UCSD Measure Density of Seawater From Scripps Pier Using Four Different Methods Lab

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Density & Salinity Laboratory Protocol

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NA = = - S - ' ( mot mot E ) ) ( mot ) mot 8.85×107×6.02×1023 5.328×10 " E! 5 ' molest ' - - I l ( mot E - -77×107 t 3.54×10-1×0.5 = = n 1.77×10-7 1.77×10-7×5=8.84×10-7 T=rV= (b) -73×10 -5×10.22×10-3 molecule - Lt . 5 ' ) (d) 111205 RENDS ] F- while → = - 214024¥02 lad"dI 214205 in - zd4d' ' ✓ ' J , k dn4 dt dNNz05 ) - d.LY#-7.-j=-dncN-Os - ta -3217102-12102 ' dt =2r= 1.77×10-7×2 2J 8.84×10-7×2 = - taek F- NzO5 r= = 411102-102 → ' re g- = unchanged is -7×10 Mol ' I S - Mol . 8-1 depends ( . - I =3 133 - on . mot 1.768×10 = T and ' Half life - First , - order P) Reaction -42=42! KA = tyz (b) A = - K U2 2×1-73×10 -5--3.46×10-55 = In = L = 3.46×10-551 RA 20028 ' s o.co/mol.dm-3,t=24h=864oosEN2Os-J-) ENDS]o= [ Naos - - . e RA 't - [ MOST = = o.co/1mo1.dm-3) 2.87×10-3 - mot 3- 46×10-5×86400 e . I E S ' P " - ' mot 5 - 16.16 ( as > ' - For Second - order Reaction DEET r=kEA5= I - [¥ ¥ = - Ha - , Half life ' - tyz know We kA= = [ -412 EAT n , - [A) , EAT KENO ] . we Fz - L - → [ FI can . = ka It - t EATO ( 5×103 mol.li' = 0.67×10-7 ) mot ! 5×103 1661.13 4h01 t mot hlfz) , - - e- 3.33×10-3 [ Fs ]= , L f- v.v apply , the = - 2h01 mot L ; ¥: = - case . - - S ( mot f- 600L " Edna"Io4=dNF4 dt simple - L I 2111027 ; EB=bz=I 50 min , IT 0.67×10-7×100×60115×103 ( Nor)= 6.67×10-3mot 21402T , t = ( TRA 't EA]o -1=155 Here - . = r= ka ak , ! -42=5×13×50×-605 (A) 0 [ NO , ] Products ' [ AI ,oo= (a) - → Atoka and CATO 2A . kbt I - . L ' ) 12=38 L moi's " - t , T.gxio-s.la = = EBT = Iss - , [A ] - - . Rb I - - , [Af 6.67×10-3 = mot L ' ' . 15×38 [ NOD ETAT = = l - 388×10-3 mot CFD = 6.94×10-4 . E ' ' mot I - (b) after Rate f = ios : K [ NOT Efz ] = = Initial f- 38×1.388×10 -3×6.94×10-4 ( 3.66×10-5 mot ' - L ' 5 - moi . Lts - ' ) ' rate : KENO DEFT = = 38×6.67×10 -3×3.33×10-3 8.44×10-4 mot Lt S - . - I ( mot 51.5 ) ' . (a) dpd = This is Here ① 2 Palo will The reaction ⑦ Run 4 time - try . Hz . The K Bitz - Half life - - - Compared , is k Rum 4 . with = ¥ 1hL excess in be can Pino tYz= Run tyz 19.2 = K Reaction r tyz = , . [A] 3 Run . ! . Pitz . with ( ⇐ order puny 3=276 Is . , K= R doesn't - Papo change) PHL to Reaction K , . to Half life approximated CATO H2 Reaction) Byo . Second ( D compared order ( First simplify to approximated be can . Palo . constant a -2k Bio Pitz approximations excess in is = Reaction some relatively f Half use No . is date Pitz trimolecular a we Run R Pri - =3 CATO . Runs K - , - KPH ) . For (b) t K fgh } - - land Run 2 K Pitz - K . 0.036,5 = ' , Pmo ' , ka - Bto - 600 ( = . RA 1×10-7 = A r df# = = (a) ( b) ( C) (d) KEAT -435 - dta#= 2B t ar - - r REATEBT dd"# = dI# DIII! dta = = - - DEF ] DI ' - - , = ; , R , k -433 , EB ] = kg EB) - - br - ; mmHg 19.2 ) trimolecular Reaction fdthI-t-dd.LI = zr - df#=cr=3r : BKEAIEBJ? ZKEAIEBJ = tyz=t÷= ' , - dd = = i CREATE B] ' 3k EAT -435 REB) daE# ; TREE] - - dtd# HEB] -1k ICE ] = - dt¥= ; ' R -437 - ta df# ddY# 5 . 3C → - ; AREA ] -435 - - = = Hgf ( mm , DIII Rs [ B) k-3 EFT REB ] = - t - KCET ; REB] Kitts - k-347 - jdd# dt# RZIEI = = i ki EB ] 1- RILE] df - k - = i REET LET 16.61 Ti . 660k = ¥ - . - e 16.68 two . = Ea RT - - DIII ' Ai - - - e + La-LA ] RT = = } dm 0176 . Es KEAT = o mechanisms . - ki = c dEd# - 0%8 competing A ✓ = EFT ka - D - - Ea and k, III - = k Ea = . , . # c Kz K2 into - moi - ' s EAT RT = EET Eff ( Equation - A. e . . Ea , I - ki t .it/4cT)Ea.zKiLTItkzlT kilt ) Ea ) . toes i - I K= Kit ka ; - " Ik C . da¥tda÷ ; , plug 720k c ( KitKat Ase Tz . dtnldktk" ' I :L:7Thm¥ , 0.068 = Kz " - - e - J/moI - Earth it - . Ea 177 kJ/md= 177×1000 . EET Ea . . t Eff z.kz ) . - - A . - e TT ) , At B F 4) step [c) EDT " KII - - . I 244 , fitting - - t 4.35 is , - e - ¥÷5/µ - .e , Effi - - . I 2×96 - kJ/mol Ink equation intercept Hi EI÷tzEa¥ z⇐r , 2Ea - = The - e - Ea .2t2Ea slope 40 - Aza;AI The N - Az e 196+2×120 From Ecj Kz - " . = TF 298.15 K ( v. d. s ) Gt H r=ki E:3? ' = Use In f ' 's Ea = (a) I :3 K2 2C → equilibrium ) in , - - k= ( K I - k= K) D ki = [A) EB] co + fast is 1 C IHA - - Te 293.45K result 1-3=290 . Ink plot , . 45K vs . YT 1-4=287.15 K . " -6480 K is ERT - 26.92 , - . 1nA - - ¥= 26.92 - 6480 , , Ea 5.38/4×104%01 - - 1-4=287 . 15k 14=77 14×77=590 't , while 141=57.2 OF (a) The - ta dEdA# From KEA ] = dh ' From a- ideal Pa Substiting get we - - eg . hft - - " - RT into dt = = PAV eg " MART - - - ( CA ) " ) [ HIRT Iii ) , Kp ( CAIR'll Kp ( RT ) ) , PA RT I dd = question lit ( Ciii ) DEAD this in Equation ta a¥( [ AT RT ) Ia Ci ) HRT = product s - kppan = gas PV AA " relationship the - for law rate RT ) Kp CA) " " " Liii ) From - eq " lil at dta# KEAT " K (b) for = order n " ( RT ) ( RT ) = K h - "" - - - reaction I , Kp - - lil 1. " Kp IA ] Kp ( RT ) relationship This nth = know we KEAT = Kp =) , . K . ' - is n valid for any (a) At B 2C → 2 , 2A and I ZA t 2B 2C Ft B t Reaction i I 2C → F → 2A 313 of G step Cc ) a , b . C . → ZA -1213 -12C + Ft B overall lb ) numbers stoichiometric are F → Ft B la ) D Ct → TZD -1 G → 2C -121) → 213+9 TF t 2A tf (b) catalyst speeds - Intermediate does I . A → B → the overall Formed - and mechanism step in appear not in consumed not appear one in in catalyst Reactant [ → Intermediate D Product → f- G → s - but the reaction up Intermediate product chemical step a the does Eq of h . a subsequent overall Reaction' (a) False (b) False (c) True First (a) r KEA ] - EAT [ A) when Reaction order = ( = t - - [ A) I Iad dt EAT oe -_ - - KAT 3255 Rae ka , X% ) EAT - . X . - complete 21=35 , [ Abe = : RA 't - ④ ( I - KA a -2 RA , for lb) 2K - - 70% [A ] [A) f. = = = R o - RA - 325 0,0013351 . [ A) = e o - - o complete e - . 0006635 ' . KA T KAI - , In 10.3 ) = 905 RA S o.IE/t3o-=hhl0.l 90% , complete ) - ka EAT , = '- 17315 ' 0.001335 0.3 CATO - For [ A) 35%1 [ A ]o= . ( (a) First - order Reaction ) - r - - - Idgaf ' KEAT = - - dta# = - - ' KATHI - l l EA ) A daft concentration - activity - = - dal N A - - AN = I . . . f!! , Inc #o e I = - d I = - it , - Adt at Ho - At = d No - - e do =dNo (t It A- = = Ito Ao - e , - . e t - - e , - - / dm - # dat foot - No = at at - - ) o (b) 226 Rd 222 → IS No Ao Ra = - - Na . glmol 226 3.7×10 " s ' ' = Mo = D= 1.39×10 t - d -42=0.693 " s - 4Hez 2.66×1021 , D= , I tyz= , 9%32×10 I -1=999 years A = = = Ao . e NIT 1600 " s years 3.15×10105 = it 3.7×1010 2.4×10 " - . e s - I ( 1. 39×10 - " x 3.15×10 " ) (c) Kd log Not Kol ) . I = A- = = = = 2×108 Ao e 16.2 -72×10 yr - = - o x -72×10 16.2 S = f I = AN game , 8.08×1022 = NA - >4 D= A ! = " Nol Kol ) . - - " ' . ooh 75 X = ) % 8.08×10 = 5 I 6-31×10155 ht - - 14.65 e ' 172×10 - ' ( K+ ) 9.45×10174 40kt, ' 9.45×1017 " 7×6.31×1015 ' 111204 ① 21402 elementary The OCT = . K Appendix kpo ki Kpo - - ki ki - Forward o . 0831 8.3141 - 0.148 ( I bar 2) . , order - 1k¥ - mot equilibrium ? ( bar dm moi - . = ' Reaction ; In . Kp ) constant on -_ 2' - K ' I I 5.97×10-3 = . constant , oh KI 2981k) - din . on RT ) Kb First -4mg equilibrium KPYCRT , = 1¥ - Ki = - , concentration - R = pressure - ( b . 09%8=4730 Hmo , , 4730151mg ) Kpo Reaction ' F- 298.15K . = phase - Rflnkpo - 12=8.314%01 From gas kf= 4.8×1045 ; 8×10615 Backward -3 ( mot dm - ' - mot i. - . ) 's dm ) Second order (a) 203 , dEdO# ( b) ddE# k [ OSTEND ki = K , [ OSTEND [ 037 K, DION dt ") t f- (d) r - - = 3 - - Kz END KLEOJEOD TO ] [ Os ] - - is I [ 02 ][ 03 END t 21420 ] [ Os ) [023-10] END I [ 023203 END i K , - I - - Kr -Lo3[ Oz] [ 023 [ OIEN'D dtd¥= I - = ( =D intermediate) K2 [ 05h02 ] , 2122203 [ 033 - [ 0] ( K Kz [ 03 [ Os) [ ODEM ] t kilos ] Kika [ Og ) [OSTEND k = I - I tkztoz) [023 END CO2 ] D slow . ; r ; KzEO3]/ END Step K [ Kioto, ] Ef:} K 2 - - 202 ) - - ki EOD EM ] ' = I dI step K - , K' Kc Eog) when - K K 1- [ 03 ][ ND [ oh (e ) - , - = ¥ Os Ot d[d¥]= at Ot M Ost M 302 → -1 is - - I EOD Off Kiki I Eon t ka 3) lenig = equilibrium - ( Oz) , fast and in equilibrium with step I 2111205 (a) 411102+02 → 11102 Ak 05 €3 11102+11103 F- 123 [NOTING ] 20 ③ 211102 steady , ① Not 02T Nose ¥ Not NO } NO, + state - , approximation . kz-LNOD-tNXO-kz-LNOT.CN#dTldtOeI=k.-LN20sT-k-iEN0D dTdN# K2 [ NOD ENG) = - Ks EMOTING ] -4103 ] KINDS] = = KENNA ki ENDS] K - - I K - - - I -44027 ENOS] (K [ NOT -41103 ] - i , Kz -41102544037 - - K - I 2kz[ NODE NOD + Zkz ddtff-kz-LNODENOD.R.cl y= EN 027111103 ) ¥ = , [ Ncos] r= (C) When dEa# K - is> = kk 2kz , [ NOD -41103 ] - - intermediate) O ¥ = KKZEN 2057 K (b) -_ ( [NO ] , Kz -4402] [ NO, ]= = k3EN0] -41103] =O k€1112057 = - 2kzEN0z][NOD -1216 ) [NOT ENG ] ENOL] [NOD [111027-41103] t=k3 - =0 ¥ , . S - I . , -12122 Step ENO, ] R= Kika K - I -12/52 2 . ENZO's ] Ethos ] step 2 is slow , step and 1 and in step - l equilibrium is fast ( C) ) Ki F- 11h05 1402 2×10301+2111024-102 → 2420+214205 K - NB ↳ + Cho ¥ NOI + 02 ④ 14024 111034 the ¥T NO 2 Neck Rate O ⑤ . ⑥ equilibrium in is ① step - determined { jog! [ 1112023 ② step , 1¥ [ NOT ! f- Kz ¥ - [NOT , Rate - determined YET [NOD t = 62 . Nos ¥ NO t , 21402 = 2=-12%14 Kz 2 ¥ [ NOT [ Oz] , - 1¥ [ NOT , NOT 02 f- equilibrium slow , " , [ Oz] ① . , ② ⑥ step = , - k= , Not 02 NOS ② . k¥16.61 REMOTE Os ] f- and ① slow , . r=¥ d 'd"f4=kzENs0z5L0z] = = equilibrium Naor 021-9211102 + " . t KEN205 ] = 16.60 16 ③ od t Oz if possible Not Ost NOL 02¥ t Ot o Only Q t 11102 ¥ Not Oct Nord Nos 11102 i - Ks ENGIENOT , EOD €3211102 KENO TEO , ] r , - - RENO ] - trimolecular EO ] . , D= Reaction 14k¥ seep 2 is rds . 16.47 2 Cr From at rate - " t -113T law rate Cr t t=kz[ CP'T [ TL"] ky - ki ¥ Crit Cpt EY! = , E Cr " , Tht + . fast , ¥ eq slow , 2Cr3t [ cist] , -422+3=42 [ Cr 'THE'T ' , . + 't TL GTL [ Cpt E) crusty ] ETL"] " equilibrium , [ Cra ] [ TL ] K2 [ CrTL5+ ] KIKI is " Cr 't TL ⇐ t TL " , Crt't Tht 't , - rectants charges 5 (rut TLH CrTL5t ¥ Cru r " , [ Tht] " = + , [ Crs ② ✓ TL + C. rut - and TL CRH - . r=R[Cru ] [713-1] , composition of the total cr2+tTL3t ① f mechanism to deforming step zcrstt TITI → = ¥1 , , slow fuse [ Crust]= [ City [Tug 16,50 20120-12111205 From law Rate Is possible 2N → mechanism NO, → 111205 → most ( o 11103 Slow . ⇐ 111034 O t . -150 111024+04 Cho NOS tool £ f- REMOS] , . 11102 t Nat 2×10301-12140241-02 mechanism to 111205 → fast to 021 zxlo-sdms.no/-!5'.a-LT=bdok.Ea--l77kJ/mol 16.61 2DI k= - A - e → Dst Iz K . I - - 177×103 Jlmol FF - 1.2×10-3 (dm ' - , 1. 2×10 mot ! s - - -31dm? mot ' - s - y ' A A = e - f- A I = - - - K - - A - e FT 60k 32.26 e 22×10 F- 720k - 8-314 Telmo " dm? mot ' - S - I 177×103 Jlmol - = = 1.22×10 o . " 0177 ( dms moi 1. s - - 's dm - mot - 1. - y s - - I e 8-314 - 720 Jlmol Eal RT - k= A. (a) ki - - A e - - e T, bn , ⇐ HT Kz , 310K -_ = Tz , - - - - A 300k Erect ¥ - ⇐ K" - ' - - 1¥ ' e , - lb ) = Jlmol 8-314 ( Lok Ea=l9kJ/mol - . - K - - Ea ) Rtn = it , In 6.5 = Hok ) Earl # E ) 145 - e = e = I -3 8-314 Jlmolik , - T2 19×1035 - kJ/mo , Ti - - = e . , Ea - - - ( - - 310k 300k ÷ jook ) - ! ¥) ERM # E) - = 6. s Density and salinity measurement Fun Fact The Dead Sea is roughly 8.6 times saltier than the ocean (33.7% salinity). This salinity makes for a harsh environment in which animals cannot flourish (hence its name). The high salinity prevents macroscopic aquatic organisms such as fish and aquatic plants from living in it, though minuscule quantities of bacteria and microbial fungi are present. Overview • Background on Salinity Conserved Elements Relation to Physical Properties of Seawater • Density by Volume (BOD Bottles) • Salinity by Density Meter (Millero Equation) • Salinity by Conductivity Probe • Salinity by Refractive Index (Refractometer) Salinity of Seawater :KDWLVVDOLQLW\: • Salinity is a measure of the total dissolved salts in solution • The average salt content in the ocean is 35 g per kilogram of seawater (parts per thousand; ppt) • Six primary constituents (conservative elements) It is highly impractical and problematic to evaporate seawater in order to make an accurate salinity measurement! For many years, chlorinity titrations were the standard procedure for salinity determination. Relating conductivity to salinity In 1978, the “Practical Salinity Scale” (PSS) was defined based exclusively on measurement of the conductivity ratio between a sample and a standard solution. Waters with the same conductivity ratio have the same Practical Salinity, regardless of the composition. Chemical processes in the ocean lead to small errors between Absolute Salinity (g/kg) and Practical Salinity (a unitless quantity). Millero, F. (1993), Oceanography, 6(3), 67. In the density lab you will use two rather complex equations (spreadsheets provided) to convert between Practial Salinity and density: The PSS-78 equation allows conversion between conductivity and Practial Salinity. The EOS-80 equation converts between density and Practial Salinity. In this course we will not distinguish between Absolute and Practial Salinity, other than to point out that there are some subtle differences that you would care about only if you were a physical oceanographer. If you are interested in further reading on this subject, visit http://www.teos-10.org/ Converts conductivity ratio to Practical Saliniy - This calculation is provided in a spreadsheet found on 6DOLQLW\5ODQG'HQVLW\.exl" You only need this portion because Rp = rT = 1 under your experimental conditions. Physical Properties of Seawater Physical Properties are “State Functions” f = c – p + 2 (Gibbs phase rule) f = independent intensive variables required to specify the state of the system c = number of different chemical species p = number of phases • Temperature • Pressure • Composition If we treat seawater like a 2-component mixture (water + sea salt), then f = 3. The most common use of a State Function in oceanography is the practice of calculating density as a function of temperature, pressure and salinity. other examples • Density • Energy • Electrical conductivity/resistivity • Sound speed • Refractive index http://en.wikipedia.org/wiki/State_function Physical Properties of Seawater Increase with increasing salinity Decrease with increasing salinity • Density • Refractive index • Freezing point ~ -1.8 °C seawater freezing point • Electrical conductivity • Temperature of maximum density • Transmission speed of sound • Compressibility • Surface tension • Solubility of non-reacting gases • Specific heat Physical Salinity Properties of Seawater of Seawater Increase with increasing salinity Decrease with increasing salinity • Density • Refractive index • Electrical conductivity • Freezing point ~ -1.8 °C seawater freezing point • Temperature of maximum density Of these physical properties, oceanographers are typically • Transmission speed of • seawater. Compressibility most interested in the density of Why don’t they sound simply measure the density of seawater directly? • Solubility of non-reacting • Surface tension gases • Specific heat Physical Properties of Seawater An equation of state describes the state of matter under a given set of physical conditions. This idea applies to all fluids, gases, solids, not just seawater! For seawater, EOS-80 provides density from salinity, temperature, pressure ρ=f(S,T,P) Also, thermal expansion, compressibility, specific heat, adibatic lapse rate, potential temperature, sound speed. For simplicity, the newer Thermodynamic Equation of Seawater (http://www.teos-10.org/) will not be used in this course. Measures of Salinity • Chlorinity (Fajan’s Method) – A measure of the content of chloride, bromide, and iodide ions in seawater – We derive salinity from the chlorinity:= Sun...
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Physical Chemistry
Fourth Edition

Robert J. Silbey
Class of 1942 Professor of Chemistry
Massachusetts Institute of Technology

Robert A. Alberty
Professor Emeritus of Chemistry
Massachusetts Institute of Technology

Moungi G. Bawendi
Professor of Chemistry
Massachusetts Institute of Technology

John Wiley & Sons, Inc.

ACQUISITIONS EDITOR

Deborah Brennan

SENIOR PRODUCTION EDITOR

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SENIOR MARKETING MANAGER

Robert Smith

SENIOR DESIGNER

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NEW MEDIA EDITOR

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This book was set in 10/12 Times Roman by Publication Services, Inc. and printed and bound by
Hamilton Printing. The cover was printed by Lehigh Press, Inc.
This book is printed on acid-free paper.䊊

Copyright 2005  John Wiley & Sons, Inc. All rights reserved.
No part of this publication may be reproduced, stored in a retrieval system or transmitted in any
form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise,
except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without
either the prior written permission of the Publisher, or authorization through payment of the
appropriate per-copy fee to the Copyright Clearance Center Inc. 222 Rosewood Drive, Danvers,
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07030-5774, (201) 748-6011, fax (201) 748-6008.
To order books or for customer service, call 1(800)-CALL-WILEY (225-5945).

