), derives liorri i ~ oiiginal \ formula ti or^, and the theory can be viewed as a Debye-Hiickel tlieury in which the linitt. (11onr.m)) size of the ions is accounted for in a fiirly rigon)ux n3annt.l. A central rchult of rhc mcxn spherical iippruximation is that
thcrjrizs h ~ more r ctlr~uen~rated electrnlyte solutions, but nlosl have mct with only linli~edauccess. One early attempt is called the Extended Debye-Hiickel theory, in which Equation 11.49 is ~nodificdto be
Thi:, expression becomes Equation 1 1.49 in Ihe litnit of small concentrations because !.I1' bttcornes negligible colr~paredwith unity in he denominator of Equation 11 3 7 in this limit. where In y2' is an electrostatic (coulomh~c)contlihutiun to In y+ ant1 In ytiS is u hardsphere (finite-size) contribution. Fur solutions of 1-1 clcctrulytcs. In y:' is givcn by E X A M P L E 11-12 Use Equariun 1 1.57 10 calculate 111 y + for 0.050 molar I.iCl(uq), and cornparc the result with that obtained from Equalinn 1 1.49. The accepted expt.rimenta1value is -1). 191.
where p is thc number density of chargcd palticlcs, d 1.; thc sum uf thc r a d i ~ rof a calion and an anion, and x = ~ dwhere , K i . given ~ by Eqiration 1 1.53. Although it is oot obvious by casual inspection. Equatiot~1 1.59 reduces to the Dchye-Hiickrl l i tnitil~g law+ Equation 11.49, in the limit of small col~centrations(Prohle~nI I (A)). The hard sphere cr~ntributionto In y, is given hy
S O L 1J T I O N: For a 1.- 1 salt such ax LrCl. I' = c., so
lny = -
l . 173(0.050)]"~
= -0.214
1 + (0.050)~'~
Although Equation 1 1.57 provides some irnprtrvenlerlt over the Debye-Auckel limiting law, st is not very accurate even at 1).050 ~uolar.At 0.200 molar, Equatior~11.57 gives -0.362 for In y, vcrsus the expzrimenlal value of -0.274. I
i
Another .cemienjpirical expression lor In y , lhat ha^ been widely used to fit experi ~ ~ i e ~data ~ t aisl
In y- = -
I.I7B[z+z I(I,/inol,L
I)'"
1 -t (I\./ r n o l . ~ - - ' ) " ~
+ C'm
where C is a parameter whose value depends upon the electrolyte. Although €quation 11.58 can he used to ht experimental In y, data Up to one molar or so, C is still strictly an adjustable parameter. 111the 1970s, significant advi~nceswere rnade in rhc theory c~felectrolyte solutions. Most of the work on thehe theories is based on a model called the pri~tritivemodel, in ivl~ichthc inns are considered I~nrdsphcrcs with charges 211 their centers and the solvent is considered a cgntirlunua rnedlu~nwith a unifur~r~ irlntivc p r ~ n i t t ~ v i t yIn. spite of the obvious deficiencies ul' (his ~rlodzl,it addresses the long-range uoulumbic interactions bctwccn the ions and their short-rangc repulsion. These turn out to be ttlajor cor~siderations,and as we will see. the primitive model can give quilt salisfactory agreement with expzrimer~taldata over a lairly large concentration range.
whcre y = n p d 3 / 6 . It1 spite of the fact that Equations I I .hO and 1 1.61 are sotnewhat lengthy, they arc. easy tu use becauhc once d has been chasen, they give In yf in tzrlns ot' (he mcll;~~-ity rm. Figure 11.9 shows experi~nentalvalues of 111y, for NaCl(aq) at 25'C and 111y, ;ls calculated frorri kqi~ation1 1.59 with d = 320 prn. Given essentlall y one atljuslable parameter (the sum (>I' the ionic ~ a d i i )the . agreement is secn to he quite gurld. We alsr) show the results fr)r the more ct~rnlt~only seen Equation 11.57 in Piguru 1 1.9.
Problems 1 1 - 1 . The de~witpof ;I glyccn,l/water solutiun thal is 40.0% glycerrll hy mas\ i\ I . 101 g - m l . ' :)I 20 C. Calculu~ethe niolalily urld the r~lolarityuiglycerol in the \r)lulion a1 20'C Calculaie the rrlr)lality ;IL U C. 11-2. (:onccnlratcd ccllfuric acid is sold as a st>luliorithat i > YX.O!'r su1t~11.i~ ncld :111d2.0'c wntcr hy Inass. Given that the density is 1.84 p-~nl:~, cillculi~tethe rlialariry clf crlncenlralrd sulfuric wid.
11-8. Derive a relaticm hetween thc 1uo1e fraclion c ~ lhe l snlkent and the mulalily o l a holutlon. 11-9. The volume uf ari aqueous sodium chlol-ide srllu~ic~n a1 25 C call he eepre\\ed as
V/mI, = 1Onl 70
+ (17.298 kg mol
')mi
+ (0.Y777
kg'.mol
')rl12
0 5 nt 5 6 mol-kg
wllcre rrr chloride.
F I G U R E 11.9 h ct>n~parisonof In y, from the mcan sphericai approximalion (tquatiun 11.59) with cnpel-imental data for N a ~ l l a q )at 25-C. The line labelled EDH is the extended Debye--Hiickel thet~iyrcsult. E q u i t ~ i ~ 1l r1.57. ~ The value of d , the sutn of the radii ul-the cation and anion, ic
t n k c ~to~ k 320 pnl.
I 1 -3. Cr)ncentratcd phouphoric acid is sultf as a solution thal is H5% phusphoric acid and 15% u.;ltcr by tnnqs. Give11 lhar thc lnolarily is 15 r11o1.L I . calculate the dznsily rllconcel~tratetl phnsphnric acid. 2 1 -4. Calcul:~[etilt 111olc fraclion (IT flucove in an aqueous solutirln thnt i~ 0 . 5 N rnrllal irt
glucose
iq
the
'
molality. Colculntc the mnlarity r>T n soluliur~thnt is 3.00m r h r in s o d ~ u r r ~
11-1 0. I K x T , ?a". nnd c" are the mole Iract~ot~, nlolnlitv. onrl mtllarity, respectively, of a solute at infinite dilulic>n,show that
where MI is the molar mass (g,mol-I) and p, is the densily ( g . r n ~ - ] )of the solvent. Note that n101e frttvtiun, molality, ntld 1no1:rl-it?are all rlirec~ly~jroportional tu each other at low cwnuclllrill~r?ns 11-11. Consider two '.;nlutionu whosc solute activities are and cry. ~vferredto the xime standard statc. Show thal the dilferencc itt the che~nicalpolentiall., o f tl~csctwo solr~tiuns is indeprrlde~ltof tllc standard slate m d d e p n d s only upon thc ralio u i l u ; . Now choose une o f these soll~tionsto he at an arbitrary co~lcc~ltration and the olher at o vcty dilute cr>ncentrnrion(csscntially infinitely dilutc) and argrle that
11-5. Shou that the relalion hctwcenmt~larityand nlolalily ior asolutiwn with a singlc solute is 11 -1 2. Uhe Fquatioos 11.4. 1 1.1 I , and the r z < o l to~f tttc previo~lstwo problems to show that where r , is t l x rncllarity, r n is the mulality, p is the density of the solulirln in g , ~ n L',- and M, i u the rrtoliu Inass (p mtjl-'1 of the solutc. 11-6. The L'KC Hanrfbook of C/~emi.v~ry rrrtd P&sic,s has tablcs of "'conuentrativc properties 01aqueous solutions" for many solutiunri. S o ~ n centries for CfCl(s) are
where 6) iic thc de11sity of the s u l ~ ~ t i oThus. n. we qcc that the three tliffrrent activity coeftciel~ts;(re related 10 nnc at~otl~cr.
11 -1 3. Use Equations 11.4, 1 1.1 1, and the reuultf of Prt>blcln 11 -l? to derive
Given that h t: density nf an aqueous citric acid (IU, = 192.12 g . In01
')
solulion st 20-C is
given by
where A i.; the rrlass pclccnt nf the solute, p is the density of the srjlutiorl. and r is the ~ilolarity.L.\irig Il!csc d a h . calcl~lnrethe ~rlrjlnlityn! cach concerrtr;~titi~!. 11-7. Uct-ive a rrlatiot~ between the Inass percenvage ( A ) of a svlutc in a solution and its rrlulality (m). Calcul;ltt the rnulillity of an aqueous silurofe solution that is 18% sucrosc by Illasc.
plot y,,,,/yl, versus r.. IJp to what concentratir~n
yz),,and y2( differ h) 2 W ?
Chaptcr 1 I 1 Solutions H:
hulirl-l
irlllial
Solutions
'1 1-24. Calculate the valuc of the freezing puinl depression tonstam lor ni~rr)bcnzene,whobe free~ingpoint is 5.7'C and whwe molar cnthalpy uf Cusion I$ 1 1.59 !d.ni(T I.
11 - 1 4 . Ihe C'RCH(~ndbmkr,SC'hemrst~onrlPhysrrs gives a table of mass pt.rcenI rlf sucrose in all aquerrus solution and its corresponding rnolanty at 25'C. Use lhese data to plot mulal~ty rcr,sus molaritg fi)r an nqrlcous sucrrlhe solution.
11-25. IJst all argument sin~ilar10 illz nnc wc useti derive Equat~ons1 1.24 and 1 1.25.
11-15. llhit~gthe d;~t;l in Tahle 11.2. c;ilculale thc octikity cueUflcic.llt of wdtrr (on a rllolc fraclinn basis) at a sucrose conccritratiu~~ of 3 (XI molal.
t l l rlr
Equotior~s11 2 2 and 1 1.23
to
11-26. Calcuklte the hoilrr~gpoint clcvatio~~ consIan1 I l ) r cyclubexi~nsgiien ~ l ~ T%&,' a t = -434li
11-1 6. llsinp tile data in Table 1 1.2, plot the aclirily uocfficirnl r)l wattr against the mule frilctiun oC wa1t.r. 11-1 7. Gsing the d a h 111 Table 1 1.2, calculntc the value or$ at e;lch value tri?tire I I -2
I O derive
and
AvaI,H = 29.47 kJ .nlol-l.
11-27. A solut1011containing 1.470g nfdichloruben~cnein 50.00 g of k n t e n r hnil+ul X0.hO C' at a prescure of 1.00 bar The bnlling point uf purt. h e t ~ z e ~is~8U.W e C'. and thc ~nuliu enlhalpy of vaporizativi~vf pure ben~cricis 32.0 kl mu1 ' I . Dctcrlnine the n~olecularIila.;s nf dichlurohen~enefrotn thcsc data.
~ 1 1 I-epruduce d
11-18. l i t the data fur Ihe ocmotic coefiicierit o f sucrose in Tablc 1 1.2 to a 41h-degree polyou1nh1and calculate the value of y,)" fnr a 1.00-rnulal solution. Compare your result nilh the clrjc obt:~inedin Exarrjplc 11 5.
11-28. C'onsider Iht Iollowtn~phnsc di;~pramlrlr a typlcnl pure si~hh~ur~cc. L:lbel the rrgiljn ctlrrcsponding to each pha~c.Illustrate how this diagrnrn chailges k)r u dllutc solutiu~lu l ' ; ~ nonvuliitile solute.
11-19. l!hir~g the data fur sucmcc given in Tahlc 1 I .2,d'tcrnline the vnluc of In y,,,' a1 3.00 ~nolal h plotting (9 - I)/m versus m and dcternlining the area under the curvc by numerical inltgration (MnthC'l~aplrrA ] rnthcr than hy cutvc fitting 4 linl. Colupare ycwr r e ~ u l twith the value g~vcrli r ~I'ahle 1 1.2. 11-2U. Lrluatiun I 1 18 ciul br uhsrl 10 dctcr~ninethe aclir ily of the solker11at its freczlng poi111
A s w n i n g that AC; i* lndcpcl~denloi~tenlpcrature.show lhat
nlierc AlUbIi(l;,) i, (tie rilolnr enthalpy rlf tusion at {he ireezing point of the purc solvent ( T i > )and ij, the difference i n thc nlular heal capacities of liquid and solid solvent. llcirlg Equation 1 I . I X, show that
~r~
',c:
'
11-21. T u k e A l u , ~ ( ~ ; {= L )6.01 k h r l ~ o l = 75.2J.K mol-',andC'; = 37.hI.K-'.1nol ti) h o w that the equation for - In (dl in the previous problem becomes
lor an aqueous solril~on.The frcclllng point drprescirlr~of n 1.95-molal aqucou~sucrose Cornpare your rehuit with srhutinn is 4.45'C. Calculate [tie value of cr, at h i s conccntra~ir)~~. the villue in Tablc 1 1.2. The valuc you calcula~zdill this prvhlern ir. for 0 C, whrreu\ thc u l u c in 'l'i~blc1 I . ? i, for, 25 C, hu! ~ h diffei-ence c ~c tairly srnall because u , docs nut vary 4 11. grwlly with tcrnpsleture (Prtlhlcm 1 1 11-22. I'hc I r e e ~ i r ~pgi n t of ;I 5 0-liiolal ;\clrl)/ I;;, I ; " , Isre Equ:$ ~ i r l r1~1.20)gives -0/(7;:>)' - e'/t l;:,)' +- . . . where U = 7;;, - T,,,,.
'
Nuw deri~orlstrotethal the h i l i n g point IIICI-ease.\ and thc fleezing prlirlr cltcl,casea ,I> ,I reqult of the dissolution of thc solute. 11 -29. A solutirm co~i~iiinlr~g lI.HO g of n plulein in I(HI luL of a suIut~~)ri t i ~ ill1 c OSIIIU~IC:pre\\llrc of 2.06 iorr at 25°C. What is the molecular mass of llie ph)~cin'!
11-30. Shnw that tht. o s n r r ~ ~prc+.run. ic of ;III iwlull\cc)lutir)nc,in he wl-i~lerl:I,
11-31. Accol-ding t r l T ~ h l c1 1.2, tile activity rll- !hc water ill ;I 1.0-rrn)lal sucrose bolulio~~ I\ O.BSR07. What external prehsure Inust bc upplied w Itic sulk~tiuna1 25.0'C to lnnke tlir actitity 01- thc water io the sululion ~ h csnrne a\ thal r r i purc water al 2i.(l C and I atln? 'Ihke the densit; o f wntcr tu be 0.997(17 E , ~ I I LI.
-
11-32. Shrm that t r , ,I: = ~ n ' ~lrlr: n 1--2 salt such a:, CuSO,, fur a 1-3 halt such as LaCl;.
~ i t drhal
u: = 1 1 : = 27,tt4yi
Chapter 1 1 /
Solutions I!: Sol~d-I-~quirl Snlulionr
47 1
Problems
11-40. Don Juan Pond ill the Wrigtlt Vallcy of An~ai-c~icn trcexs at -57 C. The ~ n a j o rhrdulc in thc pond iu Carl,. Estimate the concc,nt~-;~liun elf CaC1, in the pond watcr.
1 1-33. Vcrify the fullt>wlr~~ tablc:
Type of salt
Euatnple
1 1-41. A sululion of ~nerctlrv(ll)chloride is ;ipoor cor~duutorof eleclricity. A 40.7-g san~plc uf HgCII, is disholved in 100.0 g ot watcr, and the I r e e f ~ n gpoint uf the aolutiwn is fuunrl tu be -2.83 'C. Explain u h y HgCI, in sululion is n poor cnnductor of eleclricity.
2- 1 2-2 1-3
K,SO,
3111
MpSO,
4111
LaCI,
6m
3 1
NnjPO,
hrn
11-42. Thc hcczingpr1intofa0.25-molalaqueous solution ufMaycr's reagent, K2HgI,, isfr~und to be - I 4 I ' C . Suggest a p s s i h t e dissr)ciat~otlttactiol~t h a ~takes pliice u h e t ~K,Hgl,, is disstdvcd in w:iter
11-43. Given thc folluwing freezing-pin1 depression data, detcr~llinelhe nutnhcr of irms produced p t r formula unit when the indicated suhqtance is disfnlvcd in waler to produce i l 1.CH)-molal sulution.
S h o n [ha1 ihc gcncr.nl rev~iltf i ~ rIrr
I?
I;,
,: - 1 1
v+ t v-)n1/2.
Formula
AT/K
I, iri Equation 11.41 allows 4 + 1 as m -. 0 fur
7 1-34. Shuw I hat the inclusio~~ of the Faclor
1 1-35. Llcc L!qu;~tiunI 1 4 1 a r ~ dthc Cibhs-Jluhcni equatinn to derive Erli~alirm1 1.42. 11-36. The osn~clticctwllic~crjtot CaCl,(atl) solulit~r~s call be expressed as
'
11-44. An aqueous solutior~of NaCl has an iotlic strength nf O 3 15 ~ n o l . ~. At - what concentration rvilI an aqucous solution of K,SO, have the same ionic wength?
1 1 -45. Dcrivc the "practical" formula h r K' ~ i v e nhy Equation 11.53 11-46. Sume authors &line ionic strength in terms of mulalit!: rother !hiin molarity, in which
case
Use this expre.\swt>to calculate ;~n(lplc~tIri y- as "unction
uf tn':'
1 1 -37. Clse Cquatic~n1 1 43 IU calculate In y, for N a C l ( q ) at 25°C a=, a fuitction nC molaiity nt~dplot it verwc !TI'". Currjparv your rewltc with lhosc in Table 11.4. 11-30. 111 PI wble~li1 1-1 9, you detcrmi~~ed In y?m Frlr sucrose by calculating the area under the curve of - I versus III. Whcn dcaling with st>lutinnso f electrolytes, it is hetternumeriually to plnt (,$ -- I):JIII" versus ml!' hecauct. (yi the natural dependence I C 9 on ml". Show that
+
Show that thi.; detinition mudilies Equation 11.53 fi)r dilute solutions ro k
For an aqurclus solution at 2 5 C , where p is the density of the sulvenl ( ~ rg.mL i I).
11-47. Show that
fur all aquevps culution at 25-C, whcrc I _ i s the i o n ~ cstrength expressed in terrris of ~nolality.T&c e, tu he 78.54 and ~ h derl\ity c at water trl he 0.99707 g ~ I I . - ' .
11-39. I!\+? rllc data in 'Table 1 1 -4 lo calculntc 111y, for NaCl(aq) at 25 C by plotting ($ I ) j l l l 1 ; ? agnrmt rlrl." and determine the arcn under the curve hy nurrjerical irltegration (M;ithChap~crA). C n t n p m your values or It) y, with those you obtained in kobIem 11-37 whcrc you cnlcul:~tedIn y+ iron\ a curve-tit expression ol 4 as a p l y n o ~ n i n in l m':'.
11-48. Use the Uebye-Huckel thcory tu calculnl~the value uf In y, for a 0.010-nlolar NnCI(:lq) solution at 115 C. The experirnc~ltalvalue uf y* i i ~0.902. Take F, = 78.54 for H?Oll) at 25.0'C'.
4 72
(haptcr 1 I !Snlurinns 1 1 : 5ul1d-1lqtrirl Solutions
11-49. Derive the general cquatiun
(Hin!: Sce the derivat~onill Problein 1 1-35.} Use this result to sitow that
Show that Equation 3 can be wnttcn
d!
ah
1r $ - ( r ) ] = ~'[rt,b,!rjl
Now shu-, that the only sulution for tlr,(r)lhat is Gnitc fol- largc ~alue!,of I.
for the Dehyc-Hiickel thcury.
11-50. In Ltie Dcbgr. lliickel Iheurq.. the ivnc are rl~ndclcdas pr>in~lrlnc, and the sulvcn~is lut)dt.lecl n ct,nrinuous i~it'dium( n o ctruclurt) with H relarive perrnlrtivity F , . Consider '111 iun or typc r (I = u cution or at) anion) ait~ratcdat the origin 01- n spheric:~lconrdinatc cyclclli. Thc ~ I C ~ C I ~01 C Itli'r Z ion at lllr orlgili &ill attracl ion> rlr opposite charge and rcpcl ) the ~iuntlrerof ions (IT rqpt j ( j = a caric~nor an Ions u i ttjc same charge. 1x1 ~ y , ( rhe dniun) aituatcd at a dislance r troln Ihe cenlrill ion of tppt. i (a cntlorj or a n ,~nirm)We can irsr a Bultztnann hctrir to say that
whcrc N,jV is the hulk n u l n k r tlenxity o f ( ious and n:,,(r) is the interaclinn cnergy uf an i i'rrn with a j ion. This interaction energy will be electroslatic in 01-igin, so lei I I : , ( r ) = 4, t//, tr). wt~ercy, is the charge or1 the ion or type j and $, ( r ) is the t.lectrnstntic porcnl~aldue to the central ion of type I . A f~u~idilmmphyhics that rslaleh a cpllcrically ay~nmeri-ir, zlcctrostatic pnrari~iolt;',( r ) to a \phencally symmelnc charge densiry p, (r) is Poisson's equa~irln
I d
-
-
r'dr(
.d@ f--
-
d r )
Fx
P, 0.)
i~llerca, is the relalive pernjittivity ol Itle crllvcllt. In oui- c;r\e, p , ( r ) is ths c h a ~ g edenslty 31-c~und the central inn. First, show that
IS
where A i s a conshIlt. Use the fact that ~fthe cnllcentratiol~i \ \try small, then $,( I . ) is just Coulolnh'h law and so A = q,/.lne,,c,and
the Equation 6 ic n central resiilt of the nehyc-tfuchcl theclry. 'The iictor uC t'. Y'niod~ti~lt.h resulting Cuultlntbic putcntial, so Equation 6 is called a rr-rt.<,rrrd'L'c~uiorr~bic pnb~./l~i~?i
11-51. Use Lkjual~ons2 and h of the prekit)uc pri)hlzr>~ 10 shuw thal [he llel chiup< 111 3 y h r ic.11 ~ shell uf radius r surround~nga central ion tlt type i is
11 -52. Use the recult of the previous prublern to bhow th;it khc inwt prr>bahlc valr~cnf I - is 1:'~.
11-53. Show that
ahere c is the rnolarity of an aqueout \elution rrf a 1 -1 clcctrcdytc at 25-C. Takc e, = 78.54 for H,0(1) at 115-C. wheru C 1s the bulk nu~ribcrdun.;i~yof s p x i c s j (C', = G'1. Ll~lenrizerlie rxporientinl Icrm url;usc the ct~nililitmol~clcct~oneutral~t!: to shdw t l ~ a ~
Nrlw +uhctitutop , ( r J into Poicst>tt's equation
to
gut
for a O 50-rnr;nr aqueous sulution ufa 1-1 elcct~,olpteat 25'C. Take E , at ZSC.
= 78
54 fur H.O(l)
11 -55. How d w s the t l ~ i c k n e sof~ the ionic atrr~osphrrc.curriparc for a 1-1 clcclrulytc ,mri 2-2 electrulytu?
&i
C h ~ p l 1~1r / Solulinnr ll, Solid I iquitl Solutinn?
474
11-56. I n this problenl. we will calculale the tolal electrostalic energy of an clectroIyte solution in the Dehye-Hiickel theory. Use the equations in Problem 11-50 tu show that the number of ionh OK t y j ~111 ns"p\erical ~llcllof radii r and r dr about a central ion of type i is
Use the rcsult you go1 lor A" in the prevui~usproblem to show that
+
U Ethc ~ formula
The tolal Coulonlhic irlleraction k t w c c n thc ccntral ion of typc i rind thc ions of type j in Ihe sphci-ical {hell i u ,V, [ r ) u ,(*)37rr'rlr/ , I'wherc rt, (r) = q,q,/4rts,srr. To dctcrlninc tllc clectrostntic inter,jctibn energy u i all the ions in t i e ccolution wilh the renlral ion (01typc i ) . L',",sum ,$I '? ( r ) u'I ( r ) / lJ over all types r)f itmv in a sphericaI shell and then integrate o w r :ill \pliericul shells 10 get IJfe thc clectruneu~ralitycondition v + q ,
Cse electroneutrality to show thut
+ v-q-
= 0 tcl rewrite In yr as
11-59. Dcrive Equation 1 1.56 from Equalion 1 I .49. 11-60. Show thnt Equdion 1 1.GO reduces ro F q a t i o n 11.49 for small concentralions
NOR. using Equatiun h of Problenl 11-51),
Now argue that thc total clectrostntic energy is
\Vhy i v thcrc
.I
E:lr.tu!r o l 112
ill
lhiu eqllalion'? Wouldn't yrnt be overct>ur~tingthe eriergy
o!herwise7 11-57. Wr derived an expression Tor U" irl rht. prcvir~l~c prohlerl~.Use the Gihbs-Helmhull/. cquntiori fur 11 (Prr)hlerri 8-23) to show that
11-61. In this prohlem, we will inveqtigatc the temperature dependence (11. activities. Starling with thc cquation p , = p; RT In a , , show that
+
z,
1 f -62. I Icnlg's law h:~ysIII:II the prcwul-col-ng;~\in erluilihriutll with ;I 1lonclrctioly1cxolutirln of the ga\ i r ~n liquid is proportic~nalto the mulalily of lllc gas in the solutior~fur suAicienlly dilute solulions. What form du you think Henry'u law tiikes on for a gas such as HCl(g) dissolved in watcr'! Use the fr)llowing data fur HCl(g) at 25' C to tehl your prediction.
P,,, j 10-I bar 0.147
0.238 11-58. If we assume that the electrostatic interactions are the sole cause of the nonidealiw of an electrolyte solutiol~.thcn wc can say that
z,
is its panla1 molar where H1 is the rnular enthalpy of the pure solvent (at one bar) and enthalpy in the solution. The difference bctwwn H: and is slnall for dilutc solutions, so n , 1s fairly independen1 of tclnperature.
0.443 O.bh3 0.85 1 1.08 1.62 1.93 2.08
C1icipf~r12
478
I r h f r n ~ c ~Equilibrium ll
( \',)rles from 0 to some niaxrrnuln value dictated by thc stoichiome~ryof the reaction. l'cor cxamplc, I I 11, and n,,,, in Equalions 12.1 are equal 10 v, nloler and v, moles, respecti! ely, then 5 mill \ary frotn O to one mole. Differentiation of Equations 12.1
We shall denote the right side of Equatiot~12.5 by A,G. scr that
gi\,es
Thc ncgative signs indicate that the reactants are disappearing and thc positive signs intllvatr thi~tt t ~ eproducts are being forlnctl as the reaction progrcsqcs from reactants to products. Now let's corjsider a system cotitaining reactants and products atcowtant T and 1'. l'he Gibbs cncrxy lor this multicumponent system is a function uf T, P, n,<,n,, rr,, and rr,. whicfl wc can express tnathcmatically as G = G ( T , P,n,, 11,- n,, n,). The
total c1c1-ivativeof G is givcn by
The quantity AT(; is defined a:, rht. ctlange i n Gihha energy when the extent of re;~criot~ ctiangcs by rlne ~nnlc.The units of AT[; are then J,mrll-'. The quantity A , G has meaning o n l y if thc halanced chemical equatiou i s 5pwified. If wc assume that ;ill the partial pressures arc l o w enough that we can consider each spccies lo behave ideally, then wc can use Equation '3.33 [w, ( T . P ) = fi';'(T)+ R l ' I n ( P , / P . ) ] forthe IL,(T,P ) , in which caseEquatiun 12.6 becomes
or
where
where the suhscl-ipt n, in (he first two ptirtial derivatives stand< iOr n,, n , , n,, and n,. L'sing Equat~onsH.3 1 fur (JG/aT),,n,artd (aG/a P),.,n d G bworries
.
and
Thc quantity ArG' ( T ) is the c l i a n g in standard Ciibbs energy for the reaction hetwceri ~~tltnixed reactants it1 their clandard states a1 temperature T ntd a pressure of one bar to form untnixed products in their standard states at the sarne temperature T
nit11 sirnilar exprcssic~nsTor it,, p,, and 11,. For a reaction that takcs place at constant 7 and P , d G bccumcs
d G = )j5,dg; = p,dn,
+ / ~ ~ , d+n pydfly , t- p,dn,
I
Suhstitute Equalions 12.2 into Equalinn 12.3 to obtain
(conslant T and P) (12.3)
and il pressure OF one bar. Recauce the standard pressure P ' in Equatiun 12.9 is taken to be one bar. the Po's are usually not displayed. It must be rernernhered, however. that all the pressures are referred to one bar. and thal Q consequently 1s unitless. When the reactiun system is in equil~brium,the Gibbs energy must be a minimum with r e s p ~ to t any displacement of the reaction from its equil~briurnposition, and su Equatlon 12.5 becomes
( )
= arc;= 0 T.F
Setting A r G f 0 iri Equation 12.7 gives
(equilibrium)
480 where
and where the subscript cq emphasizes that the pressures in Equations 12.1 1 and 12.12
12-2. An Equilibrium Constant Is a Function of Temperature Only Equation 12.11 says that regardless ofthc initial pressures of the reack~ntsand produch. at equilibrium the ratio of their partial prcssureh raised to lheir respective stoichiornetric coeflicicnts will hc a fixed value at s given tcmperalurc. Considcr the reaction described hy
arz the pressures at rytrilibriurn. The quantily K , ( T ) is called the rqtiilibrilim cr>n.stant of thc reaction. Although we have used an eq subscript for enlph;~sis,this notaticm is not normally used and K,.(1') is writteri without the subscript. Fq:~ilihriu~r~-constant exptess1olls imply that the pressures are their equilihrii~mvalues, The value uf K, canriot hc etalunted unless thc balanced chemical rcaction to which it rcfers and the standard sta1t.c of each of the reatpints and products m given.
I
E X A M P L E 12-1 U'rire out tllc cy~~ilibriurn-conhtanr expre,r,io~~ ii)r thc reaction that is ~eprescr~red by the cquatior~
The equilibrium-cunstant expression lor this reaction is
Suppose that initially we have one mole of PClj(g) and no Pl'l,(g) or Cl,(g). U'liut~ the reactinn occurs tu an extent 6, there will be ( I - t ) moles of PC],($), ( moles ol PCl,(g), and 4 rnolcs of Cl,(g) In the rcaction mlxture and ths tot&[nulnber i l l moles be the ehlcnr of reaction at equilibrium, thcn the partial will be ( 1 + 6). II' wc let teq pressures of each species will be
SO l U T I C j N . According to Equation 12.12.
where P i s the tolal pressure. The equilibrium-constan1 expression is
r~llcreall tl~cpressures itre rcikrred to the sta~lclal-tlprcahore of one bar Notc thal if we had writtcn the cquarion for Ihe reanion as
then we would lloxe obtained
which is the square root r ~ f s u rprevious expi-ehuon. 'riruh. we see that the form of K,(7') and its suhsequcnl nu~ncr~cal value t i e p n d upon huw we write the chcmica!
equiition that dehcribes \he reaction.
It might appetar from this result rhat K , ( T ) dcpends upon the lotal pressure, bur this is not so. As Equation 12.11 shrlws, K , 1 T ) is a function of only the tenlperitture, and so is a crlnstant value at a fixed tetnperature. 'fhercfore, if P chntiges, then must ct~angcso that A',,(T) in Equation 12.15 retrlains constant. Figure 12.1 shows c,,, plolted against P at 200'C, where K , = 5.4. Nutc that terl d e c ~ r n ~ z1~11ifonrily s wit11 increasing P , indicating that the equilibriuil~ix ~ l ~ i f t elro111 d the product side to the reactant side of Equation 12.13 or that lcss PCI, is dissociated. 'I'his effect oT PI-csaurt: o n the posiliun of equilibriuni is an exarnple r)f Lc, Cl;ritt'lir,r:r prit/r.rl~l~. which you Icarned in general chctnistry. I,c Chiielier's principle car1 he stated as lollow,.;: I f a cherniui~lreaction at cquilibtium is subjected to a change i r ~ct~nditiorlslhut displaces it from equ~lihriunl,thdn the reaction adjusls tcrward a new equilibrium stale. Thc reaclion proceed5 in the direction that - at least partially --offset5 the change in conditions. Thus, an increase in pressure shifts the equilibrium in Equation 12.1 3 5ucI1 t h a ~thc tc~taltlun~herof moles dccreases.
1 2 - 1 . Chemical Equilihr~rlmke~uttsWhrn tlie Clitltx Ener~yIr ,I hl~nimunl
{ varies from 0 tn snnle rnaxlniilrtl value dictatcd hy the xtoichiometq of the reactron For cxaniple, i f n,, and rz,, it) Equations 12.1 are equal to v, trloles and u, inoles, rzspeut~vely.[hen { will vary froin 0 to one mole. Differentiation OF Equalions 12.1
gives
-=I , (in, = - v,dc
J n , = u,dc d r 1 , = v,dt
rcxrrnnl+
pruductr
I
The ncpativc signs indicate that the rcactants are disappearing and the positive signs intl~catrt h a ~Ihe products are bcing f0rmed as [he reaction progl-csscs from reactmts to prr~ducts.
We shall denote the right side of Equation 12.5 by ArG. so that
The quantity A,{; is defined as thc change in Gibbs cncrgy when the crtent ol'reac tion changes by one ~ n n k The . units of ArG are thrn J moi ' The quaritlty A,G has al is specified. meaning only ~fthe blilanced c h e ~ n ~ cequaliorl It w e assume that all he partla1 pressure5 are low enough thar we can c n n ~ ~ d e r each s ~ c i e to s behave ideally. then we can ure Equation 9.33 [/.L;(T, P ) = EL;(T) RT In(P, /P')] for the /A, (T, P),it1 which case Equation 1 2.6 hecomes
+
Now let's conqider a system cont~iiningreacunts and products at constant T and P. 'l'he Gibbs encl.gy fol- lhis mul~icornponentsystem is a functicln of T , P , n,, n,, n,, and n , , which we can express rna~hematicallyas G = G ( T , P , n,, n,, r a y , n,). The rotul dcrivatite ol' r; is given by
+(
d T.P,nltl.
) a%
dn,
where
I . I , E> * I
where the subscript n , in the f rst two partial derivatives stands for n,, n,, n,, and n,. Using Equation< 8.3 1 lor ( a U / i l T),,ni and (aG/a P),,n,. dG becorries
and
with siniilar expressions for pI,, 1 ~ and ~ p.,. For a reaction that takes place at constant 7' and P, dG beco~r~cs
The quantity b r G ' ( T ) is the change in s~andardGthhs energy for tlrc rcactlon between unmixed reactants In thclr ctandard states at temperature I' a~id a pressure of one bar to form unmixed products in their standard sbalrts at the same temperature T and a pressure of one bar. Recause the standard pressure P" in Equation 12.9 is taken to bc one bar, the P"s are usually not displayed. It must he remenhered, however. that all the pressures art termed to onc bar, and h a t Q conrcquentl) is unitless When the reactiun system is in equilibrium, the Gibbs energy must be a minimum with reqpect to any displacemet~tof the reaction from its equilibrium position, and
d C = )l~,d,l;
= pAdn,, t.p,u'rt, $ l ~ , d n ,
+ p,dn,
(constant T and P) (12.3)
J
Substitute Equaticlns 1 2.2 into Equation 12.3 to obtain
dG = - ~ s ~ . i ~ , ,-d < =
(l)YpU
+
+v7~~zdt
+ %pZ- L ' J L ~- v , l ~ , ) d [
(constant 7' and P)
(g)T,p
= ArG = O
(12.4)
Setting ArC; = 'O in Equation 12.7 gives
(equilibrium)
480 where
12-2. An Equilibrium Constant Is a Function of Temperature Only Lquation 12.1 I says that ~gardlcssill"the initial pressuresof the rcactarits and products, at equilibriurn rhe ratioof their pilr~ialprehsureh taiscd to their respctite .ctoichionlttric ctlcfficients will be a lixed valae at a giver1 temperature. Consider the rcaction devcrihcd by
and where the suhscript eq cmphasizes Ihat the pressures in Equatiuns 12.1 I and 12.12 art: the pressures at eqirilibrium. The quantity K , ( T )is called the equrlibriilm constarlt of rhc reactinn. Although wc h;~veused an cq suhscript fur cmphasia, this notation is i1o1 nornially used and K p ( T ) is wriltctl withc~utthe subscr~pt.Eqi~ilihriurri-constant e~pl.essionsirnply thal thc pressures are their equilibriurn values. The value of K r canuol hc evaluated l~nlessthe balanced cherliical reaction to which it rcfers and the ~tilt~ditl-d states of each of the reactanis and products art. given.
I
E X A M P L E 12-1
Wrirr nu1 thr cquil~hriurn-cunstaiilexpressitln fur the rzi~ctiorithat ic rep~esc~lted by the cquatiu~~
The equilibrium-constant expression for this reaction is
Suppose that initially we have onc mole of PCl,{g) and no PCl,(g) ur Cl,(g) Wlwrt the rmction occu~sto an extent 6. there will be (1 - 5 ) ~nolecut PCI,(g), 5 ~nolesot PC1 ,(g),and 4 nloles of Cl,(g) in the rcsctjon m~xturcand the total number of mole\ w1I1 hc (1 t ) If we 1e1 he the extent of react~onat equilihriu~n.then the partial pl-essures of each specie\ a.111 be
+
c?!,
where P is the totat pressure. The equilibrium-constant cxpressicm is
wbcrt: all thc pressurc~are rrfcrrerl tu thc ctandard pressure uT one bar Kutc tt~alif we hail writterl the equation for the reactior~as
then we wuuld have obtained
which is the squm root pT uur prcvious ekl)re\siori. Thus, we see thal [he form of R,(1') anti its subsequerlt numerical valuc depend upon how we w r i ~ ethe chemical equation thdt dcccrihes the reaction.
It triighl appear from this resull that K , ( T ) depcnds Llpon thc total pressure, t)i11 this is r l o t so. As Equation 12.11 shuws, K , ( T ) i s a func~ionof only the temperalure, and sn is a conslant value at a fixed telnperdtui-e. Therehrc, if P ch:ingcs. then <*,,I must change sn that K , (1') in Equntinn 12.15 remains constant. Figurc I?. I chows plotted against P a1 2(XkC, where K, = 5.4. Note that ttYddccleases imifonnly will1 increasing P, iridicating that the equi1ibriul-n is shifted [row 1l1cproduct qidc to Ihe rcactant side of Eqi~atir~n 12.13 or that less PCI, ia dissociated. This e r e c t of pressure or1 the pc~sitiunof equilibrium i s an example of L t I'Mrclier'.\ prinriple, which you leal-ncd in genel-a1 chetnia~ry.Le Chiklier's principle car1 be stated as li~llows:If a chenlicat reaction at equilibrium is subjected to a changc in conditions that displncus it Trt)mcquilibrii~rn. hen thereaction adjusts toward anew cquilibrium state. 'l'he reactioti prtxeeds in the direction that - at least partially -. offsets the changc i11 conditions. Thus, an increase in preshure stlifts Ihc cquilibrium ill Equarion 12.13 such that the ti~talnumber of moles decreases.
the equilibrium constant in terms of densitieq or concentrations by using the ideal-gas relation P = c RT where c is the cr>ncentrittir>n,n / V. Thus, wc can rewrite K,, as
0.0
L-
0
I
50
I
I
1U O
150
Just as we relate the pressures in the expression Cor K, to come standard prewurc P - . we must relate the concentrations in Equation 12.16 tu some standard concentration c', often taken tu be I rnu1.L-I. If wc multiply and divide each concentratiotl in Equation I 2 16 by r - , we can write 2 11 0
PI bar F I G L I R E 12.1 A plot nf the fraction of
whcrc
PCl!(g) thal ia dissociated a1 eqt~ilibriurn.c*,,, agtlinsl Iota1 pre55urr P ior lhc rcactiun gir~enby Equal~ori12.13 ut 2IK)'C.
Both K, and K < in Equalion 12.17 are un~tless,as is the I-actor ( r . ' R T j P ' ) ' y l ' r ' - \ ' 5 . Thc actual choices of P" and r.' determine the units of R to Llse in Equation 12.1 7 . If P" is taken to be one bar and u' lo be nne mol-L (as is oftten the case), then thc factor c'H?-jP' = HT/L-bar.nlo1 and R must he expressed as O.O#B 14.5 I. .bav rnnl-'. Equation 12.17 provides a rcla~ionbetween K , and K r for ideal gases. lust as we d(1n.t displ:iy the P"s in Equation 12.9 because must n f ~ e nI" = one bar, we dr~n't display the P's and r. s in Equation 12.18 because njosl often c" = one mol-1.-'. You IHUSI always be aware. however, of which refcrenve states are being used in K , and Kc when conuerting the numerical value of one tu thc othcr.
E X A M P L E 12-2 (hnsider t he assuciatinn uT potassiunl atoms in the vapor phase lo lhrm dirners
Srippvse wc start with 2 moles ur K(g) and n o ditncrs. Derive nn cxpressirlr~for K , ( T ) in tcrms uf tr4,thc exten1 c ~ ~,cactic>rj t at cquilibrium, and the prexcurc P .
S O L U T 119 N : At cquillbri~tm,lherc will he 2(I uT K , ( g ) . The lottal nr1rnht.r of ~ncrleswill be (2 spccicc will be
'
'
cq) mule%of K(g) and tq moles
cq).
The piutial prcssure u l each E X A M P L E 12-3 The valuc of K , , ( T ) (based upor] a sundarrl state uf one har) for the reaction descrihd by
NIll(g)
+ !Ii2(g) * i ~ ~ ( g )
'
1.36 x 10 at 298.15 K. Deter~rw~e thc corrcspond~rrgkdluc of K<( T ) ( h a d upon a stLinrlardstate of one n1o1.L I). IF
SO I.C77 1 0 N : I n this case, v, = 1, c, = 312, nr~dI., = I/2, so Equation 12.17 pivev
and
If 1' decreases, then t q ( 2 - Ctq1/4(1- <+J2 must decrease, which occurs by Cq decrearing I t P increases, then 6 (2- cet1)/4r 1 -- fcq)2 must increase. which rrcurs by "P truincreasing [ I I - cq)%cotn~ng smaller].
l'hc convcrsiun factor at 298.15 K i e c"RT P
--
I
We subscripted the equilibrium corlstant defined by Equation 12.12 with a P to c ~ n p h i ~ sth:it i ~ e it i s e~presscdin terms uf equilibrium pressures. We can also express
and su
(lmol.1.-~)(0083145L har mol '.K ])(298.15K) I bar = 24.79
-
K,,= K , / 2 4 . 7 9 = 5.49 x lo-'
I
I L 7. St.~rirf.~rrl C;~htr$Frat.rp~r\of Fmrrn;it~uoCan Be Uscd tu Calculate Equ~l~br~urn (:unstdnl!