PREFACE

The objective of this book is to make the concepts and methods of physical chemistry clear and interesting to students who have had a year of calculus and a year
of physics. The underlying theory of chemical phenomena is complicated, and so it
is a challenge to make the most important concepts and methods understandable
to undergraduate students. However, these basic ideas are accessible to students,
and they will find them useful whether they are chemistry majors, biologists, engineers, or earth scientists. The basic theory of chemistry is presented from the viewpoint of academic physical chemists, but many applications of physical chemistry
to practical problems are described.
One of the important objectives of a course in physical chemistry is to
learn how to solve numerical problems. The problems in physical chemistry
help emphasize features in the underlying theory, and they illustrate practical
applications.
There are two types of problems: problems that can be solved with a handheld calculator and COMPUTER PROBLEMS that require a personal computer
with a mathematical application installed. There are two sets of problems of the
first type. The answers to problems in the first set are given in the back of the
textbook, and worked-out solutions to these problems are given in the Solutions
Manual for Physical Chemistry. The answers for the second set of problems are
given in the Solutions Manual. In the two sets of problems that can be solved
using hand-held calculators, some problems are marked with an icon to indicate that they may be more conveniently solved on a personal computer with a
mathematical program. There are 170 COMPUTER PROBLEMS that require
a personal computer with a mathematical application such as MathematicaTM ,
MathCadTM , MATLABTM , or MAPLETM installed. The recent development of
these mathematical applications makes it possible to undertake problems that
were previously too difficult or too time consuming. This is particularly true for
two- and three-dimensional plots, integration and differentiation of complicated
functions, and solving differential equations. The Solutions Manual for Physical
Chemistry provides MathematicaTM programs and printouts for the COMPUTER
PROBLEMS.
The MathematicaTM solutions of the 170 COMPUTER PROBLEMS in digital form are available on the web at http://www.wiley.com/college/silbey. They can
be downloaded into a personal computer with MathematicaTM installed. Students

iv

Preface

can obtain Mathematica at a reduced price from Wolfram Research, 100 Trade
Center Drive, Champaign, Illinois, 61820-7237. A password is required and will be
available in the Solutions Manual, along with further information about how to
access the Mathematica solutions in digital form. Emphasis in the COMPUTER
PROBLEMS has been put on problems that do not require complicated programming, but do make it possible for students to explore important topics more deeply.
Suggestions are made as to how to vary parameters and how to apply these programs to other substances and systems. As an aid to showing how commands are
used, there is an index in the Solutions Manual of the major commands used.
MathematicaTM plots are used in some 60 figures in the textbook. The legends for these figures indicate the COMPUTER PROBLEM where the program
is given. These programs make it possible for students to explore changes in the
ranges of variables in plots and to make calculations on other substances and systems.
One of the significant changes in the fourth edition is increased emphasis on
the thermodynamics and kinetics of biochemical reactions, including the denaturation of proteins and nucleic acids. In this edition there is more discussion of
the uses of statistical mechanics, nuclear magnetic relaxation, nano science, and
oscillating chemical reactions.
This edition has 32 new problems that can be solved with a hand-held calculator and 35 new problems that require a computer with a mathematical application.
There are 34 new figures and eight new tables.
Because the number of credits in physical chemistry courses, and therefore the
need for more advanced material, varies at different universities and colleges, more
topics have been included in this edition than can be covered in most courses.
The Appendix provides an alphabetical list of symbols for physical quantities and their units. The use of nomenclature and units is uniform throughout the
book. SI (Système International d’Unités) units are used because of their advantage as a coherent system of units. That means that when SI units are used with all
of the physical quantities in a calculation, the result comes out in SI units without
having to introduce numerical factors. The underlying unity of science is emphasized by the use of seven base units to represent all physical quantities.

HISTORY
Outlines of Theoretical Chemistry, as it was then entitled, was written in 1913 by
Frederick Getman, who carried it through 1927 in four editions. The next four
editions were written by Farrington Daniels. In 1955, Robert Alberty joined Farrington Daniels. At that time, the name of the book was changed to Physical
Chemistry, and the numbering of the editions was started over. The collaboration
ended in 1972 when Farrington Daniels died. It is remarkable that this textbook
traces its origins back 91 years.
Over the years this book has profited tremendously from the advice of physical chemists all over the world. Many physical chemists who care how their subject
is presented have written to us with their comments, and we hope that will continue. We are especially indebted to colleagues at MIT who have reviewed various
sections and given us the benefit of advice. These include Sylvia T. Ceyer, Robert
W. Field, Carl W. Garland, Mario Molina, Keith Nelson, and Irwin Oppenheim.

Preface

The following individuals made very useful suggestions as to how to improve this fourth edition: Kenneth G. Brown (Old Dominion University), Thandi
Buthelez (Western Kentucky University), Susan Collins (California State University Northridge), John Gold (East Straudsburg University), Keith J. Stine
(University of Missouri–St. Louis), Ronald J. Terry (Western Illinois University),
and Worth E. Vaughan (University of Wisconsin, Madison). We are also indebted
to reviewers of earlier editions and to people who wrote us about the third edition.
The following individuals made very useful suggestions as to how to improve
the MathematicaTM solutions to COMPUTER PROBLEMS: Ian Brooks (Wolfram Research), Carl W. David (U. Connecticut), Robert N. Goldberg (NIST),
Mark R. Hoffmann (University of North Dakota), Andre Kuzniarek (Wolfram
Research), W. Martin McClain (Wayne State University), Kathryn Tomasson
(University of North Dakota), and Worth E. Vaughan (University of Wisconsin,
Madison).
We are indebted to our editor Deborah Brennan and to Catherine Donovan
and Jennifer Yee at Wiley for their help in the production of the book and the
solutions manual. We are also indebted to Martin Batey for making available the
web site, and to many others at Wiley who were involved in the production of this
fourth edition.
Cambridge, Massachusetts
January 2004

Robert J. Silbey
Robert A. Alberty
Moungi G. Bawendi

v

CONTENTS

PART ONE
THERMODYNAMICS
1. Zeroth Law of Thermodynamics and Equations of State
2. First Law of Thermodynamics

31

3. Second and Third Laws of Thermodynamics

74

4. Fundamental Equations of Thermodynamics

102

5. Chemical Equilibrium
6. Phase Equilibrium

132

177

7. Electrochemical Equilibrium

218

8. Thermodynamics of Biochemical Reactions

254

PART TWO
QUANTUM CHEMISTRY
9. Quantum Theory

295

10. Atomic Structure

348

11. Molecular Electronic Structure
12. Symmetry

396

437

13. Rotational and Vibrational Spectroscopy
14. Electronic Spectroscopy of Molecules
15. Magnetic Resonance Spectroscopy
16. Statistical Mechanics

568

458

502

537

3

Contents

PART THREE
KINETICS
17. Kinetic Theory of Gases

613

18. Experimental Kinetics and Gas Reactions
19. Chemical Dynamics and Photochemistry
20. Kinetics in the Liquid Phase

641
686

724

PART FOUR
MACROSCOPIC AND MICROSCOPIC STRUCTURES
21. Macromolecules

763

22. Electric and Magnetic Properties of Molecules
23. Solid-State Chemistry
24. Surface Dynamics

786

803

840

APPENDIX
A. Physical Quantities and Units
B. Values of Physical Constants

863
867

C. Tables of Physical Chemical Data
D. Mathematical Relations
E. Greek Alphabet

868

884

897

F. Useful Information on the Web

898

G. Symbols for Physical Quantities and Their SI Units
H. Answers to the First Set of Problems

INDEX

933

912

899

vii

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P A R T

O N E

Thermodynamics

T

hermodynamics deals with the interconversion of various kinds of
energy and the changes in physical properties that are involved.
Thermodynamics is concerned with equilibrium states of matter and
has nothing to do with time. Even so, it is one of the most powerful
tools of physical chemistry; because of its importance, the first part of this book
is devoted to it. The first law of thermodynamics deals with the amount of work
that can be done by a chemical or physical process and the amount of heat
that is absorbed or evolved. On the basis of the first law it is possible to build
up tables of enthalpies of formation that may be used to calculate enthalpy
changes for reactions that have not yet been studied. With information on heat
capacities of reactants and products also available, it is possible to calculate the
heat of a reaction at a temperature where it has not previously been studied.
The second law of thermodynamics deals with the natural direction of
processes and the question of whether a given chemical reaction can occur by
itself. The second law was formulated initially in terms of the efficiencies of
heat engines, but it also leads to the definition of entropy, which is important
in determining the direction of chemical change. The second law provides the
basis for the definition of the equilibrium constant for a chemical reaction.
It provides an answer to the question, “To what extent will this particular
reaction go before equilibrium is reached?” It also provides the basis for
reliable predictions of the effects of temperature, pressure, and concentration
on chemical and physical equilibrium. The third law provides the basis for
calculating equilibrium constants from calorimetric measurements only. This
is an illustration of the way in which thermodynamics interrelates apparently
unrelated measurements on systems at equilibrium.
After discussing the laws of thermodynamics and the various physical
quantities involved, our first applications will be to the quantitative treatment
of chemical equilibria. These methods are then applied to equilibria between
different phases. This provides the basis for the quantitative treatment of
distillation and for the interpretation of phase changes in mixtures of solids.
Then thermodynamics is applied to electrochemical cells and biochemical
reactions.

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1

Zeroth Law of Thermodynamics
and Equations of State

1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
1.10
1.11

State of a System
The Zeroth Law of Thermodynamics
The Ideal Gas Temperature Scale
Ideal Gas Mixtures and Dalton’s Law
Real Gases and the Virial Equation
P –V –T Surface for a One-Component System
Critical Phenomena
The van der Waals Equation
Description of the State of a System
without Chemical Reactions
Partial Molar Properties
Special Topic: Barometric Formula

Physical chemistry is concerned with understanding the quantitative aspects of
chemical phenomena. To introduce physical chemistry we will start with the most
accessible properties of matter—those that can readily be measured in the laboratory. The simplest of these are the properties of matter at equilibrium. Thermodynamics deals with the properties of systems at equilibrium, such as temperature,
pressure, volume, and amounts of species; but it also deals with work done on
a system and heat absorbed by a system, which are not properties of the system
but measures of changes. The amazing thing is that the thermodynamic properties
of systems at equilibrium obey all the rules of calculus and are therefore interrelated. The principle involved in defining temperature was not recognized until the
establishment of the first and second laws of thermodynamics, and so it is referred
to as the zeroth law. This leads to a discussion of the thermodynamic properties
of gases and liquids. After discussing the ideal gas, we consider the behavior of
real gases. The thermodynamic properties of a gas or liquid are represented by an
equation of state, such as the virial equation or the van der Waals equation. The
latter has the advantage that it provides a description of the critical region, but
much more complicated equations are required to provide an accurate quantitative description.

4

Chapter 1

Zeroth Law of Thermodynamics and Equations of State

1.1

Surroundings

System

(a)

System

Surroundings

(b)

Figure 1.1 (a) A system is separated from its surroundings by a
boundary, real or idealized. (b) As
a simplification we can imagine the
system to be separated from the surroundings by a single wall that may
be an insulator or a heat conductor. Later, in Section 6.7 and Section
8.3 (see Fig. 8.6), we will consider
semipermeable boundaries so that
the system is open to the transfer of
matter.

STATE OF A SYSTEM

A thermodynamic system is that part of the physical universe that is under consideration. A system is separated from the rest of the universe by a real or idealized
boundary. The part of the universe outside the boundary of the system is referred
to as the surroundings, as illustrated in Fig. 1.1. The boundary between the system
and its surroundings may have certain real or idealized characteristics. For example, the boundary may conduct heat or be a perfect insulator. The boundary may
be rigid or it may be movable so that it can be used to apply a specified pressure.
The boundary may be impermeable to the transfer of matter between the system
and its surroundings, or it may be permeable to a specified species. In other words,
matter and heat may be transferred between system and surroundings, and the
surroundings may do work on the system, or vice versa. If the boundary around
a system prevents interaction of the system with its surroundings, the system is
called an isolated system.
If matter can be transferred from the surroundings to the system, or vice
versa, the system is referred to as an open system; otherwise, it is a closed
system.
When a system is under discussion it must be described precisely. A system is
homogeneous if its properties are uniform throughout; such a system consists of
a single phase. If a system contains more than one phase, it is heterogeneous. A
simple example of a two-phase system is liquid water in equilibrium with ice. Water can also exist as a three-phase system: liquid, ice, and vapor, all in equilibrium.
Experience has shown that the macroscopic state of a system at equilibrium
can be specified by the values of a small number of macroscopic variables. These
variables, which include, for example, temperature T, pressure P, and volume V,
are referred to as state variables or thermodynamic variables. They are called state
variables because they specify the state of a system. Two samples of a substance
that have the same state variables are said to be in the same state. It is remarkable
that the state of a homogeneous system at equilibrium can be specified by so few
variables. When a sufficient number of state variables are specified, all of the other
properties of the system are fixed. It is even more remarkable that these state variables follow all of the rules of calculus; that is, they can be treated as mathematical
functions that can be differentiated and integrated. Thermodynamics leads to the
definition of additional properties, such as internal energy and entropy, that can
also be used to describe the state of a system, and are themselves state variables.
The thermodynamic state of a specified amount of a pure substance in the
fluid state can be described by specifying properties such as temperature T , pressure P , and volume V . But experience has shown that only two of these three
properties have to be specified when the amount of pure substance is fixed. If T
and P , or P and V , or T and V are specified, all the other thermodynamic properties (including those that will be introduced later) are fixed and the system is at
equilibrium. More properties have to be specified to describe the thermodynamic
state of a homogeneous mixture of different species.
Note that the description of the microscopic state of a system containing many
molecules requires the specification of a very large number of variables. For example, to describe the microscopic state of a system using classical mechanics, we
would have to give the three coordinates and three components of the momentum
of each molecule, plus information about its vibrational and rotational motion.
For one mole of gas molecules, this would mean more than 6 ⫻ 1023 numbers. An

1.1 State of a System

important thing to notice is that we can use a small number of state variables to
describe the equilibrium thermodynamic state of a system that is too complicated
to describe in a microscopic way.
Thermodynamic variables are either intensive or extensive. Intensive variables are independent of the size of the system; examples are pressure, density,
and temperature. Extensive variables do depend on the size of the system and
double if the system is duplicated and added to itself; examples are volume, mass,
internal energy, and entropy. Note that the ratio of two extensive variables is an intensive variable; density is an example. Thus we can talk about the intensive state
of the system, which is described by intensive variables, or the extensive state of a
system, which is described by intensive variables plus at least one extensive variable. The intensive state of the gas helium is described by specifying its pressure
and density. The extensive state of a certain amount of helium is described by
specifying the amount, the pressure, and the density; the extensive state of one
mole of helium might be represented by 1 mol He(P, ␳ ), where P and ␳ represent
the pressure and density, respectively. We can generalize this by saying that the
intensive state of a pure substance in the fluid state is specified by Ns 1 variables,
where Ns is the number of different kinds of species in the system. The extensive
state is specified by Ns  2 variables, one of which has to be extensive.
In chemistry it is generally more useful to express the size of a system in
terms of the amount of substance it contains, rather than its mass. The amount of
substance n is the number of entities (atoms, molecules, ions, electrons, or specified groups of such particles) expressed in terms of moles. If a system contains N
molecules, the amount of substance n  N /NA , where NA is the Avogadro constant (6 .022 ⫻ 1023 mol⫺1 ). The ratio of the volume V to the amount of substance
is referred to as the molar volume: V  V /n . The volume V is expressed in SI
units of m3 , and the molar volume V is expressed in SI units of m3 mol⫺1 . We will
use the overbar regularly to indicate molar thermodynamic quantities.
Comment:
Since this is our first use of physical quantities, we should note that the value of a
physical quantity is equal to the product of a numerical factor and a unit:
physical quantity  numerical value ⫻ unit
The values of all physical quantities can be expressed in terms of SI base units
(see Appendix A). However, some physical quantities are dimensionless, and so
the symbol for the SI unit is taken as 1 because this is what you get when units
cancel. Note that, in print, physical quantities are represented by italic type and
units are represented by roman type.
When a system is in a certain state with its properties independent of time
and having no fluxes (e.g., no heat flowing through the system), then the system is
said to be at equilibrium. When a thermodynamic system is at equilibrium its state
is defined entirely by the state variables, and not by the history of the system. By
history of the system, we mean the previous conditions under which it has existed.
Since the state of a system at equilibrium can be specified by a small number
of state variables, it should be possible to express the value of a variable that has
not been specified as a function of the values of other variables that have been
specified. The simplest example of this is the ideal gas law.

5

6

Chapter 1

Zeroth Law of Thermodynamics and Equations of State

For some systems, more than two intensive variables must be stated to specify
the state of the system. If there is more than one species, the composition has to be
given. If a liquid system is in the form of small droplets, the surface area has to be
given. If the system is in an electric or magnetic field, this may have an effect on its
properties, and then the electric field strength and magnetic field strength become
state variables. We will generally ignore the effect of the earth’s gravitational field
on a system, although this can be important, as we will see in the special topic at the
end of this chapter. Note that the properties used to describe the state of a system
must be independent; otherwise they are redundant. Independent properties are
separately controllable by the investigator.
The pressure of the atmosphere is measured with a barometer, as shown in
Fig. 1.2a, and the pressure of a gaseous system is measured with a closed-end
manometer, as shown in Fig. 1.2b.

Vacuum

h

(a)

1.2

Vacuum

THE ZEROTH LAW OF THERMODYNAMICS

System

h

(b)

Figure 1.2 (a) The pressure exerted by the atmosphere on the surface of mercury in a cup is given by
P  h␳g (see Example 1.1). (b) The
pressure of a system is given by the
same equation when a closed-end
manometer is used.

Although we all have a commonsense notion of what temperature is, we must
define it very carefully so that it is a useful concept in thermodynamics. If two
closed systems with fixed volumes are brought together so that they are in thermal contact, changes may take place in the properties of both. Eventually a state
is reached in which there is no further change, and this is the state of thermal equilibrium. In this state, the two systems have the same temperature. Thus, we can
readily determine whether two systems are at the same temperature by bringing
them into thermal contact and seeing whether observable changes take place in
the properties of either system. If no change occurs, the systems are at the same
temperature.
Now let us consider three systems, A, B, and C, as shown in Fig. 1.3. It is an
experimental fact that if system A is in thermal equilibrium with system C, and
system B is also in thermal equilibrium with system C, then A and B are in thermal
equilibrium with each other. It is not obvious that this should be true, and so this
empirical fact is referred to as the zeroth law of thermodynamics.
To see how the zeroth law leads to the definition of a temperature scale, we
need to consider thermal equilibrium between systems A, B, and C in more detail.
Assume that A, B, and C each consist of a certain mass of a different fluid. We
use the word fluid to mean either a gas or a compressible liquid. Our experience
is that if the volume of one of these systems is held constant, its pressure may
vary over a range of values, and if the pressure is held constant, its volume may
vary over a range of values. Thus, the pressure and the volume are independent
thermodynamic variables. Furthermore, suppose that the experience with these
systems is that their intensive states are specified completely when the pressure
and volume are specified. That is, when one of the systems reaches equilibrium
at a certain pressure and volume, all of its macroscopic properties have certain
characteristic values. It is quite remarkable and fortunate that the macroscopic
state of a given mass of fluid of a given composition can be fixed by specifying
only the pressure and the volume.*
If there are further constraints on the system, there will be a smaller number of independent variables. An example of an additional constraint is thermal
*This is not true for water in the neighborhood of 4 ⬚C, but the state is specified by giving the temperature and the volume or the temperature and the pressure. See Section 6.1.

1.2 The Zeroth Law of Thermodynamics

equilibrium with another system. Experience shows that if a fluid is in thermal
equilibrium with another system, it has only one independent variable. In other
words, if we set the pressure of system A at a particular value PA , we find that
there is thermal equilibrium with system C, in a specified state, only at a particular
value of VA . Thus, system A in thermal equilibrium with system C is characterized
by a single independent variable, pressure or volume; one or the other can be set
arbitrarily, but not both. The plot of all the values of PA and VA for which there
is equilibrium with system C is called an isotherm. Figure 1.4 gives this isotherm,
which we label ⌰1 . Since system A is in thermal equilibrium with system C at any
PA , VA on the isotherm, we can say that each of the pairs PA , VA on this isotherm
corresponds with the same temperature ⌰1 .
When heat is added to system C and the experiment is repeated, a different
isotherm is obtained for system A. In Fig. 1.4, the isotherm for the second experiment is labeled ⌰2 . If still more heat is added to system C and the experiment is
repeated again, the isotherm labeled ⌰3 is obtained.
Figure 1.4 illustrates Boyle’s law, which states that PV  constant for a specified amount of gas at a specified temperature. Experimentally, this is strictly true
only in the limit of zero pressure. Charles and Gay-Lussac found that the volume
of a gas varies linearly with the temperature at specified pressure when the temperature is measured with a mercury in glass thermometer, for example. Since it
would be preferable to have a temperature scale that is independent of the properties of particular materials like mercury and glass, it is better to say that the ratio
of the P2 V2 product at temperature ⌰2 to P1 V1 at temperature ⌰1 depends only
on the two temperatures:
P2 V2
 ␾ (⌰1 , ⌰2 )
P1 V1

(1.1)

where ␾ is an unspecified function. The simplest thing to do is to take the ratio
of the PV products to be equal to the ratio of the temperatures, thus defining

PA

Θ2

Θ3

Θ1

VA

Figure 1.4 Isotherms for fluid A. This plot, which is for a hypothetical fluid, might look
quite different for some other fluid.

A

7

C

Heat conductor

B

C

If A and C are in thermal equilibrium, and
B and C are in thermal equilibrium, then

A

B

A and B will be found to be in thermal equilibrium
when connected by a heat conductor.

Figure 1.3 The zeroth law of
thermodynamics is concerned with
thermal equilibrium between three
bodies.

8

Chapter 1

Zeroth Law of Thermodynamics and Equations of State

a temperature scale:
P2 V2
T2

P1 V1
T1

or

P2 V2
P1 V1

T2
T1

(1.2)

Here we have introduced a new symbol T for the temperature because we have
made a specific assumption about the function ␾ . Equations 1.1 and 1.2 are exact
only in the limit of zero pressure, and so T is referred to as the ideal gas temperature.
Since, according to equation 1.2, PV/T is a constant for a fixed mass of gas
and since V is an extensive property,
PV/T  nR

(1.3)

where n is the amount of gas and R is referred to as the gas constant. Equation 1.3
is called the ideal gas equation of state. An equation of state is a relation between
the thermodynamic properties of a substance at equilibrium.

1.3

THE IDEAL GAS TEMPERATURE SCALE

The ideal gas temperature scale can be defined more carefully by taking the temperature T to be proportional to P V  PV/n in the limit of zero pressure. Since
different gases give slightly different scales when the pressure is about one bar
(1 bar  10 5 pascal  10 5 Pa  10 5 N m⫺2 ), it is necessary to use the limit of
the P V product as the pressure approaches zero. When this is done, all gases
yield the same temperature scale. We speak of gases under this limiting condition
as ideal. Thus, the ideal gas temperature T is defined by
T  lim (P V /R )
Py0


V

P1
P2

P2 > P1

–273.15

(1.4)

The proportionality constant is called the gas constant R . The unit of thermodynamic temperature, 1 kelvin or 1 K, is defined as the fraction 1/273.16 of the temperature of the triple point of water.* Thus, the temperature of an equilibrium
system consisting of liquid water, ice, and water vapor is 273.16 K. The temperature 0 K is called absolute zero. According to the current best measurements, the
freezing point of water at 1 atmosphere (101 325 Pa; see below) is 273.15 K, and
the boiling point at 1 atmosphere is 373.12 K; however, these are experimental
values and may be determined more accurately in the future. The Celsius scale t
is formally defined by
t / ⬚C  T /K ⫺ 273.15

(1.5)

The reason for writing the equation in this way is that temperature T on the Kelvin
scale has the unit K, and temperature t on the Celsius scale has the unit ⬚C, which
need to be divided out before temperatures on the two scales are compared. In
Fig. 1.5, the molar volume of an ideal gas is plotted versus the Celsius temperature
t at two pressures.

0
t/°C

Figure 1.5 Plots of V versus temperature for a given amount of a real
gas at two low pressures P1 and P2 ,
as given by Gay-Lussac’s law.

*The triple point of water is the temperature and pressure at which ice, liquid, and vapor are in equilibrium with each other in the absence of air. The pressure at the triple point is 611 Pa. The freezing
point in the presence of air at 1 atm is 0.0100 ⬚C lower because (1) the solubility of air in liquid water at
1 atm (101 325 Pa) is sufficient to lower the freezing point 0.0024 ⬚C (Section 6.7), and (2) the increase
of pressure from 611 to 101 325 Pa lowers the freezing point 0.0075 ⬚C, as shown in Example 6.2. Thus,
the ice point is at 273.15 K.