T A B L E 12.1
12-3. Standard Gibbs Energies of Formation Can Be Used to Calculate Equilibrium Constants
S~andardntnlar Ciihh!, merges OC furnmarion. , l , G substnnccs at 298.15 K attd orie bar.
N(111cz that cot~lhinir~g Equatiorls 12.8 and 12.11 gives a relation between p I ( T ) , the standard chemical pc~ten~ials of the reactants and products. and the equilibrium constant. K,. In particular, K, is related to thedifference betwccn thc standard chemical potentials of thc products and reactants. Because a chemical potential is an energy (it i s lhe molar Gibbs energy of a pure substance), its value must be referrcd to solnc (arbitrary) rcro o T energy. A convenient choice c~fa zero of energy is based on the pnjczdurt. that we used to set up a table of standard molar enthalpies of formation (Table 5.2) in Section 5-1 1. Recall that we dehned the standard tnolnr enthalpy of tormiltion ol~asuhstanceas the energy as heat inwlved when one mole of the substance ia fvrmcd directly frr~rni ~ constituent s elemcnts in thcir most stable l'lmn at one bar ;und lie ttmpzrLLlurz01 ir~terest.For cxalnplc. the v ~ ~ l uute A[ H fi)r
is -285.8 kJ when 311 the species iire at 2W.15 K and one bar, and so we write A, HL[H,O(I)]= -285.8 k ~ . m o l - 'a1 298.15 K S y convention, wc a l w have that A, H '[H,(g)] = A,Hn[O,(g)] = 0 I'or H,(g) and O,(g) at 298.15 K and u n e bar. We also sel up a tdble of practical absolute cntrop~e'.;of si~hstances(Table 7.2) in Scction 79, at10 so because
Substance aciccty lcllc ammonia henrrne
hroniirie butane
c.ubon(diamond) carbon(graphite) carbon diuxrde uasl.run Inonux tdc
tlharju elllano1 tlhcnc glucuse
hydrogen bmm~de hjdn)gen chloride hpdrogcn Ruoriilc
hydrogel1 iodide we
hydropen peroxide
car] also set up a table of values c~fA,[;". Then for a reaction such as
iodine I I I C ~ ~ I ~ I ~
nlrrhanoI
we have A,G' = v,A,G
lYl+v,A,G
[Z]- v,,AIG'[A]-v,A,C;
IBI
(12.19)
'hblz 12.1 lists values r ~ fA,(;' a1 298.15 K and one bar for a variety 01- suhsvances. ; ~ n dmuch more extensive lables are available (see Section 12.9).
I
E X A M P L E 12-4 I;sing the riala in Table I 2 I . cnlctllatr A,C; (I')dnd
I
nklrugen oxide
nltmgcn dioxide dl ni lrtlgen Irlraoxide
prtq)anc SUCIIIhE
k',
a1 2'18
IS K lor
h ~ dioxide r
$111
sullclr trioxide
tctr~achluro~~~ctl~anc
Formula
.
[L~I
various
A , (; /L.I n101-
Chapter 12 I Chemical Equilibrium
Equation 12.21 gives the Gibbs energy of the reaction mixture, G , as a function of the exterlt of the reaction, <. Using d ~ evalues of A,G",2,, and d,Gc,,,; givcn in Table 12.1. Equation 12.21 becomes
and frotn Ejiluntiwn 12. I I
or
K, = 1.36 x 10
1 2 1 . A l'lot ui thc L i b b s EncrEy
'at 29R.15 K.
I
c.
'.
1 2 4 . A Plot of the Cibbs Energy of a Reaction Mixture Against the Extent of Reaction Is a Minimum at Equilibrium
where HZ-= 2.4790 kJ,mol Figure 12.2 shows G ( E ) plolted ag:linst The tninimum in the plut, ur the equilibrium state. occurs at tCq= 0.1892 ~nol.Thus, the rcaction will proceed from ( = 0 to 6 = 5C1 = 0.1892 rnol, where equilihrium is tslilblistled. The equilihriuln constant is given by
this scciion w r shall trcal a corjcrele exarnplc nf the Gibbs encrgy of a reaclioll mixture as a function of the extent of reaction. Consider the thermal decomposition of N,O, (g) to NO,(g) at 298.15 K, which we represcnl by the equatio~l 111
We can compare this resul~to the one that we obtain from A r c ' = - R T In
Strpposc we start n ~ t hone tnnle of N,O,(g) and no NO,(&. The11 as the reactlon p(xceds, 1 1 , . the numbcr of lnoles of NzO,(g). will be givcn by I - 6 and n,,: ail1 1 ' = I mol and n,,? =-0 when = 0 and that n, I 4 = 0 bc given by 26. Note that 11 nlld n,> = 2 11101 whcn ( = I mol. l b c Gibhs energy of thereastion mixture i s givcnby
,
-
(2?(AfG' l N 0 2 ( g ) l )-
K, . nr
l)(AfGr[N204(g)l) -
(8.3135 J - K I ,mol 1)(298.15 K)
If (he reaction i s uarricd out at a constant Iota1 pressure of clnc bar, then -
and
The total nunikr of tnoles in the reaction mixture is (1 have
,
P , 2 = X ~ ~ , . R ~
- 5 ) + 2 = I + 6 , and so wc
<
Thus, Bql~ation12.20 bectunes
I
According to Sectioli 12-3, we can choose our standard states such that GiIOq= h , G & and C; - = A,G&,:.
F I G U R E 12.2 A plot of the Gibbs energy of the reaction mixlure versus the extent of N,O,rg) $ 2NC), (g) at 298. I5 K and one bar.
reaction for
12-5. The Hatlo of the Rcactiori ()untient
We can also difl't.~r~tintc Equatio~,12.22 with respec1 lo $. cxplivi~i)to ubrain
+
+. t ) i r ~the firs1 logarithm lerm by P, 1 4 and 2 t / ( 1 t ) In the second logarithn~term by P,,,,. Furlhermorr. a little algebra shows that the last t ~ t term\ o atlcl up to zcru. and so Equation 12.23 becomes We can replace ( 1 - {)/{I
to the E q ~ ~ ~ l ~ ('onrlant l~r~r~rn
Equation 12.7 for this reaction schclne 1s
Realize that the pressures in this equation are not nccccsal~lyequilibrium prt.ssun.5, hut are arbitrary. Equalion 12.24 gives the value of ArG when 1 2 , lnolcs of A(g I a t prcssure f', react with v, moles of B(g) at pressure P, to produce 17, lnulcs r~t'Y(g) nt prcssure Py and 11, moles of Zrg) nr prcssurt. Pz. If all rhc prtssunhs happcn t o he one bar, thcn the Irjgarithm term in Equation 12.23 will be ztn, a i d A,<; will be cq11i11 lo ArUL; in other words, the Gibhs energy chatige will he equal to t h e st;tnrlard Gi b h ~ cnergy change. If, on the other hand. (he pressures arc the equilibriurn piesbulrs, r l ~ e ~ r A r G will cqual zero and wc obtain Equation 12.1 1. w e can write Equation 12.24 in nlrjrt: concise tor111by it~rt'oducingd t~~!:llltityCIIICLI he reurriorr quotierri Q , (see Equation 12.9)
A1 equilibrium, i)G/at = O ilr~dwc get Equation 12.11.
Wre can also cvaluate ct7,explicitly by setting Eq~~ntion 12.23 equal t o Lero. Using tlic t u c ~rtint thc last ~ w uterms in hluation 12.23 add up lo Lero, we have
and using Equation 12.1 1 for ArG':
Realize that even though Q , has thefi/,~n o I - an equilibriuin cr)nstant, the precsures are arbitrary. At cqullihrium. ArG = 0 and Q , = K,..1C 0 , c: K,,, theri Q p inus1 increase as the system proceeds toward equilibrium, which rneans that Ihe partiid prtssurzs of' (he products murl increase and those of thc reaclants tnust dccrtta%e. 111othcr wr>rdb. the reic~ionp r u ~ r c d sfrotrl 1ef1 to right as written. In ternlc ot' S , l i , it' L),, -= K,.. (her) A,G < 0, inclicati~igthat the reaction I Ssprjntaneou~frum 1t.R to riglil :I\ u r i ~ t c ~ ! . Conversely, if Q , , > K,, then Q must rlecreaxe as the reactirm proccetlh lo ey~~ilib~-i~irr~ and su the przshures of the products 111ust decreast. a r ~ t fth(7s.e of the rcirutanta rtiu;;~ increase. In tel-ms of A;G, il-Q , > K , , thcn ArC; > 0, indica~ingthat thc ~ractior)i s spontaneous from r i g h ~to left as wrilleri.
,
or tCq = 0.1892: in agrecrnenl W I I ~Figure 12.2. Problems 12- 18 through 12-2 1 ask you to carry out a similar analysis fur two other gas-phase reactions.
1 12-5. The Ratio ot' the Reaction Quotient to the Equilibrium Constant Determines the Direction in Which a ~eactionWill Proceed
I
EXAMPLE 12-5
'lhr cql~ilihriulnconsl;inr lur ttic runttion dchcrilhcd hy
15
K,.
= IO
at 960 K. Cnlculat~.I!.,(;
procecd spnraneously
and
lildiv:~tcin n l i ~ c clirtc~iam l~ lht. I-citctiot~ will
Cor
2SO,(1.0 x 10-'bar)+O2(020lharl ;= 2 SO;(] 0 x 1 0 'bur)
Chapter 12 / Chcri~icalFquilibrium
S V 1. I 1 T I O h ' : U!t. l j ~ v lc;ilculnte Ihn rrfiction qunticnt under co~ding111 Equahon 12.15.
the disswiation of N,O,(g) takcs place spontancously. The partial pressure of NIC), (g) NU,(g) incrca~csuntil equilibriun~is reachecl. 'fhc cquilibri(tr~t state is dctcrmined by the condition A f G = 0, at which point Q , = K,. Thus. initially A,G has a large negative valtte and increases lo zero a< the reautiuti goes to
these condition'. Ac-
decreases mid that of
equilibrium. We should point uut here that even h u g h A,G 0, the reaction may not uccur at a detectahlc rate. For examplc, consider the reaction given by
Note that l h e ~ equantit~csare u~litlesshccausc the pressures arc taken relative to one bar Using Equation 12.26, we have
= ( ~ . 3 1 4 ~ -mol ~ - ')(4130 ' K)In
5.11 x
)I)-'
Yhe lhct that A,(; -:O implies that the rractiun will prvcccd from left to right as aritten. This nray also hc seen from the fnct that Q , < K,.
1
12-7. Thc L'.~ri.zlicnnt an Equ~l~hrium constant with Tt.nl[>~ratbreI 5 tiivcn by llw Van't Hoff Eql~ativrr
I
12-6. The Sign nf A r c And Not That of A,.Go Determines the Direction of Rcactjon Spontaneity It i c ilnpndant to ilppreciate the diffkrence bctween A,G and ArGn.'l'hc superscript ' on AtG ett~phasizcsthat this is the valur 01- AlG whcn all the reactants aud products are unnlixed at partial pressures equal to OIIC bar: A,G" is the sranrlclrd Gibbs energ; changr.If A:G < 0, thcn K, > 1,meaningthilt thereaction wi1lproceedfrt)mreactants t o products il. all the species are mixed at one bar pdrtial pressures. IC A r c " > 0. then K , < 1, n~eallingthat 11ic rcaction will proceed frr~mproducts to reactiints if all the species arc rnixcd at nnc bar p a r ~ a lprcssurcs. The fact that A,G" > 0 does no! mean that ihe rcaction will not proceed from reactants to products if the species are niixcd under all condilions. For exarnplc, consider the reaction described by
for which Arc' = 4.729 kJ.mol-' at 298.15 K. The corresponding valuc of K , ( T ) i s 0.148. The fact that A, G" = -t4.72!, kJ.mnl-' docs no[ mean that no N,O,(g) will dissociate when we pl;tce some ot-it in a rcaclion vessel at 298.15 K. The value of A,G l o r rhe dissociation 01 N,O,(g) i s given by
Let's say lhat we fill a container with N,O,(g) and nu NO,(g). Initidly then, the logarithtn term arid A, U in Equation 12 27 will be essentially negative infiriity, Therefore,
The valur uf A,(;' at 25'C f-or this reaction i s -237 kJ per mole of H,O(I) for~ncd. Consequently, HzO{I) at onc bar and 25°C is much more stablc than a rnixturc uk H,(g) and(.ll(g) under those conditio~ls.k t . a mixture rjf Fl,(g) and 0,(g) can be kept indefinitclq. I1 a spark or a catalysl is introduced into this mixture. however, then the reaclion occurs explosively. Thiq uhservation scrves t o illustratc an impurlant point: Thr "no" o~therniorl~narr~ics is emphatic. ii'thermo+namics says that a certain process will not occur spontancously, then it wili not occur. The "yes" of thermodynamics, on the other hand, is actually a "maybe". The fnct that a process will occur spontaneously does not i l ~ ~ pthat l y it will nrcesfarily occur at a delectahIe rate.
12-7. The Variation of an Equilibrium Constant with Temperature Is Gi\:cn Ily the Van't Hoff Equation We can uae the Gibbs-Hclmoltz equation (Fq~tatior~ 8.61) AH'
to derive an equation for the tenlpernhtre dependence of K, (T). Substitute AG' ( T ) = - RT In K,.r7',l into Equatirjn 12.28 to ohtain
Note that ~f A, H " > 0 (end(~ttlermtcrcaction), the11 K , ( T ) illcreases with temperature, and if A r t { i0 (exother~ntcreaciion), the11 K , ( T ) decreases with Incrraslng tempraturc T h s is another example of Le ChPtclier'.; principle. Equation 12.29 can be intcgraied to give
12-7. The Variation uf 'In Equilibrium Constant n ~ t h Tcrnp'rdlurr
If lhc tenlperiiturr: range is s~nallenough that we can considcr A r H to he a corlstaut. then w e cat1 rvritc .
Equation 12.3 1 buggests that a plot uf It] K,.( T )versus 1/ T should he;] straight line with a slope of - ArH"/H over a sultiuier~tlysmall temperature range. Figure 12.3 sh(1u7s such I: plot for Lhc reaction H,(g) CO:(g) 7CO(g) tl,O(g) rwcr the temperature range hi10 C to 900'C.
+
+
Is Cdver by the Vdn't
t t r l l l Frludlloll
In Section 5-12 rve discl~ssedthe Lernperature variation of h r H ' .In particula~.u:c derived the equatiun
where AC; is the dlfferencc bctwcen the heal capacities uf thc products and react;]nts. Exper~rnentalheal capacity data over temperature ranges arc often presen~edas p l y nomials In the temperature. and if t h ~ sIS Lhe case. then A r l f ( T ) z,ln k expressed In the tomi (hee Example 5-1 3)
If this form for A, l l ' ( T ) is suhstitu~edinto Equation 12.29, and but11 sides itltcpl-ai~11t'd indefinitely, then wc find Lhat
The constants a lhrnugh 6 arc knuwn from Eq~~ation 12.33 and A i s an integration constant that can be evaluated from a knowledge uf K , ( T ) a1 some particular tenlperalure. We could also have intcgratzd Eqi~ation12.29 from some tznjperature TI at which Lhe vlilue of K,(?') is known ru an arbitrary temnperilturc T tu ohlain F I G L I K E 12.3
A plur of
It1
lempcrntllrt.
+
K , I T ) versus l j T furthc reaction H,(g) ( : O l ( g )+ CO(g) -tH,O(g) over the rarlge 601l'C: lo 9(H)"C.Thc circles rcpresenr cxperinlentul data
I
EXAMPLE 12-6 Givert Ihat A, H ' has arl axeriigc valur. ot - 69.8 kJ~rnul rrver the 500 K tu 7(W) K for the 1-eactiundescribeti hg
S (ILCJT 1 0N : We
IIW Equa~ir,~~ 12.3 1
ttlnperatllrc i-;lngc
with the a h w c xalucs
Equations 12.34 and 12.35 are gci~cralimtionsof Equation 11L.3I to the case whel,tl the tclnperature dependence of A,FIr ih not ignored. Equarion 12.34 shows that if In K,,( T ) is plotted against I / T, then the slope is not cunstanl, hul has n slight curvature. Figure 12.4 shows It1 K , ( T ) plotted versus I / T for thc ammonia sytltfiesis rciictir~n. Nutc thal In K,, ( I ' ) does not vary linearly with 1 / 1'. showing Illat A, H ' is tcmpcriittlrr: dclxndent.
I
E X A M P L E 12-7
Coll~ldcrthe rrilction dcccrihed by
The mular heal capncities of N,(g), I I:(:),
and NH,(g)
call bc cxpl-e\sed in Lhe 1urr11
Nutc that (he 1,cactionis exothcrr~licand so K, ( 7 ' =70n K) is le+stha11 K , ( T = i O U K).
r; [NH , ( g ) J / ~K -.I ,
lnol
' = 25.93 + 32 58 x
10-'?'
-
3.036
x 1 O-."T'
to 1500 K. Give11 that A, H' INH,(g)l = -4h.l l kJ m n l - ' at 3(W) K and that K , = 6.55 x 10-' at 725 K, derive a gener:il cxpressiun For ~ h vmiaticln c o f K , . ( T ) with ternpt.mturc in the form n i Equation 12.34. rwcr rhc temperatu~c range 3[H) K
with 1; = 300 K and A, H 'CT, -300 K) = -46. I I kJ.lnol
' and I
ono K 1 T
F I G U R E 12.4
A plot uf In K , f 7 ' ) versus 1/T for the ammonia sgllthcsi~reaction, NH,(g).
H Z ( g )+ N ~ (= ~ )
12-8. Wc Can Calculate fquilihrium Constants in Terms of I'artition Functions or
'
ArH (Tl/J,~nul =
38.10 x 10'
- 31.17T + 13.44 x
IO-~T'
Now wc uke Equatir)~~ 12.35 ~ i l TI h = 725 K and K,(T=725
-
1.965 x IO-'T'
K) = 6.55 x lo-'.
An Ilnporlant chemical application of stadstlcal thertnodynamics is the valculat~onot equilibrium constants In tarn.; u t ~nolccularparameters Consider the general homogeneous gax-phase c h e m i c ~ lreactiun
il,,A(g)
+ ~s,B(g)+-
11,
Y (g)
+ v,Z(g)
in a reaction vessel at fixed vulutne and Lemperaturt.. In this casc we have (cf.Equation
9.26)
+
d A = /lAnn,,
~
-t p > d~r l y
~
+ p,rln, (
~
(conslant T ~and V ) l
~
instead of Equation 1 2.3. Introducing the extent or reaction through Equations 12.2, however, leads to the same cortditiori for cheniical equilhriu~nas in Section 12-1,
-1his erluatiur~WRS used tu ger~eratcFigure 12.4. A1 M)O K, 111K, = -3 21. or 0.U40, in exct.llent agrcemenl with thc cxperi~nenlalvalue o f O.Wl.
K,
=
1: i s intcrzsling to cornpare the results of thia section 10 those of Scctic~n4 4 . where we dcrivctl (lie (:lausius-dapcyron eq~tation,Equation 9.13. Note that Fquntions 12.31 and 9.13 are rwentially the same because thc vaporization of n liquid can he represented hy thc "chernical equiition''
We now irltroduce statislical Ihermodynamic< thmugh the relation betweerl the che11lical potential and a pwirtitiot~funclion. I n a mixture of ideal gases, the species are independent, and so thc partiti011 function of the mixture is a pnlduct o f the partitirm funutioris of he individiral components. Thus
12-8
The vtiern~caipurer~tialof cach species is given by a11 rqi~atiorlsuch as (Prohlcrn 12-33)
where Stirling's approximiitiun lias been used for N,!. The Nl subscript un the panial derivalive indieales illat the numbers uf particles of the other specie5 arc held fixed. Equation 12.37 simply says that the clic~nicalpotential of one species of an ideal gas nli.rturt: is calculated as if the other species were not prcscnt. Thi.s, c ~ course, f is the vast for an ideal gas mixturc. 11' w e substitute Equation 12.37 inlo Equatiori 12.36, then we gct
Can Calc~ll,~teE q u ~ l ~ b r ~ l(I(mct,>nl< lrri ~n lrrnlr or P;lrt 1t.m runcr~r,n+
Using F,qu:ltiol~4.39 for the molecular parlit on funcrioi~sg i ~ c s
where we have replaced DF in Equatiun 4.39 tiy D, - t h11/2 (Figure 4.2). All the necessary parameters are given in Table 4.2. Table 12.2 gives the nurneric;il valiics of K , , ( T ) and Figure 12.5 shows In 'A plotted vercus I j T . Frotn ttie slope of the linc in Figure 12.5 we get A,H = - 12.9 kJ. mol-' colnpared lo iht: experi~tientnlvalue of ,- 13.4 k J . l ~ ~ u .The l discrepancy is due to the inadequacy of the rigid rotalor-har~t~o~liL. oscillator approximation at these 1ernperatLlreP.
For an ideal gas, tile molccula~-parlitiorl I'utictiuri is of the form , f ' ( T ) V (Section M ) so that q / V is a fiinction of temperutilre only. If wc dividc cach factc~ro n both sides of Equation 12.38 hy C"', i~nddenote thc n1111ibe1density N l / V hy p,, then w e havc
Note that K c is n function of temperature only. Reciill that K,(T) and K c( T Iare related hy (Equation 12.17)
F I G U R E 12.5
By means of Eqi~at~un 12 17 and Erlurttton 12.39, alung wrth the recults of Chapter 4. wc can calcula~eequilibr~umconstants in lerrns of njolecular paramctcn. Thia is best illustrated by means oT ex;~~nples.
The logarithm of the equilibrium constant verhuh I / 7 lor the w e c t i o ~H,(g) ~ + I:(g) + 2 1Il((g. The line is calculnted from Equatiun 12.41 and the circles are tlh: c.rpc;imental values.
T A B L E 12.2
A.
A Chemical React~onInvolving Diatomic Molecules
The valucs of K,,(?') for the reaclir~ndzrcribcd hy H2(g) calculated according to Equalion 12.41. TIK
firm 5Oi) K to I (XI0 K. The equilihr-i111rlconstan1 i b given by I
K,,(Tj
+ I?(g) e 2 21Il(g)
InK,(T)
U.
,4 Keaction Irlvolving Polyatomic Molecules
T A R L E 12.3
The Irjgarilhnl of the tquilibrium constant for the reactinn H:Ig) Ollg) + IlzO ($1
+i
As an example o f a reaction involvillg a polyutomic molccule, consider the reaction
TIK
111K,(calcj
In K,.(exp)
wllose equilibriutn constant is given by
It is allnost as corivenient to c;~lculateeach pLutiticlnfunclion separately as to substitute them into K c first. The necessary parameters are given in Tables 4.2 and 4.4. At 1500 K, thc three partition functions are (Equations 4.39 and 4.60)
Tahlc 12.3 compares the calculated vntucs or In K,,with exy~rjmenlaldata. Although thc agreement is fairly g o d . the agreement can be considerably i~npruvedby using 111oresophistjcated spectruscopic models. At high temperatures, the rotational energies of thc molecules are high enough to warrant centrifugal distortion effccts and other extennsof the simple ripid rr~rator-hartnnnicoscillalor approximation.
12-9. Molecular Partition Functions and Related Thermodynamic Data Are Extensively Tabulated
and
The factor 01.3 rxcLr1.s in yU2/ I/ hecausc the degeneracy of the ground electronic state ofO2 is 3. (Tablc 1.31. Nuticc that cach of the above q ( T . V ) / V has units of m-?. This ells us that the rztercncr state in thU (~~~ulccular) case is a concentrations of one nlolccule pcr cubic rneter, ut. that r,' = one molecule.~n-'. tlsing the values of U t ,from Table 4.2 and 4.4, the value of K, at I300 K is K c = 2.34 x 10 '. To convert to K,, we divide K r by (
~
)
l
NAPr
"
_
[!l&)(8.3145 i.mol-'.K ')(1500 K j (h.022 x 10" mol-')(I@ Pa)
to obtain K, = 5.14 x lo5, hased upon a orje bar standard state.
In the prcvious section we have seen hat the rigid rotator-hamlonic orcillator appruximaiion can hc uqed to calculatc cquilihrium constants in reasonably good agreenjerit with expclimeni. and because uf the simplicity o f the model, the calculatiorjs itlvolved are not extensive. If greater accuracy is desired, ht~wever.one must ii~cludt. corrections to the rigid rotator-harmonic oretical calculations. The thermodynamic lables that wc are about to discuss in this section, then, represent a collection of the thermody~lamicandor statistical thermodynamic properties ol' many substanccs. One of the niust extensive tabulation< of the thermochemical properties of substances is an American Chemical Suciety publication. Joumul of Phpaicnl Cfietnicrtl H c f e r ~ n rTjatrr. ~ volume 14, supplement 1, 1985, usually referred to as the JANAF Goint, ormy. ~ravy,trir force) tables. Each species listed has about n full page of thermodynan~iclspectrosct,picdata, and Table 12.4 is a replica of the entry for ammonia. Nolc [hat the fourth and fifth columns r)f thermodynamic dara are headed by - ( G ' - H ( < ) ) / 1 ' and II" - H'(7;). Recall that the value of an energy rimst be refcrrcd to some fixed reference pr~irit (such as a zero of energy). The reference point used in the JANAF tables is the standard molar enthalpy at 2'38.15 K. C:onsequently, G ( T ) and H ' ( T ) are expressed relative tu that value, as expl-cssed
12-9. T A B L E 12.4
A replica oiiht. page o1-NH,(g) data in thc IANAF tables J M A F THERMOCHEY1CAL TABLES
1598
Molecular Part~tiunFunctions and Kel.ll~dThermdynarrtir
Data Arc Extcnsivt.ly
T~hulatcd
501
by ihe headings - { G 3 ( T )- H"(298.15 K ) J / T and H - ( T ) - H'(298.15 K). FAble 12.4 gives - { G '(T) - H'(298.15 K)}/'I' for ammonia at a n111nbcr of temllcratures. The ratio ( G (1')- H"(298.15 K ) J / T rather than I G 0 ( T )- / I ' (198.15 K ) J is given because { G - ( T )- H"(298.I.S K ) J / T vurizs rtlure slow,ly with tempcrature. and hence the tables ;Ire easicr to inlerpoliitc. It is ~iotneccasary lr) specify a reference point for thc heat capacity or the entropy, as indicatcd hy lhe headings to the second arid third ctllumns. The .sixth and scvcnth uolur~~ns give ~ a l u z s of A,H' and A,G at various temperatures. We learned in Scction 12-3 that thesr data can he used to calculate valucs of A, H ' . ArG', and equilibr~urnconsta~~ts of reaction&. Because C (T) and H " (T) are expressed rclaiive 10 H ' (29K. 15 K ) in Tablc 12.4, wc must express the molecular partition function q ( V , 7.1relative t c ~a zero of energy. Recall thal in Section 12-5 we wrote q ( V. 7') as
where q U ( V ,T ) i s a molecular pwiitiun function in which the ground state cnclgy taken to be zcro. If we substitute Equation 12.46 into Equation 3.41, thcn we obtaitl
For one mole of an ideal gas,
= H r { T )=
af
PV =
is
t KT, and so Equ;ition
11.47 becomes
where H," = N , * E Because ~. y O ( ~T) , is thc molecular pa~litronfurlction in whizh l h c ground state encrgy is taken to be zero, q " ~ ,T ) is givcn by either Equation 4.57 or 4.60, without the factors of e-'~,h,,'?' ant1 e f ) ~ ' ~which ' , represent the ground state ol thc ~nolecule.Using either Equation 4.57 or J.bn, Equation 12.48 btcomec
2
J
- I
(I~ne'lrtuolccule)
( 12.4911)
12-9. Mnlerular Panition Functions and Rclatcd T h c r r r l u d y n , ~ mLl,~ta ~ ~ Are t x l ~ ~ n ~ i v Ir3t>uldlrd c~lv
Ch~pler12 1 Thpmical Equilibrium
The vi~lueobtained from Table 12.4 is more accurale than the valuc calculated from Equation 12.49b. .4t 1030 K,the ammonia moleculr ic exciled enough that the rigid rotator-bmnooic oscillator apprvximativn beginc lo become unsatiflhctory.
=~RT+Z R(-)\i~l.j
,(,
,iT -
1
(no~~linear molcuule)
(1 2.49h)
Nr11r t l tlie~i. ~ ~;II-c n o tcrms involving 0%,, , / 2 T or De/k,T in Equations 12.49 as tllcre arc in Fqu,~~ionr 4.58 and 4.61 hecauqe we have taken the energy ot the grclunri vibratrunal stale to be zero. Wc c a n u w Equation 12.49b and the parameters in Table4.4 tocalculate H"(298.15 K) If" for alrunon1:l
l ' t ~ ?very first cntly i n he liflh column i n Tablc 12.4 is - 10.045 kl.mol-'. 'Ilus value represents I J (0 K ) - H' (298.15 K), which is thc ncgative ot H (298.15 K ) H' (0 K) that we just calculated because H,' = H' (0 K). Thus, the value given by t q ~ ~ a t i n12.49h tj and the value given in Table 12.4 are in cxccllent agl.etment.
E X A M P L E 12-8 IJse Equation 12.49h and the parameters in Table 4.4 lo calculate H L ( T )NH3(g)at 1000 K and m e b x . Co~npareyour result to Tdble 12.4.
We can alst~use the data in Table 12.4 to calculate the value o f q n ( ~T. ) for amrnunia. Recall from Section 9-5 that we derived the equation (Equation 9.36)
-
-
where E, = N,,E, = Ii; and P ' I bar = 10' Pa. Equahn 12.50 is valid only for an ideal gas, and recall that q ( V , T ) j V , (M-q " ( ~ T. ) / V , ir, a l'unc~iurlul'terripxalure only for ail ideal gas. Equation 12.50 clearly displays the faw that the chemical potential is calculated relative to some Lem of energy. Because C;, = p' for a pure substance, wc can writc Equation 12.50 as
Il i s easy lo s l ~ u wthat G" .-• H i as T -+ 0 (hccause T In T H," i s also he standard Gibbs energy at 0 K.
-t
0 as T
0). arld so
According to Equation 12.5 1
H,"for
S 01.C! I l O K : Equatiorl 12.4% gives
'
H ' ( I O ( H I K ) - If,; = 42.290kl~mol Tatllc 12.4 gives
= 10' I'a. The fourth colurnn in Tablc 1 2 4 gives - {G - FJ (298. I S K ) } i T instead of -(G" - I-I,')/T, hut the first entry o f the fifth column gives H,; If'(298 I S K). ThereFtlre, h e cuponenl~alIn Equation 12.52a cat] be obtalt~edfrom
where I'
-
(G" - H i ) T
15 K)) ( H i - HU(29R.15 K)) -- (Go- H"(20X. +
e x ~ t l e n in t Equatio~l12.52a
If we sublract Equalion I from Equation 2. then
we obtain
T
T
founh column m Table 12.4
hrst entry of lifth column in Table 12.4 d i v ~ d dby T
(12.52b)
Let's use Equa~ior~s 12.51tocakcula~ey i ' ( ~ T) , for amnionia at SOU K. Suhqlilu~irlg 111edata in Table 12.4 inlo Quation 12.52A gives
The entries ~nrhc .lANAF tables for H(g) and N(g) givc 3 ,H (I)K ) = 2 16.035 L l .rnol-' and 470.82 k l . mol - I , rzqpectrvely. T h e ~ c\alum conespond to the equations
and If we suhslitute tthi~\:slue inlo Equation 12.57~1,then we obtain
If we subtrack Equation 1 from the suttl of Equaticln 3 and thrcc times Eqilatiot~2, thcn we obtain tquation 4.60 gives (Probleni 1 2 4 8 )
Thc v d u e given by Equations 12.52 is thc Inore accurate becausc Equation 4.60 is hased on the rigid rotator-harmonic oscillator approximation.
The value given in Table 4.4 is I 158 kJ.mol I. EXAMPLE
12-9
I
The JANAF tables give - I(; - H (298.15 K ) ] / T = 231.002 J mul-' K-' and H,, - f l (298 I5 K) = -8 683 ~l.rnth-' for O ? ( g )at 1500 K. IJse these data and Eq~d~r>r~\ 11.52 to calculate y"( V, 7')/1' fur O , ( g ) ut 1500 K
E X A M P L E 12-10 The JANAI: lahlcs glve A , I I - ( 0 K)lilr Hl(g). ki(&), and I(g) 10 k 28.5535 kJ,rnol- I. 216.035 kJ. lnol ', and 107. lh !d.mol-'.respectlvcly. Calculate [he value of I),, For HI@).
I
SO I I) TI C) N : The above data cml be presented 3s
and Equation 12.52~1gives
y1'(V.T I 1'
-
(h.022 x.-10'' nlol-'1(105Pa)e12!51w3 (8.313J-mul- I . ~ - ' ) ( 1 5 1 ? i ~ )
,K--~
L1!6114
,,,,(>,
I E;-:
If we suhtracl Equat~on1 from the aurn of tkiclatic)ns 2 and 3,then w e oht;~ln
The talue givcn in Table 4.2 is 294.7 kJ-nirjl-'. Labtly, the t h e r n l o d y n ~ r ~data i c in lht JANAF tahlcs can also be used to calculate halues of U,, fur molecules. Tahlc 12.3 gives A, H ' ( 0 K) = -3K.907 kJ,mol ' for Nt I, (g). 'I'he chetnicill equalion (hat rcprcscnts thih process is
f H,@)
+
N,(g)
+ NH,(g]
A , H ((I K) = -38.907 1;J.mtli-I
(I}
The thermodynanlic tables contain a great deal of therrntdy nnrnic andlor slntisticnl thermodynamic data. Their use requires a little practice, but it i s well worth the d'lilr~. Problems 1 2 3 5 through 12-58 are menrir to supply this practice.
1 2 10. Erl!~ilibr~umCurislants for
12-1 0. Equilibrium Constants for Real Gases Are Expressed in Terms of Partial Fugacities Up to ttii\ poit~tin illis chapter. w e have discussed equilibria in systetns of ideal gnsts only. In this sectirln. wc st~alldiscuss cquilihriil in systems of nonidcal gases. In Section X-8 we introduced the idca of fugacily through the equaticln
where iC(7') is the chcmical poential or thc corresponding ideal gas at onc bar. Once again to simplify the ontation we shall not display the f " in the rest of this chnptcr. Therefore. Equation 12.53can he writtcn in tlic fornl
ror~scqi~ently. we must kecp irl mind that f is taken relative to its qtandard state. In a ~~tisftll.e ( ~ ~f ; I S C S ,we would hake
Because h e ~nrhec~ilcs in n mixturc or gnscs in which the gdses do not behave ideally are IIOI iridcpendent c ~ onc f another, the parlial fugiciry of each gas generally depends tlpnn the concentrations of 311 thc other gases in the mixture. Now let's consider the general gas-phase reaclion
Ke,~lCasirj Arc: ~xpresscdin Terms of I'anial Fugacitiw
where the equilibrium constant K , is given by
Once again notice that the equilihriu~~l constant is a fullctiun of tc~npzratureonly, as dicl;~tedby Equation 12.57. The equilibrium constant defined by Equation 12.57 is called a rltt,r7nndynri1nilcquilibriuni ronrtrml. Equation 12.57, which relates Kf to A,GPis enact. being valid for real gases ax well as ideal gases. At low preshures wc can replace the partial fugitcitics by panial pressures to obtain K,, but we should e x p t thih approximalion to fail at high pressures. Thc formulas t c ~calculate partial fugacities t n ~ mequation-of-statc data are extensintjs of the ft~rmulitsin Section 8-8 where we calculated fugaciiies for pure gdses. In urder to obtain the parlial fugacities to use in Equation 12.58 we need rather extensive pressure-volume data fur the mixture of reacting gases, These data are available for the important industl-ial reactiun
Table 12.5 shows hoth K , and K j as a Kuriction of the total pressure of the rrac~iun rrlixture. Nore thnl K, is ncll a constant, b u t that K, is fairly constant with increasing total pressure. The results sliown in Table 12.5 emphasi7.e that we must use fugacities and not pressures whcn dealing with systems at high pressures.
T A B L E 12.5
The change in Gibbs energy upon cclnverring the reactants at arbitrary partial pressures to prodllcts at arbitrary parlial pressurcs is
Vdlucs of K, and K t as a fut~ctionof total prescure for the arnrr~oniasynthesis equilibrium at 450'C.
total pressurelbilr
K,/tO-'
K c /IW3
If we substitute Equation 12.55 into this equalion, then we get
Notc thd Equalion I?.% is the gcnrrulizatiun of Equation 12.24 to a system of nonide;ll p i w s . Rci~lizcthat thc values of 1he fugacities at lhis point arc arbitrap, and not necessarily equilibriutn values. If the reaction system is in equilibrium, then A,G = O ;~ndall the fugacitjcs lake on their cquilihriu~nvitlucl;. Equation 12.56 becomes
E X A M P L P 12-11
The cqu~librium constants K,, and X, can be relalt'd by a quantity K y ,such that K, = K I K , and K y has the form of an equilibrium conslallt. h u ~involving activity coefficients, y,. First derive an exprescjon for K y and then evaluate it a1 lhe variuuh pressures given in Table 12.5.
I
Chapter 12
/
c'heni~cal Equilibrium
12-1 1 . Thermodyndmic
Equilthrium Curr5l;inrs Arc txpresscd r r l lrrlns ui 4.1t ~ v t t ~ o
Now let's consider the general reactiun
5OL U TION: The relaliort kta'eer~pressure and fugaciiy is gwerl hg
11-we aubstit~~te 1h1+cxprzssirm into Equalir)n 12.58, then we tblairl
The change in Gibbs energy for converling A and B in arbitrary states k) Y and Z
iri
arhilrary states is given by A,G = v y p y
+ I I ~ J --L ~
1
1
~
1
~~
~
~
,
~
p
~
If we substitute Equation 12.59 into this equation, then wc ohtiiin where w e have used k t stmldartl\tatc ~ f' = P
--
L'\
+
I bar. Us~ngthe data in Table 12.5,
ArG = A r c ' , K I ' In
we see that
mvz
+k
QA"%
where K<
0.904
0975
The devia~~cln of K v firm unlty
IS
0.942
0877
0.688
0.496
a mcaculc of the nur~ldcaliryof the fptcrn
I
I
12-1 1. Thermodynamic Equilibrium Constants Are Expressed in Terms of Activities In t l ~ prekious t seclior) we discussed the condiliun of cquilihrium for a reaction system conqis~ingof rcal gases. Thc central result was the introduction of K , , iri which the cquilihril~tnconstan1 is expreascd in terms of partial fugaciles. In this \'zctiotl w e shall derive a sin~ilar~ x p ~ rinn s r fix g e n c r ~ equilibrium l syslerns. consi.sting of gascq. solids. licji~ids.slld/or solutions. The starting point is Equation 12.35, which we write as
whcl-c 11, is the activity of spcuies j and 11, is chcmical potential of (lie standard state. This equation essentially defines the activity, u,. Recall that we discussed two different star~dardsv;ltes ill Chaplers 10 and 1 I : a Rnoult's law sti~ndardslate, ill which 0, x , as x, -+ I , in which case p; = p;. the chemical pole~itialof purr cotriponcnt ,, ;tnd H&ry.a law standard slate, in which n , -+ ,n or a -, (*, a.; H!; + 0 or I I r,, ,+ 0. in wtlich cahc 11; is thc chemical potential of the (hypotlwdcai) corresponding ideA solution a1 unit mulali~yor unit molarity. Although Equation 12.53 is restricted to gnbcs. Fiquxlion 12-59 isgeneral. I n Fact. wc can include Equation 12.53 as a special casc of Equalion 12.59 hy defining thc activity ot- a gas by thc rzlatiot~ri, = f ; /f,'. It1 this cast, pj(7') in Equuliun 12.59 is the corresponding (l~ppothelical)ideal ga.+ at one bar and a1 the ternl,eralure of intcr-es~.Agrccing to set 4 = .fl/j;'simply alluws 11s to trcat gases, liquids, holids. (and ~olutiorls)in the sarriz notation.