1.3 The Ideal Gas Temperature Scale

We will find later that the ideal gas temperature scale is identical with one
based on the second law of thermodynamics, which is independent of the properties of any particular substance (see Section 3.9). In Chapter 16 the ideal gas
temperature scale will be identified with that which arises in statistical mechanics.
The gas constant R can be expressed in various units, but we will emphasize
the use of SI units. The SI unit of pressure (P) is the pascal, Pa, which is the pressure produced by a force of 1 N on an area of 1 m2 . In addition to using the prefixes
listed in the back cover of the book to express larger and smaller pressures, it is
convenient to have a unit that is approximately equal to the atmospheric pressure.
This unit is the bar, which is 10 5 Pa. Earlier the atmosphere, which is defined as
101 325 Pa, had been used as a unit of pressure.

Example 1.1

Express one atmosphere pressure in SI units

Calculate the pressure of the earth’s atmosphere at a point where the barometer reads 76
cm of mercury at 0 ⬚C and the acceleration of gravity g is 9.806 65 m s⫺2 . The density of
mercury at 0 ⬚C is 13.5951 g cm⫺3 , or 13.5951 ⫻ 10 3 kg m⫺3 .
Pressure P is force f divided by area A:
P  f /A
The force exerted by a column of air over an area A is equal to the mass m of mercury in
a vertical column with a cross section A times the acceleration of gravity g :
f  mg
The mass of mercury raised above the flat surface in Fig. 1.2a is ␳Ah so that
f  ␳Ahg
Thus, the pressure of the atmosphere is
P  h␳g
If h , ␳ , and g are expressed in SI units, the pressure P is expressed in pascals. Thus, the
pressure of a standard atmosphere may be expressed in SI units as follows:
1 atm  (0.76 m)(13.5951 ⫻ 10 3 kg m⫺3 )(9.806 65 m s⫺2 )
 101 325 N m⫺2  101 325 Pa  1.013 25 bar
This equality is expressed by the conversion factor 1.013 25 bar atm⫺1 .

To determine the value of the gas constant we also need the definition of a
mole. A mole is the amount of substance that has as many atoms or molecules as
0.012 kg (exactly) of 12 C. The molar mass M of a substance is the mass divided by
the amount of substance n , and so its SI unit is kg mol⫺1 . Molar masses can also be
expressed in g mol⫺1 , but it is important to remember that in making calculations
in which all other quantities are expressed in SI units, the molar mass must be
expressed in kg mol⫺1 . The molar mass M is related to the molecular mass m by
M  NA m , where NA is the Avogadro constant and m is the mass of a single
molecule.
Until 1986 the recommended value of the gas constant was based on measurements of the molar volumes of oxygen and nitrogen at low pressures. The accuracy

9

10

Chapter 1

Zeroth Law of Thermodynamics and Equations of State

of such measurements is limited by problems of sorption of gas on the walls of the
glass vessels used. In 1986 the recommended value* of the gas constant
R  8.314 51 J K⫺1 mol⫺1

(1.6)

was based on measurements of the speed of sound in argon. The equation used
is discussed in Section 17.4. Since pressure is force per unit area, the product of
pressure and volume has the dimensions of force times distance, which is work
or energy. Thus, the gas constant is obtained in joules if pressure and volume are
expressed in pascals and cubic meters; note that 1 J  1 Pa m3 .
Example 1.2

Express the gas constant in various units

Calculate the value of R in cal K⫺1 mol⫺1 , L bar K⫺1 mol⫺1 , and L atm K⫺1 mol⫺1 .
Since the calorie is defined as 4.184 J,
R  8.314 51 J K⫺1 mol⫺1 /4.184 J cal⫺1
 1.987 22 cal K⫺1 mol⫺1
Since the liter is 10⫺3 m3 and the bar is 10 5 Pa,
R  (8.314 51 Pa m3 K⫺1 mol⫺1 )(10 3 L m⫺3 )(10⫺5 bar Pa⫺1 )
 0.083 145 1 L bar K⫺1 mol⫺1
Since 1 atm is 1.013 25 bar,
R  (0.083 145 1 L bar K⫺1 mol⫺1 )/(1.013 25 bar atm⫺1 )
 0.082 057 8 L atm K⫺1 mol⫺1

1.4

IDEAL GAS MIXTURES AND DALTON’S LAW

Equation 1.3 applies to a mixture of ideal gases as well as a pure gas, when n is
the total amount of gas. Since n  n1  n2  ⭈⭈⭈, then
P  (n1  n2  ⭈⭈⭈)RT /V
 n1 RT /V  n2 RT /V  ⭈⭈⭈
 P1  P2  ⭈⭈⭈  冱 Pi

(1.7)

i

where P1 is the partial pressure of species 1. Thus, the total pressure of an ideal
gas mixture is equal to the sum of the partial pressures of the individual gases;
this is Dalton’s law. The partial pressure of a gas in an ideal gas mixture is the pressure that it would exert alone in the total volume at the temperature of the
mixture:
Pi  ni RT /V

(1.8)

A useful form of this equation is obtained by replacing RT /V by P /n:
Pi  ni P /n  yi P
*E. R. Cohen and B. N. Taylor, The 1986 Adjustment of the Fundamental Physical Constants,
CODATA Bull. 63:1 (1986); J. Phys. Chem. Ref. Data 17:1795 (1988).

(1.9)

1.5 Real Gases and the Virial Equation

The dimensionless quantity yi is the mole fraction of species i in the mixture, and
it is defined by ni /n . Substituting equation 1.9 in 1.7 yields

P1 = y1P

(1.10)

i

so that the sum of the mole fractions in a mixture is unity.
Figure 1.6 shows the partial pressures P1 and P2 of two components of a binary
mixture of ideal gases at various mole fractions and at constant total pressure. The
various mixtures are considered at the same total pressure P.
The behavior of real gases is more complicated than the behavior of an ideal
gas, as we will see in the next section.

Example 1.3 Calculation of partial pressures
A mixture of 1 mol of methane and 3 mol of ethane is held at a pressure of 10 bar. What
are the mole fractions and partial pressures of the two gases?
ym  1 mol/4 mol  0.25
Pm  ym P  (0.25)(10 bar)  2.5 bar
ye  3 mol/4 mol  0.75
Pe  ye P  (0.75)(10 bar)  7.5 bar

Example 1.4 Express relative humidity as mole fraction of water
The maximum partial pressure of water vapor in air at equilibrium at a given temperature is
the vapor pressure of water at that temperature. The actual partial pressure of water vapor
in air is a percentage of the maximum, and that percentage is called the relative humidity.
Suppose the relative humidity of air is 50% at a temperature of 20 ⬚C. If the atmospheric
pressure is 1 bar, what is the mole fraction of water in the air? The vapor pressure of water
at 20 ⬚C is 2330 Pa. Assuming the gas mixture behaves as an ideal gas, the mole fraction of
H2 O in the air is given by
yH2 O  Pi /P  (0.5)(2330 Pa)/10 5 Pa  0.0117

1.5

P

REAL GASES AND THE VIRIAL EQUATION

Real gases behave like ideal gases in the limits of low pressures and high temperatures, but they deviate significantly at high pressures and low temperatures.
The compressibility factor Z  P V /RT is a convenient measure of the deviation
from ideal gas behavior. Figure 1.7 shows the compressibility factors for N2 and
O2 as a function of pressure at 298 K. Ideal gas behavior, indicated by the dashed
line, is included for comparison. As the pressure is reduced to zero, the compressibility factor approaches unity, as expected for an ideal gas. At very high pressures
the compressibility factor is always greater than unity. This can be understood in
terms of the finite size of molecules. At very high pressures the molecules of the
gas are pushed closer together, and the volume of the gas is larger than expected

Pressure

1  y1  y2  ⭈⭈⭈  冱 yi

11

P 2 = y 2P
0

1
y2

Figure 1.6 Total pressure P and
partial pressures P1 and P2 of components of binary mixtures of gases
as a function of the mole fraction y2
of the second component at constant
total pressure. Note that y1  1 ⫺ y2 .

Zeroth Law of Thermodynamics and Equations of State
2.5

2.0

Z = PV/RT

N2
O2

1.5

1.0

0.5

200

0

400

600

800

P/bar

1000

Figure 1.7 Influence of high
pressure on the compressibility factor, P V /RT, for N2 and
O2 at 298 K. (See Computer
Problem 1.D.)

for an ideal gas because a significant fraction of the volume is occupied by the
molecules themselves. At low pressure a gas may have a smaller compressibility
factor than an ideal gas. This is due to intermolecular attractions. The effect of
intermolecular attractions disappears in the limit of zero pressure because the
distance between molecules approaches infinity.
Figure 1.8 shows how the compressibility factor of nitrogen depends on temperature, as well as pressure. As the temperature is reduced, the effect of intermolecular attraction at pressures of the magnitude of 100 bar increases because
the molar volume is smaller at lower temperatures and the molecules are closer
together. All gases show a minimum in the plot of compressibility factor versus pressure if temperature is low enough. Hydrogen and helium, which have very
low boiling points, exhibit this minimum only at temperatures much below 0 ⬚C.
A number of equations have been developed to represent P –V –T data for
real gases. Such an equation is called an equation of state because it relates state
properties for a substance at equilibrium. Equation 1.3 is the equation of state for
an ideal gas. The first equation of state for real gases that we will discuss is closely
related to the plots in Figs. 1.7 and 1.8, and is called the virial equation.
In 1901 H. Kamerlingh-Onnes proposed an equation of state for real gases,
which expresses the compressibility factor Z as a power series in 1/V for a pure
gas:
PV
B
C
 1   2  ⭈⭈⭈
RT
V
V

Z 

(1.11)

1.80

°C

–5
0

°C

2.00

0

Chapter 1

0

°C

10


Z = PV/RT

12

1.60
0

30

°C

1.40

1.20
0

30

°C

1.00

0.80
0

200

100 °C
0 °C
–50 °C
400
600
P/bar

800

1000

Figure 1.8 Influence of pressure on the
compressibility factor, P V /RT, for nitrogen
at different temperatures (given in C).

1.5 Real Gases and the Virial Equation
Table 1.1

Second and Third Virial Coefficients
at 298.15 K

Gas

B /10⫺6 m3 mol⫺1

H2
He
N2
O2
Ar
CO

14.1
11.8
⫺4.5
⫺16.1
⫺15.8
⫺8.6

C /10⫺12 m6 mol⫺2
350
121
1100
1200
1160
1550

25

He

0

H2

B/(cm3 mol–1)

Xe

CH4

Ar

NH3

–50

H2O
C3H8
–100

–150
C2H6
–200

0

200

400

600
800
T/K

1000

1200

1400

Figure 1.9 Second virial coefficient B. (From K. E. Bett, J. S. Rowlinson, and G. Saville,
Thermodynamics for Chemical Engineers. Cambridge, MA: MIT Press, 1975. Reproduced
by permission of The Athlone Press.) (See Computer Problem 1.E.)

The coefficients B and C are referred to as the second and third virial coefficients,
respectively.* For a particular gas these coefficients depend only on the temperature and not on the pressure. The word virial is derived from the Latin word for
force.
The second and third virial coefficients at 298.15 K are given in Table 1.1 for
several gases. The variation of the second virial coefficient with temperature is
illustrated in Fig. 1.9.
For many purposes, it is more convenient to use P as an independent variable
and write the virial equation as
Z 

PV
 1  B ⬘P  C ⬘P 2  ⭈⭈⭈
RT

(1.12)

Example 1.5 Derive the relationships between two types of virial coefficients
Derive the relationships between the virial coefficients in equation 1.11 and the virial coefficients in equation 1.12.
*Statistical mechanics shows that the term B /V arises from interactions involving two molecules, the
C /V 2 term arises from interactions involving three molecules, etc. (Section 16.11).

13

14

Chapter 1

Zeroth Law of Thermodynamics and Equations of State
The pressures can be eliminated from equation 1.12 by use of equation 1.11 in the
following forms:
P 
P2 

RT
BRT
CRT


 ⭈⭈⭈
V
V2
V3
2

冢 冣
RT
V



(1.13)

2B (RT )2
 ⭈⭈⭈
V3

(1.14)

Substituting these expressions into equation 1.12 yields
Z  1B⬘

RT
B ⬘BRT  C ⬘(RT )2

 ⭈⭈⭈
V
V2

冢 冣

(1.15)

When we compare this equation with equation 1.11 we see that
B B ⬘RT

(1.16)

C BB ⬘RT  C ⬘(RT )

2

(1.17)

Thus
B ⬘ B /RT
C⬘ 

(1.18)

⫺ B2

C
(RT )2

(1.19)

The second virial coefficient B for nitrogen is zero at 54 ⬚C, which is consistent
with Fig. 1.8. A real gas may behave like an ideal gas over an extended range
in pressure when the second virial coefficient is zero, as shown in Fig. 1.10. The
temperature at which this occurs is called the Boyle temperature TB . The Boyle
temperatures of a number of gases are given in Table 1.2.
Table 1.2


PV

C' = +

RTB
C' = –

P

Figure 1.10 At the Boyle temperature (B  0), a gas behaves nearly
ideally over a range of pressures.
The curvature at higher pressures
depends on the sign of the third
virial coefficient.

Critical Constants and Boyle Temperatures

Gas

Tc /K

Pc /bar

Vc /L mol⫺1

Zc

TB /K

Helium-4
Hydrogen
Nitrogen
Oxygen
Chlorine
Bromine
Carbon dioxide
Water
Ammonia
Methane
Ethane
Propane
n -Butane
Isobutane
Ethylene
Propylene
Benzene
Cyclohexane

5.2
33.2
126.2
154.6
417
584
304.2
647.1
405.6
190.6
305.4
369.8
425.2
408.1
282.4
365.0
562.1
553.4

2.27
13.0
34.0
50.5
77.0
103.0
73.8
220.5
113.0
46.0
48.9
42.5
38.0
36.5
50.4
46.3
49.0
40.7

0.0573
0.0650
0.0895
0.0734
0.124
0.127
0.094
0.056
0.0725
0.099
0.148
0.203
0.255
0.263
0.129
0.181
0.259
0.308

0.301
0.306
0.290
0.288
0.275
0.269
0.274
0.230
0.252
0.287
0.285
0.281
0.274
0.283
0.277
0.276
0.272
0.272

22.64
110.04
327.22
405.88

714.81
995
509.66

624

1.6 P –V –T Surface for a One-Component System

1.6 P –V –T SURFACE FOR A ONE-COMPONENT SYSTEM
To discuss more general equations of state, we will now look at the possible values
of P, V, and T for a pure substance. The state of a pure substance is represented
by a point in a Cartesian coordinate system with P, V, and T plotted along the
three axes. Each point on the surface of the three-dimensional model in Fig. 1.11
describes the state of a one-component system that contracts on freezing. We will
not be concerned here with the solid state, but will consider that part of the surface
later (Section 6.2). Projections of this surface on the P –V and P –T planes are
shown. There are three two-phase regions on the surface: S  G, L  G, and S  L
(S is solid, G gas, and L liquid). These three surfaces intersect at the triple point t
where vapor, liquid, and solid are in equilibrium.
The projection of the three-dimensional surface on the P –T plane is shown
to the right of the main diagram in Fig. 1.11. The vapor pressure curve goes from
the triple point t to the critical point c (see Section 1.7). The sublimation pressure
curve goes from the triple point t to absolute zero. The melting curve rises from the
triple point. Most substances contract on freezing, and for them the slope dP /dT
for the melting line is positive.
At high temperatures the substance is in the gas state, and as the temperature is raised and the pressure is lowered the surface is more and more closely
represented by the ideal gas equation of state P V  RT . However, much more
complicated equations are required to describe the rest of the surface that represents gas and liquid. Before discussing equations that can represent this part of
the surface, we will consider the unusual phenomena that occur near the critical
point. Any realistic equation of state must be able to reproduce this behavior at
least qualitatively.

T = const

2
e

f
c

P

S

ef

S
+
L

L
3

1

2

P = const
P

d

2

S+L

S

L

a
g

g

3

1
d
L+

S+

V

4

_

G

c

G

L

4

G

b

T

h

G

S
t

h

3

1
P

b

G

S+
V

G
G

L+

a

c

4
T

Figure 1.11 P –V –T surface for a one-component system that contracts on freezing.
(From K. E. Bett, J. S. Rowlinson, and G. Saville, Thermodynamics for Chemical Engineers. Cambridge, MA: MIT Press, 1975. Reproduced by permission of The Athlone
Press.)

15

16

Chapter 1

Zeroth Law of Thermodynamics and Equations of State

1.7

CRITICAL PHENOMENA

For a pure substance there is a critical point (Pc , Tc ) at the end of the liquid–gas
coexistence curve where the properties of the gas and liquid phases become so
nearly alike that they can no longer be distinguished as separate phases. Thus, Tc
is the highest temperature at which condensation of a gas is possible, and Pc is the
highest pressure at which a liquid will boil when heated.
The critical pressures Pc , volumes Vc , and temperatures Tc of a number of substances are given in Table 1.2, along with the compressibility factor at the critical
point Zc  Pc Vc /RTc , and the Boyle temperature TB .
Critical phenomena are most easily discussed using the projection of the
three-dimensional surface in Fig. 1.11 on the P –V plane. Figure 1.12 shows only
the parts of the P –V plot labeled L, G, and L  G. When the state of the system
is represented by a point in the L  G region of this plot, the system contains two
phases, one liquid and one gas, in equilibrium with each other. The molar volumes of the liquid and gas can be obtained by drawing a horizontal line parallel
to the V axis through the point representing the state of the system and noting
the intersections with the boundary line for the L  G region. Such a line, which
connects the state of one phase with the state of another phase with which it is in
equilibrium, is called a tie line. Two tie lines are shown in Fig. 1.12. The pressure
in this case is the equilibrium vapor pressure of the liquid. As the temperature is

2
G

P
c
3

1

L

L+G
4
VL

VG
V

Figure 1.12 Pressure–molar volume relations (e.g., isotherms) in the region of the critical
point. The dashed horizontal lines in the two-phase region are called tie lines. The path
1–2–3–4 shows how a liquid can be converted to a gas without the appearance of a meniscus.
If liquid at point 4 is compressed isothermally, the volume decreases until the two-phase
region is reached. At this point there is a large decrease in volume at constant pressure
(the vapor pressure of the liquid) until all of the gas has condensed to liquid. As the liquid
is compressed, the pressure rises rapidly.

1.8 The van der Waals Equation

raised, the tie line becomes shorter, and the molar volumes of the liquid and gas
approach each other. At the critical point c the tie line vanishes and the distinction
between liquid and gas is lost. At temperatures above the critical temperature,
there is a single fluid phase. Above the critical point a gas may have a very high
density, and it may be characterized as a supercritical fluid.
The isotherm that goes through the critical point has the following two properties: It is horizontal at the critical point,
⭸P

冢⭸V 冣

T Tc

0

(1.20)

and it has a point of inflection at the critical point,

冢 冣
⭸2 P
⭸2 V

0

(1.21)

T Tc

Figures 1.11 and 1.12 also show how a liquid at point 1 can be converted to
a gas at point 4 without the appearance of an interface between two phases. To
do this, liquid at point 1 is heated at constant volume to point 2, then expanded
at constant temperature to point 3, and finally cooled at constant volume to point
4, where it is a gas. Thus, liquid and vapor phases are really the same in terms of
molecular organization, and so when the densities of these two phases for a substance become equal, they cannot be distinguished and there is a critical point. On
the other hand, a solid and a liquid have different molecular organizations, and
the two phases do not become identical even if their densities are equal. Therefore, solid–liquid, solid–gas, and solid–solid equilibrium lines do not have critical
points as do gas–liquid lines.
At the critical point the isothermal compressibility [␬  ⫺V ⫺1 (⭸V /⭸P )T ; see
Problem 1.17] becomes infinite because (⭸P /⭸V )Tc  0. If the isothermal compressibility is very large, as it is in the neighborhood of the critical point, very little
work is required to compress the fluid. Therefore, gravity sets up large differences
in density between the top and bottom of the container, as large as 10% in a column of fluid only a few centimeters high. This makes it difficult to determine P V
isotherms near the critical point. These large differences, or spontaneous fluctuations, in the density can extend over macroscopic distances. The distance may be
as large as the wavelength of visible light or larger. Since fluctuations in density
are accompanied by fluctuations in refractive index, light is strongly scattered, and
this is called critical opalescence.

1.8

THE VAN DER WAALS EQUATION

Although the virial equation is very useful, it is important to have approximate
equations of state with only a few parameters. We turn now to the equation that
was introduced by van der Waals in 1877, which is based on plausible reasons that
real gases do not follow the ideal gas law. The ideal gas law can be derived for
point particles that do not interact except in elastic collisions (see Chapter 17,
Kinetic Theory of Gases). The first reason that van der Waals modified the ideal
gas law is that molecules are not point particles. Therefore V is replaced by V ⫺ b ,

17

18

Chapter 1

Zeroth Law of Thermodynamics and Equations of State

where b is the volume per mole that is occupied by the molecules. This leads to
P (V ⫺ b )  RT

(1.22)

which corresponds to equation 1.12 with B ⬘  b /RT and C ⬘ and higher constants
equal to zero. This equation can represent compressibility factors greater than
unity, but it cannot yield compressibility factors less than unity.
The second reason for modifying the ideal gas law is that gas molecules attract each other and that real gases are therefore more compressible than ideal
gases. The forces that lead to condensation are still referred to as van der Waals
forces, and their origin is discussed in Section 11.10. Van der Waals provided for
intermolecular attraction by adding to the observed pressure P in the equation of
state a term a /V 2 , where a is a constant whose value depends on the gas.
The van der Waals equation is*
(P  a /V 2 )(V ⫺ b )  RT

(1.23)

When the molar volume V is large, b becomes negligible in comparison with V,
a /V 2 becomes negligible with respect to P , and the van der Waals equation reduces to the ideal gas law, P V  RT.
The van der Waals constants for a few gases are listed in Table 1.3. They
can be calculated from experimental measurements of P, V, and T or from the critical constants, as shown later in equations 1.32 and 1.33. The van der Waals
equation is very useful because it exhibits phase separation between gas and liquid
phases.
Figure 1.13 shows three isotherms calculated using the van der Waals equation. At the critical temperature the isotherm has an inflection point at the critical point. At temperatures below the critical temperature each isotherm passes
through a minimum and a maximum. The locus of these points shown by the
dotted line has been obtained from (⭸P /⭸V )T  0. The states within the dotted line have (⭸P /⭸V )T ⬎ 0, that is, the volume increases when the pressure
increases. These states are therefore mechanically unstable and do not exist.
Maxwell showed that states corresponding to the points between A and B and

Table 1.3

Van der Waals Constants
2

D
E
c
P
A

T > Tc
Tc

C
B

T < Tc
_
V

Figure 1.13 Isotherms calculated
from the van der Waals equation.
The dashed line is the boundary of
the L  G region.