-a
Erluation 12.60 is called the Lewis equation. aftcr the grcat ~hern~udynalnicist G. N. Lewis, who first i n t r d u c d the cunuept of activity and pio~lccredthe rigorous thermrjdynamic analysis of chemical equilibria. Nole that Fquation 12.60is peticrali~ation of Equatian 12.56 to a non-ideal systetn, which tnay consist o f condensed phases ilnd solulions as well as gases. Realize that the activities at this pui~itare iirhilrary, and riot necessarily the equilibrium activites. Just as we did in Section 12-5 for (he ciiae ol. a reactinn system c)f idcal gases, we introduce a rcautiorl quotient. or an u r ~ t i rquoriefrt. i~ in this case, by a"Yfi'/
e,,= a:.: I
/
Using this nulation, we can write Equatio~lI ?.hO as
According to Equation 12.59, a, = 1 when a substance is m i t s stalldarll starc. Therefore, if all the reactants and products in a reactton mixture arc in their star~d,ird slates, then all the rr, = I in Equation 12.61 and cr, Q,, = I . giving A , G = A[G . I I thc reaction systetn 1s at eyuilibrium at fixetl T and P , the11 Arc; = 0, .md wc h ~ v e
where Q,,,,denorcs Qclin which all the activities have thcir equilihl-iunl values. analogy with Section 12-5, we dcnote Qa,,cq by K,,
111
12 -1 1 Therrnodrnamic
Equilibr~umCurlst,Inls Are ixpre5rpd in Terms of Act~\,~t~es
which we call a tl!urmodwln~?tiu t~yui~ii~riurn ronstonr. Equation 12.57 k c o m e s Ina = Equation 12.M iq cornpletelq. general and rigorol~s,and applies 10 any system in eqi~~librium. Note that Tor a ~reactiuninkoluing only gases, n, = 4, anti KU(l')=: k',( T ) ,Eyuatior~12.58,and Equalion 12.65 is rquivalcnt to Equation 12.57. Equations 1 2.64 ;land 12.65are rtiorc general than Equations 12.57 and 12.58 because the reactants car1 be in any phase, The application of this equation is best dune by example. Let's consider ii heierogeneous system such ;IS the water-gas reaction
(constant T )
( 12.hX)
For a cundcnsed phase. I'is esser~tiilllya constant over a rnndcmte pressure range, and so Equation 12.68 kcomes
EXAMPLE 12-12
t:alculntc which is used in the induqtrial prnductiotl of hydrogen. Thc (thernlodynamic) equilihl-iurn const;lnl Ctrr this cqualion i s
thc activity of C(s) in Ihe
SOLUTION: 'I'he density of coke at I(HH) C: i c ahout 1.5 p c m volutt~~.,L'. is 8.U cn~'-inol'. Fron~Equation 12-69
l n i ~= 01(7 =
A l ~ h o ~ i pwhe have clci~llwith fugacities of gases earlier, w r have not dcalt with activiiies of purc s o l i r l 5 and liquids. We nlust first chuose a standard statc for a purc cundetised phase. which we chuose to be the purc substance in its normal stale at onc bar and at the Immperature of interest. To culculate dmc activity, we star1 with
and the constant-temperature dtrivntivc of Equation 12.59
J J L= RTri It1 u
(constant T)
1-01, Nole thal the actir ily is c~scr~tially unity even nt 100 bar.
I
Accordirlg t o Exarnplc 12--12, ihe aulibity of a pure co~ldensedphasc i* unlty ;ll moderate pressures. Conrequcntly, the activities of pure solids and liquids are nor~nally not included in equilibrium constant exprcssinnq (as you may recall frottl general chemistry). For exanlple, for the reaction
the equilibrium ut~nstantis given by
(12.67)
If u,c write Equation 12.66 as
P
{constant T )
and introduce Equalior~12.67, then we haue
rl ltl u = -dP RT
-
(cunstant T )
We rluw integrate from the chosen standard stare (a = I, P = 1 bar) t c ~an arbitrary spate to obtain
II
d In m' =
L !O,
f".,,,
ri,,,,,l
* p,
Ol,,
PH>lP,
pl
if the pressures arc low cnough. Ilowever, here are cases where the activities cannot be set to u~iity,as the fl)ll(~wingExarriple shows.
1
-
v
'. and so its inular
(8.0 r~n'.mol- ' ) ( Idrnl/lOOO crn3j(90 bar) ... , -, .- .- = 0.W7d (0.082(Kdt11'.bar.~-'.nlol ')(I273 K)
K =
d
Cur111of coke at 1 0 0 bar m~d1000-C.
E X A M P L E 12-13
1-he chanpc in the stantlard molar Ciihhs energy for thc conversion of graphite into diamond i~ 2.WO kl-inol at 298.15 K. The density uC graphitc is 2.27 g.c~n-' and that oi rliamrmd IS 3.52 g.cm-I at 298,15 K. At whnt pressure wlll lhece two f o r n ~ of~ carbon be at equilibrium at 298.15 K'!
'
S O L U T I O N : We can represent the prtxecs hy the chenrlcal equation
-dP' L'R:
(constant 7 )
12-12. Uiiirrrnr-t.5 in Sulubiliry Cafr ul~t~uns Involving ionlr S l r c c ~ o
51 3
As a first approximation, we shall set all the au~ivi~y crxfficietus rqual tu iuiity :~ud
write IJsingEquatior 12.69,
~4e have
A;(;' =
[gIP -
From the followii~gset-up
I)]
('H,COOII(II~)
initial equilibrium
+ H,O(I)
'
O.l(Klrno1-L 0.100 r n o l , ~ - '- x
$
H ,O(clc{)
+
CH,C'00-
-..
-
*0
0
-
t
1
((I(,)
S ~ ~ l v i nt lg~ cexpreacirm for P givus t J = I. S l x 1 U4 bill
15 00n har
or x = 1.31 x 10 ' rnol. L-', tc~ra pH o f 2.88. This 1s thc type of calculation t h a ~i s done in general chem~ht~g. Now Ict's not set yt equal to unity. For y.. n c shall use Equ,tlion I I 57
12-12. Thc Use of Activitics Makes a Significant Difference in Solubility Calculations Involving Ionic Species where the iunic slrength I( is given by
Equi~lir~r~ 12.65 can ;ilao be applied to reactions that ~ i t k eplacc in solution. For example, let'x consirlcr the dissocialion of iin aqueous snlulion that is 0.100 molar in acetic acid, C t 1 ; C O O H ) . for which k' = 1.74 x lo-' on a molnrity scale. Thc cquatiotl for the reaction is
Being n neittral .cpeciec at il corlce~~tralirm of arnur~ti0.100 molar, thc urldissucii~led nc.t.ric : r i d has an aa~ivilycoefficic~~t of esseri~i;~lly unity and w oIlAL = r,,<. For the ions. we uac thc fact ttu11(Table 1 1.3) (ili-(J~l[,~ m('H " and s o Equation 12.70 becomes
cAc-YY
In order to calculatc ( we must knuw r,,, or c,,~. hul we cannot dzrrrmine eitlicr of these from Equation 12.7 1 because it cot~tainsy:. Wc can solve this problcm by iteration, huu7ever. We first calculate p; using tlic values of I.,,, and rAL,.that v+e ohtained above by letting y- = 1 :
or y: = 0.921. We now use this value in thc right-hand side of F,qaation 12.7 I . ;mil write
Solving for x , i ~ find e that .r = 1.365 x 10-' n1ol.L '. We now usc t h ~ svalue lo calculate a new value of (= 0.920). and use thls value in Equat~un12.7 I tn c:~lvula~c a new vaIue of x (= 1.3h6 x 10-' rno1.L-I). Cycling through once rllorz g1ve.s y' = 0.920 and x = 1.366 x 10F1 m o l . ~ - 'and . so we hnd that x = 1.37 x rnn1.L ' (11, thrcc significant ligu~es)and pH = 2.86. Thus wc see that w e calculatc a pH of
2-86 using activities and a pll of 7.88 ignoring activities. not a significant differcnce. 1:ortunxcly ~ h cnlyriad of p1.l calculations that you did in general chemistry were si~fficientlyaccurate. This is not necessarily the case for solubili~yci~lculations,a s we sh:llJ now sce. The soluhilit~;product, K*", of RaF,(s) in water at 25'C is 1.7 x 10-', and the
SO1U TI O N : The equation for the dissolution of TIBrO,(s) i c
with
L u l l ~ ny+~ 7 1 a1 lirst, we find that s = 0.0131 r n o l . ~ - ' Uf~ng . [his value of .r, we Fet
'l'hc etluilihriurn constant expression is
I< = r and y, = 0.887 for 'l'lSrO,(s) in purc water. Usirlg this value ol y, in the K c cxprcssiun girts a = 0.0148 rno1.L l -Subsequent iteratio~ls T = CI.0I49 trto1.L f'ol- rhe cnce with 0.50(1tllol.L-! KNO,(aq), w e w r i t e
IJ.;ing thc forn~ula(Table 11 - 3 ) Rrcause ,F ili much less than0.500 lnol.L-', wc iritiallp lei I< = 0.500 n1ul.1. ',which gives y., = O.bl6. Using this value In Lhe soluhility prodr~ct cxpressio~lgivcs s = 0.02 13 n l o l . ~ - l .N o w I< = 0.521 3 mu1 L and y+ = 0.612 and .\ = 0.02 1 4rrlol .I.-'.
'
Subsequent ilerativnsgjves = 0.0214 1no1.L-I. K o t ~ c cthat thesolubiI~ty01-TIBrO,1s) is ~igtlificar~tly enhanced in the 0.500 molar KNO,(aq) even though the KNO,(aql dues nut participate in the dissolutiol~rcactiott. It we had no1 included the ~ctivity cocfficient~,we would have gotten n o errect at all.
If we set y, = I * arid let s be ttie solubility of BaF,(s), then c,.?, = s and c, = 2s, and we have
Problems or s. = ( 1.7 x 10 moli .1,-'/4)"' = 7.52 x I[)-' mol, L-I. we now calculate the ionic s1rrngt.hu5irlg this value o f - s to ahlain
12-1. Expreas the cuncenlraliuns uf each species in the following chc~rlicalequalion\ in term\ of the extent of reaction, t . 'I'hc initial cortditioris art: given under each equation.
a. Using (his value of I( in Eqilatinn 1 1 . 5 7 gives y, = 0.736. Suhstitntc this value into Ilquatiut~12.55 to gel
Sn?Cl,lp) a,,
0
(7)
nu
I
b.
(1) (2)
(if ; ~ c t i b i t ycocfli~lcnfc.
I
-
EXAMPLE 12-14
Calculate Ihe solubility of TlBrO,(%) in pure water and in an aqUeOUE solulion that is 0 500 rno1.L in KN03(aq).K
'
2so,(g) n,, n,
Nl(p)
C.
= 0 0102 rnol.r.-'. Cyulitip through agatn gives y+ = 0.705 and s = 0.0107 111ol.1, '. Once nrrjre g i w s y= = 0.7(X) a t ~ dr = 0.0107 m o l . ~ - 'and {me last iteration givcc y , = (1.700 and s = 0.01X r n o l . ~ - ' to two slg~lificantfigures. Notice that in this caw thr-re is uvrr a 309 diffzrencc hetwcen calculating 7 with and without the inclusion
; I I s~ o ,S
+ Sol(!)
( 1)
*
t- 2 0 , ( 8 )
n,
2%
(2)
n,
"D
c'l,[~)
0 0
2su,(gj 0 I)
(1)
+
+
+
Oz(#)
o n~
N,O,(g)
o o
12-2. Write out the eqoilibriunl-constant expression fur the reaction that is descrihd by the quation
Cumpare your result to what you g t if the reaction is represented by
is 34.8 if the srandard rtate 15 taken to he one bar. What would rhe valuc of K, he 11for some reason the standard state were t.&en to be 0.500bar? What does this resr~ltsay alwl~t thu tlurncrical vaIucq of equilibrium conat~nts'! As~uniirjgthat we start with n,, r ~ ~ o lot r c N,O,rg) and no NO,(g), allow that 111cextent r i ~eaclirm.(LLI. at equilib~iumis given by
P l u ~try/rr,, against P given that K , = h. l at IOO-C. Is your rcsult in accord with Lc Chitel~er'sprinciple'? 12-4. In Prohlcln 12-3 yr)u plotted the cxtcnt 01-reaction at equilibrium agalnht the 1c11alp1,cssurz Ir~rt l ~ cdissuciat~unof N,O,(g) to NO,(g). Yuu found tha~ decreases as P incl-caces, in accord with 1.r CI~Btelirr'sprl~lciple.Now let's inlroduce ? I , moles ~ of an lncrt gas into the system. Assum~ngthat we s t a r with 1 1 , ~nolesr d N,O,(g) and IIO NOl(g). derive w cnpression for [, i t ; , , 111 tcrnls or P and Ihz ratio r = nqWn/n,,.As in P ~ , o b l e ~12-3, n let rLl K, = h.1 and plot Eco/fi,, versus P for I - = 0. r = 0.50, r = 1 .0,and I. = 2.0.Show that
12-9. Mncl gab-phase equilibrium conskantc in Iht. recent cherilical l i ~ e r u ~ uwere ~ c culc.ul:r~r.rl ascurrling a cvarldord alale prescure o f tme utrnosphew. Show that the curreaprlndlny equllihrium constant frlr a cpandard stale przssurr rll rme har r \ gitrn by
where Au is the sutl~of the stoichinrrletric cccfticicr~t~ of-tlh: productr 1111nusthat or tllc reactants.
12-10. IJsing the data in Tilblr 12.1, calculil(c A, G ' ( T )and K , (7') ar 25 C 1111
cZq
inlroducing an incrt gas inlu the reactlo11 ruixture at constant prefsure has the snlne effect as luu,ering the prehauiv. Whal is the efiect of introducing an i n e r ~gas into a wnction syslenl at conhtnnt volurnc'!
12-11. Calculate the value uT K L( T ) hareti uprln a one inn1 1.-' r~antlanl\tale fi,r each o i rttc equations in Pruhlc~n12-10. 12-12. Ucrive a rctotion between K ,
12-5. Re-do I'rohlcni 12-3 wiih 1 1 , / > I ( , = O.S(I and 2 0.
fz,,
t ~ ~ o l or si N,U,(g) and
n,
iuld
K . for the fullouing:
rrlnles of NO?(gl irutinlly. 1 . ~ 1
12-6. Cr,n>~clcrthe amir)orlia-synthesis reachoI), which curl bc described by
12-13. Consider the difswiatlon m x t i n ~ot~I.(g) dcscril>ccl by Suppose inillally there arc n,, moles of N,(gl nnd 3n,, molcs of H l ( g ) and nu NH,(g). Derive at) cxpreshion for K , , ( T ) in terms o f the cquilihrlum value OF the exten1 of reaction, f(.cl.a r ~ dthe pre>curc, P. IJce this expre~sionto d i h ~ lhuw l ~ ~tc4/n(,varies with P i~ntlrclate yrrur co~icl~~h~
T h e tola1 PI-eswreand the partial pressure rll 1 . r ~ ) a1 1100 C have ht.~.r~ ~ntl;~rr~rcd (1) IPL. 36.0 lorr and 28. I torr, re~pectively. thehe data tr) calculate K, (one har ctandurd hr;i[c) ~11llK (one inol-L standard state) ilt IJOO'C.
'
12-14. Show that ~ 1 , )moles of NOCl(g) and nrl NO(g) or Cl,(f), derive an expression Fnr K,, ill term!, of the equillbirum valuc of ihe extcnt of reaLtion, tCq, and the 11rcss11re.P. Grveo thal K , , = 2.()0 x I U '.calculate t L q / n , wheo , P = 0.l18U bar What is the ncw \-aluc r l i f,,<,/ti,,at equilihriurn when F = 0.160 bar7 Is this recult in accunl with I.r Chitcliel-'u prrrlciplc!
Aasuming that wc s l a t with
ideal gabes. for a reaction i~~volving
12-15. C o n ~ l d c rthe pas-phase r c a c t ~ ofor ~ ~thc synthchls of mctha~lolf r , r r ~ C:O(g) r~ and H:[F
I
I
12-8. Thc. v;ilue ol K , , at 1oW C fur the clczt~inpnc~liu~~ of carhuriyl dichlo~idu [phusgenr) ~ c c o [ . d i l to ~g
The value rlt' the cquilibrium conslant K, at 500 K is 6-23 x 10 -'. lnilially equirr~olur amounts of CO(g) and A,(&) are intruciitcetl intr, the reactirm yeseel. Dr~erminethe value uf e,/n,, at equilibrium 500 K and 30 bar.
Cllapter 1 2 / Cl~cmiralEquilibrium
11 8 12-16. Con>ldcrthe tat>cqualrons 1 COrg) t HIO(g)
* CO,(g) -tH,(g)
(2)
CH,(g)
Shuu, t h d K ,
(3)
Y
+ II,C)(E)
COtg)
+ 3 H,(g)
Kl
K,
where 6 is the extent of reaction. Given that A,C; [H,O(g)J = - 18.334 kl mo1-I al4IHM K. plot G(4) agarllst c. Differentiate G(6) with respect to 6 and show thal the rninirn~~m value of I;(< ) occur< at tCq = 0.553 Alsn slrow that
K , K1 for the surr~of lhcse two erluntionq
TH,lgl+ 2 H:O(g)
+ CO,(g) + .1H,(g)
K,
Iluw d o >nuexplain the fact that you would add thc values uf A,GL but multiply the equilthr~umcun\knnts when adding Qr~ationc1 and 2 to get Equation 7
12-20. Consider the reaction described by
a1 500 K and a total pressure of one bar. Suppose that we start with three moles of H,(g). one lnolc of N,(g), and no NH,(g). St~nu,that
Dcterrnl~~e K , . for thc reaction
12-18. Consider the reaction describetl by
wherr t is the extent of-reaction. Given that G,,, = 4.800 kJ-mol ' at 500 E; (see IBhle 12.4), plot G ( 6 ) versus 6. Differentiate G ( t ) with respect to 6 and show that thc minimum value of G ( < )w c ~ i r sat (, = 0.158. Also shor* that
at 5M K and n total pre\cure r)f one bar Suppoce that w e start with otlc rnrjlc each of Cl,(g) and Ll~,(g);iud nu RlCl(p). Shrw that
and that K,
wherc
i~the extent nf re;lctiun. Ciiven that GL ,
versur
<. DiCfcren~inteG($) with reqpect to 4.
wcur.; at Cc-
= U.544.
= -3.694 !d.tnol-' nt 500 K,plot G't,") and show thal the mi~limumvalue of G($)
Also shrw that
=
1hcr;(2
-
cYj2/27(1 - CCu)''= 0.I(I
12-21. Supposc that wc have n lrlixturc of f l ~ egases Hz(&+C02(g), CO(p), and AIO(g) a1 1 2 h I ) K , ~ i t hP =0.55 bar, Pr,, = O 2 0 bar, P,,= 1.25 bar,and I',:,=O10 bm.1~ H. the rcaction d c ~ c n b c dby the equat:otl
at equilibriuln undcr thcsc conditiorl~:lI r ~ i ~int ,w l ~ direclinn t w ~ l the l reactiun prtweed tu attain cquilibriun~l and that K, = 3trt1/(1 - -
t,)'
=
5Y
12-22. Given ttint K,,= 2.21 x 10' at 2S'C
[ ( ~ rthe
equa(inn
12-1 9. Cottcider thc reaction dcscribcd hy
at 401)0 K aljd a iota1 pressure of urlc bar Suppclsc t h a ~we start with two moles of ti,O(g) and no II,[g) or 0.1;). 5huw thnl
prcdict thc direction in which a reaction mixlure fix which PC., 0. l o bar, and PC,, = 11.0050 bar procecd~to allain equilibrium.
,, = 10.0 ha^;
=
K, at 500 K fur a gas-phase reaction doubles when thc tcnlpcrature is increased f r m 300 K to 400 K at a fixcd prcqsure. What is the value of- A r H ' lor this reaction'?
12-23. The value or
12-24. 'The value of A t t i c is 34.78 kJ.nlol-' at 1(HX) K
for the rentlion desurikd hy
Gi>en thal the valuc O F K,, is 0.236 at 800 K . eslinlate the value of K , , at 1200 K, nss~lming that h,ll is independent vf temperature.
12-25. 'l'hc ~ l l u of e A, tl is
- 12
97 bJ . lrml
12-29. The knlprratu~-edependence of the eqrlilibriuin conqtant k',, li)r Ihe rc;~vticrodcscril>cd
bv
at MI0 K fur ic given by the equation
Assuming that h , H IE it~depcndentof tempzriiture. calculate K,, at 700 K given that K, = 29.1 at 1000 K. Calcula~eIhe values of A, ti", A r H - ,and .INSfor this reaction at 52.3
12-26. The etluil~hriur~~ ~or~htaril to1 tllr: rractiun described by
12-30. At 2000 K and one bar. water vdpclr is 0.53'8 dissociatcd. At 2 1 tH) K and one bar. ~t I \ 0.88% dissociatcd. Calculate the valrie of ArII for the dih.crjcialiun of water at oor~cbar. assurning that the tnthalpy of reaction IE corIaparit o w r the range from ?(H)O K to 2 100 K.
can he expre5hed b) tllc cnlpiric;il Crlr~nula Irl
E:
=
-6.375 r 11.64 15 III(T/KI
K
-1 1790 K T
12-31. The follou lng tablc g i r ~ cthr \t:~nri;lrtl ~ n n l ; ~C;lht)\ r c r ~ ~ ~(11'g Ir>ri11:11 y irw of C'I(g) :II rhree rlil'ftrr~lttenlpcrilturch.
Ilcc this form111o to liclsrmine ArII as a furlctiun of temperature. Calculate A r H - at 2 5 C a r ~ dcomparc your result to the one you ~rhtainf r ~ m 'Table 5.2.
Use these dala to determint the value uu K , a1 each telrlpcraturc lor thc reuclirln dcsu1-ibed hy
A s s u m i ~ ~that g ArH' i s temperature indcpcnderir, determine the wlue of A, H rnlrrl lhekr data. Cumbine your results to determine A S S at cach tclnperature. Inlerpret your resultc
12-32. The Kolluu ing rxperi~ncntaldata were rlrr
't'hc ~nolarheat cap;icititer
t)f
S0,(8)
C:O,(g), ti,(g). CO(g), and HZO(g)can be ckprcsscd by
In K,. Calculate A,G make.
'.
A, H
I -3
263
* SO:(g) +
-3.007
- 1.849
O;Igl
- 1-17?
-0.591
'. and A r S ' for thic reactirln al 900 K. Sta~r.ntiy ocsurr~p~ions [hill jot1
12-33. Show Ih;it 11 = -KI'In-
y ( V . TI ,I,
tncr the tcI1iperature r;lnge 300 K to 15(K) K. Given that
at 3(X) K and that K,, = 0.695 at IOOU K, derive a general expression I'or t t ~ zrw~ationuf K , ( T ) with tcnipcratt~rcill the fwrin of Equaliur~ 12 34.
12-34. lJhe Equaliun 12.40 to cnlculatr: K ( T ) ;il 750 K For the rcactiotj descrihe~ih) tl,(p) + e 2 I i l ( f ) . Usc thc rrloIecular parameters ~ I V C I IIn Table 3.2. Co~nparzyutlr valu: r t l thc OIIC fl\,cn in k h l e 12.2 and the expcrilncntal value shu%-n111 F ~ ~ I I 12.5. I-c
(:h;~pter 12
1
Chcrrliral EUIII!I hrium
12-35. Use Ihe sti~listicalthermodynamic rnrnlula~of Section 12-8 to calculate K,(TI at 900 K, 1000 K. I100 K, and I200 K fur lhc asswiation of NaIg) to forln dirners, Na:(g) nccortiing to the equation
Usin! the B u m - O p ~ n h e i m e approxilnation r and the n~olecularparameters in Table 4.2, show that
Uornparc your prcdicrions sing this c q ~ ~ ~ ttoi othe~ ldata iri the JANAF tables.
your result at lOOn K to calculatc the Iraction of sodium alotns that form di~nersat a total pressure of otlc bar. The experirrjuntal kalues uf K , ( T ) m Llse
12-42. I!h~ngthe harnlonic ~~scillalor-rigid rillator appl-oximation, %howthat K,T)=
"'131f11~r2 (T)3;i "1
!'lot
111 K
(A) (=)(@:::):
i+, (Tl,,%r
(-> )\,;
, asainst I / 7' ~udettlrn~ioethc value of Ar H'
12-36. Llsine the data irl Table 4.2. cnlcrilatc K , at 2000 K fur the reaction rle~cribedby the equation
Ttw cxpcl-irnc~ji;ilv:~lucis 1.3 x IO..'
12-37. Using the data in Tables 4.2 and 4.4, calculatc the equilibriunl
cnrlstanl for
the water
ga$ renctit>rl
for the rcaction dcscrikd by
Usin? the wlucc OF 0 $) v,b, and U, glbcn in 'lhblc 4 2, calculate K at 500 K. I fHH1 K. 15W) K.2nd 2oW K-PI(>!In K qain.rl I j T an11rletermille the v,llue of A , H A . 12-42. Use Equatian 12.49b to calculate H'(T)- H,; for NJl,(g) from 300 K to W K and compa-e your talucr to those give31it1 'I'ablc 12.3 by plottitlg them on the same graph.
at 900 K arid 12M K. l'he cxpenrllenlal values at these two temperatures are 0.43 a t ~ d1.37. rcspcclii7c ly
12-43. ilw the JAN4F tables to calculate K, at I OM) K for h e reaction described by
12-38. lJsing~ h data c in Tables 4.2 and 4.4, calcula~ethe equilibrium conctnnt fur the reaction Compare your results tu the value given in Table 12.2.
at :(I()
K.'rhc a
10
,'
12-44. Uxc t l ~ cSANnT: ~ a h l c In s plot 111K,. vcrcuh 1 7' II.OIII4(H1 K r o 1200 K for the i-c:~c.lit>n d e s c ~ibed by
' ( w e T~hlc12.4).
12-39. Cnlculdcc tlie cq~~ilihri~un coristanl K, fur the reaction
uslng the (lato in Table 4.2 and the fact that he degeneracy of the gmund electronic stale of an i f d i n e atom i + 4 and that the degeneracy of its first excited electronic state is 2 and
that it\ e n c r ~ yi~ 7580 cln
'
The experimental values of K, are
9(K) . -
I1X)O
--
3.94 x 1K4 3.08 x lo-? Plot 17) C,, asain+l 1/ T ro determine the value
1 3 Rkl.mr)l--.
12-10. Conh~dcrthe reaction given by
1100
1200
--
l.h6 x 1W2 6.79 x 10-'
12-45. lt~ R o b l c n ~12-36 we calculated K, for the decnniposiiiun 01- CO,(g) to COtp) and n,(g) a1 2UO(l K.Uhe the JANAF tilhles to calculilte K , and compare your result to the onc that you obtained io I'robte~n12- 36.
12-46. You cnlculatcd K , at 700 K for the anlnlonia ~ y n t h ereaction ~ i ~ in Problem 12-38 IJse the data irr Table 12 4 to calculate K, and compare your result to the one that you obtained in Problem I?-?&
of ArH3. The experimcnhl value is
I
12-47. Thc JANAF tables give the follnaing dala ior I(g) a1 one bar
TiK A,G /k.I.mol-'
SOD
900
1000
34.580 29.039 24.039
1 l(M1
I2OO
18.741 13.428
Chapter 1 2 / rhem~cdlFquilibriuri> Arl cxceller~tcurve lit tu rhe plot uf In K , against l / T is given by
Show that if all thc NH ,(p) and CO,(g) rerult from the dccot~lpositior~ o i amrnt>riium c:irbam;itc. then K,, = (1/27)P ' . where P i u the turn1 prcssurc nt cq~lilihrium. 12-63, Ca1cul:lte the .cnllrbilily of t,iF(s) in wntcr at 25 C. Compare your ~ e s u l to t iIlc o r ~ c!(vu obtnitl t1y il+irrg crlncentratiuns i l l s t t ~ dof nclivll~es.Take K
'.
L!se this eupre~ciorlto dclt.rlninc hrIt a \ a iunctiun o i tc~nperalurcin the illtcrval 630 K J' 750 K. Given that
.;
A
\r)lurion that i u 0 01 50 ~nfrlnr111 MgSOlraq).
12-65. Calculate the solubility of CaF,(s) in a sulutiwn that is0.050-molar in NaF(aq). C
in the inlerx-nl 29H K
T
<
-
750 K . calc~~liite Art] '. A,S '. and hr(; :I! 208 K.
12-60, Coll\idc~the tli<socialiur~of .4g,O(s) to Ag-(s) and O,(g) acctwdinf to
Given thc following "dissociation pressure" data:
Express K,
12-64. Calculale the soll~hilityof CnP, Is) it) Take Khp= 3.9 x 10 I ' fur c&,(s~.
111
rcnm
(jf
P {in torrj antl plot
I11
K, vcrsus I ! T . .4n excclllent curvc fit to
these data is ~1vc11 hy
In K,,
Uue this exprcsqion
to
- 0,9692
t
5612.7 K
2.0953 x 10' K'
- - -1'
--T:
clcrive an equation [or A, H' from 445 K < I' < 4h0 K. Now use
~ h follor~ii~g s heat vi~plcitydata:
C.,IO,!y)J!X
-- 3.27 + (5.03 x
T ; , l A g r s ) ] / R = 2-87
10.' K - ' ) I
+ (7.55 x
C ' ; [ A ~ , O ( S ) J / R= 1.981. (4.48 x
K-'IT
lo-' K 'IT
12-61. Cihcium carbonate occurs as Imo cry~t;lllirl~ f o r m . calcite and aragonite The value of A,(; for Illc trnnrilitrrl
kJ mnl- at ZYU. The drnsity 01- calcite at 25' C is 2.7 10 g-cm-' and lhat of ;lr:lponltc i\ 1 930 g-clii-'. At whnr presrure will these two forrnr r)l CnCO, he at cquilhrium at 25 C. i u -t I .(W
11-62. Tllc decornpocitioi~of a~nrnoniumcarhatllnte. NHICOONH, takes place accorrling to
CHAPTER
13
Thermodynamics of Electrochemical Cells
Batteries produce elwtricity hecause the oxidizing agent and the reducing agent in an oxidation-reduction reaction are arranged such that the electrons that are tranafkmd from one reactant tc~the olher itre forced 10 travel thmtigh an extcmal circi~it.The hasic experinlentai sctup that rcalizch such an electron 1ra11sferis called arl eluctruchc~niciil cell. In this chapter, we will see that clcctruchc~n~cdl cc.1l.s have a number of' uwf~ul ~hermodynarriic applicatior~s.Under readily attainahlc expcri~nrrnwlcr~nditior~b.Illc voltage that a cell produces cat1 be directly related to thc Gibhs energy change ol (he untlerlying oxidation-*duc~ion reaction. This fundatnental rcliition provides us with one of thc nloat convenient and accurule methods €or determining activity coefficients of electroly tc solulions. Usirig the relation hetwe.cn ttlc cqr~ilihriumconslant and (he standard Gihhs energy change of n reaction. wc call a l w use electrouhenlicitl cell measurerllerlts 10 determine solubility prnducts and acid-dissocirt~iwcunsl;trlts. Wr: corlclude the chaptcr with a discussion of hi~llerie?,arkd Kuel cells.
Gilbert Newton Lewis wa\ horn i n Wehl Newtr)~~. Massachusetts, on 0ck)hrr 25, 1875, and died in 1946 In 1 %W,hc mcived his Ph.D. from Hnrvard IJniversity, and after spending a year stlldyitig in Gern~any.he relumed to H u ~ r u das an instr~lctur.Lewih left Har~ard111 1904 to hzuo~neSuperinkerldent of Weiphts and Mcasurev i r i the Pt~ilippi~~cs, and a year later he mover1 t o The Miisvauhusetts 111stituteof Technology. In 1912, hc acccpted the puciliun uc Dean 01 tic Cnllegc r)f Chemistry at ~ h lir~ivcrsity z of California ; ~ tBerkeley, which he chlvclnpcd illto one rjf the fine*[ ~cachirlgand rcscarch depdrtlncnl\ i n the wr)rld. Hc rcmaincd at Berkeley fr)r ~ h crest of his lift. suffering s fatal heun attack ~nh ~ laboratory. s Lewis wa\ r)ne ol Air~erica's outh~a~~ciing chernislr, certainly thc finest not tu ~zueivca Nubel Pri)e. I . C W I ~ lnndc Inany Impcxtant c~,nlrihu~~oris in chcmistry. 111 the 1920,. he ~r~troduced Lcwis forn~ulauand desurihctl ; I c.oralenl bond (which he named) irk u charcd pair of clcctrt>ns. tiis not-k or the app1icalii)n uf thrrinutlynsniic\ lo physici chewistry culminalrd ~n hi\ rluhlancling I423 text, conuthured wr-lth Mcrlc Randall. I ' l r t ~ n ) l o > ~I fZ>~ I I I I ~rrnd ~ . . ~ !he FI-L,~ Ln'ncrg~o/'Chrrr~i~-uI SU~SI(IJIC' F, I-nlrllw l l i ~ h a gcneratiun ol~cherrlistsIcarncd thrmmdynnamic~.I.ewis wu5 a rlynnrriic illd~vidunlwho was rcsponsiibc tor de\,elopment ol- niany oulstond~ngcllc~nisks,qcvcral of who~llhecame rntnlher., ul'ttle He]-kcleg f x u l t y His department protluczd ;i rern:~rkahlenurnt~erof Nnbcl Prizc winners.
13-1. An Electrochemical Cell Produces an Eleclric Current as the Result of a Chemical Reaction In this section, we will show how a hpontaneous cheinical reactlon can pnduce an electric current. Considcr the reaction
If we place a
zinc rod into a solulion or L'uSO,(ay). for exa~nplc.clcctron~will hc transfcmd from th; zinc atoms tn the copper(l1) jo~ls,producing copper atr~rnsand zinc ions in solutio~i.No clectric vun-ent is produced by this system txciiusc the rciactuntl art. in direct contact: lhc system ia shorl-vircuiied. 1C we cat) sott~ehowkecp thc rcijcr;inrs separated, howevcr, we can make elevlrons lnlrr~!Ite r i r ~ catotns t~,avclthrough a wil-c to reach the copllcr(Il) ions. A setup that allows (his IIJ be done is callcd an ~ l t , c . t ~ ~ i r - l ~ rLII 'r~~ir 523
2 P - ] is called the mrorle. Thcsc dcfi~~itinns are conveniently rcmc~nberedwith thr aid of thc mnemor~icdevice
consonants
cathode
reduction
w)wcls
anodc
oxidation
13-2. Half Cells Can Be Classified into L'arious Type5
13.1 An illustration r)f o zinc-copper elcctrucherriical cell ( T t l i ~cell i s called a nunlrl cpll.) The cquatiol~for Ihe rcacticln is Zn(s) ~ u " ( a ~+ ) zn2'(sq) Cu(s). The reaction is carried r~ut such that ttrc cleclrnns are transferred fro111the ~ i n cto the copper through an external circuit. FIGLIRE
+
+
All eIectrwhemiva1 cells conqist of two half cells. Onc of the simplcst types of halt cells consists of a metal electrode in contact wit11 a solution of its own iclr~s.Exn~nplesof 1hi.s type of electrode are the Zn"(aq)/Zn(s) electrode and the Cui2(aq)/Cu(s) elec~rode that makc up the Datiiel cell (Figure 13.1). Thc equations of the corresponding half-cell reduction reactions are
and
( , d l .An clcctmchcmical ccll for Equation 13.1 is shown in Figure 13.1. The cell ct~n.;ists of rrlds of zinc and copper (the elrcrmdes), eacli irritrrerscd in an aqueous solution containing their rcspectivc ions. Zn2+(iiq) and cu2+(aq). When the reaction drstribcd by Equation 13.1 proceeds spo~~taneously, a zinc atom in the zinc elwtrodc gibes LIPtwo electruns to the cxternnl circuit and enters the solution as ZnZ' (aq). 'L'he <pont;~rieityc ~ l - ~reaction he drives the elcctrons thmugh the cxtemal circuit lo the copper electrode, whtr? tl~cyare picked up by a CLI" (aq). wtiich deposits on the electrode as a copper atorrl. Ul11t.s~the two solutions are cclnnected electrically in some nimtlzr. the rractioll quickly " l i ~ ~ l out" e s because of a separation uf unco~~lpensated positive charge resulting from the Zn2-(aq) being produced in the ZnSO,(aq) solution and uncompensntcd negativr chilrgc rcsi~ltingl'rotn thc cu2+(aq)leaving the CuSO,(aq) solution. 'lhc i w o solutions can he connected electrically by a sulr bridgt., which consists of n snturated KE:l(aq) sululiun tliixrd with agat, a substlince that ftxrns a gcl similar to gelatin. Thc purpose of the gel is to prevent the two solutiuns from mixing while still pcrrnittirlg the passage of arl cleutric current carricd by the K+(aq) and (:I- (aq) ions. The salt hridpe therelure provides an ionic current palhwny helween the ZriSO,(aq) atld CuSU,(aq) solution<.As a ~ n ?(aq) ' ion enters rhe 7.nS04(aq) solutiotl [rum the Zrl(s) elwtmdc. ~ w o chloride inn< pass from the salt bridge into the solution. hicanwhile. ;t'; a L'U" (aq) Tot] lcavrs the CuSO,(aq) solutio~lnnrl deposits on the Cu(s) clcclrodt. two KA(aq) ions etircr thr solution. In get~cral,a11eleclrclde i s any solid on whose surface oxidation-reduction reactions lake plncc. The cleclrude at which reduction occurs [cu2+(aq) 4- 2 r - + Cu(s)l is c;~llrdthe mil,nrlrj; thc electrode at which oxidation takes place [Zrl(s) ~ n ' (aq) +
-
+
'I-he reactions may hc made to go in either directiun with n smnll change in vcjltsgc about its equilibrium value. Electrodes with this property are said to be rr.ver:rihl~.l n order to reIaIe ohservcd ccll voltages to thermodynal~~ic quantities, we will requirc that the cells that we use vt~l~sist o f rcvcrsihle electrodes. Atlother Iypr 01- rcvcrsible electrode consists of a ~nctal[such as Ag!s)] i~nda sparingly qoluhlc salt of the metal [such as Agl(s)l in contact with a xtllution of n soluble si~ltconsisting of the anion al ihr sparingly soluble salt [such as Kl(aq)]. ExarnyIes of thiq typc of electrode are the Ag(s) IAgCl{s)~HCl(aq)eleclrt,de and the Pb(s)JPhS04(s)IH,SO,(aq) electrode. The (uxidatiol~)reaction that occurs at the Ag(s) IAgCl(s)[HCl(aq) electrode may he thought of as taking place in two steps. The first step i s
which is foliowed by Ag-(nq)
+ CI
(at11
+ AgCl(s)
tc~form tbc r p a r i u l y soluble salt. The nct cquntion for the eleuirodc rcactlon is Ap(<)
+ CI
( n q ) + Ag(:l(s) t e
F ~ g u r e13 2 i s a whcmntic illustration of a qhvcr-silver chloride eleclrnde.
Ap(s) f o i l
covered w i t h ASCII s)
F I G U R E 13.2
HC1Iaq)
F I G U R E 13.3
A schematic ill~islrationtlf a silver-silver chloride electrode. 'l'he electrodc consists of metnlIic silver in crlntact with silver chloride irnrnc1,scd in a sulution o f hydmochlr~ricacid.
+
ylE
A schematic illustrationof a hydroget1 gas elsctrtnle. Thc clcctrwle r,unsists of an inen metal such as
plabnun~immersed in at, acid n,lmiun such as hydrcxhlonc acid. Ilydrogen gas is br~hhletii r ~ thc ~o ~ o l u t i u nr n e r rhc clcctrds.
r
E X A M P L E 13-1 Write the yuatior~for the elcctrodr (oxirlaliu~~) rcactiun thal occurs in the Phcs)lPbSO,(s)lH,SO,(aq) elecuode.
the oxidized and reduced states of an oxidation-reduction couple. Examples of such electrodes involve the reactions
5 C) LU -TIO N . We may picture the rzdclion as occurring in two steps, as we did for lhe react~ont h a ~cwcurs at Ihc Ag(s)lAgCl(s)lHCl(aq) electrode. The first step 1%
and followed by the formalion of the sparingly soluble PbSO,(s)
The nct squarion for the electrodc rencction is
A third type ofrevers~hleelectrode is a Fa:, electrodc. In a gas electrode, a wire or a slrip of it Jlonreiilchve~nctalsuch as p l a t ~ n u ~isnirnmcrsed In a solution through whlch a gas Is bubbled. l'he solute ol the solutlon and the ga\ have a common atorr~.'The da\sic example of a gas electrode is the hydrogen elzctr odc, Pt(s)lH,(g) IH+laq) (Figurc 13.3). -1'he equation t o r ihz ovcrall (oxidation) rracllnn that occurb at this cleclrode is
A chlorine clectrcldc wurks it1 n similar manner.
One other typc r>L reversible electrode is an uxrda~ion-reductionelectrode. Strictly speaking, all elcctrudes are ox~dat~on-reduct~un electrodes, but the terlll is reserved for elec~rodesthat consist of an Inert metal immersed in a solution containing both
The four types of electrodes discussed here do not i~icludeall possible electrodes, but most electrodes d o fall into one of thc above categories.
13-3. A Cell Diagram Is Used to Represent an Electrochemical Cell Electrochemical cells arc ofien describcd by means of a re11 di0,yrunt.For ex~~rnplc. ~hc cell diagram for the electr~uhernicalcell shown in Figure 13.1 is
The single vertical bars indicate boundaries ul' phases that are in contact, and ~ h t double verlioal bars indicate a salt bridge. TIlus, in the cell reprcscnted by the above cell diagram, Zn(s) and ZnS0,(:1q) are scpiirate phases in physical contact. aa are Cuts) wd CuSO,(aq).and a salt bridge connect5 the ZnSO,(aq) and CuSO,(:lq) solu~ions. The co~ivcntion(hat allows us Lo deduce the che~nicnlcquation corresponding to thc cell reaction from the cell diagram is that the reaction occurring at thc lett-hand electrode in the cell diagram is wrillen as an oxidation reaction arid Illat occurring at the right-hand c1ec11-odeis written as a reduction reaction. This converltion enables us
13-3. A Ccll 11 dGr,lrr I < Ils~dto Rcprc!cr>l 3n rl~r!rochcm~c,~l Ccll
Reduction takes place 3t the right electrrde, and the equatiun for the reaction is
to wite the ~ q u a ~ i of (n~ the r cell reaction unalnhigously. For the abovc cell diagram, ther~.wc have
Ztlls)
+ Zn"
Cu:-(aq) t 2
(aq) P
+ 2 P-
=+Cu(s)
(oxidation a1 the IcCt-hand electrode)
The equatiun Ibr the nct cell reaclion i s
(reduction at the right-hand cleclrodc)
+
Z l ~ ( s l 2 I l ' (aq) .-+ Z1i2' (aq)
+ H,(g)
This convenrion is easy 10 rcrnembcr because reduction and right both bcgin with r . Thc cherrlicn[ cquatiorl ctil-rcspo~ldingto t h e cell rcaction is the sum o f t tlc t*o e l e c ~ o d c
I
E X A M P L E 13-3 Writc the erluation~li>rthc eIectl-ode rt.:~c.tior!sand the net cell reaction fur the elecrrt).