Gas

a /L bar
mol⫺2

H2
He
N2
O2
Cl2
NO
NO2
H2 O

0.247 6
0.034 57
1.408
1.378
6.579
1.358
5.354
5.536

2

⫺1

b /L mol

0.026 61
0.023 70
0.039 13
0.031 83
0.056 22
0.027 89
0.044 24
0.030 49

Gas

a /L bar
mol⫺2

b /L mol⫺1

CH4
C2 H6
C 3 H8
C4 H10 (n )
C4 H10 (iso)
C5 H12 (n )
CO
CO2

2.283
5.562
8.779
14.66
13.04
19.26
1.505
3.640

0.042 78
0.063 80
0.084 45
0.122 6
0.114 2
0.146 0
0.039 85
0.042 67

*The van der Waals equation can also be written in the form
(P  an 2 /V 2 )(V ⫺ nb )  nRT

1.8 The van der Waals Equation

those between D and E are metastable, that is, not true equilibrium states. The
dashed line is the boundary of the two-phase region; the part of the isotherm to
the left of A represents the liquid and that to the right of E, gas. The horizontal line ACE that produces two equal areas (ABC and CDE ) is referred to as
the Maxwell construction. It connects the thermodynamic properties of the liquid
phase (A) with the properties of the gas phase (E ) that is in equilibrium with it.
The compressibility factor for a van der Waals gas is given by
Z 


PV
V
a


RT
V ⫺b
RT V
1
a

1 ⫺ b /V
RT V

(1.24)

At low pressures, b /V ⬍⬍ 1 so that we can expand the first term using (1 ⫺ x )⫺1 
1  x  x 2  ⭈⭈⭈.
Example 1.6 Expansion of (1 ⫺ x )⫺1 using the Maclaurin series
Since we will use series like
1/(1 ⫺ x )  1  x  x 2  ⭈⭈⭈
a number of times, it is important to realize that functions can often be expressed as series
by use of the Maclaurin series
f (x )  f (0) 

冢dx 冣
df

x 0

x

冢 冣

1 d2 f
2! dx 2

x 2  ⭈⭈⭈

x 0

In this case,
f (0)  1
df
1

dx
(1 ⫺ x )⫺2

冢 冣

and

冢 冣

and

冢dx 冣

1

冢 冣

2

df

x 0

d2 f
 2(1 ⫺ x )⫺3
dx 2

d2 f
dx 2

x 0

Equation 1.24 then yields the virial equation in terms of volume:



Z  1 b ⫺



2

冢 冣  ⭈⭈⭈

1
a
b

RT V
V

(1.25)

From this equation we can see that the value of a is relatively more important at
low temperatures, and the value of b is relatively more important at high temperatures. To obtain the virial equation in terms of pressure, we can replace V in the
second term by the ideal gas value to obtain, to first order in P ,
Z  1





1
a
b⫺
P  ⭈⭈⭈
RT
RT

(1.26)

19

20

Chapter 1

Zeroth Law of Thermodynamics and Equations of State

but this approximation is not good enough to give the correct coefficient for the
P 2 term. At the Boyle temperature the second virial coefficient is zero, and so for
a van der Waals gas
TB 

a
bR

(1.27)

The values of the van der Waals constants may be calculated from the critical
constants for a gas, as shown in the following example.

Example 1.7

Van der Waals constants expressed in terms of critical constants

Derive the expressions for the van der Waals constants in terms of the critical constants for
a gas.
The van der Waals equation may be written
RT
a

V ⫺b
V2

P 

(1.28)

Differentiating with respect to molar volume and evaluating these equations at the critical
point yields

冢⭸V 冣

⭸P



⫺RTc
2a

0
Vc3
(Vc ⫺ b )2

(1.29)

冢 冣



2RTc
6a

0
Vc4
(Vc ⫺ b )3

(1.30)

Tc

⭸2 P
⭸V 2

Tc

A third simultaneous equation is obtained by writing equation 1.28 for the critical point:
Pc 

RTc
a

Vc ⫺ b
Vc2

(1.31)

These three simultaneous equations may be combined to obtain expressions for a and b in
terms of Tc and Pc or Tc and Vc :

Example 1.8

a

27R 2 Tc2
9
 RTc Vc
P
64 c
8

(1.32)

b

RTc
Vc

8Pc
3

(1.33)

Critical constants expressed in terms of van der Waals constants

Derive the expressions for the molar volume, temperature, and pressure at the critical point
in terms of the van der Waals constants.
Equation 1.33 shows that
Vc  3b
Equation 1.32 shows that
Tc 

8a
8a

Rb
27
9R Vc

1.9 Description of the State of a System without Chemical Reactions
Equation 1.33 shows that
Pc 

RTc
a

8b
27b 2

Example 1.9 Calculation of the molar volume using the van der Waals
equation
What is the molar volume of ethane at 350 K and 70 bar according to (a) the ideal gas law
and (b) the van der Waals equation?
(a ) V  RT /P  (0.083 145 L bar K⫺1 mol⫺1 )(350 K)/(70 bar)
 0.416 L mol⫺1
(b ) The van der Waals constants are given in Table 1.3.
P 

RT
a

V ⫺b
V2

70 

(0.083 15)(350)
5.562

V ⫺ 0.06380
V2

This is a cubic equation, but we know it has a single real, positive solution because the
temperature is above the critical temperature. This cubic equation can be solved using a
personal computer with a mathematical application. This yields two complex roots and one
real root, namely 0.2297 L mol⫺1 (see Computer Problem 1.G).

We will see later that equations of state are very important in the calculation
of various thermodynamic properties of gases. Therefore, a variety of them have
been developed. To represent the P –V –T properties of a one-component system over a wide range of conditions it is necessary to use an equation with many
more parameters. As more parameters are used they lose any simple physical interpretation. The van der Waals equation does not fit the properties of any gas
exactly, but it is very useful because it does have a simple interpretation and the
qualitatively correct behavior.
The van der Waals equation fails in the immediate neighborhood of the critical point. The coexistence curve (see Fig. 1.12) is not parabolic in the neighborhood of the critical point. The van der Waals equation indicates that near
Tc , Vc ⫺ V  k (Tc ⫺ T )1/2 , but experiments show that the exponent is actually
0.32. Other properties in the neighborhood of the critical point vary with (Tc ⫺ T )
with exponents that differ from what would be expected from the van der Waals
equation. These exponents are the same for all substances, which shows that the
properties in the neighborhood of the critical point are universal.

1.9

DESCRIPTION OF THE STATE OF A SYSTEM
WITHOUT CHEMICAL REACTIONS

In Section 1.1 we observed that the intensive state of a one-phase system can
be described by specifying Ns  1 intensive variables, where Ns is the number of

21

22

Chapter 1

Zeroth Law of Thermodynamics and Equations of State

species. The intensive state of a solution containing species A and species B is
completely described by specifying T , P , and nA /nB , and so three intensive variables are required. Now that we have discussed several systems, it is time to think
about the numbers of intensive variables required to define the thermodynamic
states of these more complicated systems. The number of independent variables
required is represented by F , which is referred to as the number of degrees of freedom. Therefore, for a one-phase system without chemical reactions, F  Ns  1.
As we have seen, if Ns  1, the independent intensive properties can be chosen
to be T and P . If Ns  1, but the system has two phases at equilibrium, Fig. 1.12
shows that it is sufficient to specify either T or P , but not both, so that F  1.
Thus the intensive state of this system is described completely by saying that two
phases are at equilibrium and specifying T or P . In defining the ideal gas temperature scale, we saw that water vapor, liquid water, and ice are in equilibrium
at a particular T and P . Thus the intensive state of this three-phase system is
completely described by saying that three phases are at equilibrium. There are no
independent intensive variables, and so F  0.
Earlier we contrasted the thermodynamic description of a system with the
classical description of a system in terms of molecules, and now we can see that
the description of the thermodynamic state of a system is really quite different.
Another interesting aspect of specifying degrees of freedom is that the choice of
variables is not unique, although the number is. For example, the intensive state
of a binary solution can be described by T , P , and the mole fraction of one of the
species.
The preceding paragraph has discussed the intensive state of a system, but
it is often necessary to describe the extensive state of a system. The number of
variables required to describe the extensive state of a system is given by D 
F  p , where p is the number of different phases, because the amount of each
phase must be specified. For a one-phase system with one species and no reactions,
D  2  1  3, and so a complete description requires T , P , and the amount of
the species (n ). For a two-phase system with one species, D  1  2  3, and so
it is necessary to specify T or P and the amounts of the two phases. For a threephase system with one species, D  0  3  3, and so it is necessary to specify
the amounts of the three phases. For a one-phase binary solution, D  31  4,
and so it is necessary to specify T , P , nA /nB , and the amount of the solution.
Phase equilibria and chemical equilibria introduce constraints, and we will see in
the next several chapters how these constraints arise and how they are treated
quantitatively in thermodynamics.

Comment:
It is a good thing that this issue of the number of variables required to describe
the state of a system has come up before we discuss the laws of thermodynamics
because the conclusions in this section cannot be derived from the laws of
thermodynamics. The fact that Ns  2 variables are required to describe
the extensive state of a homogeneous one-phase system at equilibrium is a
generalization of experimental observations, and we will consider it to be a
postulate. It is a postulate that has stood the test of time, and we will use it often
in discussing thermodynamic systems.

1.10 Partial Molar Properties

1.10

PARTIAL MOLAR PROPERTIES

This chapter has mostly been about pure gases, but we need to be prepared to
consider mixtures of gases and mixtures of liquids. There is an important mathematical difference between extensive properties and intensive properties of
mixtures. These properties can be treated as mathematical functions. A function
f (x1 , x2 , . . . , xN ) is said to be homogeneous of degree k if
f (␭x1 , ␭x2 , . . . , ␭xN)  ␭k f (x1 , x2 , . . . , xN)

(1.34)

All extensive properties are homogeneous of degree 1. This is illustrated by the
volume for which
V (␭n1 , ␭n2 , . . . , ␭nN)  ␭1 V (n1 , n2 , . . . , nN)  ␭V (n1 , n2 , . . . , nN) (1.35)
where n1 , n2 , . . . are amounts of substances. That is, if we increase the amounts of
every substance ␭-fold, the total volume increases ␭-fold. All intensive properties
are homogeneous of degree zero. This is illustrated by the temperature for which
T (␭n1 , ␭n2 , . . . , ␭nN)  ␭0 T (n1 , n2 , . . . , nN)  T (n1 , n2 , . . . , nN)

(1.36)

According to Euler’s theorem, when equation 1.34 applies,
N

kf (x1 , x2 , . . . , xN )  冱 xi
i 1

⭸f

冢⭸x 冣

(1.37)

i xj 苷xi

Thus for the volume of a mixture (k  1),
V 

⭸V

冢⭸n 冣

1 T,P,nj

n1 

⭸V

冢⭸n 冣

2 T,P,nj

n2  ⭈⭈⭈ 

 V1 n1  V2 n2  ⭈⭈⭈  VN nN

⭸V

冢⭸n 冣

nN

N T,P,nj

(1.38)

where the subscript nj indicates that the amounts of all other substances are held
constant when the amount of one of the substances is changed. These derivatives
are referred to as partial molar volumes. Since we will use such equations a lot,
partial molar properties are indicated by the use of an overbar:
Vi 

⭸V

冢⭸n 冣

(1.39)

i T,P,兵nj 苷i 其

This definition for the partial molar volume can be stated in words by saying that
Vi dni is the change in V when an infinitesimal amount (dni ) of this substance
is added to the solution at constant T , P , and all other nj . Alternatively, it can
be said that Vi is the change in V when 1 mol of i is added to an infinitely large
amount of the solution at constant T and P .
Note that the partial molar volume depends on the composition of the solution. When the amount of substance 1 is changed by dn1 , the amount of substance
2 is changed by dn2 , etc., and the volume of the solution is changed by

23

24

Chapter 1

Zeroth Law of Thermodynamics and Equations of State

dV  V1 dn1  V2 dn2  ⭈⭈⭈  VN dnN

(1.40)

Dividing equation 1.38 by the total number of moles in the solution yields
V  V1 x1  V2 x2  ⭈⭈⭈  VN xN

(1.41)

where V is the molar volume of the solution and xi is the mole fraction of substance i in the solution. In Chapter 6 we will discuss the determination of the
partial molar volume of a species in a solution, and we will also see that in ideal
solutions the partial molar volume of a substance is equal to its molar volume in
the pure liquid.

Example 1.10 The partial molar volume of a gas in an ideal gas mixture
Calculate the partial molar volume of a gas in an ideal gas mixture.
The volume of an ideal gas mixture is
V 

RT
(n1  n2  ⭈⭈⭈)
P

Using equation 1.39 to find the partial molar volume of gas i yields
Vi 

⭸V

冢⭸n 冣

i T,P,兵nj 苷i 其



RT
P

Thus all of the gases in a mixture of ideal gases have the same partial molar volume. This
is not true for nonideal gases or for liquids.

Comment:
Calculus is used so much in physical chemistry that we have included a section
on calculus in Appendix D for quick reference. Since the properties of a system
depend on a number of variables, it is important to be clear about which properties
are held constant for a measurement or a process and to use subscripts on partial
derivatives.

dh

h

Figure 1.14 Column of an ideal
gas of uniform temperature and unit
cross section.

1.11

SPECIAL TOPIC: BAROMETRIC FORMULA

In applying thermodynamics we generally ignore the effect of the gravitational
field, but it is important to realize that if there is a difference in height there is a
difference in gravitational potential. For example, consider a vertical column of a
gas with a unit cross section and a uniform temperature T , as shown in Fig. 1.14.
The pressure at any height h is simply equal to the mass of gas above that height
per unit area times the gravitational acceleration g . The standard acceleration due
to gravity is defined as 9.806 65 m s⫺2 . The difference in pressure dP between h
and h  dh is equal to the mass of the gas between these two levels times g and
divided by the area. Thus,
d P  ⫺␳g dh

(1.42)

1.11 Special Topic: Barometric Formula

where ␳ is the density of the gas. If the gas is an ideal gas, then ␳  PM /RT, where
M is the molar mass, so that
dP  ⫺

PMg
dh
RT

(1.43)

Separating variables and integrating from h  0, where the pressure is P0 , to h ,
where the pressure is P, yields



P

P0

dP
⫺
P

ln



h

0

gM
dh
RT

(1.44)

P
gMh
⫺
P0
RT

(1.45)

P  P0 e⫺gMh/RT

(1.46)

This relation is known as the barometric formula.

Example 1.11 Pressure and composition of air at 10 km
Assuming that air is 20% O2 and 80% N2 at sea level and that the pressure is 1 bar, what
are the composition and pressure at a height of 10 km, if the atmosphere has a temperature
of 0 ⬚C independent of altitude?



P  P0 exp ⫺

gMh
RT



For O2 ,



PO2  (0.20 bar)exp ⫺

(9.8 m s⫺2 )(32 ⫻ 10⫺3 kg mol⫺1 )(10 4 m)
(8.3145 J K⫺1 mol⫺1 )(273 K)



 0.0503 bar
For N2 ,



PN2  (0.80)exp ⫺

9.8 ⫻ 28 ⫻ 10⫺3 ⫻ 10 4
8.3145 ⫻ 273



 0.239 bar
The total pressure is 0.289 bar, and yO2  0.173 and yN2  0.827.

Figure 1.15 gives the partial pressures of oxygen, nitrogen, and the total pressure as a function of height in feet, assuming the temperature is 273.15 K independent of height.
Comment:
This is our first contact with exponential functions, but there will be many more.
The barometric formula can also be regarded as an example of a Boltzmann
distribution, which will be derived in Chapter 16 (Statistical Mechanics). The
temperature determines the way particles distribute themselves over various
energy levels in a system.

25

Chapter 1

Zeroth Law of Thermodynamics and Equations of State
1

0.8

P/bar

26

0.6

PN2

Ptotal

0.4
0.2

PO2

10 000

20 000
30 000
h/feet

40 000

50 000

Figure 1.15 Partial pressures of oxygen, nitrogen, and the total pressure of the atmosphere as a function of height in feet, assuming the temperature is 273.15 K independent
of height (see Computer Problem 1.H).


1.

2.

3.

4.
5.

6.

7.

8.
9.

Nine Key Ideas in Chapter 1
The state of a macroscopic system at equilibrium can be specified by the
values of a small number of macroscopic variables. For a system in which
there are no chemical reactions, the intensive state of a one-phase system
can be specified by Ns  1 intensive variables, where Ns is the number of
different species.
According to the zeroth law of thermodynamics, if systems A and B are
individually in thermal equilibrium with system C, then A and B are in
thermal equilibrium with each other.
The ideal gas temperature scale is based on the behavior of gases in the
limit of low pressures. The unit of thermodynamic temperature, the kelvin,
represented by K, is defined as the fraction 1/273.16 of the temperature of
the triple point of water.
The total pressure of a mixture of ideal gases is equal to the sum of the
partial pressures of the gases in the mixture.
The virial equation of state, which expresses the compressibility factor Z
for a gas in terms of powers of the reciprocal molar volume or of the pressure, is useful for expressing experimental data on a gas provided the pressure is not too high or the gas too close to its critical point.
The van der Waals equation is useful because it exhibits phase separation
between gas and liquid phases, but it does not represent experimental data
exactly.
For a one-phase system without chemical reactions, we have seen that the
number of degrees of freedom F is equal to Ns  1. But if the system contains two phases at equilibrium, F  Ns , and if the system contains three
phases at equilibrium, F  Ns ⫺ 1. The number of variables D required
to describe the extensive state of a multiphase macroscopic system at equilibrium is F  p , where p is the number of phases.
The volume of a mixture is equal to the sum of the partial molar volumes
of the species it contains each multiplied by the amount of that species.
For an isothermal atmosphere, the pressure decreases exponentially with
the height above the surface of the earth.

27

Problems

REFERENCES
M. Bailyn, A Survey of Thermodynamics. New York: American Institute of Physics, 1944.
K. E. Bett, J. S. Rowlinson, and G. Saville, Thermodynamics for Chemical Engineers. Cambridge, MA: MIT Press, 1975.
J. H. Dymond and E. B. Smith, The Virial Coefficients of Pure Gases and Mixtures. Oxford,
UK: Oxford University Press, 1980.
K. S. Pitzer, Thermodynamics, 3rd ed. New York: McGraw-Hill, 1995.
J. M. Smith, H. C. Van Ness, and M. M. Abbott, Introduction to Chemical Engineering
Thermodynamics. New York: McGraw-Hill, 1996.
J. W. Tester and M. Modell, Thermodynamics and Its Applications. Upper Saddle River,
NJ: Prentice PTR, 1997.

PROBLEMS
Problems marked with an icon may be more conveniently solved on a personal computer with a mathematical program.
1.1 The intensive state of an ideal gas can be completely defined by specifying (1) T, P, (2) T, V, or (3) P, V. The extensive
state of an ideal gas can be specified in four ways. What are the
combinations of properties that can be used to specify the extensive state of an ideal gas? Although these choices are deduced
for an ideal gas, they also apply to real gases.
1.2 The ideal gas law also represents the behavior of mixtures
of gases at low pressures. The molar volume of the mixture is
the volume divided by the amount of the mixture. The partial
pressure of gas i in a mixture is defined as yi P for an ideal gas
mixture, where yi is its mole fraction and P is the total pressure.
Ten grams of N2 is mixed with 5 g of O2 and held at 25 ⬚C at 0.750
bar. (a) What are the mole fractions of N2 and O2 ? (b) What are
the partial pressures of N2 and O2 ? (c) What is the volume of the
ideal mixture?
1.3 A mixture of methane and ethane is contained in a glass
bulb of 500 cm3 capacity at 25 ⬚C. The pressure is 1.25 bar, and
the mass of gas in the bulb is 0.530 g. What is the average molar
mass, and what is the mole fraction of methane?
1.4 Nitrogen tetroxide is partially dissociated in the gas phase
according to the reaction
N2 O4 (g)  2NO2 (g)
A mass of 1.588 g of N2 O4 is placed in a 500-cm3 glass vessel at
298 K and dissociates to an equilibrium mixture at 1.0133 bar.
(a) What are the mole fractions of N2 O4 and NO2 ? (b) What
percentage of the N2 O4 has dissociated? Assume that the gases
are ideal.
1.5 Although a real gas obeys the ideal gas law in the limit
as P y 0, not all of the properties of a real gas approach the
values for an ideal gas as P y 0. The second virial coefficient
of an ideal gas is zero, and so dZ /dP  0 at all pressures. But
calculate dZ /dP for a real gas as P y 0.

1.6 Show how the second virial coefficient of a gas and
its molar mass can be obtained by plotting P /␳ versus P , where
␳ is the density of the gas. Apply this method to the following
data on ethane at 300 K.
P /bar
␳ /10⫺3 g cm⫺3

1
1.2145

10
13.006

20
28.235

1.7 Calculate the second and third virial coefficients for
hydrogen at 0 ⬚C from the fact that the molar volumes at
50.7, 101.3, 202.6, and 303.9 bar are 0.4634, 0.2386, 0.1271, and
0.090 04 L mol⫺1 , respectively.
1.8 The critical temperature of carbon tetrachloride is
283.1 ⬚C. The densities in g/cm3 of the liquid ␳ l and vapor ␳ v
at different temperatures are as follows:
t / ⬚C
␳l
␳v

100
1.4343
0.0103

150
1.3215
0.0304

200
1.1888
0.0742

250
0.9980
0.1754

270
0.8666
0.2710

280
0.7634
0.3597

What is the critical molar volume of CCl4 ? It is found that the
mean of the densities of the liquid and vapor does not vary
rapidly with temperature and can be represented by

␳l  ␳v
 A  Bt
2
where A and B are constants. The extrapolated value of the average density at the critical temperature is the critical density.
The molar volume Vc at the critical point is equal to the molar
mass divided by the critical density.
1.9 Show that for a gas of rigid spherical molecules, b in
the van der Waals equation is four times the molecular volume times Avogadro’s constant. If the molecular diameter of
Ne is 0.258 nm (Table 17.4), approximately what value of b is
expected?
1.10 What is the molar volume of n -hexane at 660 K and 91
bar according to (a) the ideal gas law and (b) the van der Waals
equation? For n -hexane, Tc  507.7 K and Pc  30.3 bar.

28

Chapter 1

Zeroth Law of Thermodynamics and Equations of State

1.11 Derive the expressions for van der Waals constants a and
b in terms of the critical temperature and pressure; that is, derive
equations 1.32 and 1.33 from 1.29–1.31.
1.12 Calculate the second virial coefficient of methane at 300 K
and 400 K from its van der Waals constants, and compare these
results with Fig. 1.9.
1.13 You want to calculate the molar volume of O2 at 298.15
K and 50 bar using the van der Waals equation, but you don’t
want to solve a cubic equation. Use the first two terms of
equation 1.26. The van der Waals constants of O2 are a 
0.138 Pa m6 mol⫺1 and b  31.8 ⫻ 10⫺6 m3 mol⫺1 . What is the
molar volume in L mol⫺1 ?
1.14 The isothermal compressibility ␬ of a gas is defined in
Problem 1.17, and its value for an ideal gas is shown to be 1/P .
Use implicit differentiation of V with respect to P at constant T
to obtain the expression for the isothermal compressibility of a
van der Waals gas. Show that in the limit of infinite volume, the
value for an ideal gas is obtained.
1.15 Calculate the second and third virial coefficients of O2
from its van der Waals constants in Table 1.3.
1.16 Calculate the critical constants for ethane using the van
der Waals constants in Table 1.3.
1.17 The cubic expansion coefficient ␣ is defined by

␣

1 ⭸V
V ⭸T

冢 冣

P

and the isothermal compressibility ␬ is defined by

␬⫺

1 ⭸V
V ⭸P

冢 冣

T

Calculate these quantities for an ideal gas.
1.18 What is the equation of state for a liquid for which the coefficient of cubic expansion ␣ and the isothermal compressibility
␬ are constant?
1.19 For a liquid the cubic expansion coefficient ␣ is nearly
constant over a narrow range of temperature. Derive the expression for the volume as a function of temperature and the limiting
form for temperatures close to T0 .
1.20 (a) Calculate (⭸P /⭸V )T and (⭸P /⭸T )V for a gas that has
the following equation of state:
P 

nRT
V ⫺ nb

(b) Show that (⭸2 P /⭸V ⭸T )  (⭸2 P /⭸T ⭸V ). These are referred
to as mixed partial derivatives.
1.21 Assuming that the atmosphere is isothermal at 0 ⬚C and
that the average molar mass of air is 29 g mol⫺1 , calculate the
atmospheric pressure at 20 000 ft above sea level.
1.22 Calculate the pressure and composition of air on the top
of Mt. Everest, assuming that the atmosphere has a temperature
of 0 ⬚C independent of altitude (h  29 141 ft). Assume that air
at sea level is 20% O2 and 80% N2 .