I
cherr~~cnl cell
E X A M P L E 13-2
Writc ~ t i c c q ~ n t i o ntor r thc elcctrutle rcnc~irln~ nlld the net cell reaction for the electroche!t>icolccll
Nnle that this cell (irlec not Rnr~ca \all bridge hcf~useit ha.; orily
ottc
clcctrul!liu
solutirrn S O L U T l O h': Oxidation lakes place n t the left electrdc. Oxidation of Cd{s) yicfds Cd2+{nq).so we write Cd(s)
--t
Cd" (aq)
+ ? e-
(An ilIristr:ttiorl of tl~ic1.~11is show11in Figure 13.4.) The (reduction) reactiun at thc right eleclrude is 5 0 1 LlTlOh
Oxid~tiont a k c ~place nt the left electrode. T h u ~ the , yuativn for the reactinn at the letr clectrade is -
l'he neL cell rzacrlnn is
The cell ~ ! c s c r i k dtrcw is very i m i l a r 10 thc M'c,.irt)r; .~trrr~,l
W ~ t hthe electrode conventiut~w e h a w adopted, electrons flow t h m u g h the e x t c r n ; ~ l circuit fnm left to right in the 'ell diagram if the rcacliun occurs spontaneously. The measured cell voltage is cqual to h e electrical potential of the right clecrrode nlinu5 that of thc let1 elw~rtxlr.or
F I G U R E 13.4
4 ritlc-hydl-c~pcr:gas clectrocl~crnicaiccll. Notc the H-type geomctry ur thc cell. which h~rlds the \;)It hrldrc ; * I I ccp;lralcs ~ the tu'o clcclrulytic urllutions.
Realize whcn we write 1IR -- I.; here, we are referring in the cell diagram and not to the spatial arrangement of the ccll. Bccause the reaction described by Equation 13.1 occurs spulioneouhl~(unless the ratio nf the concentrations of zinc inns t c ~copper ~ r ~ is n sentrelncly smallj, the left c l c c t r r ~ d e(as written in the cell diagram) will t;~keon
1 3 1 . The Crlrl)s trlrrgy C h a n g ~of a
Cell Rcact~onIs D~rcctlvKcldtcd tu Ihc tlr.rlr~)~i~ol~vt. I tl1r-e011 . 1 ~I. ,I1
ia said tu he reversible. The condition of rcvcrsihiliry can he exprehsed succil~ctlyhy the mathematical cxprcssiun,
F I ( ; 11 R E 13.5 A Weston ( r ~ ~ ~ d -c.ll a l - dThc cell dlilpraln 1s Cd[s)lCdSO,(ilq)(Hg?SOJ(s)lHg(l) i~idependent of the orienra~ion($[he cell in the ftguw.
electrrms a ~ i dpass them through the external circuit to the copper electrode, which will then pass them o n to the copper ions and reduce tliettl to nietallic copper. We say, then. that thc prlrtrntial o f the IeCt eleclrode is negative with re.speut 10 the right electrode. h o t c that in this case. \,\ > VL,,au thc cell voltage. -1V , is pusi tivc. A key pruprrly of a cell [hat allr~wsus tu apply the~,modynnrrlicsto elzctruchemist~y i n the ~ ~ 1 c r t r r ) n 1 c j ~ i 1 ~ (emf), c ~ f i r t ~t. ~ cwhich ~ is defined by
If a ccll ih reversible, the chemical reactiun clccw~ingwilhin the cell car1 ptoueull in either direction, depending upon the flow of current, and at the null point, the driviilg force of thc I-ci~ctiunis exactly balanced by the emf of thc putcntic~mt.ter. Becausc E is associated with a balance point of the reaction, whcrc it will gr~oibz direction or thc other with small chi~ngesabout E , one might wspect that E is relaled to thc change in Eiihhs energy for the reaction. Furthrrniort.. if- E > 0, the cell reaction p r c c c d s spontaneaurly and if E < 0. the revcrsc reacticln d t ~ sThis . concept follows from F4uation.c 13.3 and 13.4. bur the cell diaprum given by Equi~~iuri 13.2, t' :, 0. Clearly, then, rht: relation hctween E and A G ~riuslinvolvc a negative sign heciu~\e A G < 0 for the reaction associated with the ccli diagram. I ~ t ' see s just whal the relation bctv,,ccn E and A(; is. Consider a cell ill which n molcs o f electrons ;Ire (ransferred from thc left electrode to the right electrodcc. The magnitude the ckargz of one nlole of clcctrons is given by the product of tllc magnitude of'the charge on arl electron (1 -6022 x I O - ' ~C) atjd the Avngadro constant (6.0721 x 10'' mol-I), and is equal to 96 485 ~ : . r n o l - ' .'1-his quantity is callsd the Funitlr~ycortstirnr and is denuted by F.'Thus. if 11 molcs of clcctnlnh pacs ~ t ~ n i u gthc h extrrnal circilit, then the total clectric charge passing ihrougl~thc cxtcra;il cil-c~~il i.; II I.'. If AV 3 0, then the electroris will ffow from left t c ~righl (by our c o n v e ~ ~ t band ~ ) ,an amount of work equal 10 n f - AV will be done by thc ccll. II' this work is done revetsihl y, A V is replaced by E. and the electrical work done by thc ccll will he n F E . Note that iC E > 0, the cell docs work or is able to do work. Atcordin? to F.quatic111X . Ih
E q ~ ~ ~ i 13.1 i o n says (hat h e errif of a cell is (he poteritial difference of the cell rneasured i~ndzrthe ct~nditirlnt)i'rIu flow of electric curretlt.
13-4, The Gibhs Energy Change ot' a Cell Reaction Is Directly Kelated to rhe Electromotive Force of the Cell Si) f ~ r we , have no1 in~ruduvedthzrlnody rii~~nics into our discussion of electrochzrnical cclls. Such an intrriductiorj njilsl be done with some thought because the equations of rhenuodynarr~icr,a5 we have developed thcm. iipply only tu systems it1 eqililibriunl or 10 ~ v r r s i h l zpi-<~e\st.s. ?'his brings II:, to ;lrl irnpnrtant cnricept of eltctrr)che~nistry. nil~nrhy.{hiitof ;j rvvrrriblr rt.11. 1-hs enif ~X:Icell can be it~ca\urcdwith u j>otc~itintneter.;in irlatrurncr~tthat ciin be used tu r 1 1 e a ~ ran 2 electric cunrnt undcr cotiditions in which no current is allowed ti) pa+ rhrt)ugh thc ilzH (see tquntiori 13.4).l'he pntentiometer car1 he adjusted continurn~sl!, such that a current can pass ~hroughthe cell in either direction. If the emF iar strictly zpcaking, the potential dilftrence, because I i q no longer exactly rero) c h i i ~ l gonly ~ ~ >ligi~tlqwhen the current changes Irom onc riirectlcln to thc othcr, the ccll
which is the desired rclation hetween A G atid E , and is thc central equ;~tir>n of the thrtmodynamics of clcctrochemical ce1l.c. As we will sec. cicctrovhe~niui~l nieasurerrlcnts can be used to dctcm~inetherrnt~dynamicdata of ionic rcactionq thai occur i11cells. l'he relation AG = -n F E tells us [ha11the emf of il ccll tlcpeltds upon l l ~ econceritrations. or lnvrc prcciaely, upon the activities, of tlic reactant.; and 11roluctsin tlic cell reactiorj, For the general chcmical ~rcacliondescribed by
A G is given by*(~quation12.60)
537
11-5. The Standard Emf uf an E I e c t r w h ~ ' m ~Cell ~ d l Can Be Fuurld br txtrdpolalior~
Here A G = ~',p; + u , l ~ ; - v,p; - v,&, where the p; are the chemical potenlials in some appropriate and conve~~ient slandard stale Now because A G = - n F E ,
and the overall reaction associated with the above ccll diagram is
The Nernst equation for thc cmf of the ccll i s Equatior~13 7 1s called the Nertts; pqlmtroil and shon7show the emf r>t a cell dcpcnds upun Lhe autw~tics,or the concentrations, of thc qpecies participating in the cell reaction. Thc quimtity. E ' , cilllcd the standard emf of the cell, is the emf of thc ccll when the ; I C ~ Iities L nf thr products itrid rcacttlnts are equal to unity
1 3-5. The Standard Ernf of a n Electrochemical Cell Can Be Found b y Extr.~polation Bcc:iu';e rc-e ran calculate thc value of I7 for arbitrary activities once a e know E' for n ccll. it is ir~~y)urtar~t I'or 11stu be able to determine the value of E' . The following calculation illustrates a standard procedure for daing this. Consider thc cell pictured in Pipure 13.6. The cell diagram Cur this ccll is
l%ec;iuscAg(s) : I I I ~t\gCl(s) arc stdids, u-csel irAP :111d(I,,~,., et111;llto ~ u ~ i tiny l i ~ 1 1 1 : i tion 13.10. I r l addition. we regulate the prcssure of t11c hydrogen gas over the plaliuul~~ electnde such that u,,> = I. which aluounts to sctting its fugxity equal to unity. Fur a gas xuch as hydrugkn .nl ordirla1,y pressures. the fugacity is cswntially equal to the presxure. s o we can set the pressure equal to unity (one bar). The number of clec~rons transferred in the overall reaction as written above is one, so n = 1. in Fquatirjn 13.10. Finally, then, Equation 13.10 becomes
From Table 11.3, a+a- = u i and a , = y,tn, su Fquation 13. i 1 becomes Tile Ictt electrode i c a Ilydr~genelectrode, and the right electrode is a silver-silver chloride clcctrode. ,iccordit~gto our convention that oxidation occurs at the left electrode in the ccll diagram, the two elcctrode reactions are
In our discussirjl~of the Dehye-Iltick~Itheory, however, we saw that (Problem 1 1 4 7 )
A.g(s) f(>il covercd with AgCl(s)
for an aqueous solution of a 1-1 electrolyte such as HCl(aq) at 298.15 K. so we can write Equation 13.12 as
, FIGURE
13.0
:In rlrctrochctl~ic~al cell co~~sisting ol'a t~ydrogenelectrdc and n silver-silver chloride electrode.
T t ~ ccell diapra~nir)r Illis cell i s Pt(<)lH:(g)IHCl(q)lRgCl(s)lAg(~).
The replacement-of In y, by Ant':' is valid only in thc limit of- infini~edilu~ir>r~. so the procedul-c is to plot the left side c ~ Equation f 13. I 3 versus mu'?attd extrapolate to m = 0. The intercept, thcn, is E ' . Table 13.1 gives data for E versus m at 298.15 K for the cell pictured in Figure 13.6. The data in Table 13.1 are used to plot the left qidc of Equatiorl 13.13 against m'!' in Figure 13.7. The linear portion at small values of rn cxtrapt)lateq to E-' = 0.222 V.
13-5. Thr St.~r!ddrtl En11nt an E l e c l r ~ h e m ~ c Cell al
340 T A B L E 13.1
l'ht c~nf rer\u\ I I I at 298.15 K fur he cell whosc cell diagram is Ptt$)lH,(I har~~IiCl(trr~lAgC1(s)lAg(sjI
( , I ~ l5e I
1-ound by F x t r . i ~ ~ r r l a t i l ~ n
Table 13.1. Note l t ~ a the plot is less curved than the urlr in Figure 13.7, mak~ngrl~e required exIrapolation much easier. Thc plot shown in Figure 13.X i b cilllcd a Hirr,hcocl: plot and should be used [or an accuratc determination of thr value of E - .
I
E X A M P L E 13-4 Use the Iollt>wingdata to detern~lncthe v;~lue01 E' for thc cell who+c.ccll diagrani i,
S O L U 1- t 0X : Using our convent~onthat oxirlatirm takcs place ar the lcft electrc~lein a cell diagrdni, we write the two clecuode reaction\ a8
and
F I G U R E 13.7 A plrlt cjt E + ( 2 K I ' l k ' ) Inm ngalnst mi.'' at 1'38.15 K Tor the ccll whose cell d~agra~n is given by Fyua~iun13.8. The extl-apolation
tllc value of E
'.
The net wactlon is
illustrated i n Figure 13.h and or this plot to m + 0 yields
the +~andardclnf of the cell.
Although the plot in Figure 13.7 can bc extrapolated to m = 0 withrwi tnn much difficulty, we can obtain a more reliable method by using an extension of the DehyeHiiukel expression for In 6 to higher concer~tratinns. For example, we can use thc sen~ienipiricalexpres.cion given by Equal~ r j n1 1.58
Thisequiitinn is analoguus to Equatio1l13.9,so wccan~rseEquation13.13, u r k t t c r yet, Equation 13.1 6. The resulting plots using Equatiun 13.13 (plotting the lcft side against I;>)or Equation 13.1 h (plotting the left cide npainsl m l arc shown 111 F i ~ u r c13.8. Notice thal the Hitchcock plot 15 much easier 10extrapolate to m = 0, yielding a valuc. 01- E' = 0.0730 V.
where C: is an adjuh~ilbleparanw tcl: Equation 1 3.1 4 becomes
fol.
a 1-1 electrr,lyle such as HCl(nq). 11- w e substitute Equatiun 13.I5 into EquaK, wc ohlain
lion 13.12 at 298.15
Now if wc plot the leR side uf Equatiun 13.16 against ni, we can obtain the value of E" by extrapolating t c ~m = 0. Figure 13.8 shows such a plot [or the data givcn in
F I G U R E 13.8
The detcrminntiun of E by means of a Hitchcock plrlt for the data givcn in 'lhble 13. I
for a reaction in which the electnlns on each side cancel (see Plnblc~n13-57). I t we apply Equ~tion13.22 I(> the cell diagram giver1 by Equation 1.3.17. w e have
Note thal we have indicated that rcductit~nt;ikcs place at (he rinu electrode by wl.iting E , I Z I I ~(aq)/Zn(s) + 1 and that oxidatiun takes place at the hydrogenelectrode by writ ill: E 'I1l?(g)/HA(aq)J.By coupling other clcctrodcs with i i hydrogen electrode, we will obtain results similar to Equation 13.21. Fur example. if we use a AgCI(sbA?(s) electrode instead of a zinc elcctrodc, we would obtain F ~ G U R E13.9 T\\rj pltrlq
illustl-;]ling~ h cdctern~inatin~~ r)f the value of E - at 208.15 K f o ~!he cell whose crH diagr.n!nis PI(\) lH:(l har)lHRr(m)[AgBr(c)ws) from the data given i n Exanlpte 1 3 4 (;)I Thc left side ill Equation 13.13 plotted ilgalnst m I;'. (b) The left side a f F~pation13.16 ~>lr)tted njiainsl m. Both plo~scan he extrapolated to 0.0730 V, but the extrapolation is more
it i s riot pcwsihle to tileasure the voltage of a singlc electrode: only the difference irj voltngc butwecn two electrodes can be ri~easuted.If, howcvcr. wc agree tc! choose a numer~c:ll value for the standard voltage of some particular electrode, then wc can assign standard voltages tn single electrodes If we takc E" for the hydrogen eleurrode to hc zcm, we can write
13-6. We Can Assign Values of E" to Single Electrodes I'cmsider the cell whose ccll dii~grami s
and
The value of F for the corresponding ccll i s -0.723 Ihr electrode rcactiuns alc
V at Z(18.15 K. The equatiotis for
Hl(g) 4 2 ilA(aq) + 2 r -
(uxidaticln at the left clectrode)
(13.18)
~ n ' .( n q )
(reduction at the right electrode)
(1 3.19)
2 P-
-
Zn(.s)
Brcai~aeE - implies that a11 the reactants involved are in t h e ~ standard r states, wr could write Equdlion 13.27 more explicitly as
n ~ the d equatior~for the overall reaction is
If we let ArxI;-'and Arc,,(;' he the standud Gibbs energy change fur Equations 13.18 and 13.19. then A r c ' for Equation 13.20 is
Because AG' and E d are directly related by Equation 13.7. we can use Equatian 13.21 to write
E:ell = E,,
and cotltinue the proccqs by ctlupling other electrodes with a hydrogen clectrode. Thus, we will fortnally write
+ EFt,
to emphasize thal H,(g) must be at ini it fugacity (which i s essentially equivalent to P = 1 bar) and that the stmng acid corresponding to H ' (aql [fur example, HCl(aq)] must he at unit activity. Equations 19.25 and 13.26 are exan~plesof the standard reduction potentials 1ha1 we can axhign to?lectrode reactions ('ljble 13.2). We can use the entries in Table 13.2 to calculate the value of E' for a cell whose cell reaction is the sum of the two electrotle reactiotis. According to Equation 19.22
ulicre reduction rakcs place at the nght elcctlade and oxidailon t;~kciplace at the lett c1ectrt)dc We can emphasize this r~ght-lcftcunventfon hy writing Equation 13 29 as
TABLE
13.2
Standard reductinn poterr~ialsin watcr a1 2 j C. Elccll-c>dereaction
But, fi,r a pnnictilar electrode-rraulion cquatiorl, oxidatic~nis thc reverse 01' wductioi~,
K i a q j + L,-
- --tK(s)
~ a (oq) "
t,-
+2
+
SO
I?&
.--+ Cays)
--
hs+(aql r' + Na(s) E8>k,l
Llsirlg lhe Fact that
=d ; ' -
A ~ ' ~ (+a3~c i- . i\l(s~ 2 H;O!ll T 2 e - -- H,(g) r 2 OH (aq)
I
E:,= - El;,, Equation 13.30 can be written as
20'' (aq! A 2 e Zn(s) A g I S ( s , - 2 ~ -+?A@(sj+S2-(aq) FeL'(aq~+ 2 e -+ Fefs)
Tl~usthe emf rjl' the cell is the standard reduction poterltiat of the right elcct~udein the cell diagram m i n u ~thc standard reduclion potentiill of the left electrode. Consider the cell whose cell rliagr~rni s
C? [q) 7 P - -.+ CrZf(aq) ~ : d " (aq) -1 ? c -
Phil(s)
+ 2 L.
PhSO_(s) t 2 rPbBr,(sj 2 e
+
The overall cell reiiction cquntio~li s 2 Ag(.,
$
CLI" ( 1 00 rn)
+ 1 Ag' (0.100 nil + C:u(s)
(13.32)
a ttd
C'U"(~,)
+ 2 c == Cu(s)
Wc. bee from T ~ h l r13 2 that
Ag+jaq)
+ r = Ag(s) +
C L ~ (xi) " 2 e-
+ l'u(s)
E = +0.799 V
E' = t0.337 V
Pb(hj SO:-(aq) 2 Br-(aql Pb(s)
-0.364
-03583 -0 274 -{1.2hb
-0.255
Nits)
-0.30 -0.15 I
4
~ n ' iaq) ' T 2 r- --4 Sn[s) ~ h "( a q ~I- 2 u- - + I%($) Lig,ll(kj t- 2 e -+ 2 Hg(1) t 2 1-(ay) FZ' (aql + 3 1,- -+ I;c(sl 1 H - { ; q ) i 2 e . -+ H,(g) AgBr(5) t c --• 4 g ( s ) t B r - ( a q )
-0.03b O.I)(lO i0.0732
HglRrlis) $ 2 e -* 2 I Ig(1) f 2 Ur-(atll S~i'+(arl) 2 P ~n'l (aq)
+(I. t 396 +O I5
+ i\gCl!?i) + u- + A&(" + Cl-(atl) IlglC:II(>~ + 2 -+ 2 Hgll) + 2 CI-cay) Cu' ' iaql + 2 rCurs)
---P-
--
l>(sl 2 c
2 1- (aq) i l g l S O - ( > ~ 2 e --,2 Iig(1) + Sui-(aq)
Notc tha~the Ag(s)lAgl (aq) half-cell emf- is rror multiplied by 2. Thc cnlf of an electmche~nicalcell is independent of the size of the cell: emf i h an inre~lril,r,property. (See also Prohletns 13-54 thmugh 13-58.) The cmf of the cell wtiu.se cell rcuction is described hy F.quation 13.32 i s given by the N z ~ n s cql~nrion t ( ~ ~ u a t i o 3.7) n * ~with rr = 2:
-0.440 -0.408 -0.403
V?-(aq)
--.A & ( s ) + l (aq)
Agl:sb k e
and the two electnde rzaclions arc
+ Ph(s) + + -+ 2 C1 (nq) + Pb(sl 1 1-(aq)
-4
-
PhU12~.h) 1 2 L.v'I {aq)+ tr~ i : " i a -~2) P-
Cd(s)
-4
--•
-2 '125
-2.866 2.711 - 1-66 - 0 828 -0.763 - 0.7051
-
A g L S 0 4 ~ s+ lZ F- --+
---
~ r " f o r ~ , + r .---
Ilgjiaql
Fez-(aq)
+ 3 r-
Ag' i a q ) -I s
~ d ? (:q) ' -h? r
2 Ag(s) t SO:-{uq)
2 Hi(])
Ag(s)
Pd(s)
+
C)ZIg) 1 3 Ii+iaqj 4 e - -+ 2 H,0(1) Cl,igl t 2 e --,2 C1-paq)
C O " ( J ~ I+ r - + ~ u ~ ' ( a ~ )
-0.1.lh 0.126 -0.0405
+O.222,1 +0.2hX f0.337 4-0.53 15
+O.hIS5 +O.b53 +0.771 f 0.7% +-0.799 f 0.9K7
+ l.22q I 1 361
+ 1. X 1
Chaptcr 1 1 /
Therrnndvnarrlir~o i Ell~rlrochcnlir~l Cclls
:
Settillg the activities (>I.~ g ( s )and Cu(s) equal to unity and using the relati(lns I - y:ml ::y:4m' (Table 11.3)- w e =a. U , k p ~ l ~ =, N : = V I M ; = y f ~ n ' ando pet 7
13-7.
Cleclrnrllrrr~~~al Cells Tan RP l l ~ r kd1 DCterm~l~~ At tivilj
Cwfii~ient~
We set the activity of the Cu(s) and Zn(s) q i ~ a to l u~~ity and write
According to Tahlc 1 1 3, a,n,ll = y:m; = 4y:m' and u,,,,,, = y:m: -- 4y:rn'. Using the fact that y, = 0.396 fnr %11C1?(0.500 111) anti y- = 0.-119 Cur TuC'1,t I .MI In). the Nemrt equntion gives
Givcn that y , , , ,
= 0,721 at 13.100m and that y,,u,+
j3,1
= 0.456 at 1.00 m,wc have
.I.l~enegative sign iridiciiles that Equation 13.32 will be spontaneous fmm right to left ti~ldcrttie given conditions.
f X A M P L E 13-5 Use tic data in Table I ?.2to calculate thc value of the emf of a ccll whoke ccll diagram
We should point out n lirnilaticln of Equation 13.31 M o r e we go o n . It trlrns out thai 1ht.r~is always a dilferencc in electrical pjtential acrosq the interfacc of two solutions unless they have an idelltical composition. This poten~iali s called a liquid junclion porrmiul. Fr~rexample, consider the salt bridge connecting the two solutions in Figure 13.1. There is a liquid junction potential at each salt bridge-solution interface, so Q u a ~ i o n13.31 in this case shr>uldbc written as
and
where E l ' , is the sutn of (he two liquid jitnclion potentials. The magnitude of the liquid junction potential causcd by n salt bridge depcnds upon the rclativc itiobili~ies of the cation and anion that uanstitute the salt hridgc. Potassium chlrjridc is used because a pvtassiutn ion and a chloride ion are about equally ttinhile in aqtreous solution. su the toti11 liquid junction potential due tc~a potassium chloride s:~ltbridge is quite small (around a millivolt or less) and can be neglected. Generally, in designing electroche~l~ical cells with a liquid juncticln, we must take care to rnini~riizethe liquid junction potential. We will discuss liquid junction potentials in Chapter 14.
,uldthe ccll reaction i\ given by
13-7. Electrochemical Cells Can Be Used to Determine
Zn(s)IZ11C1,(0.500m)llC'i1C12(1.00 m)lCu(s) Take y,-
=
0.396 for 0.50Um 2nC$(aq) and y, = 0.419 for 1.00 nr CuCI,Iaq)
5 0 1 I! T I O N . The two clectrodc reactions are
Zn(s)
7.rl(s)
~n"(0.500m) + 2 @ -
4
+ Cucl2(1.I)() m) -
We first caiculatc {he value rlr
(left)
Activity Coefficients -t
Cu(s) -tZnCI,(0.500
from Equation 13.31,
and then uw lllc Nernst cquatirjn with n
= 2 (Equntio~l13 7)
m) In Chapter I I , we learned that activity coefficients can be deterttlined from vapor pressure measurements and from lreezing point measurements. One of the muqt convenient and accurate mcthods for determining the activity cocficients of solutio~xof electrolytes involves electroche~rlicalvelIs. I ~ t ' sgo back to Equation 13.13
which we derived ti>r a cell whose cell diagram i s Pt(s)JH,( I atn~)IHCl(aq)IAgCl(s)lAg(s). We detennincd the value OC E ' to be 0.222 V in Section 1.3-3 by plotting the lcft side of Equaticln 13.13 against nl"' and extrapolating
Chapter 1 3 !Thermuriyn,~m~r s oi tlpc trlr~lien~ic.al Cells
t o In = 0. Pirlw tha~we k n o ~the value of L ' , either by extrapolation ut B versus nr data or, rnorz conveniently, from a table o f standard reduc~ionpotentials, we can use the sainc E verrus nl data (Tablc 13.1) and Equation 13.12 tn dererrnine y , venub tn fur HI:l(aq) Sulving Equatiun 13.12 for In yL and using the first sct of data ~n Tiiblc 13.1. we get
13-7. Electmchemic~lrells Cdn Be Us~d to Det~rnmineAr tivitv Crrt.lhcicnt3
5Cl L 1J T I O R : The rzactiw at tllc Icft elec~rtale(oxirlarion) is
arid thal at the righl electrode (reductrot\) i:,
The equatioll for the net ccll reactlon i c
{ H1(Ihar) + AgRr(x)
-
1-1' (aq)
+ BrC(aq) t Ag(s)
The corresponding Nerrlkt equalion i s 01- I/+ = 0.738 a1 in = 0 . 1 2 3 8 ~rirh-kg-'.'lhe other values obtained from the data in Tablc 13.1 are plotted In Figure 13.10.
?
7
If w e set u , ~d.l q R R , . and aH? e 4ual to u n i t arid uce [he rrl;~tiunsr ~ , . uBr = t r : = y;rn(Tablc 1 i .3), w e obpain
Solving this equation for In y, glves
For the tirst set of data i n thc above table.
13.10 'I llc rucan loriic ncctivity coefficient of HCl(aq) ohluinrd lrmn Equatlutl 13.12 and the data in Tahlc 13.1 p1ut1t.d agalnkt ~nIi'. FIGURE
E X A M P L E 13-6 Show 1ha1thc Nernct cquation for a cell whose cell diagram i s Pi (s) lH,(I bar]lHBr(aq)IAgBr(s)lAg(s) can be written as
or y+ = 0.970. The other datlr are plutted in Figurc 13.1 1
Buth calculittions done in this aec~ionhave invcllvcd 1-1 electml yies. Lei's now du an cxample ir~vol\:ingZnCI,(aq), a 2-1 electr.olyte. Corisider LI cc.11 whoac icH diagram i s
The reaction at (he lefi electrode (oxidatiotl) is where y-rn 1s thc activity ofHhr(aq). Use thr. 1-esk111 uf Example 13-3 and Ihe Ir)llowing cell data ru calculate and plot y- versus , ~ r ' : ~ . A
and thal at (he right clcctrode (reduction) i s
13-8. Elcrtruchcm~c~~l Mca,~rrenler~tsCon Rr I ! ? ~ rIn l Determ~neValucs ui A r H and A,S ui Cell Rr.actlun5
The following values of E at 298.15 K versus rn can he used t o calculate the value LIT y, as a function of nl (Probleni 13-14):
'These results are plotted in Figure 13.12
FIGURE
13.11
']'he mean ionic iwkivity coeriicient of HUr(aq) plotted arainst m i i 2 .The value< rlf y-- are
obtained accordin2 lo Example 13-5.
The equation for tllc overall cell reactiorl can he written as Zn(s) t- 2 AgCl(sl - +
ZIT"!^^) f 2 C1 (aq) + 2 Ap(s)
F I G U R E 13.12 The mean iioric activity coefficient of ZnCl, (aq) plotted against m'.".
Recause the activities of the three solid phases can be set q ~ l a to l unity, the emf of the cell is given by
13-8. Electrochemical Measurements Can Be Used to Determine Values of A,H and A,S of Cell Reactions The 2 in the denclminator of the 111+~:+a:,- term results because tlvvr, electrons are trrlnsferred in the cell reaction. as we have written it. Using Table 11.3, we write ~,~~+rr= : , 4m3y:, so Equation 13.33 bccomes
Equation 13.6 serves as the bridge hetween thermndynamics and elecmxhemis~ry.We can readily express A H and A S in tertns of E by using Fquation 8 . 3 t a
and then the rclatirln A G = A H G ~ b e nE versus m dala at cuffiuientlj !ow conccnlratjons, we ccli~ldplot E f ( R T / 2 F ) ln4m3 versus m 1 and then extrapolate to m 4 O to determine E", as we did in Section 13-5 for thr cell whose cell diagram is given by Fquaticln 13.8. Altematmely, we can use the data in Tdblc 13.2 lo ohtain
-
7'AS to obtain
We can usc Fguations 13.115 and 13.36 10 calculate the values of A, H and ArS f ( ~ any r cell reaction. Fur example, the temperature dependence of E' for the cell Pt(s)lH,(g) IHCl(aqjlAgCl(s) IAg(s)
is given by
Solving Equation 13.34for 111 y, gives
55 1
1 1--9. Sulubilr~vPrtrrlurts Can Be
from 0 C to 50'0'. The equation for the cell reaurion is
thc value or 4, S
Using Equation 13.35, we have
U e ~ r r m ~ n LVIII~ c d kl~ctruchcmical( ell,
is
=
(2){96485 C.rno1-~)(1.45x lo-'' V - K
= 28.0 J . K - -111ul
I)
'
and the valac of A r H ' is
a1 75 U. arid using Equa~ion13.36. wc have
13-9. Solubility Products Can Be Determined with Electrochcrnical Cells Because AC; = - I I F L : = 0 a1 equilib~.iuni,we can use Equdlion 13.7 tu derive. ;I rrlatlon between the standard errll' and the thernicxlyna~nlcequilibrluln conatant 01 llie equallcln assoc~arzdwrfh the cell reactlcln Setring E = 0 in kqu,~~ioii13.7 giieh
E X A M P L E 13-7 l'lic cell whohc cull di;lgl-iiin is
hats a stalldart1 erni (11-0.5359V i ~ n da lerr\pcrature cr,cffuient uf 1.35 x 10 ' V K ?'>K. IS K. Ucterlninc the cell rcaction and the vulucs rlf A,(; A,H' and A, S .
.
al
.4n important upplicstion c ~ fthis equation I S ru the determir~atior~01 the solubili~y yrodl~ctso f sparingly soluhlc salts. .As a collcrre exarnplc, let's consider thc detrrminatiu~~ of the soluh~lity proclucr ol' AgCl[s). 'She t~lualiul~ far the dissolution of Agl?l(s) is
.
i~ndthat at the right cluctrodc (reduction1 i s
+
H ~ ~ C l ~ i 2s r.)
;IIIJ the cell reaction
2 Hg(1)
+ 2 CI
(aq)
IE
Ph(s) Thr value elf
P
+ Hg:Cl,(s)
4
Wc :an const]-uct ~ I ulcctrochernical I cell uhuse net ccll cqualion is E~lr~;irior~ 13.38 by i~singaAg(s)(AgCltsllCl (aq) electrode and a Ag(s) 1 Ag ( a q )elec~roclc.Thc reduclion elezlrtdc-rzaction eqtlatiolls corresponding to thebe IWO clcctrodes ~ I I - t .
'
PtK1 ,(F) -r 1 IIg(1)
w
i.i
A K G= , -,IF E' = -(2)(96 485 C>mn1-~)(0.5359V) =
-103.4 k j rnt,l-l
and Ap+(aq)
+E
-+ Ag(.c)
13-9. Snlul>ililyI'ruductr Tan Be Uctcrnm~ned~virhFlr~rrnchemiralCells
To obtain the equation for the dissolution of AgCl(s), wc subtract Equatio~l13.41 from Eqi~ation13.40. Therefore, the left eIectrode (the electrode at which oxidation takes pl:ice) it1 the cell diagntn i s lhe Ag(s) IAg- (aq) electrode. and the cell diagram is
555
The preceding two calculations of the solubility product uf a sparingly soluble s a l t illustrate the general method. Let the sparingly soluble salt l x My4A,_(S I , su that the cquation for its dissolution is M,, A , (s) I
.
+ 11, M'+ (nq) + u..
AL-(aq)
(13.42)
Notice !hat this cell ha.: no liquid jurlctin~~ and herlcc has no liquid-junction potetitial. 1-hevalue of E at 298.15 K for h i s ccll i.s given by
and its solubiIity-prod~~ct expressiorl is
According to Eqi~ation13.37.
We can conwuct an electrochemical cell whose [let cell equatioti is Equat~on13.42 hy using an M(s) lM, A,, (%)/A'-(aq) electrodc and at) M(s) IW- (aq) electrode. The net chemical equation+of the reactiutl wuurrlng in a cell w ~ t hthe cell diagram
:II
298. I5 K.Thc ~tnndardstate hcrc is a orle-molal solutiun.
is Equation 13.42, so the valuc of E' of this ccll will give the sofubilily pmduct of
M,, A"- (sJ 1
E X A M P L E 13-8 U ~ the C data in T~hlc13.2 to calcula~etllc value r)f thc suluhilitj~producl ot' Pbl,(s).
S(31 (!TI O \ . C'sing the above calculation as a guide, we use a Pb(s)lPhl,(~)II-(aq) electnxle and a ~ b ( s l l ~ b ~ +electrode, (a~) whose curresponding reduction electrode-
rcaction equations arc
E X A M P L E 13-9 Dcvise an electrochcmica1 cell (without a liquid junction) that can ht used to d c t e m i ~ ~ e the solubility product o i Hg,Cl,(s). I!se lhe data in Table 13.2 to cdculale the valuc r)l- lhc solubility prcduct uf Hg,UI,(s) at 298.15 K .
5 0 1 CJ T I U N : T t ~ eequation for the rli\uulrition ol- Hg,Cl, (s) is
Tho electrode rcactiorls that give this n e l equation are To obtai~~ thz cquatinn Ibr tile diusolu~ionof PhIz(s), we subtract Equalitln 2 from E u u a t l o ~I ~ . Therefore, h e lc,ft electrr)de(the one at which vxirlatio~~ takes placc) is the PhlsjlPb" [aq) eltctrode. and the cell diagram i\
'Ihe valuc of E -~IL 298.15 K for this cell is E
= b,',
E;
=
-0.364 V
-
(-0.126 V) = -0.238 V
Hg,CI,(s)
+ 2 e-
2 Hg(l1
-4
f 2
CI (aq)
and
We place the oxidation electrodc reaction on thc 1el1 and the reduction electrodc reaction on the right, and so wc ha\e the cell diagram
* Therefore, accardiny rn Equation 13.37 According to Tahlc 13.2,
Ch,~l)trr1 1
1
1 h~rmodyndrnicluf Elec
~rr~rlien~ical Cells
1 I - 1 U. The Uisswi.~tion(:unstantr vf Wcak ,4crds Can Be Determined vvirh
If we solve Equation 13.46 for tioti 13.45, we ub~ain
The solubilily prduct is given by Equatloll 13.34with t! = 2
CI~crrncl~~~l~ical Tells
u,,, = mH.y H and suhstitutc this rcsult into
Equn-
We clin remdrlge this q u a l i o n into the hlnn
13-1 0. The Dissociation Constants of Weak Acids Can Be Determined with Electrochemical Cells
where
Co~isidcrIhe electrcxhcmical cell
where IiA is a weak acid and NuA is its sodium s a l ~'File . equntinll.; t i ~ the r electrode reaclions are
AgCI(s) -tt, + Ag(sj t-CI ( n q )
(right)
and the equation I i j r (he net ccll rcaclion is
The procedure is 10 vary rn,,,, m,, , and m,. and to evaluatc thc left sidc o f Equ;jtion 13.47 as il function of ionic strenglh
This procedure givcs In K i as a function of lorlic strength. We now plor ( f ' I K 7 ) ( E E") ln(m,,,nr,, / m A ) against ionic strength, and thc extrapolation to zcro ionic stretigth gives 111 Ku because yl,,,y,, /y,, + 1 as I,,,+ 0. F~gurt:13.13 shows such a plot using the follouitig data for a c e t ~ cacid [HAc(aq)J,IL 298.15 K:
+
Ciiveli t h a ~the fugacily (rsse~~tially the pressure) of thc H,(g) is orit. bar. lhc cmf uf h i s crll is given by
= 0.2224 V i s thc standi~rdreduction pr,tential of thc silver-sil\er chloride electrode. Using the rclatic~no,,.ci,,. = y,. n l , , , y,, m,,-(Table 1 1.3)-we have
whcre E'
Because acetic isid is a fairly weak acid, we have neglected the a111i1llcontributiol~11ial H (aq) and Ac ( i ~ qmake ) to Im in Figure 13.13. The extrapolation of the best linear tit to the data ploued in Figure 13.13 gives - In K,,= 10.958, or K', = 1.74 x 10- '. If the H' laq) and Ac (aq) frorn the dissocintior~of HAc(aq) arc taken into nccourll, you o h t ; ~ i rK,( ~ ='I .75 x 10 ', in agrccmcnt wit11 thc ":rccptcd" vi~lucof h',' o n rnolality scale. PI-ohlem 13-41 treats the acid dissociation of- f-ormic acid. whic.11 1s about ten limes stronger 111arlace~icacid. In this case the H'(aq) 2nd .4c-[iiq) i'rorn ttlz dissociation of thc for~nicacid cannor bc ncglec~etl.
557
13-1 1. \,Ve Can Arslp Th~r~norlvnam~c Vd1ut.s In InOividual lun5 i r ~Sr~lu~ion
where T = 298.15 K , E' = 0.2224 V, and nr = I I I , , = mN3, = t r z T h e irlnlu strength is equal to Im= m,*, im = hi.Figure 13.14 i s o plot of - In Ki agnit~stCz.f5xlrapolutir)r1of a linear curre lil gives all intercepl of 11.227, or Kc,= 1.33 x 10-'.
Ionic strength I r n o l , k g - l F I C U R F 13.13 A plot 01.thc [eft side
of Equatiot) 13.47 wilh E' = 0.2224 V against ionic strength using the supplied data for the cell R(s)lH,(l bar)lHA(m,), NaA(m,). NrlCl(rn,)lAgCl(c)lAg(s).The e~tm~ulalion ol the hest lincm fit to thefc data gives - In KO = 10.958, or Kc,= 1.73 % 10 for t~icticacid.
' Iunic strcngth 1 rn01.k~-I
E X A M P L E 13-10 The value of KO fnr propat~uic acid can be dcternjined using the cell whose cell dii~pran~ iu
F I G U R E 13.14 - In K,13 given by the exprussiull in Example 11-10 for prupanoic acid, against i t ~ l i c stwngth for the cell WE)IH2(p)lHP(aq).N W a q ) , N~rlPaq)lAgCl(s)lAg(s). A lincm f i t g i w s an intercept of I 1.227, or K,,= 1.33 x lW5.
A plot of
IJw the inllnwlng data tu dctennine the value of KO at 225'C.
13-1 1 . We Can Assign Thermodynamic Values to Individual Ions in Solution The thermodynamic properties of singk ionic species are tiot measureable quantities, but we can still set up a table of single-ion therlnodynamic properties in the fnllawing manner. Consider the reaclion Z n ( h ) t.2 H ' (aq) + zn2' (nq)
+ HZ(g)
The standard emf for the associatcd elec~rochetnicalcell is 0.763 V at 298.1 5 K. I f all the species arc in their standard states
S O L ( , T I O N . We plot ~ h cleh side of Equation 13.47 against ionic strength and eukrapulatc to /.cru ionic ctwngth. Neglecting [he small concentriltions 01 H' (ilq) and P-(aql due to the rlissocialioo o f HP(:lql, the expression for - 111 Kc:is
Tile standard Gibbs energies of formation of I$(g) and Zn(s) are equal to 7cro by convention. and so wc can write
13-1 1 . We Cat1 Assign Therrr~udyna~nic V ~ l u c stn Ind~rldualIons in hulutlon
We can continue this pruccss and determir~evalues of A , F for various ions relative to h,C;-ltI+(aq)].For exalnple, consider (he I-eaction
5hl
These results suggest that we choose to set S'[Ht(aq)l = 0 ai 79K.15 K, md wc wriit:
from Equation 13.54 and
The blandnrd einffr)r the associated electrocherriical cell is 1.360 V at 298.15 K. Thus, we can wnle
from Equation 13.55. We can calculate A, H - in each case frun~the equatiun
For Equation 13.4'1, wc have Notice that the combitladun of znL+(aq)and H (aq) is rlectncally neu~ral(zn2+ 211 ) in Equation 13.50 arid [tiat the cornb~nat~on of Cl (aq) and H+(aq) is electrically neutral (CI- H + ) in Equi111on 11.52. This resi~ltalong with tnany other sini~larresults shows that H+(aq) %ill cancel fr)r any neutral salts. Consequently, we art. 211 I~hertjtu sct t112 value n t ' h , G ~ I I ~ + ( a qarl>~rrarily )l and we choose to h e 1 A,F'IllL(aq)] = O at 298. I5 K. Accepting this convention, we call wrllc
+
from which we ohtain A, H '[2n2+(aq)]
- 2A, H [H' (aq)] = - 154 k l . i i ~ c d
and fork.qudtior~13.51, wc have A - H - = -1h7 1;J.niol-' or frcrin Equation 13.50 and
We set A,H [HI (ar])] = 0 at 298.15 K. and s o we have fro111Equation 13.52. W e can detcrmi~~c A , H ' and AfS' liom lhz temperilturc variation of E u for the appropriate cell. For the cell aascxi~te(1with Equation 13 49, ( i l k - / i ) T ) , = -0 119inV.K I , s o u t ' c a n ante
and
The standard molar. entropies of the neutral species H,(g) arld L n ( s ) are 130.08 J . K .moi -' arid 41.h3 J . K - ' .mol I. rcspectively, at 298.15 K. st] Fquation 13.53 hecomes
Values of the thertnodynamic propertics of single ions are tiihulatcd in a numbcr of placcs, but one particularly valuable source is n joint Amcricarl Chemical Society and American Institute of Physics publication [Jor~rnolc!fPlljsir.rzI rrvd Chv'hrrnil.rr1RP/>T~,IIC.P Daru, vol. 11, suppl.2 (1982)], callcd the NBS Tables uj C./rc+~rr;.rllI'/~lr.rrl!o(i~"rm,nic Pn~pr,rti~s. Srlrutr.rl Vulucs{ur Inoqnnir. rind C', and C2Olgrrnir Sr t b b 1rmrr.5 it1 .S1L'i~irs. Figilrt 13.15 sl~oasa typical page of these tables. Unlike the JAKAF tahlc., (Scutiorl 1 2(I).which give data over a wide range of tcIrlperaiurzs for ahour IS00 suhs~anccs.the ahovc tables givc data only at 298.15 K, hut for several thousanrl s~~bstn~lues. LVc car1 usc the data in Figure 13.15 along with A,GLIC1-(aq)]= - 13 1.228 kl.rnr,l (frorn page 2--47 of the tables) to calculate the value of the sofithility product t ~ TICl(s) f at 298.15 K. The equation for the dissolution of TICl(s) is
'
'
The uorrespunding rcsult for Equation 13.5 1 lurns uut to be
TICl(s)
-
Tl' (aq)
f
Cl (aqj
13-1 1 , We Can Arsigl) Thcrmdvnamic Lalucs tu Indlvirlua I Ion, in Solution
and so
and so
I
TI' ' TI ' TI"
1 I,{) T>rl>a 1
E X A M P L E 13-11
Lis~rlg ttle \slues of A,G'[Ag (nqbj = 77 107 kJ.~nol-', A,C'[Br (aq)J = Oh kJ 11io1', arid A,G [ApUr(s)J = 96 40 !d. r n o l - l at 298.15 K. calculait. the value of Kqp01 AgRrO) at 298.15 K - 10'4
SO1 11 1 I0 N: Thc equation for the dissoluiion of AgBr(s) 1s
AgBr(s)
4
A g ' ( q ) 4,Rr- (aq)
and
and K I>
I
TII:l TIi.1" TICI,' I lN:l, TII'I.
-
= e - l , C - / n T - eup
-70.M7 x -
10' J,rnul
'
~ X . J IJ.K-I J ~ .t ~ l ~ ~ - ~ ) ( 2 9 s . i %
We can i11st1use tabulated values of Gibbs energics of fonr~ationof ions to calculate acid-dissocia~ionconstants. For examplc. the NB.7 Toh1p.r c!fC'lae??~iral Thrrrnorl~~im~~rir Properries give
TICI,
and
11:1:19 TICIO. 1 111
F I G U R E 13.15
I'ngc2 135 fioril t l ~ cNRS Table.; uf Chernicol 'rhern~odldytlamic Propertics IJ. Phy.~.Uhtm. Rrf 1)rrtrr.
vol. I I . \11ppl,2( I t ) K 2 j ]
at 298.15
K. Tlie equatic~nfor the dissociation of fonnic acid car1 be written an
Ch,i~>lrr1 1 i Thermodyndmirh oi
klrrlmrhem~calCells
13-12. Batteries dnd Fuel Cells Are Devices That Use Chemical Reactions to Produce Electric Currcnts
and .;(I
K = p-lrl;
!RT
= exp
[
-
21.3 x LO' ~ . m o l (8,3145 i.mol-I K-')(298.15
KT]
E X A M P L E 13-12 Given the tollowing d;~lafrr,rr~ the IVBS 7itl)lesc!f'Thr*~rtic.rrl Thrrmnd~nnnzicPrryr~*rrics, c;llculak thr: value o f rhe prrltonation cullstunl nr an~intl~rlethane at 2'18.15 K.
Batteries are devices that are cunlmonly used to provide energy for numerous ilppl~cijtiuns. Batteries are classiticd as primary if they are not rechargeable and as secundary if thcy arc chargeable. Examples of primary hawries ai-e alkaline hattcrics, commo~il y used in flashlights and toys; lithium hatterics, uscd in wristwatchrs; ~IIILIIlierclIry l>.~iteriea, used in heart paccrnaken, hcaring aids, and computerh. all of which requil-e ~ h s constunt vultagc this type o f battery pluvides. txamplea r>fsecondiryhattcrics arc lsad storage batteries used in automobilzs, and nickel-cadmium (NiCad) battcrics used ill co~dlesstools, shavers, toothbrushes, and many other device';. The cell diagrani of a mercury battery can be descrihcd by
with a sirnplitied version of lhe v t ~ r r e s p c ~ i d itell ~ ~ grc:~ctinnbeing given by
Norive that all the sp'cies in the abovc ccll rcacrion are iri cunde~isedphases. Constquenlly. there is no change in reaction concentration during discharge. which nleatls tttat the voltage (1.35 V) stays constint during discharge. This ctlnstiailt volragc, along with a long shclf lifc. is unc of thc imyurtiiot distinguishing features of a mercury battc~y. Its low atunliz miIss and slrong reducing polt.ntial givc lirhium many advantagcx iix a rttaterial lo use in batteries. For a given size, lithiu~nbilttcrica have the highest powc~,to-mass ratio of all batteries. They can also be designed t o t ~ a v elong dl-;tin r a b dnd lung lifelimes, excellent characteris~icsti)r an application such as cartliac pilcc~nakcrs arid wrislwatchcs. A lithium battery can power a wristti~;itchfor marc t l l ; ~five ~ ~ YCII[.S. The classic exarnple of a secortdary battery is a lead rlcrraye hat~cry,wt~oszccl t diagram is
5 13 1 U 'I I Cj h : The equarir)~~ tor Ihc protolraritln uf CH ,NH2(aq) is
where PbO,(s). PbSO,(s) denotes a hc~en~grrieous mixlure ot thc two stllids. U'hcn fully charged. this cell pruduces about ~ w uuolts. The 12-V hi~tteriesconin~only11acd in autt~~nobilcs coli\ist of six CCIIS in scr ics. Thc overall ccII rci~ctionupon dihchalp is
= c-L
0 ,Kl
--
--
exp
[
14.26 x 10' J mu1 --
'
I
~ s . ~ 1 a s r . m o i - ~ . ~ - ~ j ~ ~ ~ 8 . 1 ~ ~ - ~ j
During discharge, leu1 sulhte is funned at both clcctrudcs and si~lluricacid is used up. In the process of k i n g charged, the above reaction is reversed.
Chaptcr 1 1 1 Thermodynamics u;
13-12
tlpt t r r ~ c h r n i c a Lclls l
ll,{g) i 2 OH (aq) -+ 7- H20(1)
+2 t-
(13.56)
'i'he OH-(aq) i < cupplied by the concentrated NaOH(aq) or KOH(aq) si~ll~tion bathing 111~two electrodes. The oxygen is reduced at the othcr electrode according to
567
Note that the overall reaction is the combustion of hydrogen and oxygen to f u m wnter
Anuthercomniunly used rechargeable battery is a nickel-cadmium inicad) battcry. A sealed nickcl-cadmiurn battery is more stahle than a lead storage hattery and can he left jn;lclive for long pr.dods. The rwerall charge-discharge reaction can hc written as
In alrr~ostall hattcries. a lnetal in the form c~fan electrode is oxidized by an oxidizing iigcnt. Although batteries have the uor~venienceof being portablc and self-contained, they arc not huitahle for very large scale energy production. A promising il1tcrnatii;e IS a file,] ccll, which is an electrrxhemical device that utilizes the oxidation of a file1 t o produce electricity. Fucl cells d~CCerfrom batteries in thal the tucl and uxidizcr art: fcd rorltinr~ouclyinto the cell, so that its operating life i<essentially unlimited. The H: ( ; ) - 0 2 ( g )tuel cell, which has been a power source on many manned space flights, is show11schematically in Figure 13.16. Hydrogen is fed into a porous, ~ne~alljc (often ~ i i c b e l )clcctrode, whecc iil i s clxidizrd to fomm wnter acct~rdingt c ~
Rattcrics ,\rid f u ~Cells l Are Lle~,ictlsThat Use C h ~ m ~ rRcaclinrlr al in Prnducc t lrclrir Cu-rents
The currerlt tlltouph the cell electrolyte is cnrricd by the Na- (sq) and OH-.( a q ) , s t ) h~gli cr~nuentrationsare used to favm a low intcrnrtl cell resistance. The mqcu appeal ol~fuelcells is thar they are not limited by ihc inherent cfticicncy o f heat engincs. Recall from Section 6-7 that the maximum ei'liciency that we could obtdin if we burned hydrogcn tu ruri a heat cngine is givcn by (Equation 6 - 3 5 )
<
where 7;, is the tenlperature of the hot rcservior and is the telnperalure of thc cold reservoir. Thus, if we were lo run the rcac~ionin a hcal eryine at YO0 K wit11 a low-temperature rcservior at 300 K , ihe efficicncy u,ould bc
maximunl efficiency =
900K-3OK 900 K
17
- -
3
or 67%. Using the daba it1 Tablc 5.2, the maximum arnount of energy available i s (7/3)Arlf' = (2/31(2)(2X5.83 kT.mol-' = 381.07 kJ i ~ t w o m o l e s ~ ~ f t ~are , ( gburned ) in one molc of O,(g) under standard cnndilions. Thc value of E' for a half cell wllosz reaction is described by Equatio~~ 13.56 is E = U.K277 V, and that for a half cell whose rcaction is described by Equation 13.57 is 0.401 V, and so P for the ccll i.; 1.229 V. For a cell operating reversibly under standard conditions at cotwant T and P, the maximum work that can he dune by the cell is - I I ) ~ ~ which ~ , i s equal to A r G ' .
and so
Hz(€) el~lry
*'
-
I
f
021g) entry
I'
J
Electrolyte \..
. Wall
The a h v e calculalion is done fr~ra systcm under ideal condilions, but the message is clear. In practice, fuel cells can realize efficiencies closc to 90%. whcrcas heat etlgincs have efficicncirs up to abuui 30%.
Gas chamber
Porous elcctrude ,yT
,
E X A M P L E 13-13 Compare khe idcalizetl amount of work available when one mole of CH,(g) is burned in a fuel cell at 298 K and in a heat engine. with T, = HI0 K and Tc = 3W K.Calculate the value o f E for thic fuel cell. SOLU T I(> N; Thc cornhristion reaction of one molc of
methane i q given hy
F I G I J R E 13.16 cchcmar~cill~~rrratinu of a H , ( g ) - O : ( c )fuel cell. ?'he H,rg) is p;r?;sc,dover n porous electrode, wl~crcoxidalioll takes place according to Equalior~13.56. The recullir~gelectrons pas%through rhe uxtcrn;il circuit, and (lie oxidizcr [usjrrg O,(g)J is reduced according to Equation 13.57. The current is carried through thc cell hy 011 (aq).
In Equation 17.58, the oxidatirln Etate of the carbnrl atom gws firm -4 to +4. and that (11- each of the four oxygcn atoms goes from 0 to -22.Thus, n = X irr Equation 13-58.
1
Chapter 1 I / Thermodynamics ui Elcutrucbr.m~t~~l Lells
13-31. The temperalure dzpenderlce of thc standard crnf of a cell whobe cell d~agramIS Pt(sjIH,I 1 bar)]HCl(a = 1.00)1AgCl(s)~Ag(s) is given by
13-39. The value of the acid-dissuciation cvnclant of propanoic acid call be determined u4llp the cell
1132 the Iollr>wing data at ] ( P C to detrrnmine the vi~lueof K,, frlr propanoic acid at I0 C.
u h e r e Lhz tcrnpcrattlrz is in degrees Celsiuk. Write tht: as.cociated cell rcwtiun, and dztermule the slantlad change in Ciibbs energy. the cnthalpy, and the entropy at 298 I 5 K.
13-32. -rhc rzmvlaturr dcprndel~ccuC the slurrdard errlf nt n ccll Prl OIH2il bar)lHUrrlr= l.OO)llIglBr~~s)IHg(l) is giwn by
whose
T:ke L' fur {he cell to be 0.23 142 V at IWC.
cell diagram is
wllsrr I ic the Ccl91us tcrnpcrsturu. Dctvrtl~inc the chilnge in the rlarld~lrcl Ciibhs c.ilcrFy, tlic srarlcli~rdcnlhalpy. ant1 the sta~ltiarderllrupy at 298.15 K . Givcn that h I li,(g) l = 13[l 68.1 I ~ K - ~ ~ ~ o I - 5 - ~[lIg!Hr,(s)] . = 21R O l - ~ - ~ . r r l u l - ' , and 5' [llg(l)l = i~re 7h.ll2~.k-l.rl,r>l- , ~ t2~)8.15K. c;llcul;ite Ihe value of S lor Br-(aq). C u ~ ~ ~ pyour an\wer W I I ~~ h cratuc in the ,\'.8S 1i1l)lrs(IJ Chrrl~rrtrll%rrttir~cI~~ruttzrt. P~-o(,er-ficr.
13-33. 1Jse he data ~ I Tablc I 13.2 to c,llculate the value of Ihc \r>luhll~ty product of H&Cl?(s)
1 3 4 0 . The value the cell
lhe acid-dissociation constant vf acetic ;icitl cun he dctcrnlirlud us111g
at 198.15 K .
13-34. Ilse tic data III 1:1ble 13.2 to vdlculatc the value oi the soluhili~yproduct r d A&SO,isi at 298.15 K .
13-35. IJce the data at 298.15 K.
111
table 13.2 trj c;~lculatsthe v;ilue of lhe suluhility pnxluct n t PbSO,(sj
13-36. Givelithat dlcstanrlard rzdl~ctionpc)tential oi 2 D+(aq) ~ a l c u l a t ethc valuc of the equilibrium cvnstanl uf
+ 2 t-
-
Use the i'ullowing data at O'C to dcternlillc the value WE K,, Cur acetic acid : ~ t (PC. ' h b c L fm the ccll to be 0.23655 V at 0 C.
L),(g) is -U.OG34'
13-41. The \ a l u ~01 lhc acid-dissociatlo~constant c~fft>rr~ucacid can Lw dctcrn~ined~ h i n g the cell
ClO-(aql -t H,0[1)
+ 2 c-
+ U1 (aq)
+ 2 OtI
(aq)
E' = (1.40 V
colculatr. the vnlr~enl' the acid-discncialinn constant of HUIIXaq). Hint: k'rx~n [lie fact that h',&= I MI x 10 ".
d to u x
Determi~wt ~ l ccquation for the uverall reactiorl nt this ccll and write thc corrz$pt>ndir~g Memst cqwalrnn Now use tilt rclation
13-38. Givrrl the rrriu~:lir,~~ ele~~rt~Clt'-rt;~ctiun diitii at 298.15 K to show that = calculate the balue ct'thc acid-dissocintion constant of 13C102(aq).
~HF~YH,,"~cI-YcI-
-
F
YF~
k'
Ill
K,,
Chapter 13 / Therrn~d~nam~cs ul Clectru~h~mical Cells
where E' ic the stat~dardentf of the cell
Show that this t q ~ ~ o t i ocan n be writt.cn a?
Ws)lH,(g) IHCNaq) IAgCl(s)lAg(s)
Using rhe relation Llescribc hum tllc value of
% call be detcrlninctl by plotting show that
sg:iins~ionic strcr~glhillid cxtl-;~pc>lating to /.crt>.U~tlikcthe cusr uf prop:lnr)lc acid prcsentcd in Exmrlyle 13-10, the HI (aql and FoC(aq) frvm thc diswciation o f Ht'o(aql cannot he ipntrred. Thr way to prrlcced is as follows. First, calcuhk tlje values of In Ki and ncglecling the H7(arl) and Fo (;iq) fr.olu the dissociatiotl of formic acid. Plot In K: against I",, and ubtnir~a preliminary value uT KL>by extmptdaling to zero ionic strcl~gth.Now use Ihe preliminary \aluc of K O I r ) calculate nt,,, using
uhtl-e mHIvand mGdS,>ilrz thc stoichiometric concentrations 01- HI:u(aq) and NaFdaq). Realizing t h a ~m,,,will be fairly small, ncglecr m , , , with rcspecl 10 nr,,, and mN3, and
wrile
'1'11e valut. of ; . Ican k eestim;~te~l usin? Erluotion 1 1-56 on
;I
rnolnlity rcsle
Using !he above prucrdure and the lollowing dntn at 25'C. calcul:~t' the value of Crjr tornlic acid at 25 C.
K,
13-42. T t ~ value e c ~the f discnciatiun cor~stautOF water. Kw, can be determined u ~ i n g the cell w
P I ( E ) ~ ~ ~ : ( ~ ) ~ K OKCI(aq)lAgCl(s)lAg(s) H(R(~I~ Show that thc emf rll. this cell is given by
(1)
Usin: the fi~lluwingdala fi)r t t ~ cccll givcl~by Equation I with I I I , , , = 0.01W)rnol,kg-', determine thc value of Kw at each ternpemture by plotting thc left side of the abuve equalirm again5t ionic strcngth and extrapolating to zero. Take E" 10 k given by
Now plot In hry against 1/ T , curve f t ywu~rcsult to
where
vl,,,
15the
partial molar volume of the HCl(aq) at mnlality m. -kt,, i a [he p,ir.rl:ll
md TdGCI are the molar vnlumej of Agl s) molar volume of H?(g) at pressure I>. and and AgCI(s). k g u c that at rel~tivclylow prehxlires
;111duse Bqualitm 12.29 to dctcr~ninethe value o f A r H ' for the dissociatirln of water as a quadratic polynonliai in T. The experimental value of A,H" at 2S'C is 55.9 H .nlol-I. 13-43. L'cc ~ h data s in [ablc 13.2 tu ualculatc tire valr~eof thc frrluhility produci r \ l PbCl,(s) a1 298.15 li,and compare your rcsult wilh the value that you obtain uslng the data in the bL3S lirhIes uf Chrmicrrl Th~r~nt)d?.lrurnic Propertirs [ A Phj.s. C h ~ l nDohi, . vol. If, suppl. 2 ( 1982ij.
13-44. U.rc t l ~ tiah c in'l'ahlc 13.2 to calculate the kaluc uf the solubllity pruduct of HgzSO,(s) XI 298.15 K, and compare your rewlt with the vnlue that yo11 o b t a i ~using ~ the data in the NBS 7ii blcr r!f r ! i ~ r r ~ iThcr~(~tyi~clnnlr c~~l Pro~~c,rfirr [J. Ptiys. Ulwm. ncrtr~.vol. 11. suppl. 2
where Ej is the emf of the cell at pressure P,
13-53. Consider lhe cell whose cell dingrim is
-
Show that A rV x - V , , ,. Using a vlrial equation (11' state throi~ghthc wcond virial ccotlilicient (Equation 2.23). shuw that (see Prubler~~ 13-52)
{1982\(.
13-45. G I V Cthat ~ A,C; [ ~ h ~ + ( a q=) ]-21 43 kJ rr~ol I , A,G' [~O:-(a~)l = - 7 4 53 kJ mol ' ar~dA , G [PbS04(s)J= -81 3.4 k) mol-' at 298 15 K, calculate the valuc o f the mlubility h ollc obvulned in P r o b l c t ~17-35. ~ p r o d u ~ clf t PbSO,($) Cornpare your rcsul~& ~ t the 13-46. Clse the dam in Figure 13. IS along with A I G L [ R .taql] r = - lI13.96 kJ ~ n o l - 'tocalculatc thr value of the solubilily product of ttlalllum(1) bl-on~ideat 298,IS K.
-
Now use thc data in Table 2.7 and Figure 2.15 to cstitnate the value of B2,( T I B , , ( T ) / R T fur H,(g) at 2S5'C.[The vulue of B;,(I") from numerical tables is 0.4 l h.1 Pillally, use tllc lilllnwing data at 25'C Tor the above cell to calculatc the value ot E a\ a function of pressure, and cornparc the result o f yrlur calculatio~lwitti the c-upcri~r~ent;il -cult5 graphically.
1 3-47. I!5s t hc data rn kKS 7i1blr.5 r!l ( 'hr+izic.ulY henno~t~ntlt~nr Yroprr-fi~,~ 1.1 l'lfys. Clrrt~~. il,ttr~,vol. 11. ~ u p p l .2 (1982)J lu c,ulcuIate thc value of the solubility product of barium sulfatc at 2921.15 K.
-
13-48. Given that AIC; [HSO;(aq)l -755.9 1 kJ-mu1 and thal A,G' [ ~ O : - ( a ~ )= l -744.53 kl.nlnl-', calculate thc value of the second acid-dissociation conslant of sulrui-ic acid a1 298.15
K.
13-49. l l s e the data ill ,VRS T,lbl~x1 , f C'l~rrnir~cll Il%ermorl'~ntil~ii Pml~rr-liraIJ. Pliys. C'hcm. UCIIU, vol. 11. suppl. 2 (1982)] to calculatc the u l u e of the first acid-dissociation constant d a r c e n i c acid. H,AsO,(aq) at 298.15 K.
1 3-50. IJse thc dava in NBS Yirbks rd Ctierrricul T/ler~noclrnurnic Prr,l,rrfirs [J. I Chrm. Zlufu, vnl. 11. siippl. 2 (198?)] to calculate the value of thc f i r ~ base-protonatinn t cr~nshnt of hydrilzillc at 298.15 K. 13-51. Cr)rupalc tllu idcaliied alnnullt ut work avail;iblz w11ct1one rnc~lrof prrlpanc is burned ill ii turl cell at 2'18 K and in a heat cl~ginewith 1 , = %)(I K and 7: = 300 K. Calculate the valuc 01- E ' for this tuel cell. 13-52, Dcrive a g e ~ ~ u rrcl;ilir~n al for rllc prussure rt~.pcnderlr.u<)I-theemf ol (1.11 : ~ zunstlnt t terllpcraturc. Show that for thc ccll whose dragram ic PI(&)IHz(g)IHCl(aq)l.4gCl(s) IAg(+)that
i111clcctn>cher~lical
The next jive prob1rm.r involve ~~olrultrrrrfg !he vc~lur( B E -fora rlt'~.fr~dr W U C I ~ U thilr ~ ~ IS (I rombinutinrl cg'urhrr el eat rod^, renrritms. 13-54. Somerimes II rri~ghthe nccecsary to calculate thc viluc of E' ior a reduction electrode reaction that is a coinbination uf othcr reduction electrode rcnctionc. For cxample. cons~clrr the lwo reduc1ir)n electrode renctirlns l . K H 1 ( a t l ) + M r ~ O , r a q ) + 5 r -4H,Cl(l)+Mn'1faq)
+
2. 4 H+(aq) + MnO, (s) 2 e- + 2 HLOtl) + Mn2+(aq) We can use these ddla to calculate the raluc (]I' F' for 3. 4 H ' (aq) MnOj (aq) + 3 e - + MnO,(s) + 2 H,O(I)
+
F.]~l49lV E; = 1.20K V
W e hrst ~lntt.that Equaliu~l3 resulls from subtracting Equation 2 frnm Equation 1. It ir important I(>reali~ethut E , i~ not equal to Ey - E;. howwcr. because E" is an i ~ t r ~ n a i vproperly. r The standnrd Cibhc energy change for Equatiot~3 is given by AG; = AG; - A[;,. howevcr. Using this fact, uhow thnt
18-55. Given l t h t
calculatt. the raluc of E ' for
13-56. Ciiten t h ; ~ ~ ~;'(nq)
+ 2 e-
-
Cr(s)
B' = -0.9 1 V
calculate ihe value uf E' for
13-57. 'Ihe p r e ~ i o u sthree problems rlevc!op the idea that you must use value< of AG' In cnlculntc v;lluzs of 8 - ul- reduction electrode reactions from the val~lcsof E irotn other
clcclrrxlc 1,cactiuns. When the cambination is such that the electrons or1 each side of the rquutiu~~ c,~ncel. hnwevrr, we do nth have to use AG" as an intermediate quantity and can write E Y ~= ,, k ,, 1 E , , dirzc~lyas we did 111 muation 13.22. In this p~oblern,we derivc Equation 13.22. Consider the ~ w reduction u clectrude reactions
1.
A+tiIY
2. B
+ n, r
-
X
-+Y
Ei,md
Ei.A~
Show that the comhrnntion in which the e l e c w n s cancel
3.
~r~,\+?l,
Y + n, B + n , X
is
13-58. Crmsirler thc reduction electrode reactiot~
Using the fact that the dissociation constant for H,O(I) is K y = I .(XI8 x 10 calculate the value nf E" for
l4
af 29X. 15 K1
Nonequilibrium Thermodynamics
To this point, we have discussed the thenuodynamics only OF systems in cquilibrii~~n.
Lars Onsager wah born i n Oslo, Norway or1 Novr~nher27, 1'103, and dlcd in 1076. Hc received a dcpree in chemical crigincering I-iun~the Nurweg~anlnstitule nf 'I'echnology in '1.1-undheimin 1025, altcr which ht: wcnt to Zuricl~to discusc with Peter Debye a flaw he had discor*eredin the Dchyc-liiickel theory. Oncagcr was a graduate ~tutlentwith Dchye fro111192h lo 1'128, dunng which time he dsxeloped the Onsagcr linliting law for the crrnductivily ot dilute ~ v l u ~ i of n~~s electl,olytes. In IY28, he c m i ~ r ~ t ctod he llnitell Sla~ehand spenl livc years all dssoc.ialz in chcmi$trg ut Brown University. where he developed tlh: Onsager rcciprucal relations, which ttw Norwegian Ir!
;$nilIiihhs postdt~cluralfellow at Yale Univctsity. Wher! it was dicuvcrud thi~the twd never rcceived hic doctorate. Ihe dcpartmerlt pcrs~latlctltun1 to hubmil n lhesi\ Tre a Yale ductvrarc. Rather than subnu1 a tl~esisun t1lc reciprt~alrclationh, he chose inslead a tn,~therr~atlcal topic. In 19-15,he was npporn~edas the J. Willard Gihhc Professor or Theoretical C:hc~~~i.;try and rernairlcd at Y:ile untll his rtllirznicnt in 19711. Hc became a CIS r i r i ~ e rl ~r ~191? H? taught r)~ilyel ;dude courses at Yalc hccilucr I Ihis ~ ~-e]luIaliun as a rlitlicult lecturer. Studt~~ts uflen jokil~glyrrferrrrl lr) his aotislial rucchanicc C I I ~ I L F ac ~ S -'Advn~lce~i Norwcgin~~ 1 and 11" He spenr his remaining years a i the Centcr fol- Thet~,cticalSludle!, at the lJn~\,el-c~rq. uf M ~ a ~ n i at Coral Gables. 1;lorida. working in hiophysicb. Onsager had a wide range oi lntcrests in theoretical chcmiclry and physics incluriing thc dcvelopme~ltof the thzrr~lndqna~cic theory of irrevers~bleprwesscs. {Insager receivcd the Nobel Prize for cherrlistry in I%+ tor the disctlvery of the recipnlcal rclntirlnh hcnriog his riartle, which are fundamenla1 l r ~ the r rhemli~dynamicsof irl.cvclcihle processes.
Realize that thermodynamics allows 11st o derive relatir,nships hetwcrn measurable propcrtics such as temperature, pressure, and heat capacity. Nole thal these are properties of systems in equilibrillrn, There arr.rrlany measurable properl1c.sof n o n c r l u i l ~ h r i ~ ~ l ~ ~ systems such as (herma1 conductivity, electrical conductivity, ;inJ diHuaiun c u c f t i c i e ~ ~ t ~ . Nole 1ha1 lhesr: prupcrtlcr ~nvolvcP I - < ~ c s s that c s arc the result of differerlces in tetliperature, clcctrjcal potcntial. and concentration, respectively, so that huch systerrls are nut in equilibrium. We want to investigate whether we can exterld our 1reiIlmen~of' thcrmodynam~~,s s o that wc can dcrivc relationships hetween various nonequilibriurn properties. Ii so happcns thar w c call dn this with the introduction of n few reaso~~able, 3dClitio1ji11 yrihtulatts. Th15part ot thcrmt~dyntlrnics is called t!otrt,r/uilit)ri~~~n tl~rr~-rtrtirlj~trutr~ir~~ ur the ihermrd7nrrrlair.r of ip-rr~i,rrbiblu pruut.sxrs. 111 the first 1u.o sectior~:,of this chap~er,we will I-ezxaiiiinc llic Sccontl I .;IN of - 1 - h r r m r ~ l y n a ~ nWe i c ~begin therr because the cntropy, although strictly an equilibriutu therlnudyniimic prupcrty, clcarly has a "dircctiunnlity" to it became tiS 2 0 fur ally natural p r ~ c s it1 s all isolated systell~.I n the subsequent two sections, ue will iritroduce rime into our thenmdqnamic equations and discuss (he linear Hux-forcc reliltions tll;tl :ire ~ e n t n 10 ~ l noncquilihrium therrnudynaniics fur systems that are no1 too far I'I.OIII ctlr~ilibriun~. The mobt celebrated result oT these ~ w sec. o tioils ;IW the Onsager rzcipl-ocul rclatlot~s,which :he will see are the keys for deririrlg the relalionships I,ttween various norieyuilibriun~ prclperties. U'e present our first practical application c ~ fnoncquilihrium thermrdyn~tnivsin Section 1 4 5 untl show how various appar~*ntlydisparil~c. t.lwtr;lkinctiv qumritics arc actually related to each other. Then. it1 Sec~iori1M.we u h t : a \irnple zhrrnicaI k~neticscheme to ahow the fundan1c11t:dbahis of the Onsnper reciprocal relations. 111 ~ h next z few sections, wc usc noncquilibrium tllcrmodynnmics to derii~requatidn\ IT the magnitude of the liquid junction potcntial, which we sti~tcd i r Chapter ~ 13 1s !he eleiltrical potential thar results whencvcr two difflcre~~t clcclroly~c solutions are brtbught Into contact. All the systems w e discuss in thc first nine hcctions 581
I'haptcr 14 / Ncrn~qu~l~hrir~rn Thermndvnamirr
arc ~ l i s ~ t ? r i t i l lS~Y ~ S ~~~)I u T I~ inS , the sense that they consist of two subsysrcms at different trlrlpcratL~res.chcn~icalpo~cnliuls.nr electrical potcr~ti:rls in curitact with c x h othcr. In Scction 14-10, wc eatcnd our approach lo cnntinuous systems, which are those in which letupcraliirc, chemical potential. or electrical potential vary slnoothly from onc region to anr~lhcr.In particular, we will sludy ditfusion in binary and ternary solutions. We discuss a few gencral principle'; of tloneq~~ilibriunl thennodynamics in the laqt two sections. In Scction 14-1 1. we show lhat the entropy production is a minimum when a systcjn i s i n a steady stntc near equilibriunl as cumpmd with an eq~rilihriumstate, a( which point the enunpy production is zero. In the last section, we briefly discuss c.xtcn.;ions of nr~nequilibriurnlhermodynamics to systems that are tihr from equilibnurt~.
14-1. Entropy 15 Always Prod~rcedi r ~a Spuntancu~iProcecs
Comhininp Equations 14.3 and 14.4 gives
which is Ulausius' qtatement of the Second Law rlf Therlnodynanlics. We can illustrate these ideas ctncl-ctclq. with ttle following examplc Consider a two-compartment system as shown in Figure 14.1. tach compartment i s in equilibrium with a hent reservtjir at different temperatures 7; and T,, and the two conlpartrnents are separaied b j a npid, heat-conducting viall. The total changc of c ~ ~ e of g gcorttparl-
~nent1 is
14-1. Entropy Is Always Produced in a Spontaneous Process In our discussiorl of the Second Lnw of T h e m d y n a m i c s in Section W, we showed t t ~ u t(Equation (1.17)
dS > 0 d.S = O
(sp>ntancousprocess in an isolated system) (reversible procew in an isolated system)
(14.1)
Itrluations 11.1 apply only lo processer that takc place in isolated sysrcms. For other types of systelns. we found it convenient to view rlS as consisting of lwo parts. OJIC part of d S is the entropy creatcd in the system by any spontaneous processes occurring within it. and thc othcr part is the change i n entropy resulting from the cxchangc of encrgy 35 heill between the cystem and its surrounding\. These two conrributic~ns liccuut~tfur thc cnlire change in the cnlmpy. We denoled the past of d S that i s created hg any sporttailenus process by d.Sl,. This quantity is always posidvc. We denoted the par( of ti5 (hat rcsults f-ron~thc exchange of cnergy as hent with the surroundings by rlSpxrh. This quantity is given by 6 q / T , where T is the tetnpcrature o f the surraundings; it can he pos~live,~~cgativc, or zero. The quantity bq will be 6qrc+i C the exchange is carried nut reversibly and 6qImif it is carricd out irreversibly. I'huh, we can write for 17r7r prclcess (Equation 6.18)
For ;I revcr~ibleprocess, dS,,
where 6,y, is the energy as heat exchanged with its reservoir and Siql is the enel:! as heat rxcharlged with cornpartlnent 2. Sirnilwly,
Essentially hy definition
Now each compartment is ill equilibritlm with its hcat reservoir. so
= 0 and Sq = 6qm, so that we have
I-rrr an irrevcr~lhleor spontaneous process, d.Tl,, r 0 and Sq = 6yIm,so that
F I G U R E 14.1
A two-cornparln~entsvstetn with each compartment in contact with an (e\sentially infinite) heat reservoir, onc at tempenlure T,and the other at tctnperature T,. The two cornpmmctlts are separated by a rigid, hcat-conducting rvall.
Chdptcr 14 1 Noncqu~l~briu~n Tlt~rmodynamirs
14-2. Entropy Always Increases When There Is a Material Flow from a Region of Higher Chemical Potential to a Region of Lower Chemical Polcntial
(cotistant V 1
Consider the isolaled two-cumpartment ,gstcm shown in Figtlre 11.2. t a c h of the two compartrnenl:, is in equilibrium individually, but thc two compartments arc 11ot in equilibrium with each other. Thc two compiwtrnenls arc separiltzd by a cliathcrtr~al
LJsing the Lict that S = S,+ S,,wc obtain
+
(See Equatio~i6.15.) Bi11 d U , = Sy, = Seql S,ql (Equaticln 14,h) and d U , = Sty, 6,qz (Equation 14.7), so Equation 14.9 becomes
-+
(heat-conducting), Rcxible, permeable rvdli: therelilrz. cnergy hcat. volumc. alld mus1 ]:+ye triatler can flow !'rum one cornpartmen1 tc) thc other. Each cnn~parl~nrnt enough that the transprlrt of energy or matter h ) i n one cumpartlnent 10 the other duec not app~tciahlydisturb the equilibrium state oi each compartinent. The ~olalcntrupy or the two-compartment systcm is 8 = S -t S,, with S,= 6 ,(U,, C', . l a , ) and S, = .S2(I!,, V2.n , ) . Rccause each conipanmen t is in equilibrium irrdividually. we can w;,itz
Finally, Equation 14.8 allows us to write
-
d U , P; dIV l TI
7,
p dL n
l
(14.13)
TI
with a similar equation for dSz. The evaluation of the derivatives in Equation 13.13 follows from the equation dU = TdS - P d V pdn. Therefore,
+
where
i s thc entropy rxchrmged with the reservoirs (surroundings) and
is the eiuropy producer/ within the two-cu~npartrncntsystem. 'The condition dS,, 2 0 implics that 6(q , must be positive if T, > T, . Physcially, this means that energy as heat Rows from conipartment 2 to compartmenl 1 if T: > T,, as we should expect. Conversely, 6 T?.Note that the dirzclinnaiily of the heal How is dictated by d%,, 2 0. The valuc of dSexFI, is arbitrary in the sense that it car] be positive. n g ~ t i v eo, r z r o and has nothing ro do with the direc~ionalityof the heat flow. The separation uf d S into the sum of dSc>lLl, and rlSpR, is Cundamental. We will sec that nonequilibriun~thermudyna~nicsih based upon the facl t h a ~rl.SprK, > O fur any spontaneous process.
Rigid,
adiabatic u j l l F I G U R E 14.2
An isolated two-cnnlp~rtrnentsystem in which the twu colnpurtlnents iue wpararcd by 3 diathemal (heat-conducting), flexible, pcrmeahle wall. Each colnpartrnent is at uquilihri~~ln individually, hut the two cumpdrtmcnts are not in cquillhrium witt~e ~ c h othur.
Chapter 1 4 / N U I I ~ U I ~ I ~TTl ~ ~L e rI r~nT~Id ~ i i ~ ~ n i c ~
506
where w e huve written dSp, instead d S because the two-compulme~ltsystem is isolated. T h c twrl-compartrncnt system in Figure 14.2is isolated. so U,
+ Uz = conatant dI:', = -dllz
V, -+ V, = constant
rlV, = -dV,
n , 4- n, = constant
ddn, = -dn,
(14.15)
,
Note thal rln must be negativc if p , > p,, meaning that rnattcr will fluw rrrrn~con]partment I to compartment 2. as e x ~ c t e d . We call relate Equation 14.19to Equation 14.18 in the following way. Because the V = V, 1', is fixed, we can write
+
for the two-compartment systcm, and so
Substituting Equation 14.15into Equalion 14.13gives
(constant T and V )
To make the complete connectiun heiween Equations 14.18 and 14.1'1, substitute Quation 14.19into the above equation to obtain
Thc first tern) in Equation 14.16is essentially Equation 14.12. I ~ t ' look s at the second term. Fquation 14.16 says lhat
I ' C I S ~=, ~- d A = - d n , ( p 1
-
p2)
>O
in agreement with Equation 14.18.
if 1; = Tz and 11, = f i 2 .Thus, d V , > 0 if PI > P,, which means t h a ~the flexible wall t+-illmove to the right in Figure 14.2i f P, > P,, which is what wc should expect. The third tcmi in Equation 14.16 has to dn with matcrial flow, ur diffusion. If we lake T, = T, = T and PI = P2 for simplicity, then Equaticln 14.16 i s
I
E X A M P L E 14-1 Suppsc a two-compartment syslem is encloscd by flexible. diathcrmal walls and is immersed in a heat bath at temperature T and pwssure P.Dcrive the analog rlt Equation 14.21 forthis system. Assunie that the wall sepwatjng the w o compartnlcnts allows only rrlatter to flow from une compartment In the other.
Kecall that (replacing activity by concentration) SOLUTION: $Vestartwith
lijr a solution of conccnlration c, or [hat y depends iogarithnlically on the cc~ncentration. 'rhos. Quatior1 14.I H says thar d n , < 0 if p , > p,, which means that matler will flow frt~rnuorrlpartment I to compartment 2 if the concer~trationof the dilfusing species i~ grr:itel- i n cunlpartlnent 1 than in cornparlrrlent 2, as we know from experience. S o far w e have becn treating an ixolated (two-compartment) system in this section. Suppose instzad that the ~wo-cn~npartment system is in contact with a heat bath at ternperature T, so that T, = 1; = T. Furthermore, for simplicity, let the wall separating the two compartments he rigid so that V , and V, are fixed. In this case. the twu-compartment systcm is at constant I ' and constant V l and V,. The appropriate thermodynamic state functic~nunder thexe conditions is the Helmholv, energy, A = A ( T , V = V , V,, n), aod the governing equatin~risd A = pdn at constant T and 1'.Thus, we have for the Iao-cornparttrle~ltsystem,
+
dC = - S d T
+ V d P + pdn
= pdn
(ton\tani T and P)
Becawe the pressure is held comstant, we can wrile (see Equurioti 14.20)
d.Spd =
TdS-rlH T
-
dG
- --
T
(consrant T and P )
potential dimerencc is called elertroo.~mosis.For a given system and diaphragm, the elecrmo~moticpressure (EOP) is delincd as
where .I,, i s the volunle flow across the diaphragm. The electroostno~icpressure is the ~in~l,ortionnlity constant relating A P to A @ at the point whcn thc volulrle Row has ccaqed.
Now coucidcl. rhe opposite cxpcrin~entwith thc same system in which the electrodes arc sllort circuiteti so that A111 = 0 and the solution is forced thrrlitgh the diaphragm by n pistun (Figurc 14.5). I n this expc-riinent, therc is an eleclric current through the diaphragni us well as a volume Ril\v, J , . It turns out that the magnitude of the electric current is directly proporlional to the volume flow. We define a quantity called the srrerrmi~lgcllrreni. SC, by the relation
F I G U R E 14.5 A schem;itic diagrnnl of an apparatus to meawre ~trcamingcurrelit Each comparime;lt containc idenlical solutions r ) f an electrvlyte and the two conlpartments art. separated by a pllrnur diaphragm. .4nammeter nieasurcs the eleciric current that accr~nipatliesthe flu~dHnw uar~setl by the piston.
with a pressurc rlifmtlce A P , a ~ c ~ l ~ diffcrence age results. If this voltage difference is measured with a potet~tiomcterso Lhilt I = 0, then A$ turns OUL to be proportiunal to A P. This fact allows us to dcfine a strr,nming pnicniiul, SP, by
The stiranling cu~rent1s the pn~por[ionah(yuunstant relating the current 1 In the vrllume flow J , . It is found empirically that the tnagnitude of the electroosmotic pressure differs only in sign fron~the streaming current, so that
EOP = -SC
Similarly, &hen a potential rlifferencc i v applied across the diaphragm under conditions in which A P = 0 (horizt)ntal tubc), there is a volume flow. J,, as well as a current flow,I. It rurns out that J,, is proportiunal to 1 under these cot~ditions,and vie dchne a quantily ~lrurmo~rmotiujlow, EOF, by
We will see below that Equation 14.36 results from the Onsager reciyrucal relatiutls.