1.23 Calculate the pressure due to a mass of 100 kg in the
earth’s gravitational field resting on an area of (a) 100 cm2 and
(b) 0.01 cm2 . (c) What area is required to give a pressure of 1
bar?
1.24 A mole of air (80% nitrogen and 20% oxygen by volume) at 298.15 K is brought into contact with liquid water,
which has a vapor pressure of 3168 Pa at this temperature.
(a) What is the volume of the dry air if the pressure is 1 bar?
(b) What is the final volume of the air saturated with water vapor if the total pressure is maintained at 1 bar? (c) What are the
mole fractions of N2 , O2 , and H2 O in the moist air? Assume the
gases are ideal.
1.25 Using Fig. 1.9, calculate the compressibility factor Z for
NH3 (g) at 400 K and 50 bar.
1.26 In this chapter we have considered only pure gases, but
it is important to make calculations on mixtures as well. This
requires information in addition to that for pure gases. Statistical mechanics shows that the second virial coefficient for an
N -component gaseous mixture is given by
N

N

B  冱冱 yi yj Bij
i 1 j 1

where y is mole fraction and i and j identify components. Both
indices run over all components of the mixture. The bimolecular
interactions between i and j are characterized by Bij , and so
Bij  Bji . Use this expression to derive the expression for B
for a binary mixture in terms of y1 , y2 , B11 , B12 , and B22 .
1.27 The densities of liquid and vapor methyl ether in
g cm⫺3 at various temperatures are as follows:
t / ⬚C
␳l
␳v

30
0.6455
0.0142

50
0.6116
0.0241

70
0.5735
0.0385

100
0.4950
0.0810

120
0.4040
0.1465

The critical temperature of methyl ether is 299 ⬚C. What is the
critical molar volume? (See Problem 1.8.)
1.28 Use the van der Waals constants for CH4 in Table 1.3 to
calculate the initial slopes of the plots of the compressibility factor Z versus P at 300 and 600 K.
1.29 A gas follows the van der Waals equation. Derive the relation between the third and fourth virial coefficients and the van
der Waals constants.
1.30 Using the van der Waals equation, calculate the pressure
exerted by 1 mol of carbon dioxide at 0 ⬚C in a volume of (a) 1.00
L and (b) 0.05 L. (c) Repeat the calculations at 100 ⬚C and 0.05 L.
1.31 A mole of n -hexane is confined in a volume of 0.500 L at
600 K. What will be the pressure according to (a) the ideal gas
law and (b ) the van der Waals equation? (See Problem 1.10.)
1.32 A mole of ethane is contained in a 200-mL cylinder at 373
K. What is the pressure according to (a) the ideal gas law and
(b) the van der Waals equation? The van der Waals constants
are given in Table 1.3.

Problems
1.33 When pressure is applied to a liquid, its volume decreases.
Assuming that the isothermal compressibility

␬⫺

1 ⭸V
V ⭸P

冢 冣

T

is independent of pressure, derive an expression for the volume
as a function of pressure.
1.34 Calculate ␣ and ␬ for a gas for which
P (V ⫺ b )  RT
1.35 What is the molar volume of N2 (g) at 500 K and 600 bar
according to (a) the ideal gas law and (b) the virial equation?
The virial coefficient B of N2 (g) at 500 K is 0.0169 L mol⫺1 .
1.36 What is the mean atmospheric pressure in Denver, Colorado, which is a mile high, assuming an isothermal atmosphere
at 25 ⬚C? Air may be taken to be 20% O2 and 80% N2 .
1.37 Calculate the pressure and composition of air 100 miles
above the surface of the earth assuming that the atmosphere has
a temperature of 0 ⬚C independent of altitude.
1.38 The density ␳  m /V of a mixture of ideal gases A and
B is determined and is used to calculate the average molar mass
M of the mixture; M  ␳RT /P . How is the average molar mass
determined in this way related to the molar masses of A and B?
1.39 Figure 1.13 shows the Maxwell construction for calculating the vapor pressure of a liquid from its equation of state.
Since this requires an iterative process, a computer is needed,
and J. H. Noggle and R. H. Wood have shown how to write
a computer program in Mathematica (Wolfram Research, Inc.,
Champaign, IL 61820-7237) to do this. Use this method with the
van der Waals equation to calculate the vapor pressure of nitrogen at 120 K.

Computer Problems
1.A Problem 1.7 yields B  0.135 L mol⫺1 and C  4.3 ⫻
10⫺4 L2 mol⫺2 for H2 (g) at 0 ⬚ C. Calculate the molar volumes of
molecular hydrogen at 75 and 150 bar and compare these molar
volumes with the molar volume of an ideal gas.

29

1.B (a ) Plot the pressure of ethane versus its molar volume in
the range 0 ⬍ P ⬍ 200 bar and molar volumes up to 0.5 mol L⫺1
using the van der Waals equation at 265, 280, 310.671, 350, and
400 K, where 310.671 K is the critical temperature calculated
with the van der Waals constants. (b ) Discuss the significance
of the plots and the extent to which they represent reality. (c )
Calculate the molar volumes at 400 K and P  150 bar and at
265 K and 20 bar.
1.C This is a follow-up to Computer Problem 1.B on the van
der Waals equation. (a ) Plot the derivative of the pressure with
respect to the molar volume for ethane at 265 K. (b ) Plot the
derivative at the critical temperature. (c ) Plot the second derivative of the pressure with respect to the molar volume at the critical temperature. In each case, what is the significance of the
maxima and minima?
1.D (a ) Express the compressibility factors for N2 and O2 at
298.15 K as a function of pressure using the virial coefficients in
Table 1.1. (b ) Plot these compressibility factors versus P from 0
to 1000 bar.
1.E The second virial coefficients of N2 at a series of temperatures are given by
T /K
75 100 125
150
200
250 300 400 500 600 700
B ⬘/cm3 mol⫺1 ⫺274 ⫺160 ⫺104 ⫺71.5 ⫺35.2 ⫺16.2 ⫺4.2 9 16.9 21.3 24

(a ) Fit these data to the function
B ⬘  ␣  ␤T  ␥T 2
(b ) Plot this function versus temperature. (c ) Calculate the
Boyle temperature of molecular nitrogen.
1.F Nitrogen tetroxide (N2 O4 ) gas is placed in a 500-cm3 glass
vessel, and the reaction N2 O4  2NO2 goes to equilibrium at
25 ⬚ C. The density of the gas at equilibrium at 1.0133 bar is
3.176 g L⫺1 . Assuming that the gas mixture is ideal, what are
the partial pressures of the two gases at equilibrium?
1.G Calculate the molar volume of ethane at 350 K and 70 bar
using the van der Waals constants in Table 1.3.
1.H Plot the partial pressures of oxygen, nitrogen, and the total pressure in bars versus height above the surface of the earth
from zero to 50 000 feet assuming that the temperature is constant at 273 K.

2

First Law of Thermodynamics

2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
2.10
2.11
2.12
2.13

Work and Heat
First Law of Thermodynamics and
Internal Energy
Exact and Inexact Differentials
Work of Compression and Expansion
of a Gas at Constant Temperature
Various Kinds of Work
Change in State at Constant Volume
Enthalpy and Change of State
at Constant Pressure
Heat Capacities
Joule–Thomson Expansion
Adiabatic Processes with Gases
Thermochemistry
Enthalpy of Formation
Calorimetry

In this chapter we begin to emphasize processes that take a chemical system from
one state to another. The first law of thermodynamics, which is often referred to as
the law of conservation of energy, leads to the definition of a new thermodynamic
state function, the internal energy U. An additional state function, the enthalpy
H, is defined in terms of U, P, and V for reasons of convenience.
Thermochemistry, which deals with the heat produced by chemical reactions
and solution processes, is based on the first law. If heat capacities of reactants and
products are known, the heat of a reaction may be calculated at other temperatures once it is known at one temperature.

31

2 .1 Work and Heat

2.1

WORK AND HEAT

Force is a vector quantity; that is, it has direction as well as magnitude. Other
examples of vector quantities are displacement, velocity, acceleration, and electric
field strength. In this book vector quantities are represented by boldface italic
type. The magnitude of the vector is represented with lightface italic type. Force
is defined by
f  ma

(2 .1)

where f is the force that will give a mass m an acceleration a .
Work (w) is a scalar quantity defined by
w f ⭈L

(2 .2)

where f is the vector force, L is the vector length of path, and the dot indicates
a scalar product (i.e., the product is taken of the magnitude of one vector by the
projection of the second vector along the direction of the first). If the force vector
of magnitude f and the vector length of magnitude L are separated by the angle
␪ , the work is given by fL cos ␪ .*
The SI unit of force is the newton N, which is equal to kg m s⫺2 . The SI unit
of work is the joule J, which is N m or kg m2 s⫺2 . The differential quantity of work
dw done by a force f operating over a distance dL in the direction of the force is
f dL.
Since pressure P is force per unit area, the force on a piston is PA, where A is
the surface area perpendicular to the direction of the motion of the piston. Thus,
the differential quantity of work done by an expanding gas that causes the piston
to move distance dL is PA dL. But A dL  dV, the increase in gas volume, and
so the differential quantity of pressure–volume work is P dV.
Work w can be positive or negative since work may be done on a system or a
system may do work on its surroundings, as shown in Fig. 2.1. The convention on
w is that it is positive when work is done on the system of interest and negative
when the system does work on the surroundings. (As we will see later, a similar
convention is applied to heat q ; q is positive when heat is transferred from the
surroundings to a system, and q is negative when heat is transferred from the system to the surroundings.) Thus, the differential of the PV work done on a system
is given by
dw  ⫺Pext dV



(2 .3)

where Pext is the external or applied pressure.
Work is often conveniently measured by the lifting or falling of masses. The
work required to lift a mass m in the earth’s gravitational field, which has an acceleration g , is mgh , where h is the height through which the mass is lifted.
*Since this is our first contact with vectors and matrices, we want to note that they are represented
by boldface italic type. They may have units like other physical quantities. Sections D.7 and D.8 of
Appendix D give information about the mathematical properties of vectors and matrices.


We use dw rather than dw as a reminder that work is not an exact differential (Section 2 .3), and so
the value of its integral depends on the path.

Vacuum

Vacuum
m
h
System

m
System

w is positive
(a)
Vacuum

Vacuum
m

m

h
System

System
w is negative
(b)

Figure 2 .1 (a) Work is done on a
system by the surroundings. In this
case the stops are pulled out, and
the system is compressed to a new
equilibrium state. (b) Work is done
on the surroundings by the system.
When the stops are pulled out, the
system expands to a new equilibrium
state.

32

Chapter 2

First Law of Thermodynamics

The work done by a system in lifting a kilogram 0.1 m is
w  ⫺mgh  ⫺(1 kg)(9.807 m s⫺2 )(0.1 m)  ⫺0.9807 J
where g is the acceleration of gravity. The negative sign indicates that work has
been done by the system.
The total work w on a system when there is a finite change in volume is obtained by summing the infinitesimal amounts of work given by equation 2.3:
w ⫺



2

Pext dV

(2 .4)

1

To make this calculation of the work for a finite change in state, Pext must have a
definite value at each volume.
If the expansion or compression of a gas is carried out very slowly, the pressure throughout the gas will be uniform and equal to Pext (within an infinitesimal
amount) and the maximum work of expansion (negative) or of compression (positive) will be obtained, as we will see in Section 2.4. When a process is carried out
in this way, the pressure given by the equation of state can be used in equation 2.4.
Such a process is said to be quasistatic. When the gas is allowed to expand rapidly
or is compressed rapidly, the pressure is not uniform and so such a substitution
cannot be made.
The integral in equation 2.4 is called a line integral because its value depends
on the path. Line integrals are discussed in greater detail in Section 2.3. In the
quasistatic case Pext  P and the pressure is a function of temperature and volume, and so equation 2.4 should be written
w ⫺



2

P (T, V ) dV

(2 .5)

1

In an ordinary definite integral, the integrand is a function of one variable. Later,
in Section 2 .4, we will replace P with nRT /V for an ideal gas and integrate equation 2 .5 at constant temperature, but now we want to take a more general point
of view and consider the two processes in Fig. 2.2. The state of a mole of gas can
be changed from (2P0 , V0 ) to (P0 , 2V0 ) by an infinite number of quasistatic paths,
but we will consider only the two paths shown. In the upper path, the pressure
is held constant at 2P0 and the gas is heated until it reaches 2V0 . Then the volume is held constant while the gas is cooled until the pressure reaches P0 . For this
path, w  ⫺2P0 V0 . In the lower path, the volume is held constant at V0 and the
gas is cooled until it reaches P0 . Then the gas is heated at constant pressure until
it reaches 2V0 . For this path, w  ⫺P0 V0 . Note that in both cases the work is
the negative of the area under the path. This example shows that w depends on
the path.
When work is done on a system that is thermally insulated so that there is
no exchange of heat with the surroundings, the thermodynamic state of the system is changed. This type of process is referred to as an adiabatic process. Joule
performed experiments in 1840–1849 showing that the change in state of water in
an adiabatic process is independent of the path, that is, whether the work is used
to turn a paddle wheel (Fig. 2.3) or is dissipated by an electrical current flowing
through a resistance or by the friction of rubbing two objects together. Since a

2 .1 Work and Heat

33

Isobaric step
(heat flows in)

Initial
state
2P0

Isochoric step
(heat flows out)
P0

0

0

V0

2V0

Figure 2 .2 For the change in state of a mole of gas from (2P0 ,V0 ) to (P0 ,2V0 ), the work
done on the gas depends on the path. By the upper path, w  ⫺2P0 V0 . By the lower path,
w  ⫺P0 V0 . For a clockwise cyclic process, w  ⫺2P0 V0  P0 V0  ⫺P0 V0 . In the cyclic
process, the gas is returned to its initial state, and so ⌬U  0  q ⫺ P0 V0 . Thus, q  P0 V0
and heat is absorbed by the system in the cyclic process.

given change in state of the water in the calorimeter can be accomplished in different ways involving the same amount of work, or by different sequences of steps,
the change in state is independent of the path and is dependent only on the total
amount of work. This makes it possible to express the change in state of a system
in an adiabatic process in terms of the work required, without stating the type of
work or the sequence of steps used. The property of the system whose change is
calculated in this way is called the internal energy U. Since the internal energy U
of a system may be increased by doing work on it, we may calculate the increase
in internal energy from the work w done on a system to change it from one state
to another in an adiabatic process:
⌬U  w

(in an adiabatic process)

(2 .6)

In words, the work done on a closed system in an adiabatic process is equal to
the increase in internal energy of the system. The symbol ⌬ indicates the value of
the quantity in the final state minus the value of the quantity in the initial state;
⌬U  U2 ⫺ U1 , where U1 is the internal energy in the initial state and U2 is the
internal energy in the final state. If the system does work on its surroundings, w is
negative and, furthermore, ⌬U is negative (i.e., the internal energy of the system
decreases) if the process is adiabatic.
Although equation 2 .6 provides a way to determine the change in internal
energy of a system, it does not provide a way to determine the absolute magnitude
of the internal energy of the system. However, the internal energy can be fixed
arbitrarily for some given equilibrium state of the system, and equation 2.6 can
be used to determine the internal energy with respect to that reference state.
When equation 2 .6 is applied to a system of arbitrary size, the internal energy
is an extensive quantity, but in working problems we will often deal with molar
quantities and express the change in molar internal energy ⌬U in J mol⫺1 .

Figure 2 .3 Joule heated water by
performing work on it, in this case
by rotating a paddle wheel, and
found that the temperature rise depends only on the amount of work
done on the system.

34

Chapter 2

System
in state 1

First Law of Thermodynamics

Heat
reservoir

(a)
Heat-conducting wall

System

Heat
reservoir

(b)

System
in state 2

Heat
reservoir

(c)

Figure 2 .4 (a) A system in state 1
is insulated from the heat reservoir.
(b) The system is brought into contact with the heat reservoir through
a heat-conducting wall. (c) The system is then insulated from the heat
reservoir and is found to be in
state 2 .

A given change in state of a system can be accomplished in ways other than
by the performance of work under adiabatic conditions. A change equivalent to
that in the Joule experiment may be obtained by immersing a hot object in the
water. We should not say, however, that the water now has more heat any more
than we would say it has more work after it has been heated with moving paddle
wheels. In other words, heat and work are forms of energy crossing a boundary.
After the experiment, the temperature of the water is higher, and it has a greater
internal energy U. Heat is transferred when there is a gradient in temperature, as
shown in Fig. 2.4.
Since the same change in state (as determined by measuring properties such
as temperature, pressure, and volume) may be produced by doing work on the
system or by allowing heat to flow in, the amount of heat q may be expressed in
mechanical units. When Joule was doing his experiments the unit of heat was the
calorie, which is the heat required to raise the temperature of a gram of water
1 ⬚C, from 14.5 to 15.5 ⬚C. Joule was able to determine the mechanical equivalent
of heat, which is now known to be 1 calorie  4.184 kg m2 s⫺2  4.184 J. Now
we find it more convenient to express heat in joules and to define the calorie as
4.184 J. A joule of heat is the amount of heat that produces the same change in a
system as a joule of work. The dietary calorie is actually a kilocalorie.
Since heat is an algebraic quantity, it is important to adopt a sign convention.
The convention is that a positive value of q indicates that heat is absorbed by the
system from its surroundings. A negative value of q means that the system gives
up heat to its surroundings. The change in internal energy U produced by the
transfer of heat q to a system when no work is done is given by
⌬U  q

(no work done)

(2 .7)

In words, the heat absorbed by a closed system in a process in which no work is
done is equal to the increase in internal energy of the system. Or, put another way,
if no work is done, the heat evolved is equal to the decrease in the internal energy
of the system.
It is important to understand that it is the measurement of work in the surroundings that makes it possible to determine the quantity of heat q transferred
to a system. In Section 2.5 we will find that there are a number of different kinds
of work, and each of them can be readily measured by measuring the raising or
lowering of weights in the gravitational field of the earth.

2.2 FIRST LAW OF THERMODYNAMICS
AND INTERNAL ENERGY
Since the internal energy of a system can be changed a given amount by either
heat or work, these quantities are in this sense equivalent. They are both usually
expressed in joules. If both heat and work are added to a system,
⌬U  q  w

(2 .8)

For an infinitesimal change in state,
dU  dq  dw

(2 .9)

2 .2 First Law of Thermodynamics and Internal Energy

The d indicates that q and w are not exact differentials, as discussed in the next
section.
Equations 2.8 and 2 .9 are statements of the first law of thermodynamics. This
law is the postulate that there exists a property U, referred to as the internal energy, (1) that is a function of the state variables for the system and (2) for which
the change ⌬U for a process in a closed system may be calculated using equation 2 .8. The first law is not restricted to reversible processes.
This mathematical form of the first law seems obvious to us now, but prior to
1850 it was not obvious at all. Before 1850 the principle of conservation of energy
in mechanical systems was understood, but the role of heat in this principle was
not clear until Joule’s experiments led to equation 2.8.
If ⌬U is negative, we may say that the system loses energy in heat that is
evolved and work that is done by the system. The first law has nothing to say
about how much heat is evolved and how much work is done except that equation
2 .8 is obeyed. In other words, the entire decrease in internal energy could show
up as work (q  0). Another possibility is that even more than this amount of
work would be done and heat would be absorbed (q ⬎ 0), so that equation 2.8 is
obeyed. Although the first law has nothing to say about the relative amounts of
heat and work, the second law does, as we will see in Chapter 3. Since the internal
energy is a function of the state of a system, there is no change in internal energy
when a system is taken through a series of changes that return it to its initial state.
This is expressed by setting the cyclic integral equal to zero:



dU  0

(2 .10)

The circle indicates integration around a cycle, that is, where the initial and final
states are the same. The cyclic integrals of q and w are not generally equal to zero,
and their values depend on the path followed.
The first law is frequently stated in the form that energy may be transferred in
one form or another, but it cannot be created or destroyed. Thus, the total energy
of an isolated system is constant.
The internal energy U of a system is an extensive property (Section 1.1); thus,
if we double a system, the internal energy is doubled. However, the molar internal
energy is an intensive property. We will use U for the extensive property and U
for the intensive property.
The quantity of heat transferred to an object can be calculated using q 
⌬U ⫺ w , where w is the measured quantity of work done on the system. The
change in internal energy ⌬U in the process can be calculated from the quantity
w of work required in an adiabatic process (see Section 2.1).

Comment:
The statement of the first law of thermodynamics in mathematical form was a
great achievement, and actually did not occur until after the statement of the
second law. A key idea is that the quantity of work required to produce the same
temperature rise in the system as an unknown quantity of heat can be used as
a measure of the quantity of heat; thus heat is measured in terms of joules, just
like work.

35

36

Chapter 2

First Law of Thermodynamics

Now that we have established that the internal energy is a state function, we
want to be sure that we know how many variables have to be specified to describe the state of the system. If the system involves only PV work, the internal
energy of a mass of a pure substance can be described by a mathematical function of T, V, and n or T, P, and n ; these functions are represented by U (T, V, n )
and U (T, P, n ). We will give the precise form of these functions only for ideal
gases because the functions for real substances are very complicated. The internal energy of a homogeneous binary mixture can be specified by a function
U (T, V, n1 , n2 ), U (T, P, n1 , n2 ), or U (T, P, x1 , nt ), where x1 is the mole fraction of
substance 1 and nt is the total amount of material in the system. Thus the description of the extensive state of a homogeneous mixture of N species requires N  2
variables, one of which must be extensive. The intensive state of a pure substance
is determined by two intensive variables (T and P ), and the intensive state of
a homogeneous binary mixture is determined by three intensive variables (T, P,
and x1 ). Thus the intensive state of a homogeneous mixture of N species is specified by N  1 independent intensive variables. In Section 5.4 we will discuss the
change in this rule when chemical reactions are involved and are at equilibrium.