E X A M P L E 15-3 Stlow that the clestroosn~olicprclsure and the streaming current have the same unils. SOLUTl(1h: The S1 units of A P arc ~ , r n . =J,m-' ' and thosc uf A $ J-C-', so that the units o f EOP are
The S1 units (if I are C - s - I . and thost. o f
JV
UC
The streaming ptential and elecltoostnotic tluw are relatcd to each other by
V=
are m3.5 ', so that the units of SC are
fhus the llnitr nE EOP and SC are the same.
Then: ale other pairs of eltctrnkinetic quantities that are relatcd to each other. Fur example. if the fluid is forced through the diaphragm shown in Figure 14.4 or 14.5
Equations 14.36 and 14.39 are just two of a number or relations hetheen various electrokinetic cluantitier We will 11uwshou how Equations 14 36 and 1 4 . 9 result frum the On~agerreciprocal relations. If h e apparatus illustrated in Figure 14.4 is at a fixed temperature and pressure. then the appri~priatethermodynamic equations are
Chdp~er14 / Nonequilihri~lrnThermodynamics
i +6
The Dnsa;r+ K e ~ i p r u c Kclationr ~l Are Haserl url thc Princ~pleot II~td~lrd Hdldn~c.
a h e r e p.,, i s the chemical potet~tiitlof the cation, / I , - i s the cl~ertiioalpcltcntial of the anion, p,, i s the chendcal potential of the water, and is the electrical potential in compartment I . 'The Ii~sttwo terms in Equ;ltion 13.40 rcpresmt the electrical work involved in changing the nurtlher c ~ fmoles of cation hy d r l , . - and the nutnber of moles of-aniot~by drr,-. if we use an equation similar 10 Equation 14.40 l i l t compartment 2,
The correspnrading linear fl ux-force x1arirm.j arc
thcn
We can riow use Equations 14.46 lo pro\c F q ~ a t i o n s14 3h and 11.39. The elzctruoslnotlc presaure (Equation 14.34) i:, obbainetl by letting J, = 0 in Equat~on14.16~ and solwng fur A P / h @ to obiain
+,
IJqing the facts that dn,+ = - d n ,
Jn,- = -dn, , and dn,, = - d ~ 1 ~gives _ The streaming current (Equallon 14 35) is obta~nedby setullg A $ = 0 in t q i 1 ~ 1 1 1 o n ~ 14.46 and dividing Equation 14.4hh by 14 46a lo obtdrn SC = --Ll,,
Exatnple 14- 1 showcd tl~atrlSiIK,= - d C / ? - for a system a t constant 7 and P, s o Equation 14.41 van hc writtell as
(14.4Xj
Lvr Erlualions 14.47 and 14.48 show that EOP = -SC bccause L,, = L,,; lhus ploying the validity of Equation 14.36.
w h c ~ cA F , =I"+ - - - p I , , = p:- .--J L , , A/L* = pZu- p n , and A$ = $,. Now divide Equation 14.42 by dr to get
l?r, EXAMPLE 1 4 4 Pruve that tilt srrcar~lingputenlid. SF! is lhc ncgnti\c of elec~ronsint~tic Row. EOt:
whcrc. .II = d n , , / d r , . I = d n , . : d t . and Jw = dtrla/ d t . The flow of the neutral 1-1 elc.ctroly~eacross the diaphragm i s ,l>= J+ = J the clcctric current across it is I = r J+ - e./ , and the chertdcal potcntiid of a I -- 1 rlectrolyt:: is p. = !I I p - . Suhhtituting these lhrec rclirtiol~s~ I I ~ Equiitior~ (I 14.43 gives
.
+
h l o s ~electrokinetic experiments arc carried out such that the crtnccntratiar~*)I' the solution is the samc in botl~ .partments. In this casc, there is n o conceritratinn dspcnder~ceof A / l , or A p *. s r ) (lquation 9.9)
*-here i sbands for s sr w. Suhstituring A l l l = 1'(A P into Equatiutl 14.43gives
The tcm1 in parentheses is sirxtply thc total \'r)lumt- Row, J , , acnls, rhc diaphragm, su we finally havc our basic erlli-opy production equation for eleutroklnetic phencbmena:
SCI L U T I Cl h : 4~.corrlingtu Guatioll 11.37, SP = (Arl,; LIP),.^,. Lct I = 0 irt Equatioo 13.4hhl~~gt15P = - L i b ~L,,.ElcctronrmolicHowisgivcn hy EOF = (.i,.!II,,=,, (Equalion 14.38).Let AP = (1 ~ I Equalion5 I 14.46and diwdc orlc t)y the ollier lo ubtain EOF = ( J , : ' I ) = I.,,iI Therefure, SP = -EOF heci~uwI .,,, = L , .
,,,-,,
.,,.
,
Problerris 14.. 11 to 14-15 have yo11 derivz relaiic~nshctwccr~other electrokinetic quantities.
14-6. The Onsager Reciprocal Relations Are Bascd on the Principle of Detailed Balance 111this section. ~2 will usc a chcmical kinetic ccheme to gain some insight to the r)ripi11 of the Onsager reciprocal relations. OE (he lniany applications of nonequilibriun~the,,to i~ariousprtlccsacs. the applicatin11 i o cliemicul kincrics is exceptionally rnrxiy~arni:~ Ijmitcd in practice because of thc requirement that the Ilux-forcc rcl;~tionsbe fincar. This lineari t j requtrtment mcans that chemically I-eilcting >ystems rnilht be ulosr. 10 eq~litibriun)Tttr he f~wmalismof r~oriequilihriumthcl-mudyr~atr~ics 10 he LIPPIICL~~IC. Nevrrthzlrrs, wc nil1 sce in this section how the 0n.cagc1.rcciprociil rcli~lit)nsariw hg studyi[q 'L u ~ n p l cchcmical reactiotl.
14-6. TheOnsag~rReclprucnt Relations 4rc H,lfcd nn thc Pririrrple of neta~lcdH d l , ~ n r ~
Let's start with a reversible first-ordcr elementary chemical reaction described hy kxv
X+. Y
+
where we ha\ e used the fact that [XI [Y]= constant. Equation 14.53 is our fundamental entropy production equation. We cat1 write Equation 14.53 as
k,, 'rllis reaction i s tcw simple to involve the Onwger reciprocal relatjon~,but we can use it to introduce sonle not;itiotl. The ratc oT this reaction is given by
where J i~
where [XI and [Y1 are concentra~ions.We can define the flow of !hi%reactirlrl fmm left lo ri$t hy
ih called thc nj'jinig. Thc linrir Rux-fnrcc relalion is
At cquilibriu~n.J = (1 and Equation 14.49 givcs
]XItq
and
+
A giwn in Equation 14.ih is rafid onl? for small values of A/ R where K i~ the molar gas conqtant. This mcans that [XI and IY] n~usthc cloqe to their equilibrium values because { i , = pI at equilibrium. koblern 14-18 shnwi that
u,=WI-~Y[,~
(14.50)
+
Note that a, +a, = O because [XI [Y]= [XIcq [Yjq. Using t h ~ sresuit, Equatiorl 14.49 cat1 he rxpresred in terms of' a, by (Problem 14-16)
Now let's apply r~u~~cqi~ilihrium thcr~nodynamicsto this rc:+clion. For cnwsriience only (scc Problcln 1 6 1 7 ) -wc take the reaction system lo he isolated. Fmm the quation
Re
Equation 14.49 and v.hcrc
EuperimentalI>, the linear relation betwccn J and
Now define
n,=~~I--
~ I V C I hy I
if L Y ~ ~ L<<X 1I and ~ ~ a,/[YlrCI << I . If we substitute this expression for A into Eq~ration 14.56. wc obtain
If we compare Equation 14.57 with Equalion 14.5 I , we see that I. = k y x ~ y l , ~ , ~= !R k,,l.rllly,iK
cee th~lt Ly dn, riSrc, = - -Rnx - IT T
II'cvc divide Eqi~ation14.52 by I/', we obtain
(cnnslant I; ~ t Vd )
E X A M P L E 14-5 Consider the elerncntary chemical rcaction descrikd by
Show [hat *c affinity in this case is given by
A=
/Ax +fly - ILL -
T
SOi. IJTIO h : The now of this reacliot~f m ~ uleft tu right is given by
11-6. Thu 0nsagc.r R c ~ ~ p r u cRclatiuns al Arc B a d 011 the Prwnciple of Det~wledBaIdnce
Tlesc three affinities are not independent. however. because -4, = -(Al sequently. the enimpy production equa~ionuoiisis~sof only iwo [errnu
Thc basic ther~x~odyr~anil rqu~lionfur ll5 is
+ A?).Co11-
and sv Sp" = Divide bv 1/
!iydnx pudn, ~ ~ : , d -7dt - Ttl[ - y -;-
n ~
The associated linear flux-force e q u a l i o ~ ~are a
tu ohtain
At equ~llbrtu~n A, = .Ai = 0 and so J ,
-
J , = 0 and J?
--
J, = 0 , or
Equalion 14.63 i~ an interesting rzsull. Aucordirig to illis etluaiiori, the condidon of tliennodynmic equilibrium dws not require thnt all the flows vanish. only thnt thcy beequal. Therefore. accordillg to Equation 14.63, thc rcaction may circulatc in onc dircction or the othcr indcfinitcly. This notion, hr~wevcr,viulatcs a fr~iidamenral p~-uiciple of naturt, which cays that under equilibriuru cottditions. any ~llolecularprtlccss i~nrlits rzver,e prtlcess must puke place, on the average, at the satrle rate. l'hz Lonhequences of this principle to thc reaction schcnic i n Equatiul~14.58 is that cadi 01 ihe indivitlual flows in Equation 14.59 must be zero. Thus, nut only nlust these flews bc cquiil tu each other a1 tquil~brium,hut they individilally rrjusl he equal to Lero. I'his priiiciple is callcd the prit~ripicr:rlfdcrailt:dhulnnc-r,. The condition of detailed halance
Now let's 1ooL at Lhc cIcn~cnritryreac~ionscheme described by
requires that
This is ihe reaction scheme used by Onsager tu glean the basis o f the wiprr,i-aI relattc~ns.We can write the kinetic eili~ationsaa
J , = k,,IXI .I?
=
k,, [Yl
.I, = k,,lZI
-
A-,,LYj
-,
k,,lZ]
k,,[Xl
Oi11y tuu of thrrce kinctic cquatiutis are iridspendcnt hccausc [XI ctlrizlant. Thc atfinjucs associated uiih ihzse ihrec flows are
- 1Y 1 + 121 =
E X A M P L E 14-6 S h o ~ that tke principle of det;~iletlbilldil~ereql~ii-esl h a ~k,, k,,k,,
=:
k,,k,; k , ,
l : i t c i ~ ~dlwb c t r,csul;physically.
5 0 1 11 Il [:I N : S i ~ i rEqualion t 14.65~Irlr lY)rq/[X]tq.Equatiur~IJ.hSh trlr, 121,,~[YJ,,,, and Equ:>tir~n14.b5~-fur IX1eq/IZ]c,lanti multiply the three results lrlgrlher tcl gel kxrkYLkL3 = ~ u ~ Z Y k , ) rThus, . the prndlict of the thrcc ratc corlqtants In one direztioo isequal ro the prucl~ctufrhe thrzz rate cunslants in the olhcr tlireclit>r!.Note that thc prir~ciplzot detailed h l a n c e say$ t t ~ the t six ratc constants arc not indupcndcllt.
Chapter 1 1
;h u n e q r r i l i t ~ r ~ ~r'ttr~rmdvndmir~ rri~
+
14-7. Elcctruch<:m~calPotet~tial~ Play the Role nf the Chemical Porpnlinls
603
Just as we did for the case X Y , we consider the kinetic system described by Equatior~14.58 to bc near equilibrii~m.In this case
# d Q being performed. The First Law of Themrdynamics f'or an clectrochemicnl system h o m e s
Tlic third Ilux. ./,, is a linear cnrnbinatin~~ of J, aiid J, and i s given by (hoblern 1 4 2 O j
111Fquation 14.70, thc sccnnci tcnn on thc right rcprescllts pressure-~~nlume work, the
third tenn represents the work invvlvcd in changing the colllposition o f the system. and the fuurth Term represents electrical work. The total charge of the systenl is givcll by
Thc two independent affinities are (Problem 14-21)
z, i s the valencc. n, is the nurnhcr of molcs of the j t h species. and F is the Famday constant. Substituti~~g Equation 14.71 into Equation 14.70 gives where
C'ornhining E q ~ ~ a ~ i o 14.67 r ~ s and 14.68 givcs
Equation 14.72 stiows that the work involved ill changing the cot~ipositiunof an elecwochetriical system is composed of two parts, a chemical pan, p,dn ,, and w electrical part, zjI;$dni. We define a quanlity, fi, , called (he c l c c t m ~ h e r n i r n i ~ o t ~ n t i a l o f c n m p r ~ ej ~, rby~
fi,]
= Vj
H w e compare Equations 14.69 with Equations 14.62, we sec that
"
Zj"l),
Using this definition of ,E,. Equation 14.72 becomes tlU = TrlS
I I' we ucc thc detailed balance condi~ionsky,[YIcq= kZY[ZIFI,and k,,[X Itq = k,,[Zlq (Equations 14.65b arid r , l . then L , , and L,, becolne L 1 2=
kzx [ Z J W K
and
kxlzl
-
PrlV
+
ji,dn,
( 1 4.741
The electn~vhemic.alpotential plays the snme role in electroche~nicalsystclns tliitt the chrrtiical pjte~itialplays ill sysie111sc0111po~eduf 11euln11species.
Lz, = R
Thus, w e sce that (he principle of detailed balance is the basis of the Onsager reciprocal relat~ons.Onsager later proved the validity of reciprocal relations more geocrally u i n g
I
E X A M P L E 14-7 Show that thc condition for chemical equil~hriumfor the general wactirln described hy
14-7. Electrochemical~PotentialsPlay the Role of the Chemical Potentials for Charged Systems in Different Phases Btiore we dixcuss electrocl~en~ical systems in the next few sections. it 1s cnmcnient to irl~rnduvea quantily called the elecrmchrmiuaIpotential. If a reglon has an electrostatic ~wtc~itial I), cliang~ngthe charge by an amuunt d Q will result In an amount of work
where each species may be charged and in different pha~esat different electnc pLenl~als.
t
14-8. The hlagnrft;dcuf the
S01.U [ION, Rrut subtract d ( T S ) ~ dG =. -Sd7
raddl d ( l ' V ) to Equation 14.74 to obtain
+ VclP +
fi!rln,
L~quddJunctionPotental Del,e~wlr I;pon
'Tr.jrw,llortN I I ~ T I ' I ~ ~ ,
400 rnmo1.L-I and 20 mmol-L-' If
w e acsume t h ~ thc t nct~vitycocficicnts of thcse soluhons art: about the same, then the Nernst pctlent~alat 25 C is
(14.75)
I
Yole thar
which i s typical uf rcsting-state potentials in biological membranes. which is a p c n c r ~ l i z a to~f ~ E~ y u~a t l o ~9~26. JUSL as we did in he beginnin r>T ChapIer 12, we intrvduce the extent ufreactiun toobtain Equatio~ls12.2, and thcn sir~~ply tcvllijw t h t nrgulricnt to Equation 12.6 tn nhtairl
<
I l ~ cnnditlor~ e for eq~tilitxlu~n is 1hul (BGjnt:),
,= 0. ant1 su we have
derived in earlier chapters can be rzwi-~trenin tenns 01. to include electn)clie~nicalsystems. IVec;~napply Eqi~alion14.77 lo Lhecase i n which twr~U ~ U ~ O eU lSc ~ t r ~ l ys~lutions lc a n ~cparatcdhy a men~hranethat ia permcahlc to onc of thc ions. If wc dcnote the two solulions by u and P and the pcrmeahle iun by i . thcn Equ;ltic>n14.77 bccomcs I t turns out that all the equntiona involving 11 that we
1 6 8 . The Magnitude of the Liquid Junction Potential Depends Upon Transport Numbers Co~lsiderthe electmclietr~icalcell whose cell diagram is (Figure I 4 . h j
The vertical dotted line betweerl the NaCl(u,) n ~ t dthe NaCl(u,) sepreuprlrs the jur~ction hcrwcen the two solutiot~s.The silver-silver chloride electrode reacr s ~xversibty with the chlot-irlc ions. and thc rcaction thar tnkcs pliicc at thc right clcctrodc (scductionl is described hy
A~("]-AgCl(h)
electrodes
\\.sing the fact ihar p, = p;
+ RT lna,.
wherc (a, is thc activ~tyot the inn I , wc have
F I G U R E 14.6
schematic lllustratiur~ni thc ccll wtlosc cfll diagram is Ag(s)l AgCl(ailNaCl(r~,) f NaClin,)l!\gCI(s)1 A s ( $ ) Thc +ilr,er-silverchtoridt. electndc ru;jclc rcrersihly w i ~ h[he chloride ions. The I U , ~NaC'lfuq) solutions niay be scp~ratetlby a prmw ~iiaphrilpnl to ~nirliruizcttlcir ~r~ixing, 1 . 1 net ~ rcactioll of this cell ic NaCl(u,) 4 NaCl(rr:). and the cell I!, called :iconcentration cell.
\I
'I hus. we see that an eleclric pt)tenlial aurohs thtr lnemhriinc results. This potentiill differer~uecounleracts the tendency of the i ions tc~equalize their concentratic>ns LIII i t ~ 1w0 t sirlcs of thc r n c l n h ~ ~ ~This n c . potellti;ll difference is often c;illcd lhe ,Vvrn.\t oolc.trtir;l, especially i n biophysics. A number of biological mernbranea are ~ n u c hnlorc pcrmcablc to one particular ion than to others. For example, squid axon nerve cell nienihranes in their resting state are altnust zxclusively permeable lo pot~ssiumiclns. Tlic concentrations of the potassium ions on the two sides ol' the membrane art ahour
Poruu+ d-aphragtrl
where thc electron. e-, is in the Ag(s) electrode. The reaction that takes place at the lctt elzctrodc(oxidaticm) is described by
Now if we let the electrical potential that an ion experiences i n electrode campartmen1 Or I j be $" or $&.then (Equation 11.73)
'I'hc equatiot~h ~ the r nvcrall cell I-eactinn is given by the sum of these two cquations
Solving for A$! and using Equation 14.85 gives us
'I'hus, thr clrlving iorce ol the cell is due oniy to an activity chnngc, or a conccntratiun change. Such a cell is ualled a mncenrmtint~cell. If wc apply Equa~ml14,77 to the cquations for the two electrode rcnutions given hy abovc. we Ilnvc
In Equatiorh 14.86, Aft i s the liquid junction potential and E is the emf o f the ccll. We w7iII low develop the noncquilihrium thennndynnniic cquntionsfor this (isothermal) systelrl. Following thc rcasoning that wc used in previous secticms, we may wri~t:
and
where a and B dcnrlte thc solution clcotrode compartments. Using the fact that the B electrodes are sirnib, we can write p,*,, = @ ,! and pi,, = p,, so if we subtrucl f;.rlu;uion 14.8 1 T~.omEquation 14.80, we have
where J (.I 1 is the llow oT cations (anions) attd Jy is the flow of w:atcr ~no~cculcs. M'e will fir->I [I-anstimnEquaiion 14.87 into one involvitlg the Aow of neutral snlr, J<, rather than thc individual ions. Fur concreteness, we will assume that lhe elecirudes react rttvetsibly with the anion (as in Figure 14.6). In this case, the fow ok thc c;ititln is thc ramc as thc Row of neutral salt because the caliun is no! produced or removed at the clcctrudes. For a 1-1 sali such as NaCI, J5 = .I, when the electrodes react reversibly with the CI (aq) iorls. But for an arbitrary salt, My A " , J, = v.+ Jh (Problem 14-32). I According to Equation 11.32,
Brcduse the elect~nnsare in qimilar (s~lrerelectrode) phascs in the twu electrode uonlpartmenl5. thc chrnlical p a n of thc eledrochcmical potential cancels in fi; fi: , Lirlds d
-
Suhjtihrting J-
= 11,
J, and Equation 14.88 into Equation 14.87 gives us
where fl and 4"re h e electrical potcntials of thc electrons in the silver electrodes. 'She diHerence in these potentials is the electromotive force of the cell, E . Thus, u7c sec from Equations 14.82 and 14.83 that the emf of the cell shown it1 Figure 14.6 i s given by Introducing the electroneutrality condition z+v+ Equation l4.M gives us lPmblem 14-33)
+ t-v..
= 0 and Equation 14.85 into
PI-crblem 14-3 1 helps you sh?w that
T S ~= ~ ,J , A f l q + I E + J w A f ~ , where if tlrc elcclrodcs react revcrsihly with a zl-valent ion.
r, the electric current passing across the liquid junction,
(14.90) is given by
11-8. Thc Magnitutle c j l rhr I iquidJunct~onPotenual Ilr[>rrid:tpuri Tr,in>porr V ~nberi
N o k thal the force that drivcs ihc clcctric currrnr across the liquid junc~iot~ is the emf of the cell, and not the liquid junction potential. Equation 14.90 is~i'tquite our desired form for T s ~ , ~The ~ , . total vr>luine flow is givcn hy
'v_
Rut >> V,, especially in a dilute solution. so J , z Equntin~l14.90 in the form
about M)S: of the ionic curretit. 'The fraction of the electric current carried by an ion of !ype j, I ; / I. is called the trrmsport n~tmbrra i d is denort.d by
+
~ ~Thus,7we can ~ write .
Experirncntally, a concentru~ioncell such as that depicted ill Figure 14.6 is run such that J,,, the vc>lume flow from one cotnpartment to the other, is zcro, so uur hasic eritropy production equation ir
For a binary salt. t+ t- = 1. Tmnspon numbers can be deterniithed by experimental methods (see Problem 14-28). U'e can wntc Equation 14.97 in terms oFt- = I+,/I.
Sulving Equation 14-99ror
r+
varlzty of
gives
The afsociated phenomenological equations itre
- "-L
Ll2-
L,,
The emt of the cell is determined untlcr cunditions of zero curxnl flow. so if we sci I = 0 in Equations 14.94. wc find that
il
v + z ,F
IT we substitute Equation 14.100 inlo E q ~ ~ a t i c14.95 ~ n and substitute t h a ~rebult inlo Equation 14.H6, we ohwin
+
The rntio L , J L , , has a nice physical interpretation. To see what this is, f i r s t divide 1, by 1 u~Equations 14.94 with All, = 0 to get
w h e ~ ewe havc used the O n s ~ g e reciprocal r relation in going from thc second to the third ratio. But using the relation J.+ = rJ- J, (the electrodes react rwersihty with the anirli~l.w e her
'1-I,c
I ,it it) I I ,'I in ~ ~ u iEtl rt~14.97 i s thc fructir~nof the total inni; cLlrrrnt thal is hy the cetinr. S o ~ ~ t r i l tu r y what you ~lliphlhave thought, the cntinns and anions do no1 neressarily c;irry tht s;lmc alnoilnr of currrnt heuousz they do no1 rhrcessarily tnmc .iI thc same speed. For exun~plz,the chloride ions in n NaCl(ay 1 rnlutit~nw r y
cal.1 led
Last, we use b p s = IJ+ A,?+ u- Afi-, electroneuual~~y ( v , + V _ Z _ = 0), and the tact that r, -t r . = I ( 1 , and t are thc fractln~~s of current camed by the uatlona and anions) tu pct (Probletn 14-35)
If we use the rzlatic~nsA p , = R T I n ( ~ + , ~ / u)+and , , A p - = HT In(u , , / m Equation 14-101 becon~es
,
,), thcn
Let's apply Equation 33.102 tci n 1-1 rlcctmly~e.If w e replace the aur~vitic.: by rimcentratinns ill Equation 14.102, we habe m , - ? = r . , -l = t.,, c r - , ? = r . - , = (,., .*milrr = (. ,, = r > , , and so t i + , = (,+, , =
,
,
14-9. Tbe L~quid]unction Pofential Is kvell Approxirr~dledby thc Hedwsun Equat~on
Equation 14.10.1for this cell is
Note that Equatiorl 14.103 says that thc magnitude of the liquid junction potential deper~ds.upon r - I + in this case. It turns our that t,+ a t , , . , so we expect that the liquid ;unction potential should k quite small for a concentration cell whose cell diagrain is -
+
whcrc we have uSBd the fact that I,, I,,-= I in going from thesecrlnd line to the third Line. If we assume that fM+ is constani and integrate Equation 14.105 from electrode compartment I to electrode compartmenl2, we gct
EXAMPLE 14-8 Llsc Equation 14 101 and the data k l o a to c~timatethe liquid junction pulential at 25 C for the cells
.4f(c)lAgCl(h)lh1CI(rl) ! MCl(r,)lA~Clfs)lAp(s). I,,
SOLUTION:
u,/rnol .L-'
M ' = Na'
0.392
0.010
M' = K+
0.4'4)
M+ = H+
0.H25
0.010 0.010
For Na',f,x+ =0.392,so!,,-
u2/rnul~~-' 0.0050 0.WSO
0.0050
where we have used h e relalion a: = a , ~ a , , - in going from the first line to the second line. Before we can use Equation 14.106, we must deal with the (nonexperimental) ratio a,,- ,!ti, . According to the Debye-Hiickel theory, or its simple extensio~i(see Equation 11-57),
= 11.608. Therefore,
,,
(8.313 ~ . m d - ' . Kl)(298.15 K ) 0.0050 InA$ = (O.608 - 0.392) 96 4 ~ c5 mol-' ' 0.0 10
.
Iny = Thc values fur K' and Ht are -0.36 m V and + I 1 .h rnV,'.respectivcly.Theexperimental values are es~rmaterlto be -3.68 ntV, -0.33 mV, and + I 1.13 mV. r e s p t i % e l y .Note Ihaf the magnitude of A$ for KCl(aq) i s aboul a factor of 10 smaller than the others.
I
1.173i;?(Zr/r n ~ l , L - ~ ) ~ ' * I
+ ( l r / ~ n o l . I .I)"'
Note that rhis equ~tionsays that y,,, = y,,-, so n , + = a,- in each compartment (Problem 14-29). Jf we accept this result, then u7ecan substituteac1- = (U,,U,~ )"' = (a:)'!' = a , into the second term on thc right side of Equation 14.106 to obtain
Equation 14.101 i s oftcn written in the differential form
Equation 14.104 is a fundalnental equation for the liquid junction pr~tenddWe will apply Equation 14.104 lo a number of special cases in the next section.
14-9. The tiquid Junction Potential Is Well Approximated by the Henderson Equjtion Once again, we comider a cell whose ccll diagram is
We used Equat~on14.107 (with activities replaced by coricentralions) in Example 1+X to calculate A @ fur several concentration cells. We can also use Qualion 14.104 to calculate rhe values of the emfs of the concentration cells in Example 1 4 8 . We start with the first line of Equation 14.105 and use = 1 - t,,. tn obtain
!,-
14-9. The I i q u ~ dlunctiun Porentiai
+
whcrc we hale used E q u a ~ i o11.32, ~ ~ which says that p, = v+jr I v p . Because the electrodes react reversibly with the C1 (aq) ions. we now use Equatior~14.84 in the differential form
Is Well hpprux~mdtcdby the Hendrrmn Eq~ral~orl
613
ion in a dilute gaq will bc accelerated by an electric ficld, but an ion in a sr,lut~on~ u c h as an aqueous sulutic~nwill quickly (of ihe order of 10 s, see Problem 1439) curue to a constant vclncitp as the viscous drag on thc ion balances the torce arising flnm (he elcxtric field. If wc let v, be the (cljnstant) drift \-clocity r d t h e jlh ion, thcn thc mobility of the j th ion, it,, is defined by the equalion
''
to write Equarion 14.108 as
If we iIssikme thar 1,- is constant and integrate from electrode cornparlment I to eleclrorle cnmpanment 2, we obtain
whcrc we have used the fact that u = u:. Recall that the electrodes are rci~crhibll:wilh respect to C1 (aq); it they were reversible with respect to the cation, (hen t,,, would bc replaced by -1, ,. tn Equahon 14. I09 (Problem 1+36)
I
E X A M P L E 14-9 Use Equaliori I1.1(W and the d;~t;~ in Example 14-8 to calculnte the vallle nl' E ior
I
each cell in Example 1 6 8 .
where & is the electric field strength. Thus. the mobilily is the drift velocity of an ion in a unit electric field. Because the units of v, are mes-' and thuse o f f are V . ~ n - l ,(he SI units of mobility, u,, art: m2.v-'.s-'. For a 1-1 electrldytc such as NaCI, the current density, j. is given hy
Note that the utiits of j are (mnl.dm 3 ) ( ~ . m o l - ' ) ( r r i - s - 1=) A .m-', whcrc A stnr~rls for amperes. Thc first term of the right side uf Equation 14.1 11 detiotes a current Hux of cations, and thc second term d e n o ~ e as currrnt Rux of anions. The ca~ionsand anions are moving in opposite directions under thc influence of E. but they have opposite sign<, st) thc electric current llux is In thc sarnc direction. Mubilities arr uu\tonlnl.ily laken tc~he posiiive quaritities and Equation 14.1 1 1 is written as
Note that j . = r+:+I;id+E and ,I- = c 1,given by 'l'hc values fur K ' anrl Ht are 17.45 ntV and 29.38 mV,respectively. 'rhc cxperilnental values m eslimated lo he 13.5 mV, 16.81 rnV. ant1 28.3 niV, rcspectivcly. The small discrepanc~cs bctween the calc~~laterl arid [he cxpcrin~clltalresrllts cum? Irotn tllc assumption that t,- I \ coristant.
I F u t', co the cation transport tl11nlht.rr is ,
for a single I - , I electrolyte. For a mixture of 1 - I electrrilytcs, I, becornes (Prohltin 14-40)
To this point, we have considered only cclls in which the two solutio~lsin contact rorilairled the samc salt. 1xt'?r now consider. cells of thc type
Refnrc we can integrale Equation 14.104 for any case other lhan two solution< of the same salt i n contact, we inust cithcr kntlrv 01-assume the relative concentrations of the
ions with recpect to cach other, in addition to triaking some assumption concerning single-ic~nactivities. There are a nuttther of intcgration schemes for hli~ation14.104, but we will discuss only onc of them, which is due lo P. Henderson. Rcfnrc we can preaerit Henderson's intcgration scheme, w e niust introduce the mahiliry of an ion. An
for thc ith ion. We are itow I-eadyto derive the so-callcd Henderson equalion forthc liquid junctiu~l potential. 'To carry out the integrations in Equation 14.1W, we must make assurnptiuns about how [lie I] ?rtd the u, vary across thc liquid~jrinctioit.We will tirst ns
Chapter 1 4 / Nonequilibrium Thermdynamics
the liquid junction is a mixture of HCl(aq) and NaCl(q) and that the composition of
14-9, The l.irruid JunctinnPnrential ts Well Approximated by the Henderson Equat~on
and so
the solution varies as
c l = (I
-x
)
and
CFiaCl = X r ?
where x varies frum 0 to 1 from electrode compartment 1 to electrode compartment 2. Nule that r,,, = r , and r,,B, = O when x = 0 and that c,,, = 0 and c,, = c2 ahen .r = I . Thc wriable u is the proportion of the solution in electrode compartment 2 that makes up the sulutinn within the liqu~d junctiun. Generally, at a position within the liquid juncriotl where the mixing fraction i% x, the concentration oC ion j will tx
where (;,,
and r,,; are the compositic~nsof iun j in electrode comparlments 1 and 2.
lrspectively. For the HCl(r.,) NaCl(c2)liquid junction that we discussed above,
rNs+(x) =
r-,
f
cNJ1 (x)
,,+("C,-~-~CI
t,,
= I .Thc
2.21 I
- 0.901~
numerical results are plotted in Figure 14.7.
T h e mobilities nl variouq ions at 298.15 u;10"
m Z . ~ -. s l- l
K at infinite dilution. Ion
~/lf.-m ~ '.V-[,s"
-1)
(14,115)
= r N a - , z=~CIX
r , , (*)=r.,,
+
[ ) . S191
-
T A 8 L E 14.1
ton
c,,,(*) = r . , , + , , ( I - x ) = c , ( l
tYa+
(0.1W)(5 19)r - I) 9011-
2.2 1 1
I )X
= cI
+ (cl - c l ) - ~
We must also make assumprions about how the transport numbers vary with solution composition. Ibdo this, we simply substitute Equation 14.114intoEquation 14.1 13 to obtain
Br- (aq) CH,COO-(aq)
8.09 4.23
Cl-(aq) 1-(aq)
7.91 7.96
NO; (aq) OH-(aqj
7.40 20.6
where we ascumc that the mobilities are constants. When these approximations are used i r b Equation 14.104, tbc ~ s u l t i n p equation for A$ is called the Henderson equation.
E X A M P L E 14-10 Use Equations 14.115 and 14.1l h and the mobilities listed in Table 14.1 to plul the transpurr numbers of t l l N a ' (4'1). and C1-(q) as a function of mixing fraction x
for the liquid junctlon IiCl(0.050 M) 1 NaCI(O.1W M).
5 0 LUTlO l\l : The denominator ul-Equation 14.116 is
Mixing fraction F I G U R E 14.7
Na+(aq), and C1-(aq) within the liquid junction NaClIO.100M) plotted against the mixing fraction according to thc
?he transport numbers of H'(aq),
HC1IO.QSO MI
Hendersor~integration scheme.
L.,el'.r
f 4-10, The Flux-Force Relations for I_'oritinuuusSystcms
apply these approximations to a cell such as
lntlolve Gradients of Therniodvnamic Quantities Rather
T h a n Differences All the systerns wc havc discussed in this chapicr consist of two conlpartments, in which some ihzrmodynamic property such as temperature or concentratitm chilnges 1atI1e1,abrupt1 y from one cf>mp;lKtnent to Ihc other. In lhc vocabulary ill-nnr~cquilihrium ~hern~odyr~arniva, we say that such systems arc discontinuous. \Ve can irnaginc systclns such 3%a tnetnllic bar in which the rzrnperature vane< slnuothly from onc cl~td to the other. a>r a aolution in which the concentr;ltiol~varies so~oothlq'h-rlrn rlnc rcgiur~ lo another. We say that systems like these arc continuous xystclns. 111 [his cectiun. we will show how llonequilihr~umthenriodynamics can be forlnulnted to treat co~ltinuous systems. For simplicity only, we will consider systemc in which the pmpcrties v;lry in only one direction. which we take to hc along ihe .w axis. We now assumc that we can subdivide the system into slices of thickncss Ax = .k, - A , (Figure 14.8). which i., large enough tlrat therrriodynamic ~ i l r i i t h lsuch ~ ~ as 1.. P, and p llave well-dcfitictl local values hut small enough that these local valucs are the '.;amcveryuhere within thc rcgion of th~vknessA x . We furthcrrrlnrt: assume thai heh he Inca1 tlicr~nodyr~u~i~ic quantitie:, satisfy tliz same theniindynarriic equatior~athat we havc idcrived t i ~ requi11bnuin systerns. These two assumptions are formali7cd by the Posrularr r$ Lorul Eqrrilibrhm, 1-his pustulate obviously places restrictions on the systems we ca11trt.;lt. hut eaperie~lceshows that it is a good approairnaiio~~ for rliarly systems. Most syuetus r l f chcmical interest vary rather smoothly on a rr~checuIarscale. with nntable exceptions bcing systems involving phenumcnn such as shuck waves or exploaior~s. Ler's consider Eq~ratiun13.23,
(I;) = c,,. (.r). I f we substitute Equation 14.114 into Equation 11.1M and integrate, then we obtain (Prt~hlem1441)
because c,,
ill agreement with Equn~ion14.107. Nnw let's consider a liquid junction such as HCl(uf. KCl(r). I n this caiz, h e Hcndcrson equation yields (Prohlcm 1 M 2 )
I
E X A M P L E 14-11 Ljsc the data in Thlr 14.1 lo calci~l.lte the valt~cof A $ a1 25 1: for thc liquid jt~ncliun
SO 1.U I'IO N , IJclr~gtilt
mobilities in Table
13.1. Equatio~~ 14.117gives us
in g o d agrutnlent with the expcrirrlerlv~lcaluc of 31.2 V. w
1
There art: other integratirm schemes for Equation 14.104, but Henderson's is relatively simple and gives satisfactory agreement with exycri~nentalvalues.
F I G U R E 14.8 The geo~netrjused to dcrive Equation 13.119. The pusitively dirccrrd fluxr5 J,, and J,: are rhvwn poiuting iri the pusitive .Y dircction,
14-lil. Th? Flux -Force Relations for Cor~tinuousSystems
which ;~pplicsto the discontinuous system shown in Figure 14.3. We now apply this equation to any subrrgio~~ of volume A Ax and divide Spnvlby IJ = rl Ax to write
)
n ACjijT) + -1 ( d1 -AX
11
A
cdt
whcrr A ( I , / l ' ) = I/TZ - l / T l and A ( p / T ) = P,/T: sn~all,wc write Equation 14.1 1 K as
AX
- /L,/T,.
(14. 1 181
-
K e a l i ~ i nthat ~ AX is
61 9
E X A M P L E f 4-1 2 Discuss the unla irr Fquat~on14 123. S O L U T I U Y : Thc utjitsof o = $,,<,/V are J.K-' .rn-'.\-',
~ ~ ~ h e u nuif t q = 7.n The unitsofthe $'nremol-rn-'.s-' audofrhc J p , ;;).I a r t 1.r n r j l - ' . n ~ -' . s o thc u1ii1~O I ~the right side 01 Eqrvutirl~l14.123 are ( r n n ~ ~ n ~ - ' . s - ~tnti1-' ) ( ~ . .ln-I I = J - m - 3 . s - 1 , in agreement with the units of 0 = T o . arcJ ni
'
':
We call give aphyslcal Interpretation to the 1 : in Equation 14.123. If wc use the fact that a flux. J, , is equal to o, v , , where I#,is the veloclty of compnent i (Prohlem 1 1 4 ) , then the J," can be expre5sed as
whcre
The qu:\nli~ies.I, and are calledflures. Note h a t Jri and Jfl in Equation 14.120 difler l i , ~ w nJ , ; and $ in Eqlwtioo 14.24 hy a factor of i / A . The derivatives ill Equation 14.1 19 ;we called gradients of I / T and -11 / T. respectively. As an example of the application of nonequilibrium thermodynamics to cont~nuous systetns, we will consider diffusion in an isothermal system (iso~hermaldiffusion). In this case, we have J),
Thus, we we thal J y i s the fluh uf sulutc 1 rclative to that of the holvent (component I ). Let's a ~ n s l d e ar binary colulion, such ap qucrose in water or sodium chlorlde in water. In xuch cnseu, there is only one independent Row and Equation 14.123 is
with just one Ilux-Force equation whcre cP simply stands for 7 ' n and rl is the number of compoilenrs in the system. 'Chc chemical potentials i n Equation 14.121 are not independent because of the Gibhsnuhem equittloo (Eyuatiun 10.10), which we can write as
Equariot~14.125 can be written in l e m s o f a gradient in the concentration rather than n gradient in the chemical potential by usit~gthe relation I L , = p z ( T . P) RT Inn, and assuming that al = uZ. This gives
+
We will use Equation 14.122 to eliminate the chemical putentidl of the solvent in favor oC the others. As usual, we l e ~ thc chemical ptential of the solvent k fi,. SO
Equation 14.125 now becomes
Substitute this ~esultinlo Fquation 14.122 to obtain
Equation 14.126 has been known experimentally since the 19th century and is called Fick'.r ICIW ~ f d ! f l l i s i nIt~ .i s usually written as
flux = -U-
il c.
a ,T
1+ I
where D IS the dljfusilm coeficienl.Thc negalive sign in Equations 14.12h anJ 14.127 sirnpty means that the directian of the diffusional flux is uppositr: the concentration gatdient. By comparinp Equiit~ons14.126 iiod 14.127, wc sce that D
0. Thc Flux-Fxce Relations for Continuous Systemr
with L , , = L ,:. In an ideal solution, p , dcpcnds only on r? and 11, dcvnds only o n r,, but in a gcncral nt~nidealtcrnary solution. p I and EL^ dcpcnd 011 both c: and (, (P~.ohlern1-5) Therefore, we have
L , , HI= L C:
Note that because L,? > 0 (Equation 14-35), L) is an intrinsically positive quantity. The phenomenological coefficients L?? v q approximately linearly with concentration, so Equation 14.128 says that diffusic~ncoefficients should be fairly independent of conccntritlnn. 'The plots c ~ fthe diffusion cuefficicnts r,T LiCl(;iq), NaCltaq), and KCl(aq) at 25'C against the square root of the concentration in Figure 14.4 show that the diffusion coefficients have only a sligt~tdependence on conccntrariotl. Lct's now discuss ;I bolution of tu7usolutes in watcr (a ternary systcmj. Irl [his case, tyuation 14.123 become%
where pl, = a p ; / a z i .Equations 14.1 30 bccomc
wlth the two flux-force reliltions where the D,, arc defined by comparing Equations 14.133 and 14.134. Note that the reciprncal relations L i z = L,? dn rlnt lead 10 Dz, = U , , , nor does LZj= O lead to 3,= 0. Only when hoth J L ? , = p,? = 0 and L2; = (I will 112, = 0.
EXAMPLE 1 4 1 3 Use the f o l l ~ ~ t i nexperimelltal g data fur a NilCI(O.25O M)-KCliO.250 hI) ~ctllulic>rl nl 25'C LU k e r ~ l Lhe j Onsager recipmcal relation%.
1.. = 1A)!!t C --. D.;p,% -= -= -3 !4211,2 - P L , P J 2
arid
L;,
=
n , : /,, ~ n,,P,, F;#>j - @:>P3;
Substituting dic e~pc~~inlcrltal vnlucs for the vario~~s quantities g i v o w
F I G L I K E 14.9
A plol t ~ the f d~tfusionc t d i i c ~ c r ~of l s LiCl(dq). NuCl(atl), and K('l(oq) at 25 C ag-ins1 the hc1ual.e root r d Ihe corlccr~tratirro.showirig that the rlil-ru\~on ccefticients arc fairly independent
KTL,,ihT-c~nis ~ l ~ l agrcc i h
' = -7.5
x 10.'
arid
RTL,,/M.CIII'.S-~ = -7.3 x lo-'
wirh each ut11c.r within exveri~nentalenur.
14-1 1. A j!cady State 15
14-1 1. A Steady S t a t e Is a State of Minimum Entropy Production
3
Stare of hlinimu~n[n4rnpv Prt~rlurt l o r l
SOt.UllON: Weset Jm =OinEquation 14.134a andsolve for Xn toget
In thcse linal two sections. we will prore some genenl results of nonquilibrium ~hermnodyniin~ics. Firsl, we will prove a result involving steady-stale systems. We will see that thc steady stale plays a sirnilar role in nonequilibrium rhwlr~odynamicsto that which the equilibrium state plays in equilibrium thermodynamics. For s~mphcity. let's consider the cirnple system shown in Figure 14.3. This systetrl is described by Erluations 11.24. 14.25, 14.26. and 14.27,
Thus, we see that the steady-slate force, X,,,is not equal to zero even though its arqociakd flux. Jn. i s . Substitute this result into Equation 14.134 to get
' " = ( L ' L:L. , nn- ~ ) x ~ I f we klrbstitute this result into Equation 14.131. wc obtain Spnd=
with
(
I-l2{,
,;> "
Li,)
I,",,
The inequality here follows from Equation 14.33.
Suhsliruting Equation 14.133 into Equation 14.133 and using L," = Idfit: gives
= 0 and all the fluxes arld forces are equal to zero: thc prrlperties At cqtlilihriu~n. of thc qystcnl arc uniform and (here al-c n o fluxes. At 21 steady stale. the fluxes and forces in Equation 14.133 do not changc with time and %r, > 0. One way tu maintain n steady state i l l Figure 14.3 is lo use very large cumparttllcnts (strictly speaking. itllinitely larger so thal the flux oC energy and rriatter fmm onc campartment to the orhvr does n o 1 aller thc lcnlpernturc or I t ~ echclnical potellrial i r ~eithcr corthparttncnt. Now 1cr.s considcr the case in which only X u is hcld constant. T h ~ sm5e can he achieved expzrirnentally by using a system o f two relatively small crlmpartmcnts in cnch of which {he chemical potential changes as molecules flow from one ro the other, hut where one corrip;i~tmenlis in contact with a heat b a l l at temperature TI and the other cotnparrment is in utlntact with a heat bath a1 tempcralure 1 ; . Contrav to what y o u n ~ i g l thiok, ~t the ct~ncentrations(or chemical potentials) of the t w o compartments du not cqualize, evcn thuugh lnolecules can pass from anc compartmcnr to thc nthrr. In
'fhc folltnving cxample derives this result in ternls of equations.
E X A M P L E 14-14 Detcmiinr. the values u i Xn, J,:, and for the syslem described ahuve, when hcld t ~ x r r l l >Xn ~ ~ict allowcd to ;~djuhttn its steady-state value.
s~,,,
X ,is
Let's hee huw S7TG,varies with X m kccping
X,!fixcd. To do this ~nathematically,
w e differentiate Sprndgiven hy Quation 14.135 with respect to
X nto obtain
But thc right side here is equal to 2J" by Equation 14. I B h . so wc have
uscd the fact that Jn = 0 at the steady slate. A second derivative of with recpecl ru Jin gives ( a ' ~ ~ ~ \ , / a=x2Lnn : ) > O, so w e see that the rate of cntropy production i a a minimum at a steady state. Equation 14.136 has a nice physical interpretation. At steady state, thc unrcstrained li~rcewill adjust itself so that the rate of etltropg pmduction is a minimu~n. Equation I I . 1 3 h is an exarnplc of the I'ri~lciple of Minimum Entropy Prcxluctiun. An equilibriirm stare i s n statc uf zcrn entropy production; a steady sllite is n xtatc of ~ n i ~ ~ l r n uentropy ni production. In a sense, a steady state plays the same role in nunequilihriun~thennodynamics as an eq~~ilibriurr~ slate plays in classical (cquilihriun~)thcrmodqnarnivs. For tl~zilca~e trea~edin Example ]&Id, the rate of entropy production is a minimuni. Thus, if we hold 2; and T, conswnt and let cncrgy flux artd niaterial flux take place until a steady start: is reached. then s will decrease to its minimum value vr~iislstcnt p-' with the fixed ( n u n ~ ~ rthermrdynainic i) force X , .If we were then t o let X,, v ~ n i s hby removirig he contact uf the cumpartments with their respective heat baths, the system xvnuld come tn equilibriu~n,when Sprtd= 0. where we havc
We can rllustrak these results pictorially 111 Figure 14.10, which shows mSprntirlue to decrease until thc point Spn, = 0, where the vystem is now in a state of cquilihriurn. Problem 14-50 shows that a steady state, once estahlist~ed,is stable with respect tu small fluctuations in the rlolifixed forces. In closir~gthis stction. WE will prove the Principle uf Minirtlun~Entropy Production more generally. In a .cteady state, ccrtain (~herrt~odynamic) forces are held fixed, and the properties of the system do not change with time. It is obsenred experimentally that il' the forces X I . X I , . . . X, ;ICCheld I~xed(by contact will1 external hathc, say), then Ihe other forces X , . . . , X,,will eventually attain constant values that may nr may no1 be rero. The tluxzs J,, I?, . . , .I, will attain constant nonzero values, and the liuxes JL+, , . . . , J,, will vanish. This is essential1 y the definition of ri steady state. Now SpI, is given by
,, .
.
14-1 2 . The C lansdorff-Prigogi ne Inequality Appl ics to Systems That Do Not Necessarily Have Lincar Flux-Force Kelations I n Section 14- 1 1 , we showed that thc cntrnpy pn)cluclion in a steady Wtc i x a minilnu~rr with respect to the set of forces the1 are not fixed. We proved it, however, only under the conditioiis that ( I ) the flux-fi~rcerelations are lincar. (2) [he Onrager reciprwal cocfficicnts are illdependent of tirlle. rel~tionsare valid, and (3) the phent~rr~erjolugical Thus, our principle of minimum entropy production in steady st;~tcsis n r l t universally \alld. In this final section. we will discuss a mort: general condition for thc nature of ;l steady- statc. The total entropy prilductiun in a continuous syhtem i s g i v t . ~hy~
where the J( and X I are the thermodynatt~icfluxes and forces, ~ s p e c ~ i v e lThe y . tirllc derivative of P is
.
which we ~ r i ~ast .
.
. X n vanable. Problem 14-52 helps you a ~ t hX I , X : , . . . , X , fixed attd X,,,, attair~sa minimu~llvalue a ~ t h respect to each of the no~lfixcdtorces, prove that X , , , . . . . , Xn.Realize t h a ~the Principle of Min~muniEntropy Production is limild to cases in ~ h i c hall the fluxes are liriearly rclatcd tc~the forces.
The Belgian chcn~istsPeter Glansdurff and Ilya Prigogine have sl~uwn(hat
whereas nothing can be said about the sign of iI,P/i)l. This mean5 that for syatclus that are far from equilibrium, a stationary state does nut necessarily correspond to a srate of niininlum cntrupy production. Mie will prove Equation 14.I41 for the exarnplc of the one-rlirne~jsiutialticat fluw i n anlirnlv. bar of length 1 . In this case. n = 1 arld J2 = J,., the flux of energy as heat, and X, = a ( I j T , l ? x . the s ~ d i e noft thc reciprocal lenlperature, and su Equation 14.141 becomes F I G U R E 14.10
A picturial illuhtration of the principle of mini!num entropy ~>roduction. The quadratic s~~tiilce i s Sp, given hy Pqua~ion13.135plotted apirlsl X I . and X,,. Wc st,i~-t at the puilit PI wlth the coordinates X i , and X,', and then allow X,, to vary. 'The recult is that the hyhteirl will 111ovcto the poirjl P! at the niii~imurnof thc parilhnla ti)rmcd hy the intercertion of the quadrit~c surfi~ccant1 lhe plxlc Xi, = constarll.
Tn get from the srcuntl term to the third terrn. u.c have uxcd the i';~clth31 cross partial second derivativz~%re equal (MathChaptcr D). N o w ir~tegr;~te by parts to obtain
Chapter 1.1 / N o n ~ < ~ ~ l i l i t >Tl~rrrrtr>rlvn;brni# r~u~li s
626
If the tetnpertlture i s fixed 31 the twu ends of thc bar so that a ( l / T ) / a t = 0 a1 x = O and x = I, then the lirsi term on the right side of Fquation 14.142 vanishes and we have
Wk obtain the inequality because the integrand in Equation 14.147 is always positive (see Problem 340 or 8-58),
I
E X A M P L E 14-15
Prow the inequality in Equation 14.141 for the case cyf one-d~mensionalisothermal diffus~onin a binary st)Iution.
We can writc a J U / 3 x in terms of 3 7 ' / 2 t by referring tu Figure 14. II. The difference between the flux of energy at x Ax and that at x , J , (x iA x ) - Ju (x), is equal to the change in energy it) the volume A A x . If we let u he the energy density. then we
SO L UTI 0 N I In this case, we see fi'orl~Equation 14. I24 that
+
have ;uld EO
there is o~ilyone flux and ivrce, with J, = ~ , datid X, = a ( - ~ ~ , / T ) / a = x S~rbstitutc.Il and ,Yl into Equation 14.141 and intcprak by part5 to
i-. 1 :Ticd/i.;d.r). get
'l'he ~ ~ e g a ~ai vi ~e noccurs hccau
The first term on the righL vanishes if we
maintain h e chemical
pcllentials (coticenlr;~-
tions) 6hed at each cnd, so that
We can writc Au as hruomes
AT where p is the molar density, and so Equation 14.145 By [he same argument that w e used to dcrive Eqrratio~~ 14.145 (Problern 1d-531,
a J,:
-=
ax
- aT --PC,%
If we suhslitute this rcwlt into Eqiiatiori 14.143, we find that
Eql~adon14 149 simply expresses thccr~nscrvatio~~ nl Lhe nunher nl~diffusi~y molecules Substitule Equation 14.119 into Equalirm 14.148 tn uhtain
"'I -
at
(2)
(3) (3P;
L, T.",
and so Equation 14.151) hecolncs
F I G U R E 14.11
The geontctry used tu derixe Equation 14.145.which repreuenk the consemation of energ!.
The inrilualiry occurs because T and (;lc-2/illrI),,,, ties
are inlrinsically positive quarltl-
Chapter 1.1 i Nonequilihrium 1 herrnndynan>icr
N r ~ l ethat in both proofs of Equation 11.141, wc did not assume a lit~carflux-force rel;itiun, and so we never assumed the validity of the Onsager reciprocal relalions. Problems 14-55 and Id-Sh have ~ O I prove I the I;lansrlorff-Prigoginc i n c t ~ t ~~a lt yfor thc chcn-lical rcaction s c h c n ~ cX Y and thc trin~lgularschetne given by Equation 11.58, rcapccrively. Konequilihriurr~thenriodyna~riicsas we have prehenterl in (his chapter I S restricted 10 procehses that are near equilihriuln, have linear flux-force relations, and obey thc Oncagcr reciprucal rclatiuns. Although these ;ire not serious restrictior~s for marly cummo~~ly studied processes such 3 s heat flow and diffusion, there are a nurrlher o f i n l potrant cases f o r which h e restricticms art: unavoep~ahle.Noleworlhy examples abound in biophysical arid phgsiolugical systerrjs, in which rapid r n e ~ a h o l i cprocesses mil other hiocliernical reacticlris play c e n ~ r a lroles. I:or~siderahle research has Iocusserl on extending ntlncquilihrium thermodynamics to qystems far from cquilihriuln, particularly i11 biophysics.
+
14-11 . In this problem. w e will show thnr ue call use vilriuus linear ct>rrihi!~at~ons of floxcs and force:, and $till presene the Unsilgzr reciprocal relations. In man) applicalior~, ,>I ~ ~ o n e q u i l i b ~ ithrrmudynamics, tin~ certain lincar flux-force rrlatir!ns ;Ire rimre c r m \ z ~ ~ i c [ ~ t lhan olllers, ~ r l dthc ~ c s u l tof this prohlsm say\ lhst wc can urc ;my convenient linc.,lr cotnbinations that we wanr. W e wilt prnvc tluf rcsult for r>oly a special ~ d s e hut , the ~ . c h u l t is general First sti~rtwilh S,Tr, = .I, X, + .I,X, with
+ l,lzx: J. = L,, X I + L,,XL
Jl
= Ll,,Y1
To keep the algebra tu a rlnnrttlutn. dcfine ilrv fluxes by
W ~ ~ C T r~ C ' mid 11 a r t constants. No* stdvc rhcsc two equation.: f o r .I, and .I: and s ~ ~ b s t ~ t ~ r r c them into S to obtain
Problems 14-1. Cunsicltr a two-co~npart~nunt systcm it1 colltact with a hcnt bath at tcrllpcratbrc T, so rtiat 7,= 7,= T. Let the two-cunipartr~le~lt sysleln be surmunded hy rigid, tnipernieahte wall.;, hut let rhc will1 separiiling rhc rwo companmcnts bt. permei~hlci ~ n dRcnible. Shuw thi~td A = rlrr (11 . - - ;I ..) - rl C', ( P, -- P, ) 5 0 N r m sliuw thal
which szrlea to tlefirle X ; and Xi. Now define Ihe phe~ln~ncr~olngical coeffizirnla !W,, tq
,
Conwrt thcsc equstions into the h r m ut Equatio~~s 1 , nlld show thui ,PI,, = ,I.1:, follrrr\s from L , , = L , , . 14-2. n~\cussthe physical meaning
t d
each term
on the right side uf Equatioll 14.12
14-3. Ehtenri Equalion 13 21 11, include a flexible wall between ths two comp,~~-trnzots. 14-4. Show I ha1 the 1uo Iernl!,
c)ri
[he right hide
r>l- Sprlrl =
J I , XI, + J,, X,, have I he 58me units
ah
14-5. Show thul Fick't law in the iurrr~J,: :x - 11
, IS small.
14-7. I t thc
\ ~ C ~ Z I iInI
/I
11,
- / d l can
he wrilren
ah
.I,,
t,,
-
&he11
14-12. The didgunal Irrrr~hi ~ Equation i 14.36 v;in be directly r c l a ~ ~tod expcrilncrllally rllr~iconducmrtrr. ( J , / A P),rL,) is c q u d to L , . surahle quant~ties.Show that the ~nrr~horiir~ut Shoiv hat the rirmr fr-icol ( ' o r d u c t i ~ t 1~J~, ~~,~ I / J ) ~is~ ~. -q ~u ,ato l L,:.
,.
14-1 3. 1-ih: r r r :nr~tl~lerrrr~o.smmti~~,flo~i~ i r dutinzd as ( J , / A I ) ),,,._,,.and tlw clrrn,rit is delined a!, ( I / A P ) , $ o . Show lhilt lhew quantitlus iuc sqrlal.
w t otlil
.rrrt',r~r~i~rc
14-1 4. lr~stcado f writing the fluxec as lirlzar.cr~nlbi~~atiuns uf I he Irlrceh. wv car) wrirc the force\ For twu fwccs and turofiuxe5. w e t~avc a:, linear ctrnlbtnarlnns of the
F~gurc14.3 tlac a Hexihle *all helween [he two crlrnpurlmenlc. then Sl> , ( ,
-
J[,X,:
+ .(,XmZ+ J , X , .
0
J , = d Vl ! O r iind X,. = ( f', / T , ) - t I'?/ T1).Write uut the linear flux-f(,tcerrlatiuns fm this system. Show that L , , , and L,.,. h a w ttlc E ~ ~ I units. I~C where
14-8. P l o w that L , , L : ? > L ; : ~ 14-9. PIOYCthat L2:=- 0 and that LC,L,! >
Lf!if .kpm, =
XI I.,, X,X,.
14-1 0. Provc that the largest phcnnrr~er~oiogical ct~erf~cienl musi he one of the diagonal ones.
arhe~c h = L , I L ,,, - - L l l L 2 1 Nole . rocdlrzhllhrn& L , , = L ? , .
that
R I i = I(,, as a consequence
r)r tllc Onsafer 1rr.ip-
14-15 . Shri* 1 hat Ihc lsvo electrokinetic qualllilics ( A $ !.I, ), ,, (.\rrorid
Chapter 14 1 Nonequilihriun~Therrndynamics
630 14-16. Derive Equation 14.5 1 .
14-17. At thc beginning rll-Section 14-6, w e discussed thc chemical eyu;llir!r~X + Y and took ~ h t reaclitm . cystem t o be isolated. Derive the same firral resulr col~sideringthe reaction sq'stcln to he held at a fixed ten~peralureartd v01unle.
-+
I L ~ 1 1 y)/ T isequal to R ( ( a , / [ X I r q ) ( m y / l Y lcq)J = Kuxikx, ky,)/.k,,[Y]~q whure u, = [ X ) - [XI*\I and when the reaction systcm is near equilihriutn
74-18. Shnw ihnt A = (
0.2rX1 M and 0.020 M in KCl(aq). Assume that the activity cuemcients in the two solutions are the samc. 14-25. Consider a membrane that is permeable to sodium ions. Show that the Gibbr energy required ro transport one mole rlf srxlium ions across the ~nernbranei.: g v e n hy
14-19. Consider the elen~cutarychemical renclion described by where A$ =
$? - ~ 1 Take , A$ and calculale the valuc of AG.
rt)
he 70 r11V- u N,,. :/tiv., ,, to he 10, and T lo 'PC 37 C.
Shrw that t11c affinity in this caqe if givcn by .A = (u,p, -f- v y v U- V y P y - ~ ~ l * ~ l ! T
11-21. Show that A, = R.I, / k,, lX1c,Iand that A, = R J 2 / k,,[YlCq fur the triangular rcaction sctrclrrc discirssed in Scut~on1 4 4 when it is near equilibrium.
14-26. In r h ~ cproblem, we will discuss the Dor~nunrflecf, which occurs fur s y t e n l s like the one shown in Figure 11.3. Tho solutiuns are ceparated by a membrane thal if penneablc to $mall ions but not to pulymers or protein ~nolecules
14-22. I>ihcufr why the rcilctinn cchemc
If wc ignore any small transport of water mrrlecules, the conditloll for equilibrium ~ c r o c s the mctnbrane is 14-23. 'l'l~isprr>hlrni illuctratcs , u ~ alternate derivation nf the Orlsager reciprucnl relations for the triangular kinetic ~cIrelncd i s c ~ ~ s s eind Section 1 4 6 . First show that A, given by Brluatio~i14.hO car! bc writtcn as
Now
Lice
Equnt~orls14.68 and the fact thal A, = - ( A ,
Show that if we replace activilies by cunccntrations. then
these two conditions
give us
+ A:] to derive The pofe[itial,A ylr, is called the Drmnrm pofcnfiuL Show that c,+,,r ,,-,, = c., Now show that electroneutrality girea r
and that 11-21. Cnlculale the value of the Irankn~enlhranepotential at 298,15 K ol a membrane thal is pcrnmcable cmly lo potossiu~nions if the solution on the twn qidcs of thc tneinbranz are
+
+
=c
and
cK+,?= crl-,? = ('
,?L?~
,,
= r2.
Uce ~hewequaliunslo verliy the entries in the last four c o l u m ~ in ~ sthe follou~ngtable (all concentrations me mol L and the temperatulr is 298.15 K).
'
c ~ f4g+(aq) must have migrated uul c~ithcclcctrode cr~rnparl~rieril. Ttlc fraction ul'~hzt r ~ r a l current flow (0.0 I00 faradays) carried by the Ag+(aq) ions is
Similarly. at Ihe otherelectrode,the drcrrilhr o l llh: mnoullt of Ag- Pay) is(O.I(KH) - 0.0947) mules = O.iKl53 moles. But in the ahser~ceof ~uigratiun,thc passags of 0.0100 taradiljc would tesult in a decwnw of 0.0100 mules. \(I O.MW7 moles mu>[ haw rn~gratedi n t o the elwtnide conlpartnlent. Su w e cee oncc again t h a l
I t is
oftende\irahle 10 suppress the Dwn~raneffect and the abo\,c table shows rhal this can
tx done by adding relatively high conccr~tratinr~a of ~ a l t . 14-27. Considcr tllc ccll whew ccll d~agrarr~ I!,
Wrilt lhe equalinns fur the electrude reactions and the overall rcact~ono l this ccll. Would yc?u call thic a concrntrfition cell? 14-28. 'Fhi5 problem illustri~tesa melhod for dercrrlulung transpo1.r ~ l u ~ n b c cnl)c.r~rr~er~tally. rc Corlsidzr rlle schematic diagram below.
The tuhe i i filled with AgNO,(aq) and here are silver electrcdes at each end of the t u k . When a current is passed through the ccll (cluctrolysjs), the reaction< at tIie twtl elrztmdes arc deacrihed hy A:(>) -* Ag'(uq) e - and Ag+(aq) e - + Ag(s). Thus, if olle mide of chiirgc ronc faraday) is passcd th~nugllthe solul~rln,then one rnnle ul .4r+iaq) iivr mill Irrr~r~ a1 ulie clcclrutlc and one molt will be removed at the other. 'Thc current between the elrctl-des is carricd by Ag+(aqj ions movirjg tn orlu direction and NO;(aq) ion:, nioving In thc other dircctiurr. Btcauac thc .Ag+ (q) ionndr) not carry all Ihe current, they r l o nrlt inure away t h r ~ the j elrctnjdr a1 n h i c h they are h r m e t l as t'asl as lhcy form Situjlarlj. thcy do not a l ~ i v eat [fie r>t!wr slcctrcxie a5 fasi as they are rrmove-ed.'l'he~cfarrs.Ag4!arl) ions ilccu~nulatc~lrok~o(l the electrode at which thcy arc produccd and me dc~,lr.tdaraurld rll' clectside at which thc) iirc rclrlc)vcd. ~~tjc:, Cr)nuidr~-~ l duc c ~ r o l p i hcell ~lictui-ctlall>rnr.1 1 ) be divided I I I I ~two ~ l ~ ~ ~ i c~)unes par,lrnentr, and cuppuce lhal each electrode cunlpartlnent initially coorain, 0.1000 r ~ ~ o l of ApNO,(atl j . Now silppose that 0.0100 fwadnys arc pn~scdthrough the A&NO;(aq) xolulion, and we find that thcrc are O.IU53 n~oIu!,ot AgNO,(aq) i11 the electrtlde c;)mpsrtment in uhich Agl (nq) is producd arld 0.0947 rrlolts in t t ~ ecrlirlpartrnent in which 4gttaq) is removed. If there Rere no rriigralic)n of .lg ' ratlj ions fi-om one electl.xIc c a ~ ~ ~ p a n : n e n l trl the olher. the pkdshagc uf (1.0100 fiiradoys would rccult in an irlurca~edl- 0.0100 mnlc!: A g ' ( ; ~ q )in thc clcctrodc culrjpoi,rnlt.nt in ullich ~ ~ + ( isaprt>rluccd. ~ ) The ob+er:ed incrcasc. howevrr, is rmly 0.0053 mules: thei-efore, (0.0100 - O.(H153) nlolcu = [I O O l i IIIOICS
+-
Suppose now that 638 Care passed through n AgNO,(aq) sr)lutlon and that 3.312 g id' AgNO,(aq) are found in une electrode curnparlmetlt and 4.602 g in the cjther. Calculate the transprl number uf Ag+(aq) in AgPiO,(aq). Aasurtlc that the initial unlc>urltof AgNO,(arl) in the rwcl cornpillirnenl ic ttlc same.
14-29. Show [hat thc lkhye-Huckel thtory says that
u , . = u,
fnr ;i 1-1
Seu Equulioll 1 I .JH
14-30. Use Eyuat~rln14.103 and the fullr>w~rlg data to calclilak ihr ~niuesof the liquid lullctlo~~ potentials nl 298.15 K and compare your results with the exprimental ~alucsgiven belo&. 1Jse ctmncet~trationsin place ol-activities.
-
+
14-31 . Show that E =
AiL,lz,
F for anelectrode tt~arreacts rekcrsibly wit11 rill ion of valc~lcc
14-32. Shuw that Lhe flux of an ion o i a ualt M , , , dL. is given by J , = I:, J< or J _ = J 5 , akre J, is the flux o f neutral sait, as ling u s ncither ion ih pnlduccd ur removed in art electr~dcrcmion. 14-33. L)ed\s Equatiorl 13.W from Erlualirlr~ 14.89. 14-31. Shrw
that
I t c rolunlr flow defirled by Lrbu;ltion 14.92 has units of 1. +
14-36, b h o ~that E q u a l i o ~14.109 ~ becomes
if the eleztrdes react reversibly with the cation inskad
uT the
anion.
'
14-37. Use Equalior~14.109 and the data below to calculate the value of the emf at 298.15 K of cach ccll given. Compare your resulls with the experimental values. Assume that the activity c~fficicntcare unity in each c a w
14-43. Use the data in Table 14.1 to calcu[ate the value of A $ at 2SC Tor the tillowing liquid junctions. Compare ynur resutts wllh the gi\'c11expcrill~entalvalues. All concenlrdlions are 0.0100 mo1.L I .
Junctiun
A+!mV
Jul~ction
A $;mV
14-44. Pnwe t c ~yourstlf thnt a ~naterinlflux. J . is equal trt r71. whew r , is cnr~ccntmticv~ ;it~dt ' i s vzlncily. 14-38. We clai~nedin Section 14-9 that an ion in solution In an electric field will quickly come tn n cotlstant velocity h e c a u ~ cof the viscous tirag on the ion from [he solvent mnohules. L e t the viscous drag be linearly proportional to the velocity of the ion but in the oppwitc direction. Then Ncwton's equation lllr thc motion of the ion is
14-45. Shuw Fnra mixture of two 1- 1 clcctrclytes. 2 and 3, that the Debye-HiickcI tIicory gives
E v a l ~ t eand compare the magnitude of c, = 0.0010 M. 0.010 M, and 0.Itl M .
where f is thejriu;ior~constant. Show that if the ion is at rest initially, thcn the solution lo this equation is
= ( 8 / r , / 3 r ; ) , and p,? = (a@,/ac,), ar cl =
1 4 4 6 . U tider what conditions do ( n / c , / J r , ) , and ( a r ~ , / a c , ) , equal zero?
1447. U x the fulluwing experimental data for a NaCI(0.250 M)-KCl(0.500 M) snlut~nnat 25'C lo verify the Onsagcr reciprocal relatiuns.
nz,= 1.35 x
Nnte that whcn f r / r r ; >> 1. the11 v has a constant value v = z e E i f . We can obtain this result by letting I; iw a constant in newt or^'^ equation above. The mohility is &fined as u = I I ! L so . u r e ree [ha[ rr = Zu/ f.
14-39. A Famr)us cxprcssion for the rriction constant is f = 6agn. where q is the viscosity or t i l t ct)lvctlt ntld u is the rndius o f the ion. This expre~sionf o r f is called Sfokes's lau.. Given that the viscosity o f water IS 8.9 x 10 kg-rn-'.s-l a1 2 S C , estimate h e magnitude o f f for an ion such as Na'(aq). Now estimate f/na and use the result of the previous problem to show thnt I! attains its steady value in about 10-l2 s to 1 0 I' S.
10.' crn2.s
'
U?, = 0.013 x 10-\rn'.s-l f i , , / R T = 5.251 M-' p , , : R T = 1.119 M-'
D,,= 0.22 x
'
cm2.s-[
D,, = 1.86 x 10 cm2-s-I p n / R T = 1.129 M-' prr/N7' = 3.105 M - '
1448. Use the fulli>iving experi~ncntaldata for a NaCl(0.500 M)-KC1(0..50n M) sulut~onat 2 5 C to verirj the Onsager reciprocal rclations.
'
14-40. Show that the transport numher of ion i in a mixture of ions i s g ~ s m by
14-49. Prove- t11ar
q,,,given by Equa~ioh14.135 i s necessarily a nli~lirnurn
at
X,; = X n : 0.
14-50. Tn thlr pnhlerr~.we will prove that a steady state, once eslablished, i s stablc with respect si~lallfluctuations in the \anable forces. We do this here only for a system with two fluwar: and tw? forces. (The fnlluwing prclblem give< a more general treatrnenr.) Let X, be t i x d and XI he uariablc, so that J, J 0 but J , = 0.Nuw lel the variable force X, vary from its steady-state value by an arnnurit 6 X,. Show that 6 4 = L2,6X2. Because L,? > 0, howevar, dJ2 tnust lwvc the same sign as SX,. Now argue that if 6 X , > 0.then 6 J2 > 0 means that X, is made smaller, thus causing X 1 to g back toward its steady-state value. to
14-41. Use the Henderson intqration scheme tu derive Equation 14.107 (with a c t i ~ i l i ere~ placed by conccntratiuns) fur a liquid junction hlUl(c,) ! MCl(c,).
14-42. Use the Hcndeaot~integration scheme to derive Equation 14.1 17 fur a llquid junction
N w argue 1 1 ~the t same is true if JX, i0.
Chapter 1.1
: Nonequ~l~hrium TIternmnd\,nam~cs
14-51. In this problclri, wc will develop a gcncral prnnf nbuul thc rtahility ot t11c stcad) state for a system near erluilibrium. Let XI, X : , . . . , X , be tixed furczs and X , - , , . . . . Xn be variable fnrcc~,so that . I , , = I,, I = . . . = J), = 0. Nnu, let o ~ l uof thc bariahle fnrccs, X,,. change siighllp by d X n E ,s o that X S + X: 6 Xm,with L 1 5 ) I I C: n. Show that S J,,, = L ,q,,,b X ,,z. Arguc as wc did in thc prcviuus pr,oblcrn that thc sign of 4 JSjcwill always tie such that [he flun .Irfi reduces X,,, hack to its cteady-
,
+
+
14-55. Prove till: Cilansduril-Prigogitle illcquality fur the chctnicnl scheme X
2 Y. Dn 111)1
assume t h d t the sysler11is Ilem equilibrium.
14-56. Prrne rhz Ciln~~sdorff-Prigi~ginc incrluality for ~ l l zcherrlical schcnlt given hy tirl11.1tion 14.58. Dv nod a,sun)e that ~ h re;tctiun c i \ near cquilibritrm.
14-57. Conihii~c.Equations 14.127 and 14.139 to derive Ihr st>-calltddiffusirw cqun11011
14-52. 111 lhih prnbleln. we prow that
.
rltl,ii~lsa lninimum value with respect tu each of the nnnfixed forces, X,,, . ... X,,,in a steady state The mathematical condition thrit be a nliniinum iand not a n1;txiinum ur at) lnf ccticm poitit) i? thal
untl
Nuw show that
The solutions trl this equation give the ct)ncc~~tration of ;l diffusing \ulntance at any prrinl and tiine. starting trom some intitial dihtr~butionof crmcentrution. l'lls ~nlutioniu lllc diffusi4,n equatloli if all the diffuring substance is inillally lncntcd at the uripin ic
where 4 1s ttw total cuncerltration, initially Incated at the origin. F i r r ~show that ( , ( I , r > does indeed hat~kt!,the diffusion equation. Plot c ( u , t)/r;, versus x for inclrasing vnluc\ of Dr and interpret the result physically. What do you chink tlie follow inttgml equals?
14-56. We can lllterpret c ~ A r- ,) d r / c , ,in the pl-etm~ous problztn as the pruhabil~tythat a diffir\irig particle is lucated k t w e e n x and x -td.x at time t (hlatt)C:hapter B). Firs1 $how t t l ~ t <.(I, I ) / < - : , is normalized. Now calcr~lalelhc avcrnge distance that the p a t i d e will he fn~ltld
Iron1 thc ar~gin.1nterprt.l your result physically. Now show [hat and
Th: root-mcan-squale (i~stancetrnvcled by a difrusirlg particle ic Now shun. thal
The cuntlitiun Ihat .in,d be a min~mum1s that I,,, z L:,, u hiich i h t.x;~ctly ~htlbamr a\ Equation 14 32. xvhlch 1s a con,equence ul thti pu,iilve hlinlle na[ure rub Spml{see Problcln 1 4 8 ) .
14-51. Use the approach of Proble~n8-57 to repartition n 2 = nM1,,*b e t w e n two initially identical con~partt~lct)ts to show that {i12.4!;)n;),,,,,n > O and that idp,/,Jvr, I,,, = ~ - 1 ( i ~ ~ (2, 7 )/, ,~ >c 0~ )
'
Given th.1 U = 10 crn2.s-' for all ion tliltusing in walcr, calculate tllz viiluc rjl lo-'' s, 10-' s, and 10 s. ~ ~ S C U your E S result%.
I,",,a1
14-59. Cirnsidzr a clowd circuit consi\tirlg of two dirferer~t~nctalliccon~luctorsu-heie the junctions be4wot.n the dissimilar metals are tnaintained a1 uniI~rcnt teinpcl-alures, Ti ~ I I L I 7:. It turn\ OUI that a voltage wilh a resulting electric c u n r l ~ twill he g c r ~ e ~ a l eilld this circuil. This eticct was firs! r)b.cerved by T.J.Sezhrckin 1822 ;utd 1s now knuwn ,I> lhr .Sc~ht,rI. c f l d ~ rThe . ~ i ~ r r ~ h p ~ l l clnf d i l l in p lhc circuit ( m c ~ s u wlth d a potrntiumrter) dcpcnds U ~ I I rhc tctnperatilres al the two ,juoctiuns arid upon the Iwo n~ctals.'Thus, !f unz junctlon 1s liked ar a kn4,wn tel~~peri~lure, tt~cnrhc ternpt.rdlllre at tllc other ltlnctic>r>can he J e t r ~ - ~ ~ ~ i r ~ t . ~ l by mca~uringthe emf. Such a device is called a tlr~~r.ttrot.o~tplc. ~ltldis uhed rxtrnlr~vclyto nltasurr Ivrnprature. [,el's now cons~dzlthe inverse ~ l i f t ~whcl-e t, lhe trro jul~ztiuiicart. at Ihc snme temprrature a ~ t dd n e l ~ h currcnt c is mainlained across the junctirr~~s bctween the lwo nwtals. 111 this caw, thc :empcralure at nnc junction will increase and the ternperaturc at the other w ill decrr-we,ui>lts
Chapter 14 /
638
Nuncq~r~l~hriurn Tlicrniutlrr~arr~~r-s
eleclric current and changcs sign when the direction or the current is reversed. Thc evolution nr ahsorpl~or~ olenergy as hcrtt when an electric current passe\ across the junction uf lna dissimilar rnetals i s called the Peltter efir!, after its discoverer, J.C.A.Pe1tier. We detine the Pelticr cnefficient IT by 11) the
J,: = X I
(AT
Answers to the Numerical Problems
= U)
wherc .I, i~ t l l ~R ~ I X of ellergp i ~ heat s r ~ - i n - ' . s - ' a) r ~ dI is thc flux of thc elcctric current ( C - m 2 - s -). l'hc cntropy pnxluctiun etl~raliunlljr a ~t\ern~ocouple can bc written as
whcre 4 i s d ~ electric e potential in the circuit. Show that both sidcs of this equation have tllc somc units. Write out the two linear flux-inrct. relations uld show that
1-4U. 1.3h x 10" revolution-s
1-1. 1.50 x 10''s ]:5.00 x l b c m '; 9.93 x 111-'' J 1-2. 3 x 1 0 " s I : 1 x 1 U r m ; 2 x IO-".~ 1-3. 0.67ct11-': D.UI5 m: 1.3 x 10 23 J
1-41. I / c ~ n - ' = ZO.HOh(J 1) - 0.0018(J I)': B = 111.4~3 crn-I: fi = 0.00044 c ~ n 1-42. I / c ~ n - '= 3 . 8 4 5 4 J 1) - 2.555 x IO-'(J 4-I)'; = 1 9227 crn-'; fi = 6-39 x lo-" cm-' 1-43. l h . 9 2 9 c m ~ l : 3 3 . 8 1 9 c m - ' ; 50.753 cm- I; 67.631 ern-'; 84.477 cm-I; 101.28 cm ': 118.04 v m - ' 1-44. N 2 : 3 . 2 , l:CIH,: 3 . ? . 7 ; C 2 H , : 3 , 3 , 12: C,H,: 3.3. 18: C,H,: 3. 3.30 1-45. 11CN: 3 . 2 . 4 ; C D , : 3 . 3 , 9 ; S 0 3 :3. 3. 6 : SF,: 3, 3, 15: (CII1),CO:3, 3, 24 1-46. 8.845 x 10-"' J = 53.27 kl .nlol. 1-47. 1.871 x ll1-ly I = I 12.6k ~ . m o l - ' 1-48. The center of mass lies 7.2 pm beyond the central nitrogen atom. I = h.645 x 10-'"g.1n~; I j = 0.42 I 3 urn-'
1-4. Z x 10-li J
F., i u 1neasurt.d under [he conditiori of IIO currcnt flow) This equation r,clating ttlc Sccbcck cffcccct and the Peltier effect ha\ been known experitncntally since thc riliddle o i [he I KOh. Not until Onsager's formulation of the reciprocal relations waq i t derived currectly, h(!wrvt.r. There were scvcral previous derivations, none uf which was correct, ant1 some r l i which were quite biz=. (Keca11 that the emf.
'
Chapter 1
1-5. ( a ) 1.07 x ID1*: (b) 5.41 x 10" photons: (c I 2.b8 x 10'' p h o t o ~ ~ s 1-6. 1.338 x 10" s ';3.139 x 10-Iq J 1-7. 1.70 x 10'' photon ,s-'
1-8. 2.178 69 x 10-IXJ 1-9. 3 1-10. 2 1-12. 13.398 r V 1-14. 1.282 x IU. ' 111 1-15. 52.917 72pin 1-1 B. Thc final equntion illustrates conservatinrl of cncrgy. 1-21. 9.IN 432 x 10 '' kg 1-22. 2.178 638 x lO-I3 J 1-24. k = 2 4 e ' : 479 ~ . t n - '= 479 kg.s-' 1-25. 0.01KI jrn" 1-26. y = - 6 1 1 ~ ~ ' 1-27. i= 0.1119hl; 25 = 56.59 cm-' 1-28. 4M ~ . m - '1.27 ; x s 1-29. i: = 9 63 x 10'' s-I: r:, = 3. I9 x IO-." J 1-30. 83.9 N.ri1.': 1.20 x 10-"s 1-31. iCk5 jcm ' = 2!38R.7!: - 5 1.56Sv(t1+ 1) 1-32. 1051 Scnl-!: lih7.?1cn1-' 1-33. 1 = 21h9.0~111-':? i s = 13.0cm-I 1-36. 1 = 3.348 x LO ' kg-m2; Rp = 142.; pm 1-37. 798ROcc1-' = 955.6kJ.rnol 1-38. 113.0pm 1-39. 305.5 pm
-'
+
+
'
+
'
Chapter 2 2-1. 2.98 x l ~%atni.3.02 x 10"at 2-2. 7.39 x 10: torr. 0.972 atm 2-3. 1.00 atm 2-4. -40'
AnsweE tn the Nu~r~eric-C! Prublvms
2-6. 3.14 x lo4 molcculcs, I .X5 x 101\cm2 .rno~-] 2-7. 44.10 2-9. v,-, =0.77.y, ~ U . 2 3
Chapter 4 3-10. ( E ) = Nk,T
3 3-11. ( E ) = - N i , , T 2
ON' --
v
1
2-1 0. 2.2bar. 2.2bar 2-11. TIz 2-12. 62.3539 dm' . t o r r . ~ - .rllol-' ' 2-15. 0.11499X d~ri'-nlul-I; 11.03XhS dm' -mol I ; 0.01663 dm' .rnol-] 2-1 6. 353 bar; 8008 bar; 438 bar; 284 bar 2-1 7. I03 1 bar; 41 1 bar 2-18. 10.00mvl,L !; 10.28 ~ n t d . ~ - l 2-1 9. 1570 b x ; -4250 bar 2-21. K-K: 345 bar; P-R: 129 b;u 2-22. 0.07073 L,mol l. 0.07897 ~ . r l l o l - ' , 0.21671,-mol I : 14.14 m o l . ~ - ' ; 4.h IS r11oI I , , 2-23. l l - l 2-40. 1 kJ vs. IOU kJ 2-43. Yes 2-44. - 15-15 cn!' mul ' 2-45. -60 c~n!-mu1 ' 2-52. (C.m)'(lu3)i[C:.&',kg-',nl-')(nl") = kg ln2.s.l = J 2-54. 5.86 x 1 0 - ' V J ~ n " :1.35 x IO.-" I , m 6
-',
'
MathChapter 6 B-I.
4-5. f,IT = SoOK) = 9.0 A 10-"; f2(.T= IOIHIK) = 4.9 x IO-~'; f,(T =ZO(KIR) = 2.2 x lo-"
3-18. Cbis a f u n c t i o ~of~ T' = T / Q , , whcrc (-1, = h v / k, 3-23. Pa) h (h) 9 ( c ) 12 3-24. 9 total. 3 allowed 3-25. h nllowed lerrns 3-26. 27 total, 1 dlnwcd 3-27. 10 allowcd tcrnis 3-28. 1.94 x lo-" 3-29. 0 092 8-30. 142(3 3-31. (1.286 3-35. f, = p - h b " % T (1
-
e
":**'
);
0,9999,
[.(I1 x 10-" I003 x 10-l2 3-36. 1.000.0.!4Yh2,0.Y65[)
+
4-7. Uc = L)] f R(-jk* CO: 10S3 !d-tl~ol-I;NO: 638.1 k ~ . m o l - ' ; K,:54.1 kl.mnl-'
MathChapter D D-1. K = l i P D-2. u = 1 / T D-4. ; ~ k , 7 = i n ~ 7 ' 1-11. f,,, = P --).,-:J = 1.01 x I O - V O ~ H ~ ; D-7. O ; n / v 2 :~ A / ~ T ~ ' ' \ ;B!) v 0.0683 for CIZ: 0.358 for I?; etc. D-9. 0; 0,. - 3 ~ / 4 ~ ' ' ' l ' ( l+ ' 8) 0-10. exact 4-1 2. 87.4 K. 4?.7 K 0 . 1 1 . inexact; cxuct 4-13. 90r IU 4-8. 6332 K, 4462
K
+
4-14.