2.3

EXACT AND INEXACT DIFFERENTIALS

The internal energy U is a state function, like V, because it depends only on the
state of the system. The integral of the differential of a state function along any
arbitrary path is simply the difference between values of the function at two limits.
For example, if a system goes from state a to state b , we can write



b

dU  Ub ⫺ Ua  ⌬U

(2 .11)

a

Since the integral is path independent, the differential of a state function is called
an exact differential. The quantities q and w are not state functions. The integrals
of their differentials in going from state a to state b depend on the path chosen.
Therefore, their differentials are called inexact differentials. We will use d instead
of d to indicate inexact differentials. In going from state a to state b the work w
done is represented by



b

dw  w

(2 .12)

a

Note that the result of the integration is not written wb ⫺ wa , because the amount
of work done depends on the particular path that is followed between state a
and state b. For example, when a gas is allowed to expand, the amount of work
obtained may vary from zero (if the gas is allowed to expand into a vacuum) to
a maximum value that is obtained if the expansion is carried out reversibly, as
described in Section 2.4.
If an infinitesimal quantity of heat dq is absorbed by a system, and an infinitesimal amount of work dw is done on the system, the infinitesimal change in
the internal energy is given by
dU  dq  dw

(2 .13)

2 .3 Exact and Inexact Differentials

where the d is used with U since dU is an exact differential and d is used with q
and w because they are inexact differentials. In other words, U is a function of
the state of the system, and q and w for a process depend on the path.
It is interesting to note that the sum of two inexact differentials can be an
exact differential. To illustrate this point further, we consider the path from
a to b in Fig. 2.5. We may define the path by a curve y  y (x ) connecting
a and b .
The differential dz  y dx is not an exact differential,



b

dz  z 

a



b

y dx  area I

(2 .14)

a

because this area depends on the path between a and b , as may be seen from
Fig. 2 .5.
The differential dz  y dx  x dy is an exact differential. Since dz  d(xy ),



b

d z  ⌬z

a





b

d(xy )  xb yb ⫺ xa ya

(2 .15)

a

The reason dz  y dx  x dy is an exact differential may be seen from
Fig. 2 .5. The integral of dz from state a to state b may be written



b

a

dz  ⌬z 



b

y dx 

a



b

x dy  area I  area II

(2 .16)

a

The sum of these areas is independent of the shape of the curve (path) between a and b . If 冮 dz does not depend on the path taken between the points,
then dz is said to be an exact differential. Thermodynamic quantities like U,
H, S, and G (all of which will soon be introduced) form exact differentials,
since their values are dependent on the state variables and not on the path by
which the system got there. There is a simple test to see whether a differential
is exact.
For a system with just two independent degrees of freedom, the total differential dz of a quantity z may be determined by the differentials dx and dy in two
other quantities x and y . In general,
dz  M (x, y ) dx  N (x, y ) dy

(2 .17)

where M and N are functions of the independent variables x and y .
To show the test for exactness we now consider a function z that has an exact
differential. If z has a definite value at each point in the xy plane, then it must be
a function of x and y . If z  f (x, y ), then
dz 

⭸z

冢⭸x 冣

y

dx 

⭸z

冢⭸y 冣

x

dy

(2 .18)

37
b

yb
II

y
ya

a
I

xa

x

Figure 2 .5 Path of a system in
going from state a to state b .

xb

38

Chapter 2

First Law of Thermodynamics

Comparing equations 2.17 and 2.18, we find
M (x, y ) 

⭸z

冢⭸x 冣

(2 .19)

y

N (x, y ) 

⭸z

冢⭸y 冣

(2 .20)

x

Since the mixed partial derivatives are equal,

冤 冢 冣 冥  冤⭸x 冢⭸y 冣 冥
⭸ ⭸z
⭸y ⭸x



⭸z

y x

(2 .21)

x y

then
⭸M

⭸N

冢 ⭸y 冣  冢 ⭸x 冣
x

(2 .22)

y

This equation must be satisfied if dz is an exact differential. It is Euler’s criterion
for exactness. This relation is also very useful for obtaining relations between the
derivatives of thermodynamic functions.
To illustrate the use of equation 2.22 let us reconsider the differential dz 
y dx . Since M  y and N  0, (⭸M /⭸y )x  1 and (⭸N /⭸x )y  0, so that equation 2 .22 is not satisfied. Therefore, dz  y dx is not an exact differential. On the
other hand, dz  y dx  x dy is an exact differential: M  y , N  x so that
(⭸M /⭸y )x  1, (⭸N /⭸x )y  1, satisfying equation 2.22.

Example 2 .1

An inexact differential and an exact differential

Suppose
dz  xy 3 dx  3x 2 y 2 dy
Can z be a function of the state of a system (i.e., a function of x and y )? The partial derivatives of z are
⭸z

冢⭸x 冣

 xy 3 ,

y

⭸z

冢⭸y 冣

 3x 2 y 2

x

The mixed partial derivatives of z are
⭸2 z
 3xy 2 ,
⭸x ⭸y

⭸2 z
 6xy 2
⭸y ⭸x

Thus dz is not an exact differential, and so z cannot be a function of the state of the system.
However, consider the total differential of z ⬘:
dz ⬘  2xy 3 dx  3x 2 y 2 dy
Can z ⬘ be a function of the state of a system? The partial derivatives of z ⬘ are

冢 ⬘冣
⭸z
⭸x

y

 2xy 3 ,

冢 ⬘冣
⭸z
⭸y

x

 3x 2 y 2

2 .4 Work of Compression and Expansion of a Gas at Constant Temperature
The mixed partial derivatives of z ⬘ are
⭸2 z ⬘
 6xy 2 ,
⭸x ⭸y

⭸2 z ⬘
 6xy 2
⭸y ⭸x

Since the mixed partial derivatives are equal, dz ⬘ is an exact differential, and so z ⬘ can be
a function of the state of the system. In fact, z ⬘  x 2 y 3  const.

When the differential of a physical quantity is inexact, it may be possible to
use it to define another physical quantity that has an exact differential by multiplying by an integrating factor. For example, if the differential of the physical
quantity f (x, y ) is given by
df (x, y )  y (xy  1) dx ⫺ x dy

(2 .23)

df (x, y ) is an inexact differential because



⭸[ y (xy  1)]
⭸y



⭸ (⫺ x )
⭸x




 2xy  1

(2 .24)

 ⫺1

(2 .25)

x

y

However, multiplying equation 2.23 by the integrating factor 1/y 2 yields
df ⬘ ⬅





df
1
x
 x
dx ⫺ 2 dy
2
y
y
y

(2 .26)

The mixed partial derivatives are



⭸(x  1/y )
⭸y



⭸(⫺x /y 2 )
⭸x




⫺

1
y2

(2 .27)

⫺

1
y2

(2 .28)

x

y

The function f ⬘ is given by
f ⬘(x, y ) 

x2
x
  const.
y
2

(2 .29)

Integrating factors are useful for obtaining exact differentials from inexact differentials and in solving first-order differential equations.

2.4

WORK OF COMPRESSION AND EXPANSION
OF A GAS AT CONSTANT TEMPERATURE

Since work done in compressing a gas is positive, we start by considering the compression of a gas at constant temperature using the idealized apparatus shown in
Fig. 2 .6. The gas is contained in a rigid cylinder by a frictionless and weightless

39

40

Chapter 2

First Law of Thermodynamics

m
P2
h

m

P1, V1, T

P1

P2, V2, T

V2
(a)

(b)

V1
(c)

Figure 2 .6 Compression of a gas from P1 , V1 , T to P2 , V2 , T in a single step.

piston. The cylinder is immersed in a thermostat at temperature T, and the space
above the cylinder is evacuated so that the final pressure P2 is due only to the
mass m . The gas is initially confined to volume V1 because the piston is held up by
stops. When the stops are pulled out, the piston falls to the equilibrium position,
and the gas is compressed to volume V2 . The pressure of the gas at the end of the
process is given by
P2 

mg
A

(2 .30)

where g is the acceleration due to gravity, and A is the area of the piston. The
amount of work lost in the surroundings is mgh (Section 2.1), where h is the
difference in height, and so the work done on the gas is
w  mgh  ⫺P2 (V2 ⫺ V1 )

(2 .31)

Since V2 ⬍ V1 , the work done on the gas is positive. This is the smallest amount
of work that can be used to compress the gas from V1 to V2 in a single step at constant temperature. The work done is given by the shaded area in the P –V plot of
Fig. 2 .6c. Notice that the pressure used in calculating the work is not the pressure
of the gas but the external pressure determined by the mass m , cross-sectional
area A, and acceleration of gravity g .
However, we can carry out the compression with less work if we do it in two
or more steps, as shown in Fig. 2.7. We can compress the gas in two steps by first
using a mass m ⬘ just large enough to compress the gas to volume (V1  V2 )/2 in

P2

P2

P2

P1

P1

P1

V2

V1
(a)

V2

V1
(b)

V2

V1
(c)

Figure 2 .7 Compression of a gas from P1 , V1 , T to P2 , V2 , T in two, three, and an infinite
number of steps.

2 .4 Work of Compression and Expansion of a Gas at Constant Temperature

the first step, and then using the larger mass m for the second step. The work lost
to the surroundings is given by the shaded area in Fig. 2.7a. It is clear that by using
more and more steps we arrive eventually at the diagram in Fig. 2.7c, which shows
that the minimum amount of work is required in the limit of an infinite number
of steps. In this case the pressure is changed by an infinitesimal amount for each
infinitesimal step, and the work is given by the integral of equation 2.3 at constant
temperature.
w 

冮 dw  ⫺ 冮

V2

P dV

(2 .32)

V1

In case c it is not necessary to distinguish between the external pressure and the
gas pressure because they differ at most by an infinitesimal amount.
The work w done on a gas in an expansion can be determined using the idealized frictionless piston arrangement shown in Fig. 2.8. The gas is initially confined
to volume V1 because the piston is held by stops. When the stops are pulled out,
the gas expands to volume V2 . The mass is chosen so that P2  mg /A; in other
words, this is the maximum mass that the gas will raise to this height. The work
gained in the surroundings is mgh , and so the work done on the gas is
w  ⫺mgh  ⫺P2 (V2 ⫺ V1 )

(2 .33)

This is the negative of the largest amount of work that can be obtained in the
surroundings by the expansion of the gas from V1 to V2 at constant temperature
in a single step. The work done on the gas is given by the negative of the shaded
area in Fig. 2.8c.
More work can be obtained in the surroundings by using two, three, or an
infinite number of steps, as shown in Fig. 2 .9. The largest amount of work in the
surroundings and the largest negative work on the gas are obtained in the limit of
an infinite number of steps. The work done on the gas in the limiting case is given
by equation 2.32.
The work obtained in the surroundings in the single-step expansion is clearly
not great enough to compress the gas back to its initial state in a single-step
compression; this is evident from the shaded areas in Figs. 2.8c and 2.6c. (The
subscripts are different in the two figures because we have followed the usual convention of labeling the initial state with a 1 and the final state with a 2.) However,
the work obtained in the surroundings in the infinite-step expansion is exactly the

m

m

h

P1

P2, V2, T
P2

P1, V1, T

V2

V1
(a)

(b)

(c)

Figure 2 .8 Expansion of a gas from P1 , V1 , T to P2 , V2 , T in a single step.

41

42

Chapter 2

First Law of Thermodynamics

P1

P1

P1

P2

P2

P2

V1

V2

V1

V2

(a)

(b)

V1

V2
(c)

Figure 2 .9 Expansion of a gas from P1 , V1 , T to P2 , V2 , T in two, three, and an infinite
number of steps.

amount required to compress the gas back to its initial state by an infinite-step
compression; this is evident from the shaded areas in Figs. 2.9c and 2.7c.
The infinite-step compression described by Fig. 2.7c and the infinite-step
expansion described by Fig. 2.9c are referred to as reversible processes. These
idealized processes at constant temperature are reversible because the energy
accumulated in the surroundings in the expansion is exactly the amount required
to compress the gas back to the initial state. This can also be seen by applying
equation 2 .32 to the gas for a complete cycle from P1 , V1 , T to P2 , V2 , T and back
again. (Note that here we are using the same subscripts for the expansion and the
compression.)
wcycle  ⫺



V2



V2

P dV ⫺

V1

⫺

V1



V1



V2

P dV

V2

P dV 

P dV  0

(reversible)

(2 .34)

V1

Another important point about reversible processes is that they can be reversed at any point in the process by making an infinitesimal change, in this case
in the pressure. Thus, a reversible expansion or compression requires an absence
of friction, a balancing of internal and external pressures, and time to reestablish equilibrium after each infinitesimal step. When these conditions are not met
the process is irreversible, and the system and its surroundings cannot both be restored to their initial conditions. Remember that these are isothermal processes,
and there is heat flow that we have not talked about.
All real processes are irreversible, yet it is possible to approach reversibility
closely in some real processes. Heat may be transferred nearly reversibly if the
temperature gradient across which it is transferred is made very small. Electrical
charge may be transferred nearly reversibly from a battery if a potentiometer is
used so that the difference in electrical potential is very small. A liquid may be
vaporized nearly reversibly if the external pressure is made only very slightly less
than the equilibrium vapor pressure.
The concept of a reversible process is important because certain thermodynamic calculations can be made only for reversible processes. For processes in the
chemical industry, the greater the irreversibility, the greater is the loss in capacity
to do work; thus, literally every irreversibility has its cost.

2 .4 Work of Compression and Expansion of a Gas at Constant Temperature

Example 2.2 Work of compressing an ideal gas at constant pressure
Two moles of gas at 1 bar and 298 K are compressed at constant temperature by use of a
constant pressure of 5 bar. How much work is done on the gas? If the compression is driven
by a 100-kg mass, how far will the mass fall in the earth’s gravitational field?
w  ⫺P2 (V2 ⫺ V1 )
 ⫺P2

冢P

nRT



2



 ⫺nRT 1 ⫺

nRT
P1

P2
P1





 ⫺(2 mol)(8.3145 J K⫺1 mol⫺1 )(298 K)(1 ⫺ 5)
 19 820 J
h⫺

w
mg

 ⫺(19 820 J)/(100 kg)(9.8 m s⫺2 )
 ⫺20.22 m

Now we consider the work required for a reversible isothermal compression
of a gas and the work that can be obtained from a reversible isothermal expansion.
In a reversible process, the change is accomplished in infinitesimal steps. Such a
reversible process is often spoken of as one that consists of a series of successive equilibria. Since the gas is at its equilibrium pressure (within an infinitesimal
amount) at each step in the expansion, we may substitute the pressure given by
an equation of state into equation 2 .32 and integrate it. If the gas were allowed to
expand rapidly, the pressure and temperature would not be uniform throughout
the volume of the gas, and so such a substitution could not be made. If the expansion is carried out reversibly at constant temperature for an ideal gas, the external
pressure is always given by P  nRT /V. Substituting in equation 2.32, we obtain
wrev  ⫺



V2



V2

P dV

V1

⫺

V1

nRT
V2
dV  ⫺nRT ln
V
V1

(2 .35)*

since the temperature is constant.
In integration the lower limit always refers to the initial state and the upper limit to the final state. If the gas is compressed, the final volume is smaller
and wrev is positive. The positive value means that work is done on the gas. The
isothermal expansion of one mole of an ideal gas by a factor of 10 yields wrev 
⫺(1 mol)RT ln 10  ⫺5229 J at 273.15 K.
*ln represents the natural logarithm, and log represents the base 10 logarithm; ln x  ln (10) log x 
2 .303 log x .

43

44

Chapter 2

First Law of Thermodynamics

Since for an ideal gas at constant temperature P1 V1  P2 V2 , the reversible
work is also given by
wrev  nRT ln

P2
P1

(2 .36)

The equation for the maximum work of isothermal expansion of a van der
Waals gas is obtained by using equation 1.28:
wrev  ⫺



V2

V1



 ⫺nRT ln

Example 2 .3



nRT
an 2
⫺ 2 dV
V ⫺ nb
V
1
1
V2 ⫺ nb

 an 2
V1 ⫺ nb
V1
V2





(2 .37)

Work of reversible expansion of an ideal gas

One mole of an ideal gas expands from 5 to 1 bar at 298 K. Calculate w (a) for a reversible
expansion and (b) for an expansion against a constant external pressure of 1 bar.
(a )

wrev  nRT ln

P2
P1

 (1 mol)(8.3145 J K⫺1 mol⫺1 )(298 K) ln

1 bar
5 bar

 ⫺3988 J
(b ) wirrev  ⫺P2 (V2 ⫺ V1 )  ⫺P2

冢P

nRT
2



nRT
P1



 ⫺(1 bar)(1 mol)(8.3145 J K⫺1 mol⫺1 )(298 K)

冢1 bar ⫺ 5 bar 冣
1

1

 ⫺1982 J
More work is done on the surroundings when the expansion is carried out reversibly.

2.5

f

Figure 2 .10 Idealized experiment
for the determination of the surface
tension of a liquid.

VARIOUS KINDS OF WORK

There are a number of ways that work can be done on a system, or a system can do
work on its surroundings, other than PV work. If a system has a surface, surface
work may be involved. If the system is a solid, there may be work of elongation.
If the system involves electric charges, there may be work of transport of electric charge from a phase at one electric potential to a phase at a different electric
potential. If the system is in a gravitational field, there may be work of transport
of mass from one height in the field to another height in the field. If the system
involves electric or magnetic dipoles, there may be work of an electric or magnetic field in orienting these dipoles. Here we consider only surface work, work
of elongation, and work of transport of electric charge.
Let us consider the work required to increase the area of a surface. The force
f required to increase the area of a liquid film, as illustrated in Fig. 2.10, is
f  2L␥

(2 .38)

2 .5 Various Kinds of Work

where ␥ is the surface tension of the liquid and L is the length of the movable
bar. The factor 2 is involved because there are two liquid–gas interfaces in this
experiment. The surface tension is force per unit length and is usually expressed
in N m⫺1 . It is a temperature-dependent quantity in general. The surface tension
of water at 25 ⬚C is 71.97 ⫻ 10⫺3 N m⫺1 or 71.97 mN m⫺1 .* The surface tensions of
liquid metals and molten salts are large in comparison with those of other liquids,
as shown in Table 6.6. The work required to move the bar in Fig. 2.10 to the left
by a distance ⌬x is
w  f ⌬ x  2L ⌬ x␥  ␥ ⌬ As

(2 .39)

where ⌬ As is the change in surface area (2L ⌬ x ). This is the amount of work
done on the liquid system. According to this equation, surface tension is equal
to the ratio of work to change in area, so it can also be expressed in J m⫺2 . The
differential of surface work is given by
dw  ␥ d As

(2 .40)

Surface tension arises from the fact that the molecules in the surface of a liquid are attracted into the body of the liquid by the molecules in the body. This
inward attraction causes the surface to contract if it can and gives rise to a force in
the plane of the surface. Surface tension is responsible for the formation of spherical droplets, the rise of water in a capillary, and the movement of a liquid through
a porous solid. Solids also have surface tensions, but it is harder to measure them.
Two other forms of work are more familiar, so we do not need to discuss them
in detail. When a piece of rubber is stretched, the differential work done on the
rubber is given by
dw  f dL

(2 .41)

where f is the force and dL is the differential increase in length. When a small
charge dQ is moved through an electric potential difference ␾ , the work done on
the charge is given by
d w  ␾ dQ

(2 .42)

These differential work terms become a part of the first law if surface, elongational, and electrical work are involved:
dU  d q ⫺ Pext dV  ␥ d As  f d L  ␾ dQ

(2 .43)

This is an important equation because it shows how surface tension, surface area,
force, elongation, electric potential, and electric charge come into thermodynamics.
The variables involved in work form conjugate pairs of intensive and extensive variables, as shown in Table 2 .1. If both the intensive variable and the
extensive variable are expressed in SI units (see symbols in Appendix G),
the work is expressed in joules.
In this section and the preceding one we have considered processes in thermostats without saying anything about the quantity of heat flowing into or out
of the gas. Now it is time to talk about the flow of heat that accompanies such
processes.
*In older literature surface tensions are usually expressed in dynes cm⫺1 . In these units the surface
tension of water at 25 ⬚C is (71.97 ⫻ 10⫺3 N m⫺1 )(105 dynes N⫺1 )(10⫺2 m cm⫺1 )  71.97 dynes cm⫺1 .

45

46

Chapter 2

First Law of Thermodynamics
Table 2.1

Some Conjugate Pairs of Thermodynamic Variables
Intensive
Variable

Extensive
Variable

Differential
Work

Pressure, P
Surface tension, ␥
Force, f
Potential difference, ␾

Volume, V
Area, As
Length, L
Electric charge, Q

⫺P dV
␥ dAs
f dL
␾ dQ

Type of Work
Hydrostatic
Surface
Elongation
Electrical

Example 2 .4

Calculating other kinds of work

(a) A piece of stretched rubber exerts a force of 1 N. How much work has to be done on
the rubber to stretch it one centimeter? (b) The surface tension of water is 0.072 N/m at
25 ⬚C. How much work has to be done to increase the water surface by one square meter?
(c) A mole of electrons is transported across a potential difference of 1 V from the positive
electrode to the negative electrode. How much work is required?
(a ) w  f ⌬L  (1 N)(0.01 m)  0.01 J
(b ) w  ␥ ⌬ As  (0.072 N m⫺1 )(1 m2 )  0.072 J
(c ) w  ␾ ⌬Q  (1 V)(96 500 coulombs)  96 500 J

2.6 CHANGE IN STATE AT CONSTANT VOLUME
The quantity of heat q can be measured by determining the change in temperature
of a mass of material that absorbs the heat. The heat capacity C is defined by the
derivative C  dq /dT, but dq is an inexact differential because heat is not a state
function. Therefore, the path has to be specified; for example, we may consider
a constant-volume path or a constant-pressure path. First we consider changes in
state at constant volume.
When a system changes from one state to another at constant volume, the
change in internal energy U may be calculated from the heat q evolved and the
work w done on the system by the surroundings. For a chemically inert system of
fixed mass the internal energy U may be taken to be a function of any two of T,
V, and P. It is most convenient to take it as a function of T and V. Since U is a
state function, the differential dU is given by
dU 

⭸U

冢 ⭸T 冣

dT 

V

⭸U

冢⭸V 冣

dV

(2 .44)

T

The first term is the change in internal energy due to the temperature change
alone, and the second term is the change in internal energy due to the volume
change alone. Since the differential of the internal energy is given by dq ⫺ Pext dV,
if only pressure–volume work is involved, then
dq 

⭸U

冢 ⭸T 冣

V



dT  Pext 

冢⭸V 冣 冥 dV
⭸U

T

(2 .45)

47

2 .6 Change in State at Constant Volume

If the change in state of system X takes place at constant volume, it may be represented by
X(V1 , T1 ) y X(V1 , T2 )
In this case equation 2 .45 reduces to
dqV 

⭸U

冢 ⭸T 冣

dT

(2 .46)

V

Since the change in temperature and the heat transferred are readily measured,
it is convenient to define the heat capacity CV at constant volume as
dqV
dT
⭸U

⭸T

CV ⬅

冢 冣

(2 .47)

V

This equation may be applied to a system of any size, but frequently we will be
concerned with the intensive thermodynamic quantity CV, which has the SI units
J K⫺1 mol⫺1 . Since the heat capacity at constant volume is readily measured,
equation 2.47 may be integrated to obtain the change in internal energy for a
finite change in temperature at constant volume:
⌬UV 



T2

CV dT  qV

(2 .48)

T1

This is illustrated in Fig. 2 .11. Over a small temperature range CV may be nearly
constant so that
⌬UV  CV (T2 ⫺ T1 )  CV ⌬T

(2 .49)

In principle, the quantity (⭸U /⭸V )T may be measured in an experiment devised by Joule. Imagine two gas bottles connected with a valve and enclosed in a
thermally isolated container, as shown in Fig. 2 .12. The two bottles constitute the
system under consideration. The first bottle is filled with a gas under pressure, and
the second is evacuated. When the valve is opened, gas rushes from the first bottle
into the second. Joule found that there was no discernible change in the temperature once thermal equilibrium had been established, and so dq  0. No work is
done in this expansion since Pext  0, and so dw  0 and dU  dq  dw  0.
Since the temperature is constant, equation 2 .44 becomes
dU 

⭸U

冢⭸V 冣

dV  0

(2 .50)

T

Since dV 苷 0,
⭸U

冢⭸V 冣

0

(2 .51)

T

Thus , Joule concluded (incorrectly) that the internal energy of the gas is independent of the volume. However, this method is not very sensitive because of the
large heat capacity of the gas bottles relative to the gas. Equation 2.51 actually

C

T2
q=

C dT
T1

T1

T2
T

Figure 2 .11 The heat q absorbed
by a substance when it is heated
is equal to the integral of C dT
from the initial temperature T1
to the final temperature T2 . If an
amount n is heated at constant volume, C  nCV , and if an amount
n is heated at constant pressure,
C  nCP .