N:: 0.32%:
4-16.
* 2114
H:: 9.45%
3-41.
tqiactiori of Fraction uf atoms. I(HK) K atoms. ?5MK
,
4-22. Utkk = 227.6 K: C3t,cc,, = 325.9 K; - 5 + A ~-!!b h , l 3?>1K/I q:1m Y.803 at 5!HHI
+,
K
4-23. 2. I. 12, 24.2. 1
,
MathChapter C C-1. 1.25 :-:lOK'%,197 x I 0 ' 5 . O 467'A
C-2.0?498,U4?9%.
.,492S
4-24. t-lvl = 2-1-11 K : Ov,,, = 3016 K; i4v , , , A = 1 0 3 K; I-),,,,, = 3765 K; C,:R = 6.21
4-25. 2: 1.JbO r H)-" kg.ln2; 1.7U2 K ; 2XJZ K:3649 K;4715 K; IU49 K; Rh3 3 K;4.33R
4-28.I = h.74t x 11U (:Im = 0.547K.
kg.rn2.
4-29. see 'I'ablc 15.4: 5.30R Chapter 3 3-7. -hyB1.I)
4-39. (a) 3K!2 (b) 7 K / ? IC) 6 K !dl 13K/2 (c) 121I
,,= 4480 K; O ,,,,,,,= 5464K; U
4-34. U,, (4
?,,,,,
n,,HD
= 42.6 K; = 64.4 K
Chapter 5 5-1. KE = 9.8U kl; 11 = W.3 rn.s I; 27.1 C 5-2. 15 bar; 3000 J 5-3. 28.8 bar; 3 60 J 5-4. 4.01 kJ 5-5. - 1.73kl 5-6. I I ..IkJ 5-7.+325 J: +309 J ; they d ~ f f tbccause r 711 i:, il path fl~nction 5-9. -3.93 kJ.mo1 5-10. -3.92 kJ-invl 5-12. V,=11.351.; V2=22.70i-. T 2 = lO90K: AII = 1 0 . ? k l - ~ n o l - ~A; H = 17 0 k ~ . m o l - I ;q = 13.6 kJ; u. = -1.40 kJ 5-13. 418 J 5-19. T, = 226 K. 11: = -898 J 5-20. 519 K 5.21. 421 K 5-22. q , = 122.9 kJ-mol-I. AH = 122.9 kJ-mol I . AU = 113 2 k~-rr~all-'. u: = +9.77 k ~ - m r l l - l . q , = 113.2 k ~ - m u l - I , At1 = 112.9kl-mu1 I, AIJ = 113.2 k ~ . r u o l - ' w , =0
'
Answers tu the
'
5-23, h,U = 288 3 kJ in01 5-24. 74.6 kg 5-25. 2rl5 K 5-26. 1340 kJ 5-35. A , H = 416 hJ 5-36. A, I I = 521 6 kJ 5-37. A r l l = +2 9 kJ 5-38. A , H"[iructose] = +1244.3 kl,rnol 5-39. rnt-thnnol. -22 7 k~.g-'. N,H,(I) = - 19.4 k 1 . g 5-40. 32 5 kl 5-41. (a) -44 14 k l , cxolhermic ( b l -429.87 kJ, cnotherm~c 5-42. 47 X kJ mol-' , 44 0 kl mol-' from 'I',~ble 19.2 5-43, l J h '31d kJ.mol '
Kurr~eric-a1 Prvblems
6-9. 19. I 3.~-' ; positive because the gas is expanding
-
6-1 0. qwU= Pl(V,
Answ~rslo :kc Numcr~calProblems
7-10. Suhslanct:
-
I',1;
Pentane Hrxme Hcptilr~t.
' 6-1 3. AS = 37.4 J K-' 6.14.
AS = 30.6 J K
'
mol-'
'
Lthylcllc nsidc Reiisz~le Dl ethy l ether Z;.trnct~lnrnrr~elhane Mercury Bromine
-
h-17. A S can ht. posithe or ncgative for an ~ ~ o t h e r mpmcesq. al AS = -5 76 1 K-' 6-10. A S = 217.9 1 K
' 58.7
Penrdne Iiehane
6-19, A S = 4 4 O J . K
Hzptanz
Ethylcnc chide Bcnzc~~c Uicthyl cdler Tctmchlnr~~r~~erhane
73.5 77.6 32.0 35.7 4h.3 13
Mercury Brorriinc
7-13. 1'12 05 J K '-mu1 7-16. 223 2 J , K - ' ,rnr,l- ' compared tu 223. I l - K 1no1-' 7-1 8. 23? R J li- mol 7-20. 19h.7 J ,K-' ,rnl>l-' 7-21. 139.3J.K-' ,anal-' 7-22. 272.6 J - K - -ulcl-' 7-23. 27'4.3 J - K - -n~oL-' 7-24. 1 5 1 7 3 K .mol-'; residual entropy 7-2 5. 185.6 J . K- .t ~ ~ a l - I 7-30. 212.8 J , K- . mnl-I 7-31. 159.9 5.1(-' .rnul-I: residual entropy
E-5. Each urnhe her in a row is the sutn of the two number\ above i t . E-6. K4 E-7. 1.12 x 10. vs. 0.0194 in Table J.1
'
6-41. 191.6J.K '.niol-' 6-42. 2 13 , 8 J - K - ~,mul-~ 6-43. 193.1 J . K-' ,mol-'
6-45. 21% at 1 atm:
Chapter 6
6-2. d :!y
+
i:
6-6. ym, = l : ' ~ , , ( T ) ~ i l
-L
.
...
at 25 atm
' ' '
'
C,(T)~T
V P,~~Y:AS=KITI-~ -
VI
6-8. 5-76 J .K '; positive kcause h e gas is cxp;~nrling
Chapter 7
7-2. 37.5 J.K-' 7-3. 192.6 J.K-' 7-4. 3 . 7 5 J-K-' 7-5. 44.51 J . K - '
7-32. 153.1 J - K '-mol 7-33. 3 5 . 4 J.K ' .mol- ' 7-34. 113.: I . K-': mnl-' 7-35. S i H , , 29s t 5 K j = 130 7 J - K ',mrll I : S'(D2,298.l5K) = I43 9 ~ ~ K - ' . m n l - ' : S3(H1>, 29XK) = 143.5 J.K-' ,mcll-'
' '
'
Chapter 8 8-1.
'
E-4. b
7-36. 253.5 1 . K ,rnol-'. The experimental value is 253.7 J.K-' ,rnr,I-'. 7-37. 234.3 J - K-' ,niril-' The ex~t.rirnental value is 210.1 J,K-' .rnol-'. The dil-fewnce i h due Ir) rt..;idual mtrr)py. 7-38.- 172.7 J . K - ' . 1no1-' 7-39. -49.6 5 K -mul-' 7-40. (a) CC), (h) CH,CH,CH, ( c ) (:H,CH,CH2CH,CH, 7-41. (a) D 2 0 (b) CH,CH:OH rc) CH,CH,UH?CH,NH, 7-42. (d) I. la) > (bl > ( c ) 7-43. ( c ) > (h) % (d) (a) 7-44. ~ranrlmionaltin hr~th 7-45. 239.5 J - K-' -mu1 7-46. 188.1, J K -mu1 7-47. ( a l 2 . 8 6 J . K '-lnol (h) -242.9 J - K - '.mril-' (c) -112.01.K-' , ~ n o l - ' 7-48. (a) -332.3 I . ~ - ' . n l n l - ' Ib) 252.66 1. K - ' .nlnl-' ( c ) - 173.0 J - ~ - ' - r n u lI
'
09'C) = 0;
A , , a p c ~ 70°C) 5 = 0 33 kJ-mol Avmpc(~ O5C ) =
-':
-0.44 kJ-mu1
'
8-2. A w , , I ~ ~ ( ~ 0 . 0= 9 '0; C)
~ ~ ~ ~ ~ ( = 7 435.9 5 . 0 J.rnol ' ~ )- I ; ~ ~ ~ C ( 8 5= . 0-429.8 ' ~ ) J.rnol-I; no 8-5. PV = RT and P(V - b ) = R7' 8-7. - 0.050R kJ.rntll8-8. R 8-9. 7.87 x I O - ~d n ~ ' . h u - ' . ~ - '= 0.7H7J . K - ' .rnql-' 8-1 2. fmrn data: 155.6 bar; from van der Waals: -4.80 har; from Redlich-Kwong: 161.1 bar 8-13. -0.M52 kl . ~ n o l 8-16. ( i + F p i 2 ~ = )T 4.47 x lV4 dm3.rnol-'.K-'; 2, = 25.27 J - K I-mol 8-17. 138.8 J-mrll 8-19. V and U 8-20. 0.01561.K ' . ~ n o l
'
-'
'
'
'
Anrvvers tu tht~Numerical
Chapter 13 13-1. (a)
+ 2 HI(aqj
Prublems
-
PbJ,(c) I H2(ri ib) Cu(s) + 2 A g U l O , ( q ) + 2 Ag(9) + Cu(ClO,),(aq) (c) 2 ln(s) 3 C,d2+(aq) + 7Cd[<) 2li1"((aq) Ph(s)
+
(ti)
+
+
Sri(s) + 2 Ag+(aq) 4 2 Ag(\) SI?' (aq) 13-2. (a) F W s ) + Hg,SO,(?) + 2Hg(l) 4 PbSO,ls)
(b) H , g IIg,Cll(h) + 211g(l) $. 2 HCI(aq1 (c) Z ~ ( F ) HgO(s) + Hg(1) ZnO(s) (d) Cd(s) 4- HgLSO,(s) --+ 211gII) t CdSO,(aq) 13-3. (a) 110erect; (b) decreaw: ( c ) lncreasz; (d) decrease 13-4. (a) increase; (b) no cflcct. (c) decrease: (d) decrease; ( e )no effect; If) increase 13-5. (a) 8; (b) 4; (c) 2; i d ) 20 18.1 0. (a) yzm': (bj 4y:rn'; (c)4y;m'
+
12-49. 2.443 x 1oT2m-I 12-50. 3.84 x 1 0 ' ~m-' versus 3.86 x I 0l5 111-~ 12-51. 1.87 x 1 03' m-' verru.: i 9 1 loJ5 1 1 1 ~ ~ 12-52. 5 66 x 10" TI)-' \ersu+ 5.51 x 10" m 12-53. D,, = 427 8 kl,rr~ul-I 12-54. 11, = 1 fA2 kJ . m(h ' 12-55. D. = 1598 k ~nlol-I . 12-56. K ) , 0.53 1.1; therttorc K, at 503 har 12-57. Rr must bc smaller thmi K, at one bar 12-58. In a 1.08 at 100 bar 12-59. A 7 H ' = 159.2k~.mt>l-': AvS- = 2 1 7 . 7 ~ - m o l ,-K~ - ' : A,& = 91.3kJ.m01-~ 12-60. A r t ] ' = 3 l . h 7 k ~ . t n u r ' : A,S" = 96.511.rnol-'.K I : A,G' = 2.YlOkJ.n~ol 12-61. 38(W) h~ 12-63. 0.05 l n ~ o l , ~ - '
-
'
id)
+
+
y$i2
13-11. 0.2223 V 13-12. (1.984 V 13-13. 0.358 V 13-14. 0.834; 0.745: 0.704; O.hSh: 0.587: 0.521; 0.450: 0.400 13.1 5. U.788: 0.877: 11.863;0.894: 0 . W ; 0.927; 0.94U 13-16. 0.966: 0.95 I: 0.930; 0.9n6; 0.R7tr:
0.838 13-17. -0.268 V, The H,(g) electrt~deis positive. 13-18. 0.4738V. The Phcs) electrcde is positive. 1 3-19. -0.48 1 V,l ' l ~ eZ,nis) electrode is
positive. 13-20. H,{g)lHl(aq)llz(+)IPt(s0.5335 ); V 13-21. Pb(s)lPKl,(aq)lPbCI,(h)lPb(q); -0.140 V
Answc~sto thc hu.neric,>l Pruhlerns 13-22. Znisl lZnCl,(aq)l IHCl(aq)lH,(g); +0.7h3 V 13-23. -0.21 V 13-24. 0 41 9 \.:; yes 13-2.3. -O.?P? V: no 13-26. 1.326 V 13-27. 0.7 12 V 13.28. 1."+9V 13-29. HZ(p)lH,SO,(aq)IPhS04(~)IPb(s); -0.3s3 v 13-30. (1.0732V. U.981, 0.975. 0969.0.963. 0.955 13-31. H,(gI ? AgCI{r) + 2 4g(s) 2 tICI(aq); E ' = 0.2224 V; A , f T = -42.92 hJ-mrll I : A,S' = -124.6 J . I I I ~ I - - ' . ': K 4 , H - = -80.0: k ~mnl-I . 13-32. Ill(g) - lig,Ur?(s) + 2 Hg(1) -1- 2 HBrrrrq); ArG0= -26.96 kJ.nlol-'; A,S' = , 29 7 I - m o l - ' . K - [ : A . f f ' = -33.8 kJ-mu1 I; S 1Br-(nq)] = 83.5 J-mol .K-I 13-33. 1.21 x 13-34. 1.16 x lo-' 13-35. 1.40 x I U - R 13-36. 1.30 13-37. 8.R x 10 13-38. 9.4 x IU-' 13-39. I .?3 x lo-' 13-10. l.h5 x 10 ' 13-41. 1.78 x lo-' 13-42. Thz values of Kw at O"C. t OC. 20-C, 30'C. W C . T O T , and 6U'C me 1.15 x 1u-IS, 2 94 x 10 1 5 , 6.85 x lo-". I .45 x 10- 14, 2 92 x 10-1; ,445 1 o 14. and 9.60 x 1 O-.'",respectively. The correspondirhg balues uf A r H nare (in kJ.rnul I) we 63.88, h0.33, 57.26, 54.67. 52.58. 50.97. and 49.81.The value nf K* at 25°C crnrl-s pur to be 1.01 x lo-' and the value of A,H' comes out to be 55.W k J mul-'
+
+
"
13-43. 1.85 x 13-44. 7.90 x 13-45. 1.64 x 13-46. 3.70 x 13-47. 1.08 x 13-48. 1.01 x
lC1-I and 1.71 x 10 lo-' and 7.99 x 10 10 I O-' 1 (I-"'
"
-'
'
1K2 13-49. 5.65 x 10 13-5U. 9.85 x 10-' 13-51. 148U kl versus 2 1 O X k t ; F,'" = 1 .OY V 13-53. B?,. = 6.1 x 10 atm-' (estrnlatcd). or 5.39 x I fl a m (using tnblualcd
'
'
values)
13-55. 0.852 V 13-56. -U.74 V Chapter 14 14-24. 59 nlV 14-25. 12.7 kJ.mol-' 14-27. H 2 ( P lI + Hz(P?); the cell ir a concentration cell 14-28. 0.47 14-3U. -3.95 mV; -2.31 mV; -0.356 mV; -0.208 mV; f I 1.6 mV 14-37. 13.93 m k 4 1 . 7 8 mV: 17.45 mV: 52.36mV: 29.42 mV; 88.26 m V 14-39. TAe the radius o f the Na+(aq) ion to be 100pm. Then / - = 1.7 x 1 0 - I k g - s I. Take the mass of the NP-(aq) iori tn be 100 amu. Theri J!nt = 1 x 1013s I . 14-43. 26.9 ntV. 26.9 mV; 31.2 ~ I V -4.37 ; rnk -4.37 mV: -5.01 1nV 14-45. p2,/j~?I= -0.013 for c, = r , = 0.0010 M; -0.043 frlr r , = r, = 0.010 M; -0.15 for r , = c, =0.10 M 14-46. For an ideal solution. 14-47. R T L , , / M ~ c ~ ~=~-9.9 s x lo-' and R T L , , / M . c ~ ~ . s -= ' -9.5 x lo-' 14-48. K ~ ' L , , / M . C ~ ~=. ~-1.5 - ~ x 1N6 and RTL,,/M.cm2.s-' = -1.5 x 14-57. The integral i s equal to zero. 14-58. ( x ) = 0; xn,,
'.
Il lustration Credits
CHAI'TEK-OPENIhG PIiOTOS Chapter 1. Cuurtery [IT A IP Em~lioSegk Visual 4rchlves. NT.LMcggcrs Collecliun. Chnprcr 2: b e r l i n p h h n e s I.ahoratory, Leiden. (:ndnc~ynt rhc Callcch lirchi\es, Earnest \VdTson rollcctiuri. Chapter 3: Courtcsh ut Juhrl Simm (memorial stonc) md Univtrrsit) of Vienna. Coulte~y of All* Niel.; Bohr l.ihratv Ch,iplcr 4: Courlt.;j c ~ tIIC Berbeley Chi~pler5: C u u r l t > j at .41P Ern1110Segre V~>ual Archil.63. Phys;r.> 1i1da.v(:allectinn. Chap~rr6: Cr>urlt>yk b f AIP t.:1111lioSege Visual A:chi%e.. L . J ~ ~ l'ollecuon. E Chapter 7: K ~ k s n > u ~ r uvonr m dt. Gtsch~edznis >an de X.tuur~~ete115chapp"n~ I,e~dcri. (:oulrcsy oi AIP Elnillo Scgrc Visual Archi\?,. Chdpter K: Photr>:rdph by I+.l.ange, courtesy of AIP Enlillo Scgri Viaudl Archives. Chapler 9:Ccurtehy c:f ALP Kmil~oScgre Visual Ar
C11i1p:cr 10: Rcprin I cd ct>urle+yrrf The Huncroft I,ih:;lry, Un1etl51ty(11 Califr>rniu,Berkeley C'halwr I I : i l k h y t ) Pnaograph by Francis S~mon.ctjllncsv :\IP Er~~iliu Segrk Viauul Archive!, l'hapter l?: Courlc5r. of :ZIP Et~lrlluScgri. 15su::l ~ k ~ h i v eW> , Cddy Cullecliun Chiper 13 t<eprintcii with perm~usiontmni the Jr>~iu~iirl :t rhrmk,crl Ei!,rr.rr!iun, Vul, 61, Nn I . ilDli4l. C ' ~ ! p y r i ~I?) t ~ I1984 by lunn nr T11cl~iicalF ~ I I C U ~ I ~Oc I. I. . AIT,:~JC~II ~ I I c ~ I . I LS,~; II! C I C I ~ , Ctiupl-r :1 C o ~ r l c +~i y l"il< Uniwrjily L ~br'iry.\l,i:.~r,c;ipIs .rnd Archives.
TEXT FIGLIKES Fig. ?, 10: W ~ t hpcr~riirsior~ Irorl~C~uug-.lenSu. Indu.rrrra1and D~gf~lreri+lg I'l~mristrc Rr~rrin-h38. 803 (1936). Cupyripht 194h Americim C'hemic;~lSocie~p Chapler 5 quote: With permihhion of Open ('curt '1.1-adeB Academi~Honkh. a divhitin ot Caws R~bli\l~ing Crhml~aliy.P c n ~ IL, . tr.c)rllrllI>r,rrEin'insrri~rPhili).>ophcrS<,irorrsI. tdircd hy P.A. Sch~llp.copyr.igl111973. Fig. 8.1 I : Rcl>r'jnretl!$ it11 l>crrni,hii~nt ~ n ~Hr .lH . Nzwron, birlii\rrrtrl clod Er~girrrrr-lvg (71vnli\lrj R<,\i,idrth 27. 302 t 1935). Cnpyriphr 1935 4111cri~,un Chcrrllrul Sr:~ict!. Fig. 9.14: Wltl~ycrrrlisriic.irlcrr~dCh~mrtrrlR&I ~ f z cl)rrtir v 17, 1531 (1988). Cupy~ighl195Y A~nzrican ('hernical Swizly ,311dthe 41iierrfuli lnsl~turc of Pllysics for thu Kation;ll Irlatitulc u1 Science and Tt.chn\>logy Table 12.4:W~rllpcl,nllss!otr K ~ O I I I J(~r,,rrrrrrl of P l z ~ . ~ ~ crrlzd . u i C l t r ~ n ~ c He?;.rl,r~rc tll U~rtci14. sllpplc~nznt1 (1985). (hpyrjpllr 14H5 An>rricnii C'hcnii~llSocicr! ; ~ I I L I thc A~llcrlcnrlltlsllrl~lcuf Phy\ic> lur ~ h c Nii~ir)nalI i ~ ~ t i ~ru~( Science tf . i~nd Teclu~ology. Fig. 13.15.Wrlh pcrrr~iaaiu~i Urmn Joclr~~t~ul r$ I'l~yvi~-ol a~lclI'l~entrrnlH ~ f i , ~ , ( ! Dutcl mt.~ 1 I. supplelnrr~t2 (19811 Copyright 1982 Amcricerl C hcrrjicul Sucic~yand Ihe Anicricun Insrilu~cuE iJhyaic9 f r ~ihc. K;lriuri;~lI r ~ ~ l i l ~illr tScicncc c ar~d Tcchnulopy.
l ndex
A absolute entropy. 279 activity, 4 I 0 if.4431. 455R con(lensmi phase, 5 10 dilute uolutiun. 443# tclllyeralurt. dependence, 475 iictivitp coeffiiicnt. 41 I activity quolienl, 504 adiabatic cumpressibility, 344 adiabatic exyansiun. l96fi, 222 adiabatic fli~tnetctnperature, 225 adiabatic prwess, 194 Ictnpcritrure change. 196 affinity, 599 a~tlplitudc.10 at~gularv e l t ~ i ~ 19 y, a!iharmr>n!c terrnq. 33 anharmonicit) cojlstarkt, 14 anodc. 5 3 1 a\vmmetric top. 27, 1 6 5 8 atmt>sphere.atm (unil), 5 1 average. 96
azeotropc. 407
B Ral rner Furmula, 4 BaIn~crscrics, 4 bar iun it). 5 1 battery, 5 h5 Berthclot equation. 335 binary ~ i l u t i v n 38s . binonlial meuicient, 230 binrmnisl expansion, 230 binomial series, 230 B A r Irequcncy condition. 3 Bohr rad~us.3 1 bojIir~cp i n t elcr.~tint~ constant, 450 boiling point elevatiotl. 450 Bolr~mannconstant, 7 3 Boltz~~latln tac~or.105
Boltzmann, 1,urfwig (bio). 1W Roltzmann statistics, 123 criterion. 1241 boson, I21 Ruyle temperature. 7 1 Boyle's law. 197
C calcite-aragonite equiIibrium. 527 canwnical ensemble, 106, 107J C'arnot cycle, 2 5 8 8 cathude, 530 cell diagram. 533 Celsius scale. 52 centrifugal distortion conrtant, 36 chen~icalaffinity, 599 chenlical potcntid. 361f, 3938 partition [unction, 369fl Clapeyrun equation, 362 classical harmonic oscillator, 32 cIas';icnl thermndynamics, 2 Cluusius4lapeyron equation, 365 8 Clausius, Rudol T (hio), 236 Clauaiuc, statement of second law, 248,583 cwfficient of thennal expansion. 182, 3 11 coexistctice curve, 61, 35U, 4UY cdIigative properties, 4.188 compressibilily, adiabatic, 344 isothermal, I81 con~pressibilityfactor, 54 con cent ratio^^ ccH. WS,606 consulate temperature, 408 cunvergencc tcst, 136 corrcction for nonideaiity. 320 correspondi~~g statcs, law of, 67#. fugacity. 330 Rcdlich-Kwong, 68 second virial coefficient, 75 van der Waals equation. 68
coup1111gcoefticient, 590 critical consrant<, h51 and Pcnp-Kobincrln paranlctcrs. 89 Kcdlict~-Kwrillg pxarneters. 648, 88 crllrcnl exponent. 383 #. crlllr ul upnlcscrnce. 35h critical point. 60, 63. 354 C I O , ~c ~ ~ I I I c ~ 590 L'~~, c r ~ h ~equation c sl,~tc.611 #: cyclic process. 240
D D;lltorl's law of panial prescurch, 86 Daniel ccll. 5 3 0 d c B m ~ l j cwavelength. 3 I L)eh~<-HuckeIK , dchne(1. 46O047 1 Dcb!c--1Iuckel theory, 459 8 , 3 7 2 8 IIehy1. Putcr (biu). 438 Lkbyc 7 ' ' law. 278 1)eh) r Lcrnpcraturz, I 39. 27K L)eh>r thetlr) c l t hcat capacity, 134, 278 dc.gc.~~cracy, 7 rlrgree r)r frcedom. 2 1f f . 35 1 dcrhative. 175 plr[lLi1.175 total. 178 ddililcd balance, princlplc of, 6U1 diffukion ctlzfticicl~t,620 d~ffils~on equatinn, 637 C ~ I ~ I W I - ~ force, I O ~ 778: d i ~ s r l c ~ a l ~ctlnhtants on ;1110ctectrt*Ilcn~jcal cell,. 556 fL ~i~\u)ci.~~iibli crwrg). I7 distill.~tic>n,400 D ~ I I I Lcffcrt. ~ I 63 1 I)uni~anpntcntjnl. 63 1 d~,oplct.vapor precsurc. 381 8 nulong and Pctit. 116. 130
E L~llcteinmodel rd otorl~iccrystal, 115. 130, 140 Llllstein temperalurc. 14 I elcct~icalwork. 602ff elec~rocheri~icnl cclls. 529 a l ~ ddicswi;ilic>n corlsrits, 55bJ aiui holuhjllty product. 5 5 3 8 clcc~r~>chernlcuI p>tcotial. 603 elrctrkjtl. 530 clcvt~ubinrlicphrnrlrnclla. 593 /j elcvtro~n;lpneticspsctrum. 1[) ~ I E c ~ ~ ~ I I I I oforce. ~ I ~ ' ~ 5% elr;ir\ln r,olt, cV, dcfincd. 3 1
clcctronic energy, 8 electronic partit~onfunction, 143,# electronic temperalure, 171 elec~rr)rlarrlosis.594 . clccrrous~nr~lic lluw. 595 clcct~'oosmoticprzscurt., 593 rrrii, i 3 h and tnllwlpy, 55 1 anrl rnlropy, 551 anri Ciihhc erlergq, 537 pressure deprdericc, 575.576 cndr,thermic. 207 cncrgy
elzclrtinic. 8 8 trans1;itiulnal. 7 vthralio~lai.13 >irial expanciol~.339 cnsuntble, I06 en,errlhlz a v c r q e . 112 cr~thnlpy,200 precsu1.c cizpeodsnce, 31 5 temperarure dependence, 21 7 virial expansion, 339 enthalpy ar~dcruf. 551 entropy. 240 and di+ordcr,249 8. 288,u. altd emf. 55 1 arid phnsc transitiori. 277 anrl \tillis~icalthcr~nudynamicr,?491J: corl-scticln tor ~~orlrdc,~lity, 31 9 8 1~i1.tltio11 func~ion.2hIl.jl., ZX I ,g v~rlalc x ~ p l ~ s i o339 n. enlrulpy cxclla~~ge. ?hh, 107. 581 cnlropy r l t rllixir~:. 252. 255 fl cntrupy product lor^. 147. 266. 3 7 . 5x2 fl and disur.dei. 2 4 3 8 . 2KX I!. arid h ~ d t(low. 2hb and partition runcliurl, 28 1 and ijrlal cxpat~siun.339 rate of. 588 etll~ilihriur~~ n. 4x1 ~,rjr~vcr
excess Gibbs energy uf mixing, 418 excess thermodynamic qwnr~ties,436
ground state, 3
excited stale. 3 cxulhermii, 207 cxtendd Dcbyc--Huckel thet)ry. 464 cntcnvvt quantity. 5t1 exlerIl of reactlnrl. 177.-181 "g4:. 4 8 6 8
H
F Faraday colistanr, 53;
fcnnlon, 121 F~ck'slaw of tlihsion, SS9,619 First Law of Thertllndynamics. 192 flame krnperdture, adiabatic. 225 fluh, 589. b l 8 flux-towe relation$, 5903 force constant. 10 force, d~crmdynamlc.,5 KY Frjurier's law of h u t conduction, 589 fractional djstillatior:. 400 frcc~ing-po1r11 depre.;hirln. 448 .$j irre~ing-pt~inl clcpressi rln cun&nt, 449 friction cwstant. 641 fuel ccll, 5Mff fi~p;~i.ity, 325 8 cquilibriu;r~curlst;lnt, 5416ff. laam tnf c t rehpunding ~ states, 330, 331 Wlich--Kwong equation. 338 van der \ G a l s , 337 virial rxpan+ior~.?Zh Cuga~itjcoefficienl. 316 fundamcl~titlv ~ b r ~ t i nfrequency, ~~al 15, 151. 251
G Ealluna function, 235 E;ausq~wdjsrrjhulion. 99 ,& petmzlric series, I36 Giauque, William I biok, 112 Fihh6 enerp?. 304 and emf, 53: ,~nrlextent ~11're;tclim.3.168, 518 ar~l:przs>urc.321 -6 nnn-i'- V w-r~rli.? 106 3 virial expa~~sion. 3 39 Gihhs-Duhen~ equurion. 3 9 0 8 Gibbs c n e g p of mjxing. 41 8 Ciibbs-Hcltd~oltz :quaIio~i.323, 335 Chhbs, J MJlllard ihiwl, 348 Gihhs phase rule, .Y8 Glansdori-Prigoqce ineyualiry, 6 2 5 8 gntiicnt. thcrniody~larnrc,61 K g r ~ p h i ~ c d i s m n - rquilihium. ld 511
tiurd sphere pulentlal, 82 second virial coefficient. 83 harmonic oscillator. 11 heat. 186 r n o l c c u l ~irltcrprctatio~j.14%11. tleot capacity, I 14. 202 fi Uehye thetlrq, 139. 278 Eirlateln, I 15. 130, 140 prehsure deperidence, 3 34 vrnal exI,arlsiorl. 339 volume dependence. 333 heat CIIEIIIC, 150 elriciencg. 259 heat of cumbustion, 107 hcat of fusiuri, 205 heat uf reactiull, 210 hcliun~-ncon lnscr, 3 1 Hvlrnl~oll~ cncrgy, 302 virial expansion, 340 Ilelmholtz. IIerlnann von (bio), 3W Her~deranricqualion, h 13jf tlellry's law, 403 Hcnry's law constarit. -103 Henry's law ctarldard \tale. 4 14.440
hertz. 30 Hcss's In% ZOX, 2 15 Hildehrand. Jr1t.1 [hro),3x8 Hitchcr)ck ptrjt. 541 homogeneous functicln, 426 Ilooke's law, 10 hydrogen atomic \pectruln. 4 hydrogen elc~,trl,~lr. 532 H~ickel.Ench (hlo), 438
1 ideal gas eq~t;~tiun, dl) ideal rubkt bmd. 342 ideal solution, 395 8.401 inequalily c > i C'ldusiu\. 24H. 583 iticxijct dlffer~er~t ial, IBO, 192 i r i tiurzd active. 24 infrai-ed illactive. 24 lrr~eg~atirlg factor, 240 intensive quilntily, 50 i~~trrmolrcrilar interacrion, 7 3 8 ir~tci~lal pressure. 333 i u ~ n calniospt~erc,462 ionic niuhi tity, 61 2,@ ionic strength. 460 lonizatiorl etlctgy, 3 1
1sol:ited 9ystem. 245 isutherm, 60 isothermal comprrssibili~y.181. 22 1 . 312 iteraliot~,39
I JANAF tablev. 449 Joulc. James (hit>), 184 ~ o I I I ~ J? ( U I I ~ ~ 2) . Jr~ulcThorncon cuefhcienl. 227, 3.10 fl JouloThomson cxpeiimcnt, 226fi Joule-Tllomsc>n ~ n v c ~ s i otclnpilrature, n 311 K, kelvin. K (unit). 5 1 l'H8pital'c rulc. 139 law of cnrreuporidii~gstate<.h7jJ. 75 f i ~ ~ a c i t330 y. Rcdllch-Kwr~np equation, 68 secontl eirial coefficicnt. 75 tar1 dcr Wanis cquali(xrt. 68 law r l i rectilinear diametrrs, 375 lead storage battery, 565 I< Chfitelier's principle. 48 1,525 1.e111lard-Jor~cs parameters. 751 Lennard- Jui~e+ ~wtentinl.74 Ievcr rl~lc.347. 4W 1,ewis cquation, 509 Lewih. GilberI N. (bio). 528 I~ncnrnunequilibrium thermodynamics. 591 liquid junctir~npotential, 547, MU, 6 1 0 s Iwal cquilibrium, 617 I .c>ndond i ~ p e r s ~ oforce, n 77,ff, 8 1 I yman seriek. 4 M
hlaclaurin seriel. 138 Margu1t.s equalion, 43 1 Maxwell equal arca construction, 61, 382fl hlaxwell rclntion. 3U8. 7 191 ntcaii. Oh rrleau ior~icactivity. 455. 4561 mean ionic acrivity cneflicic~~t, 456 Incan ir)r~rvmrjlnlity. 455 rrtcan sphcrlcal approximation, 363,fi incchaoicnl co~ldrictancc.629 mellit~gp i n 1 curve. 350 mclcuqbbattery. 565 1* microwave spcctroscupy, 21 rninirr~umentropy production, priilciple oC, 622 J]: ~nixing.entropy of. 256 ~nnhili~y, 61 18
rilolality, 440 molar pas co11stant, 53. 54t mrdmity. 44 1 lnolecular in~erpretatiot~ of heat, 1 9 8 8 tnoieculu i~~terpretatiun of wurk, 1 9 8 8 molecular partition [unction, 120 ~uurnentu~n o f inertia. 14. 25fi Moore's tables, 8 Morse potenlial. 33, 34 n~ultinumialcoefticient. 231 ~nullinomialexpansinn, 231
N natural variable, 316# Nernst equation. 538 Ner~lstpotential, hOS Nernst, Walthcr (hill). 272 neurtr)n, N (unit), 10 Newton -Raphson method, 3 0 8 ~ ~ i c k e l ~ a d m i uhalkry. rn 566 nonequilibrium therrnody~~amicc, 5818 nonideul sdutiun. 401 nun- P- V work, 3M normal boiling point. 352 norn~almelling poinl. 351 t ~ o r m dmode. 24 nurllcrical integration, 4 2 8 0 cwld l unction, 1 0 1 Ohm's law, 590 Onsagcr, tars (bio). 5S11 Onstiger reciprocal relatlorls. 591 f l orthobaric densiiy, 355 t)srrtotic crdficieot. 444.457 osmotic pressure. 45 1fi overtone%,34, 35
indepettdcrit, irtdistlnguishable particles, 1 27
linear p>l ydtrrmic molecule, 166 molecular, 120 nrmlinear plyatomic molecule, 167, 282 rotational, 156 8 translatiorial, 143 ff van der lVaa1i equation, 1 1S vibmtional. 152fi pascal, P il (unit). 513 Paschen scrics. 31 path runtlion. 192 Peltier effw-.638 Peng-Robinson equation, 57 period of ixcilla~io?,32 pzrn~i~tatinn. 129 phasc diagram. 350 waler. 357 phase rulc. 34% p t ~ e t ~ n r n c ~ ~ o l ocoefficicnt, g~cal 590 phrlil,n. 3 Planck cunstunt. 3 Planck, >lax (hio). 1 Prnwm's c q ~ ~ s t l o472 n. pr~lan~iihili~p. 7q polarizabi lit? vr>lume 79. H[lt postulate of local equilibrium, 617 putentiornett% 537 pressure-cornpsition diagram. 396 8-429. 130 pressure tinil-. 5 1 f primitiw model. W principle r>Cdttaild balance, 601 principle of minimum entmpy production. h22 ,JJ prubability. detinrd. 95 pure heat transfer. 168
u?;ldatic>n-reduction clectrrde, 532
R
P parlinl deriwtive, 175 partial ~nolarqunntite.;, 3#7# pafiial lnolilr tolurne. 7848 partial pressure, 86 partilio~~ funclinn. 1098 and chelnical pntctltial, 365'8 arid energy, 1 10 and cntropy, 2 M # and equilibrium constanl, 4958 atomic. 282 diatu~nicmolccule, 159, 282 electronic, 144$, 28 l ff: independent, distinguishable particles, 120
Raoult's law. 1Wff Raoult's law. negative deviations, 402 Raoult's larv puqitive deuiarinn. 402 Kaoult'b law $tandad slate, 414.439 rate of entrap? production, 588 ratio tcst, 136 rrac1it)n quolletlr, .LEN !( rcciprvcal relation\. 591 j 1 rcctilinem diameters, 375 Rzdl~cLKtl.otrg?onstants. 581 RedlicbKwunp equatir~n,5 7 reduwd mass. I I. 19, 3: redwed parameters. h7 regular soultion. 419. 4 3 2 8
molecular Lheory, 41 Y f l residual entrnpy? 289 reverse osmusis. 453 reversible cell, 536 reversible e l e c t d e , 531 revcrsible process. 189 reversible wurk, 184 rigid relator. 1R rotatio~~al constant, 20, 1 5 , 25t rwtational &@es of freedom, 23 rotational panition function, 1 5 6 8 nltatirmal temperature, 151r, 156, 1 6 3 rubber band. 342 rubber elasticity. 342 S ~ a lbridge. t 510 Schriirlinger equalion, h screened coulomhic polential, 473 Seconrl r a w r ~ Thermndynamics, f 248, 249 Kelvin's statement. 260 second cclitral Inolnent, 97 second eIectrut!~moticRow, 629 secrlnd electnlc>smolicpressure, 629 second moment. 97 sccond streaming current, 629 second streaming potential, 629 second virial coefficient, 70 Seeheck effect, 637 selection rule, 14, 19 anharmonic oscillator, 35 harmunic oscillator, 14 npid rotator, 19 semipermeable membrane, 45 1 silver-silver chloride eleclr&, 531 Simon cquation, 374 Simpcon'f rule, 44 s i n g l e 4 e c h r d e pntenlialh, 542x Sl units. 2, 50 solid-gas coexistence curve, 350 solubility product, 5 1 4 , 5 5 3 8 and elecclmchernical cell, 5533 solute, 439 soluliun activities, 4 4 3 salvent. 439 spherical top. 27, 165 spontaneous process. 301 g square well potential, 83 second virial coefficient, 83 stability. 343 fi slahility conditions, 3 4 3 8 , 385 standard boiling point, 352 standard deviation, 97
Index
\t;tndard emf, 538 standard ttntrrlpy of reaction. 289,R ctandard Gibhs energy oi'fornation, 484jf..
4HSr stariclard mcliing polnt. 35 1 standard riiol,v er~ihalpyoi for~rlalion,2 1 I , 114t sti~ntlardn~oldrentl-opy. 2 7 9 8 , ?X4t, 2 x 5 8 standard reaction enthalpy, 2 10 statidartl wductiori potentialc, 544, 5451 standard state, Iicnry's Inw. 414, 440 standard slale of a gas. 3 1 9 8 alanrlard ctate o f a solution. 41 3 ~ 3 hrandard stotc. RaouIt'h law, 114,434 stntc lunctlon. 1x9 station;try stute. 3 ?i~i~tis[icul ttlem~udyria~nioa, 2 stzildy \[sic. 622J. Stirlirtg's approximalir)n. 232 Srokc'h law, 634 strea~nir~g curmnl. 594 alrearr~ingpotential, 595 suhlirnarion, 353 surrounding*, 1% syln~r~ulric top. 27, 165 splrimetry n u m k r , 15'1, I h?r system, 186 T
Ihylra expanaiun, 33,141 tzmpcrature-composition diagram, 3 4 8 8 , 429. 430 rhcnnocl~crni\try,207 thcrrnacr>uple,h37 thertr~crdyn~mic energy v~rlnlexpansirln. 334 \-olume dc.pendence. 310 thcr~ric~tiynnm~c rqullihriuui constant, 507, 5 10 thcrn~tdynamic1i)rcr. 589 ttrer~l~r)d) namics of ~rseversiblrprocch.\es, 581 Th~rriLAWof Ctier~nodynan~ica. 275 fl, tic IIIW.39h tom (rlnir), 5 1 lotal derivative. 178 Iraycctrjry. 33 lru~~hlarior~al degrees of freedom. f3 Ir:~njlillirmalcnelgy, 7
[ransport number, expcrimtntal determination. 631 irdpczoidal approximnt~on.43 tl-iplc point, 350 I'roulun's rule, 142
v van dcr Wlnls cnnhtants, 55, 5ht and cri~icalcrmstants. van der Waala eiluatiun. 55, 1 18 vnll der Wralh. Johannea (biu), 48 vm't t4ofirqu:1don. equilibrium constant, 401 ff.
viin'~11011'. Ji~ct~hus (blo), 476 vapul p c c w r e and external prcscure. 38 1 vilprr? pruacure. droplet, 381 J. \arlancc. 97 ~ ~ h ~ a t ~ tdegrcex l n u l of frcedo~u,23 vihratlunal C I I F T ~ Y .I3 vihr,~~ional parlition function, 152 fl, r i h ~ ; ~ t ~ i ltrnlpcralure, nal 15 I!. 152. 1621 vi~ialctwlliclent. 70 virlal equulion 01-state, 70 v ~ ~ iexp;lnhlun al encrgy. 339 enthnlpy. 330 cnlropy, 339 Glhbc energy. 334 hcal c;rpacity. 339 H r l m h o l ~cncrgy. ~ 340
W walr, 31 wave furictiut~,6 u,avcrlumber, 5 Wcalon standilrd cell. 535 wrjrh. 18h clc~,lrlval,h02J. rnuximurrl available, 303 111c>lrclkirinterpretation. 198.ff: nun-Y-V? 3Uh Z zcrr I-poirlt energy. 13