48

Chapter 2

Gas

First Law of Thermodynamics

Vacuum

Figure 2 .12 Joule’s experiment
in which a gas expands into a vacuum. Joule found there was no discernible change in temperature and
concluded that (⭸U /⭸V )T  0. We
now know that this applies to ideal
gases but not to real gases.

applies only to an ideal gas. The molecular interpretation of this relation is that
there is no interaction between the molecules of an ideal gas, and so the internal
energy does not change with the distance between molecules. On the other hand,
the internal energy of a real gas depends on the volume at constant temperature,
but the second law of thermodynamics is needed to derive an equation that can
be used to obtain (⭸U /⭸V )T experimentally.*

Example 2 .5

Heat absorbed in expansion of an ideal gas

Calculate the heat absorbed and the changes in internal energy for the two expansions of
an ideal gas described in Example 2 .3.
(a) According to Joule’s experiment, ⌬U  0 for the reversible isothermal expansion
of an ideal gas. Therefore,
qrev  ⌬U ⫺ wrev
 0 ⫺ (⫺3988 J)
 3988 J
(b )

qirrev  ⌬U ⫺ wirrev
 0 ⫺ (⫺1982 J)
 1982 J

Thus, more heat is absorbed by the gas in the reversible isothermal expansion.

2.7 ENTHALPY AND CHANGE OF STATE
AT CONSTANT PRESSURE
Constant-pressure processes are more common in chemistry than constantvolume processes because many operations are carried out in open vessels. If
only pressure–volume work is done and the pressure is constant and equal to the
applied pressure, the work w done on the system equals ⫺P ⌬V, so that equation
2 .7 may be written
⌬U  qP ⫺ P ⌬V

(2 .52)

where qP is the heat for the isobaric (constant pressure) process. If the initial state
is designated by 1 and the final state by 2, then
U2 ⫺ U1  qP ⫺ P (V2 ⫺ V1 )

(2 .53)

*In equation 4.112 we will find that for a van der Waals gas
⭸U

冢 ⭸V 冣

T



a
V2

This should not be surprising because van der Waals added a /V 2 to the pressure to provide for intermolecular attractions. The quantity (⭸U /⭸V )T is often called the internal pressure. It has the dimensions of pressure and is due to intermolecular attractions and repulsions. The internal pressure
changes with the volume because as the volume is increased, the average intermolecular distances
increase and the average intermolecular potential energy changes.

2 .7 Enthalpy and Change of State at Constant Pressure

so that the heat absorbed is given by
qP  (U2  PV2 ) ⫺ (U1  PV1 )

(2 .54)

Since the heat absorbed is given by the difference of two quantities that are functions of the state of the system, it is convenient to introduce a new state function,
the enthalpy H, defined by
H  U  PV

(2 .55)

q P  H 2 ⫺ H1

(2 .56)

Equation 2.54 may be written

In words, the heat absorbed in a process at constant pressure is equal to the change
in enthalpy. For an infinitesimal change at constant pressure
dqP  dH

(2 .57)

where dH is an exact differential since the enthalpy is a function of the state of
the system.
When pressure–volume work is the only kind of work (electrical and other
kinds being excluded), it is easy to visualize ⌬U and ⌬H ; in a constant-volume
calorimeter (Section 2 .13) the evolution of heat is a measure of the decrease in
internal energy U, and in a constant-pressure calorimeter the evolution of heat is
a measure of the decrease in enthalpy H.
The enthalpy H is an extensive property. Therefore, for a homogeneous mixture of Ns species involving only PV work, the enthalpy can be specified by Ns 2
variables, one of which must be extensive. The intensive state of a homogeneous
mixture of Ns species involving only PV work can be specified by Ns  1 intensive
variables.
Changes in state at constant pressure are of special interest in the laboratory,
where processes generally take place at constant pressure. For a chemically inert system of fixed mass, it is most convenient to take enthalpy H as a function
of temperature and pressure. Since H is a state function, the differential dH is
given by
⭸H
⭸H
dH 
dT 
dP
(2 .58)
⭸T P
⭸P T

冢 冣

冢 冣

If the change in state of a mole of X takes place reversibly at constant pressure, it
may be represented by
X(P1 , T1 ) y X(P1 , T2 )

(2 .59)

For such a change equations 2 .57 and 2 .58 may be combined to obtain
dqP 

⭸H

冢 ⭸T 冣

dT

(2 .60)

P

Since the change in temperature and the heat transferred are readily measured, it is convenient to define the heat capacity at constant pressure CP as
CP ⬅

⭸H
dqP

⭸T
dT

冢 冣

P

(2 .61)

49

50

Chapter 2

First Law of Thermodynamics

Since the heat capacity at constant pressure is readily measured, this equation may
be integrated to obtain the change in enthalpy for a finite change in temperature
at constant pressure:
⌬HP 



T2

CP d T

(2 .62)

T1

Comment:
The enthalpy H is not entirely new because it is defined in terms of U, P, and V.
From one point of view, the enthalpy is redundant, and so why is it introduced?
The answer is convenience. Use of H is more convenient in considering
measurements and processes at constant pressure, and U is more convenient in
considering measurements and processes at constant volume. Later, in Section
4.2, we will formalize this process of changing independent variables by use of
Legendre transforms.

2.8 HEAT CAPACITIES
Values of CP at 25 ⬚C for about 200 substances are given in Table C.2, and values
for 298 to 3000 K are given for a smaller number of substances in Table C.3 in
Appendix C. The dependence of CP on temperature is shown for a number of
gases in Fig. 2 .13. In general, the more complex the molecule, the greater is its
molar heat capacity, and the greater the increase with increasing temperature.
Power series in temperature may be used to represent CP as a function of
temperature:
CP  ␣  ␤T  ␥T 2

(2 .63)

Parameters for several gases are given in Table 2.2. The change in enthalpy with
temperature at constant pressure is then given by
H2 ⫺ H1 



H2

H1

dH 



T2

C P dT

(2 .64)

T1

100
80


CP

60
40
20

400

600

800

1000 1200 1400 1600 1800
T/K

Figure 2 .13 The effect of temperature on the molar heat capacities of several gases at
constant pressure. Starting with the highest curve at 1200 K, the gases are CH4 , NH3 , CO2 ,
H2 O, N2 , and He. (See Computer Problem 2.B.)

2 .8 Heat Capacities
Table 2.2

Molar Heat Capacity at Constant Pressure as a Function of Temperature from
300 to 1800 K: CP  ␣  ␤T  ␥T 2  ␦T 3


⫺1

JK
N2 (g)
O2 (g)
H2 (g)
CO(g)
CO2 (g)
H2 O(g)
NH3 (g)
CH4 (g)


⫺1

mol

28.883
25.460
29.088
28.142
22.243
32.218
24.619
19.875

⫺2

10

⫺2

JK


⫺1

⫺5

mol

10

⫺0.157
1.519
⫺0.192
0.167
5.977
0.192
3.75
5.021

JK



⫺3

mol

⫺1

10

⫺9

0.808
⫺0.715
0.400
0.537
⫺3.499
1.055
⫺0.138
1.268

J K⫺4 mol⫺1
⫺2.871
1.311
⫺0.870
⫺2.221
7.464
⫺3.593

⫺11.004

Source S. I. Sandler, Chemical and Engineering Thermodynamics, 3rd ed. Copyright 1999 Wiley,
Hoboken, NJ. This material is used by permission of John Wiley & Sons, Inc.

H2 ⫺ H1  ␣ (T2 ⫺ T1 ) 

␤ 2

(T ⫺ T 21 )  (T 32 ⫺ T 31 )
2 2
3

(2 .65)

JANAF Thermochemical Tables and Stull, Westrum, and Sinke, in The Chemical Thermodynamics of Organic Compounds, give values of HT⬚ ⫺ H298
⬚ for various temperatures so that H2⬚ ⫺ H1⬚ is readily calculated for the substances listed.
The superscript indicates that the substance is in its standard state; standard states
are discussed in Section 2 .11.

Example 2.6 The change in molar enthalpy on heating
Using data in Table C.3, calculate the change in the molar enthalpy of methane in going
from 500 to 1000 K.
H1000
⬚ ⫺ H500
⬚  (H1000
⬚ ⫺ H298
⬚ ) ⫺ (H500
⬚ ⫺ H298
⬚ )
 (38.179 ⫺ 8.200) kJ mol⫺1
 29.979 kJ mol⫺1
(Note that this is not the change in the enthalpy of formation of methane; see Problem
2 .29.)

The relation between heat capacities at constant pressure and constant volume can be derived using equation 2.45 at constant pressure (P  Pext ), where it
can be written as

冤 冢 冣冥

d qP  C V d T  P 

⭸U
⭸V

dV

(2 .66)

T

Dividing by dT and setting dqP /dT  CP , we obtain

冤 冢⭸V 冣 冥 冢⭸T 冣

CP ⫺ CV  P 

⭸U

⭸V

T

P

(2 .67)

51

52

Chapter 2

First Law of Thermodynamics

The quantity on the right-hand side is positive so that CP ⬎ CV. The two terms
on the right-hand side may be interpreted as follows: P (⭸V /⭸T )P is the work
produced per unit increase in temperature at constant pressure, and
⭸U

⭸V

冢⭸V 冣 冢⭸T 冣
T

P

is the energy per unit temperature required to separate the molecules against
intermolecular attraction.
Equation 2.67 takes on a particularly simple form for an ideal gas because
(⭸U /⭸V )T  0 and (⭸V /⭸T )P  nR /P . Thus,
CP ⫺ CV  nR

or

CP ⫺ CV  R

(2 .68)

This relationship may be visualized as follows. When an ideal gas is heated at constant pressure, the work done in pushing back the piston is P ⌬V  nR ⌬T . For
a 1 K change in temperature, the work done is nR (1 K), and this is just the extra
energy required to heat an ideal gas 1 K at constant pressure over that required
at constant volume.
We will see later, in equation 4.120, that by use of the second law the difference between CP and CV for any material may be expressed in terms of the cubic
expansion coefficient ␣ and the isothermal compressibility ␬ (see Problems 1.17
and 1.18). The values of CP and CV for liquids and solids are nearly the same.
Thermodynamics does not deal with molecular models, and it is unnecessary
even to discuss molecules in connection with thermodynamics. This is one of the
strengths of thermodynamics, but it is also a weakness because thermodynamics, by itself, does not provide the means for predicting the numerical values of
thermodynamic properties of particular substances. We will see later that kinetic
theory and statistical mechanics do lead to quantitative predictions of thermodynamic properties.
Kinetic theory (Chapter 17) shows that the molar translational energy of a
monatomic ideal gas is 32 RT . The translational energy Ut is independent of pressure or molar mass for a monatomic ideal gas so that
Ut  32 RT

(2 .69)

According to equation 2.55, the molar enthalpy of a monatomic ideal gas is larger
than the internal energy by P V (or RT ), so that
Ht  32 RT  RT  52 RT

(2 .70)

Thus, the translational contributions to the molar heat capacities of monatomic
ideal gases are expected to be
CV 

CP 

冢 冣
冢 冣
⭸U
⭸T

⭸H
⭸T

 32 R  12 .472 J K⫺1 mol⫺1

(2 .71)

 52 R  20.786 J K⫺1 mol⫺1

(2 .72)

V

P

Tables C.2 and C.3 in Appendix C show that values of CP for monatomic
gases are constant at 20.786 J K⫺1 mol⫺1 independent of temperature, except for

2 .9 Joule–Thomson Expansion

cases where electrons in the atom can be excited to low-lying levels [see especially
O(g)].

2.9

JOULE–THOMSON EXPANSION

A gas flowing along an insulated pipe through a porous plate that separates two
regions of different constant pressures may be heated up or cooled down. This
Joule–Thomson expansion is shown in Fig. 2 .14, where P1 ⬎ P2 . To push one mole
of gas through the porous plate, work amounting to P1 V1 has to be done on one
mole of the gas by the piston on the left. Work amounting to P2 V2 is done on the
surroundings by one mole of gas pushing the piston on the right, and so the net
work on the gas is
w  P1 V1 ⫺ P2 V2

(2 .73)

Since the pipe is insulated, q  0, and
U 2 ⫺ U1  P1 V 1 ⫺ P 2 V 2

(2 .74)

U2  P2 V2  U1  P1 V1

(2 .75)

H2  H1

(2 .76)

or

Thus we see that there is no change in the enthalpy of the gas in a Joule–Thomson
expansion.
The Joule–Thomson coefficient ␮JT is defined as the derivative of the temperature with respect to pressure in this process:

␮JT  lim

⌬P y 0

⭸T
T 2 ⫺ T1

⭸P
P 2 ⫺ P1

冢 冣

(2 .77)

H

The Joule–Thomson coefficient is zero for an ideal gas, but for real gases (⭸T /⭸P )H
is positive at low temperatures and negative at high temperatures; that means
that a cooling effect is obtained below the inversion temperature and a heating effect is obtained above the inversion temperature. Below the inversion
temperature, this effect can be used for refrigeration, but we will not discuss it

Porous plate

T1, P1

T2, P2

Insulation

Figure 2 .14 Joule–Thomson expansion.

53

54

Chapter 2

First Law of Thermodynamics

further. The inversion temperature for nitrogen is 607 K, for hydrogen 204 K, and
for helium 43 K.

2.10

ADIABATIC PROCESSES WITH GASES

In Section 2 .4 we discussed the work of compression and expansion of gases in
contact with a heat reservoir. Now we consider the compression and expansion
of gases in isolated systems. No heat is gained or lost by the gas, so the process
is adiabatic and the first law becomes simply dU  dw . If only pressure–volume
work is involved, dU  ⫺Pext dV. If the system expands adiabatically, dV is positive and dU is negative; thus, if the expansion is opposed by an external pressure
Pext , work is done on the surroundings at the expense of the internal energy. The
relation dU  dw  ⫺Pext dV applies to any adiabatic process, reversible or irreversible, if PV work is the only kind of work involved. If the external pressure is
zero (adiabatic expansion into a vacuum), no work is done and there is no change
in the internal energy for all gases. If the expansion is opposed by an external
pressure, work is done on the surroundings and the temperature drops as internal
energy is converted to work. Integrating yields



U2

dU  ⫺

U1



V2

Pext dV

V1

⌬U  U2 ⫺ U1  w

(2 .78)

where w is the work done on the gas.
For an ideal gas, the internal energy is a function only of temperature and
so dU  CV dT (equation 2.44). Thus, when an ideal gas expands adiabatically
against an external pressure, the temperature drop is simply related to the change
in internal energy. If CV is independent of temperature for an ideal gas in the
temperature range of interest, then



U2

dU  CV

U1



T2

dT

T1

⌬U  U2 ⫺ U1  CV (T2 ⫺ T1 )

(2 .79)

Since q  0, then ⌬U  w and
w 



T2

CV dT  CV (T2 ⫺ T1 )

(2 .80)

T1

where the second form applies when CV is independent of temperature. This relation applies to the adiabatic expansion of an ideal gas with CV independent of
temperature whether the process is reversible or irreversible. If the gas expands,
the final temperature T2 will be lower than the initial temperature T1 , and the
work done on the gas is negative. If the gas is compressed adiabatically, it will
heat up.

2 .10 Adiabatic Processes with Gases

When an adiabatic expansion is carried out reversibly, the equilibrium pressure is substituted for the external pressure, and so for an ideal gas
C V dT  ⫺ P dV  ⫺

RT
dV
V

dT
dV
 ⫺R
T
V
If the heat capacity is independent of temperature, then
CV



CV

T2

dT
 ⫺R
T

T1

CV ln



V2

V1

(2 .81)

dV
V

T2
V1
 R ln
T1
V2

(2 .82)

This equation is a good approximation only if the temperature range is small
enough so that CV does not change very much.
Since CP ⫺ CV  R , equation 2 .82 may be written
␥ ⫺1

冢 冣

T2
V1

T1
V2

(2 .83)

where ␥  CP /CV. By use of the ideal gas law we can obtain the following alternative forms of this equation:
(␥ ⫺1)/␥

冢 冣

(2 .84)

P1 V ␥1  P2 V ␥2

(2 .85)

T2
P2

T1
P1

Thus, when a gas expands adiabatically to a larger volume and a lower pressure,
the volume is smaller than it would be after an isothermal expansion to the same
final pressure. Plots of pressure versus volume for adiabatic and isothermal expansions are shown in Fig. 2 .15.
Example 2.7 Reversible adiabatic expansion of a monatomic ideal gas
Figure 2 .15 shows that when one mole of an ideal monatomic gas is allowed to expand adiabatically and reversibly from 22 .7 L mol⫺1 at 1 bar and 0 ⬚C (at point A on the graph) to a
volume of 45.4 L mol⫺1 (at point C ), the pressure drops to 0.315 bar. Confirm this pressure
and calculate the temperature at C. How much work is done in the adiabatic expansion?

␥


P2  P1

冢 冣

T2  T1

冢 冣

w 

V1
V2

T2

冮T

1

V1
V2

 (1 bar)



5
2R
3
2R

22 .7 L mol⫺1
45.4 L mol⫺1

␥ ⫺1

 (273.15 K)





5
3

5/3



 0.315 bar

22 .7 L mol⫺1
45.4 L mol⫺1

2/3



 172 .07 K or ⫺101.08 ⬚C

CV dT  32 R (172 .07 K ⫺ 273.15 K)  ⫺1261 J mol⫺1

55

Chapter 2

First Law of Thermodynamics
1.0

A

0.8
Isothermal
0.6
P/bar

56

B

Adiabatic
0.4

C
0.2

0.0
20

25

30

35

40

45

50

V/(L mol–1)

Figure 2 .15 Isothermal and reversible adiabatic expansions of one mole of an ideal
monatomic gas.

2.11

THERMOCHEMISTRY

The quantity of heat evolved or absorbed in a chemical reaction or a phase change
can be determined by measuring the temperature change in an adiabatic process.
Since very small temperature changes can be measured, this provides a sensitive method for studying the thermodynamics of chemical reactions and phase
changes. If the temperature rises when a reaction occurs in an isolated system,
then in order to restore that system to its initial temperature, heat must be allowed to flow to the surroundings. Such a reaction is said to be exothermic, and
the heat q is negative. If the temperature falls when a reaction occurs in an isolated system, heat must flow from the surroundings to the system to restore the
system to its initial temperature. Such a reaction is said to be endothermic, and
the heat q is positive.
Since the enthalpy is an extensive property that is a function of the state of
the system, its differential (at constant T and P ) can be written in terms of the
partial molar enthalpies of the species in the system (see equation 1.37 in Section
1.10):
Ns

dH  冱 Hi d ni

(2 .86)

i 1

where Ns is the number of species and Hi is the molar enthalpy of species i . When
the temperature and the pressure are constant, equation 2.57 yields
N

dH  dqP  冱 Hi d ni
i 1

(2 .87)

2 .11 Thermochemistry

Now let us apply this equation to a system in which a single chemical reaction
occurs.
To connect heat absorbed or evolved with a chemical reaction, it is of course
necessary to know what the chemical change is and to have a measure of its
amount. To discuss the thermodynamics of chemical reactions, we will find it convenient to represent chemical reactions in general by
Ns

0  冱 ␯i Bi

(2 .88)

i 1

where the ␯i are the stoichiometric numbers and the Bi are the molecular formulas for the Ns species involved in the reaction. The stoichiometric numbers, which
are dimensionless, are positive for products and negative for reactants. Thus, according to this way of writing a reaction equation, the reaction H2  12 O2  H2 O
would be written
0  ⫺1H2 ⫺ 12 O2  1H2 O

(2 .89)

The reason for using this convention is that it makes it easier to write thermodynamic equations for chemical reactions.
The amount of reaction that has occurred up to some time is expressed by the
extent of reaction ␰, which is defined by
ni  ni 0  ␯i ␰

(2 .90)

Here ni 0 is the amount of substance i present initially, and ni is the amount at
some later time. Since n is expressed in moles and ␯i is dimensionless, we see that
the extent of reaction is expressed in moles. The concept of extent of reaction is
important because it provides a connection between the amount of reaction and
a particular balanced chemical equation. We will also use the extent of reaction
later in calculating equilibrium compositions.
Equation 2.90 shows that dni  ␯i d␰, and when we substitute this relation
into equation 2.87, we obtain
N

dH  dqP  冱 ␯i Hi d␰

(2 .91)

i 1

Dividing by d␰ gives
⌬r H 

⭸H

冢 ⭸␰ 冣

T,P



dqP
 冱 ␯i Hi
d␰

(2 .92)

The quantity ⌬r H is the reaction enthalpy.
The reaction enthalpy is the derivative of the enthalpy of the system with
respect to the extent of reaction. This is perhaps easiest to visualize for a very
large system for which we can write ⌬H /⌬␰  ⌬r H. If one mole of reaction
occurs, ⌬␰  1 mol and ⌬H  (1 mol)⌬r H. To know what a mole of reaction is,
we must have a balanced chemical equation, since the way an equation is written
is arbitrary with respect to direction and with respect to multiplying or dividing
by an integer. Thus, the enthalpy of reaction for 2H2  O2  2H2 O is twice that
of the reaction H2  12 O2  H2 O. To distinguish between the extensive property

57

58

Chapter 2

First Law of Thermodynamics

⌬H and the change in enthalpy for a specified chemical reaction, we will write
⌬r H for a reaction. It is evident from equation 2.92 that the reaction enthalpy has
the SI units J mol⫺1 . Here the mol⫺1 refers to a mole of reaction for the reaction
as written. An overbar is not used on ⌬r H because the subscript r indicates that
the mol⫺1 unit is involved.
As a further specification of the states of the reactants and products, we will
usually consider reactions in which the reactants in their standard states are converted to the products in their standard states. When substances are in their standard states, thermodynamic quantities are labeled with superscript zeros (actually
degree signs). Thus, if reactants and products are in their standard states, equation
2 .95 becomes
N

⌬r H ⬚  冱 ␯i Hi ⬚

(2 .93)

i 1

T

HERMODYNAMIC STANDARD STATES

The standard states that are used in chemical thermodynamics are defined as follows:
1.
2.
3.
4.

The standard state of a pure gaseous substance, denoted by g, at a given temperature
is the (hypothetical) ideal gas at 1 bar pressure.
The standard state of a pure liquid substance, denoted by l, at a given temperature is
the pure liquid at 1 bar pressure.
The standard state of a pure crystalline substance at a given temperature is the pure
crystalline substance, denoted by s, at 1 bar pressure.
The standard state of a substance in solution is the hypothetical 1 of the substance in
ideal solution of standard state molality (1 mol kg⫺1 ) at 1 bar pressure, at each temperature. To indicate the standard state of an electrolyte, the NBS Tables of Chemical
Thermodynamic Properties (1982) use two symbols. The thermodynamic properties of
completely dissociated electrolytes in water are designated by ai. The thermodynamic
properties of undissociated molecules in water are designated by ao. The thermodynamic properties of ions in water are also designated ao to indicate that no further
ionization occurs.

Lavoisier and Laplace recognized in 1780 that the heat absorbed in decomposing a compound must be equal to the heat evolved in its formation under
the same conditions. Thus, if the reverse of a chemical reaction is written, the
sign of ⌬H is changed. Hess pointed out in 1840 that the overall heat of a
chemical reaction at constant pressure is the same, regardless of the intermediate steps involved. These principles are both corollaries of the first law of
thermodynamics and are a consequence of the fact that enthalpy is a state function. This makes it possible to calculate the enthalpy changes for reactions that
cannot be studied directly. For example, it is not practical to measure the heat
evolved when carbon burns to carbon monoxide in a limited amount of oxygen, because the product will be an uncertain mixture of carbon monoxide and
carbon dioxide. However, carbon may be burned completely to carbon dioxide

2 .11 Thermochemistry

in an excess of oxygen and the heat of reaction measured. Thus, for graphite
at 25 ⬚C,
⌬r H ⬚  ⫺393.509 kJ mol⫺1

C(graphite)  O2 (g)  CO2 (g)

The heat evolved when carbon monoxide burns to carbon dioxide can be readily
measured also:
⌬r H ⬚  ⫺282 .984 kJ mol⫺1

CO(g)  12 O2 (g)  CO2 (g)

Writing these equations in such a way as to obtain the desired reaction, adding,
and canceling, we have
⌬r H ⬚  ⫺393.509 kJ mol⫺1

C(graphite)  O2 (g)  CO2 (g)
CO2 (g)  CO(g)  21 O2 (g)
1
C(graphite)  O2 (g)  CO (g )
2

⌬r H ⬚ 

282 .984 kJ mol⫺1

⌬r H ⬚  ⫺110.525 kJ mol⫺1

In this way an accurate value can be obtained for the heat evolved when graphite
burns to CO.
These data may be represented in the form of an enthalpy level diagram, as
shown in Fig. 2.16. In addition, this diagram shows the enthalpy changes that are
involved in vaporizing graphite to atoms and dissociating oxygen into atoms at
25 ⬚C:
⌬r H ⬚  716.682 kJ mol⫺1

C(graphite)  C(g)

⌬r H ⬚  498.340 kJ mol⫺1

O2 (g)  2O(g)
C( g) + 2O( g)

716.68
kJ mol–1

1215.02
kJ mol–1

C(s) + 2O( g)

498.34
kJ mol–1
C( s) + O2 ( g)
–110.53 kJ mol–1
CO(g) + 1 2 O2 ( g)

–282.98
kJ mol–1

–393.51
kJ mol–1

CO2 ( g)

Figure 2 .16 Enthalpy level diagram for the system C(s)  O2 (g). The differences in level
are standard enthalpy changes at 25 ⬚C and 1 bar.

59

60

Chapter 2

First Law of Thermodynamics

2.12

ENTHALPY OF FORMATION

Since absolute enthalpies are not known, enthalpies relative to a defined reference
state are used instead. The defined reference state for each substance is made up
of the stoichiometric amounts of the elements in the substance, each in its standard
state and at the temperature under consideration. These “relative” enthalpies of
substances are called enthalpies of formation and are represented by ⌬f H ⬚ . Since
the same reference state is used for reactants and products, the same ⌬r H ⬚ is
obtained as would be obtained with equation 2.93 and absolute enthalpies. Thus,
standard enthalpy changes for reactions may be calculated using enthalpies of
formation as follows:
N

⌬r H ⬚  冱 ␯i ⌬f Hi⬚

(2 .94)

i 1

Note that the enthalpy of formation does not have an overbar because the subscript f, for formation, indicates that the mol⫺1 unit is involved.
The enthalpy of formation of a substance at a given temperature is the change
in enthalpy for the reaction in which one mole of the substance in its standard state
at the given temperature is formed from its elements, each in its standard state at
that temperature. If there is more than one solid form of an element, one must
be selected as a reference. For thermodynamic tables at 25 ⬚C the reference form
is usually the most stable form of the element at 25 ⬚C, 1 bar pressure. Thus, the
reference form of hydrogen is H2 (g) instead of H(g), the reference form of carbon
is graphite, and the reference form of sulfur is rhombic sulfur. For thermodynamic
tables that cover a wide range of temperatures, different reference states may be
used in various temperature ranges. In any case, the enthalpy of formation of an
element in its standard state is zero at every temperature.
From reactions given previously we can see how the following enthalpies of
formation at 25 ⬚C are obtained:
⌬f H ⬚ /kJ mol⫺1
CO2 (g)
CO(g)
C(g)
O(g)

⫺393.509
⫺110.525
716.682
249.170

These enthalpies of formation should be identified in Fig. 2.16. Enthalpies of formation ⌬f H ⬚ at 25 ⬚C for some 200 substances are given in Table C.2. These values
are from the NBS Tables of Chemical Thermodynamic Properties (1982). Table
C.3 gives enthalpies of formation from 0 to 3000 K for a smaller group of substances from the JANAF Thermochemical Tables (1985).
The values of enthalpies of formation given in these tables come from four
sources: (1) calorimetrically measured enthalpies of reaction, fusion, vaporization, sublimation, transition, solution, and dilution; (2) temperature variation of
equilibrium constants (see Section 5.5); (3) spectroscopically determined dissociation energies (see Section 14.3); (4) calculation from Gibbs energies and entropies
(see the third-law method in Section 3.8).

2 .12 Enthalpy of Formation

You may calculate ⌬r H ⬚ for any reaction for which the reactants and products
are listed in tables, but the reaction will not necessarily occur spontaneously in
the direction written. The question as to whether or not the reaction can occur is
answered by calculations based on the second law of thermodynamics.

Example 2.8 Calculating the standard enthalpy of reaction at constant
temperature
What are the standard enthalpy changes at 298.15 K and 2000 K for the following reaction?
CO2 (g)  C(graphite)  2CO(g)
Using Table C.3, at 298.15 K,
⌬r H ⬚  2⌬f H ⬚ (CO) ⫺ ⌬f H ⬚ (CO2 )
 2(⫺110.527 kJ mol⫺1 ) ⫺ (⫺393.522 kJ mol⫺1 )
 172 .468 kJ mol⫺1
At 2000 K,
⌬r H ⬚  2(⫺118.896 kJ mol⫺1 ) ⫺ (⫺396.784 kJ mol⫺1 )
 158.992 kJ mol⫺1

So far we have talked mainly about the reaction enthalpy at 298.15 K. To
calculate the standard enthalpy change at some other temperature, given the value
at 298.15 K, it is necessary to have heat capacity data on the reactants and the
products. Since enthalpy is a state function, we can use the paths indicated below
to calculate the standard enthalpy change at any desired temperature.
reactants
298

冮T


Q



298

T

products

q冮


T
298 CP,prod dT

CP,react dT

reactants
⌬r HT⬚ 

⌬r HT⬚

⌬r H 298


products

CP,react dT  ⌬r H298
⬚ 

⌬r HT⬚  ⌬r H298
⬚ 



T

 ⌬r H298
⬚ 



T

298

298

(2 .95)



T

CP,prod dT

(2 .96)

298

(CP,⬚prod ⫺ CP,⬚react ) dT
⌬r CP⬚ dT

(2 .97)

where
⌬r CP⬚  冱 ␯i CP,i⬚
i

(2 .98)

61

62

Chapter 2

First Law of Thermodynamics

The reaction heat capacity ⌬r CP⬚ does not have an overbar because the subscript
r indicates that the unit mol⫺1 is involved.
If data are available on the heat capacities of reactants and products in the
form of power series in T (see Table 2.2), ⌬r HT⬚ may be expressed as a function
of T as follows:
⌬r CP⬚  ⌬r ␣  (⌬r ␤ )T  (⌬r ␥ )T 2

(2 .99)

where ⌬r ␣  冱 ␯i ␣i , and so on. Substituting in equation 2.97 gives
⌬r HT⬚  ⌬r H298
⬚ 



T

[⌬r ␣  (⌬r ␤ )T  (⌬r ␥ )T 2 ] dT

298

 ⌬r H298
⬚  ⌬r ␣ (T ⫺ 298) 

⌬r ␤ 2
⌬r ␥ 3
(T ⫺ 2982 ) 
(T ⫺ 2983 )
2
3

 ⌬r H0⬚  (⌬r ␣ )T  (⌬r ␤ /2)T 2  (⌬r ␥ /3)T 3

(2 .100)

In the last form of this equation the constant terms have been added together
to obtain a hypothetical enthalpy of reaction at 0 K, hypothetical because the
power-series representations of CP are for a limited temperature range. Within
this temperature range equation 2.100 does represent the standard enthalpy of
reaction as a function of temperature.
The JANAF tables give standard enthalpies of formation at a series of temperatures, and so these values may be used directly to calculate enthalpies of reaction.
Some values from the JANAF tables are given in Table C.3.
Some thermodynamic tables give values of HT⬚ ⫺ H298
⬚ to assist in the calculation of ⌬r HT⬚ for a chemical reaction or phase transition:
HT⬚ ⫺ H298
⬚ 



T

298 K

CP⬚ dT

(2 .101)

Depending on the table, ⌬r H ⬚ for phase transitions in the intervening temperature range may be added to the right-hand side of the equation.
Example 2 .9 Calculation of the bond energy of molecular hydrogen from the
enthalpy of formation of hydrogen atoms at 298 K
What is the value of ⌬r H ⬚ at 0 K for the following reaction?
H2 (g)  2H(g)
The calculation using ⌬r H ⬚ (298 K) illustrates the use of H0⬚ ⫺ H298
⬚ from Table C.3:
⌬r H ⬚ (298 K)  2(217.999 kJ mol⫺1 )  435.998 kJ mol⫺1
H2 (g)

298 K



Q

q



H298
⬚ ⫺H0⬚ 8.467 kJ mol⫺1

H2 (g)

2H(g)

⌬H298
⬚  435.998 kJ mol⫺1

H0⬚ ⫺H298
⬚ ⫺(2)(6.197 kJ mol⫺1 )

0K

2H(g)

⌬r H ⬚ (0 K)  (8.467  435.998 ⫺ 12 .394) kJ mol⫺1
 432 .071 kJ mol⫺1

2 .13 Calorimetry

63

Alternatively, this value may be calculated from the enthalpy of formation of H(g) at 0 K
in Table C.3:
⌬r H ⬚ (0 K)  2(216.035 kJ mol⫺1 )  432 .070 kJ mol⫺1
This value is often referred to as the H H bond energy. In Chapter 11 we will see how
this value can be calculated theoretically; there this energy is referred to as the dissociation
energy D0 .

Comment:
The concept of a thermodynamic cycle will be used in many ways. The important
idea is that it may be easier to measure changes in a thermodynamic property,
such as enthalpy, along three sides of a cycle than along the fourth. For example,
it is easier to make calorimetric measurements of enthalpy changes at room
temperature and use heat capacity measurements to calculate the enthalpy change
at a higher temperature than it would be to make the calorimetric
measurement at the higher temperature.

2.13

CALORIMETRY

Heats of reaction are determined using adiabatic calorimeters; that is, the reaction
or solution process occurs in a container, which is immersed in a weighed quantity of water and is surrounded by insulation or an adiabatic shield that is kept at
the same temperature as the calorimeter so that no heat is gained or lost. A simple adiabatic calorimeter operated at constant-pressure is illustrated in Fig. 2.17.
Thus, ⌬HA for this adiabatic process is zero. When a certain amount of reactants
R are converted completely to products P in a constant-pressure calorimeter, the
changes in state involved may be represented as follows:

R(T2) + Cal(T2)

∆ H(T2 )
=

∆HR

R(T1) + Cal(T1)

∆H

Stirrer

Thermometer

Insulated cover

P(T2) + Cal(T2)

0

A

∆ H(T1)

∆ HP

(2 .102)

P(T1) + Cal(T1)
Dewar flask
A

The adiabatic container, thermometer, stirrer, and weighed quantity of water are
represented by Cal. Since the enthalpy is a state function, the enthalpy change for
the actual process may be written two ways:
⌬HA  ⌬H (T1 )  ⌬HP  0

(2 .103)

⌬HA  ⌬HR  ⌬H (T2 )  0

(2 .104)

B

Figure 2 .17 Adiabatic calorimeter
operated at constant pressure. A
reaction between solutions A and B
is initiated by rotating the reaction
vessel around the axis indicated.

64

Chapter 2

First Law of Thermodynamics

Since the heat capacities of the reactants, products, and calorimeter may be assumed constant over the range T1 to T2 , these equations become
⌬H (T1 )  ⫺⌬HP  ⫺[CP (P)  CP (Cal)](T2 ⫺ T1 )

(2 .105)

⌬H (T2 )  ⫺⌬HR  ⫺[CP (R)  CP (Cal)](T2 ⫺ T1 )

(2 .106)

where these CP ’s are extensive properties. Thus, the results of the calorimetric experiments can be interpreted to obtain ⌬H for the conversion of a certain amount
of R to P either at T1 or T2 . The heat capacity term in the first equation can be determined by using a calibrated electric heater coil and measuring I 2 Rt with only
products present, and the heat capacity term in the second equation can be determined with only reactants present. In this expression I is a constant current that
flows through resistor R for time t .
Once ⌬H has been determined in a calorimetric experiment, ⌬r H for a balanced chemical reaction can be calculated using ⌬r H  ⌬H /⌬␰.
When a reaction is carried out in a sealed bomb (see Fig. 2.18), no P V work
is done, and the first law may be written ⌬U  qV. Thus, the change in internal
energy for the reaction is obtained. When the reaction is carried out at constant
pressure, the first law may be written ⌬H  qP. Chemists are usually more interested in ⌬H than ⌬U because chemical reactions are generally carried out at
constant pressure. If ⌬U is determined in a bomb calorimeter, the value of ⌬H
may be calculated using equation 2.55:
⌬r H  ⌬r U  RT

冱 ␯g

(2 .107)

where 冱 ␯g is the sum of stoichiometric numbers of gaseous products and gaseous
reactants. Remember that stoichiometric numbers are positive for products and

Thermometer

Stirrer
Ignition
wires

Adiabatic
wall

O2

Bomb in which
sample is burned

Figure 2 .18 Adiabatic bomb calorimeter for carrying out combustions at constant
volume.

2 .13 Calorimetry

negative for reactants. In writing equation 2 .107 in this way we are ignoring the
volume change due to solid and liquid reactants, because this is negligible in comparison with the change in gas volume. We are also assuming that the gases are
ideal.

Example 2.10 Calculating the enthalpy of reaction
from the heat of combustion
The combustion of ethanol in a constant-volume calorimeter produces 1364.34 kJ mol⫺1
at 25 ⬚C. What is the value of ⌬r H ⬚ for the following combustion reaction?
C2 H5 OH(l)  3O2 (g)  2CO2 (g)  3H2 O(l)
⌬r H ⬚  ⌬r U ⬚  RT

冱 ␯g

 ⫺1364.34 kJ mol⫺1  (8.3145 ⫻ 10⫺3 kJ K⫺1 mol⫺1 )(298.15 K)(⫺1)
 ⫺1366.82 kJ mol⫺1
This is the quantity of heat that would be evolved at 25 ⬚C and a constant pressure of 1 bar.

Example 2.11 Calculating the molar internal energy of combustion
In an adiabatic bomb calorimeter, the combustion of 0.5173 g of ethanol causes the temperature to rise from 25.0 to 29.289 ⬚C. The heat capacity of the bomb, the reactants, and
the other contents of the calorimeter is 3576 J K⫺1 . What is the molar internal energy of
combustion of ethanol at 25.0 ⬚C?
The change in the state can be written
C2 H5 OH  3O2

2CO2  3H2 O

冦  other contents冧 [T  25 ⬚C, V ]  冦  other contents冧 [T  29.289 ⬚C, V ]
for which q  0 (adiabatic) and w  0 (constant volume V ) so that ⌬U  0. This change
in state can be written as the sum of
C2 H5 OH  3O2

2CO2  3H2 O

2CO2  3H2 O

2CO2  3H2 O

冦  other contents冧 [T  25 ⬚C, V ]  冦  other contents冧 [T  25 ⬚C, V ]
冦  other contents冧 [T  25 ⬚C, V ]  冦  other contents冧 [T  29.289 ⬚C, V ]

⌬U1
⌬U2

where ⌬U1  ⌬U2  ⌬U  0, so that
⌬U1  ⫺⌬U2
or
⌬r U  ⫺

(3.576 kJ K⫺1 )(4.289 K)(46.0 g mol⫺1 )
0.5173 g

 ⫺1364 kJ mol⫺1

When a solute is dissolved in a solvent, heat may be absorbed or evolved; in
general, the heat of solution depends on the concentration of the final solution.
The integral heat of solution is the enthalpy change for the solution of 1 mol of

65

66

Chapter 2

First Law of Thermodynamics

solute in n mol of solvent. The solution process may be represented by a chemical
equation such as
HCl(g)  5H2 O(l)  HCl in 5H2 O

⌬sol H ⬚ (298 K)  ⫺63.467 kJ mol⫺1
(2 .108)

at 1 bar pressure, where “HCl in 5H2 O” represents a solution of 1 mol of HCl in
5 mol of H2 O. As the amount of water is increased, the integral heats of solution
approach asymptotic values.
The solution of liquid acetic acid in water to form an aqueous solution in
which undissociated acetic acid is in its standard state is represented by
CH3 CO2 H(l)  CH3 CO2 H(ao)

⌬sol H ⬚ (298 K)  ⫺1.3 kJ mol⫺1

(2 .109)

where the ao indicates that the ions do not dissociate further.
When a solute is dissolved in a solvent that is chemically quite similar to it
and there are no complications of ionization or solvation, the heat of solution
may be nearly equal to the heat of fusion of the solute. It might be expected
that heat would always be absorbed in overcoming the attraction between the
molecules or ions of the solid solute when the solute is dissolved. Another process
that commonly occurs, however, is a strong interaction with the solvent, referred
to as solvation, which evolves heat. In the case of water the solvation is called
hydration.
The importance of this attraction of the solvent for the solute in the process of
solution is illustrated by the dissolving of sodium chloride in water. In the crystal
lattice of sodium chloride, positive sodium ions and negative chloride ions attract
each other strongly. The energy required to separate them is so great that nonpolar
solvents like benzene and carbon tetrachloride do not dissolve sodium chloride;
but a solvent like water, which has a high dielectric constant and a large dipole
moment, has a strong attraction for the sodium and chloride ions and solvates
them with a large decrease in the energy of the system. When the energy required
to separate the ions from the crystal is about the same as the solvation energy, as
it is for dissolving NaCl in water, ⌬H for the net process is close to zero. When
NaCl is dissolved in water at 25 ⬚C, there is only a small cooling effect; q is positive.
When Na2 SO4 is dissolved in water at 25 ⬚C, there is an evolution of heat because
the energy of hydration of the ions is greater than the energy required to separate
the ions from the crystal.

Example 2 .12 Calculating the enthalpy of neutralization
in aqueous solution
Calculate ⌬r H ⬚ (298 K) for the following reaction using Table C.2 .
HCl in 100H2 O  NaOH in 100H2 O  NaCl in 200H2 O  H2 O(l)
⌬r H ⬚ (298 K)  ⫺406.923 ⫺ 285.830  165.925  469.646
 ⫺57.182 kJ mol⫺1

For dilute solutions it is found that the heat of reaction of strong bases, such
as NaOH and KOH, with strong acids, such as HCl and HNO3 , is independent
of the nature of the acid or base. This constancy of the heat of neutralization is a
result of the complete ionization of strong acids and bases and the salts formed

2 .13 Calorimetry

by neutralization. Thus, when a dilute solution of a strong acid is added to a dilute
solution of a strong base, the only chemical reaction is
OH⫺ (ao)  H(ao)  H2 O(l)

⌬r H ⬚ (298 K)  ⫺55.835 kJ mol⫺1

When a dilute solution of a weak acid or base is neutralized, the heat of neutralization is somewhat less because of the absorption of heat in the dissociation of
the weak acid or base.
Since for strong electrolytes in dilute solution the thermal properties of the
ions are essentially independent of the accompanying ions, it is convenient to use
enthalpies of formation of individual ions. The sum of the enthalpies of formation
of H and OH⫺ ions may be calculated from
H2 O(l)  H(ao)  OH⫺ (ao)

⌬r H ⬚ 

55.835 kJ mol⫺1

H2 (g)  12 O2 (g)  H2 O(l)

⌬r H ⬚  ⫺285.830 kJ mol⫺1

H2 (g)  12 O2 (g)  H(ao)  OH⫺ (ao)

⌬r H ⬚  ⫺229.995 kJ mol⫺1

The separate enthalpies of formation of H and OH⫺ cannot be calculated, and
so, to construct a table of enthalpies of formation of individual ions, it is necessary
to adopt an arbitrary convention. Enthalpies of formation of aqueous ions in Table
C.2 are based on the convention that ⌬f H ⬚  0 for H(ao). In other words, by
convention,
1
2 H2 (g)

 H(ao)  e⫺

⌬f H ⬚  0

where e⫺ is the electron, which is assigned ⌬f H ⬚ (e⫺ )  0. This electron is not dissolved in water, but is a formal electron required to balance the equation. Therefore, the enthalpy of formation of OH⫺ is given by

1
1
2 H2 (g)  2 O2 (g)  e

 OH⫺ (ao)

⌬f H ⬚  ⫺229.995 kJ mol⫺1

On the basis of these values for the enthalpies of formation of H and OH⫺ , the
enthalpies of formation of other ions of strong electrolytes may be calculated.
From the enthalpy of formation of HCl(ai) it is possible to calculate the enthalpy of formation of Cl⫺ (ao).
1
1
2 H2 (g)  2 Cl2 (g)

1
2 Cl2 (g)  e


1.

2.

 H(ao)  Cl⫺ (ao)

⌬r H ⬚  ⫺167.159 kJ mol⫺1

 Cl⫺ (ao)

⌬f H ⬚  ⫺167.159 kJ mol⫺1

Eleven Key Ideas in Chapter 2
In thermodynamics work w is a signed quantity, and it is positive when work
is done on the system of interest and is negative when the system does work
on the surroundings. The work in a process depends on the path, even when
the process is reversible. The differential of the work is represented by dw
as a reminder that work is not an exact differential.
The change in the internal energy U of a system in an adiabatic process is
equal to the work done on the system. This provides a way to determine the
difference ⌬U in internal energy of two states of a system.

67

68

Chapter 2

First Law of Thermodynamics

3.

4.

5.
6.

7.

8.

9.

10.

11.

A system can also undergo a specified change in internal energy by absorbing heat or evolving heat q . Since the change in internal energy can be expressed in joules, the quantity of heat can also be expressed in joules.
According to the first law of thermodynamics, (1) there exists a property U
that is a function of the state variables of a system and (2) the change in internal energy for a closed system can be calculated from ⌬U  q w . However,
the first law by itself does not provide any information as to whether a given
process will proceed in the forward direction or the reverse direction.
The mathematical test for whether a variable is exact or inexact is whether
the mixed partial derivatives are equal or unequal.
When a gas is allowed to expand, the maximum work is obtained when the
process is carried out reversibly, that is, when the process is carried out with
an infinite number of infinitesimal steps.
In addition to pressure–volume work, there is surface work, elongation
work, electric charge displacement, and other kinds of work such as work
of electric and magnetic polarization, which are not discussed here.
The heat capacity at constant volume is defined as the partial derivative of
the internal energy with respect to temperature when the volume is held
constant. The enthalpy is defined by H  U  PV , and the heat capacity
at constant pressure is defined as the partial derivative of the enthalpy with
respect to temperature when the pressure is held constant.
A cooling effect is obtained in the adiabatic expansion of an ideal gas,
and the maximum cooling is obtained when the expansion is carried out
reversibly.
The change in standard enthalpy in a chemical reaction is equal to the
summation of the standard enthalpies of formation of the reacting species,
each multiplied by its stoichiometric number in a specified chemical equation. If the standard enthalpy of reaction is known at one temperature, its
value at any other temperature can be calculated if the molar heat capacities of the species involved are known throughout the temperature range
involved.
Calorimeters are useful for determining standard enthalpies of formation
of species, but we will see later that these values can also be determined in
other ways.

REFERENCES
M. Bailyn, A Survey of Thermodynamics. New York: American Institute of Physics, 1994.
K. E. Bett, J. S. Rowlinson, and G. Saville, Thermodynamics for Chemical Engineers. Cambridge, MA: MIT Press, 1975.
H. B. Callen, Thermodynamics and an Introduction to Thermostatistics, 2nd ed. Hoboken,
NJ: Wiley, 1985.
K. S. Pitzer, Thermodynamics, 3rd ed. New York: McGraw-Hill, 1995.
J. M. Smith, H. C. Van Ness, and M. M. Abbott, Introduction to Chemical Engineering
Thermodynamics, 5th ed. New York: McGraw-Hill, 1996.
D. R. Stull, E. F. Westrum, and G. C. Sinke, The Chemical Thermodynamics of Organic
Compounds. Hoboken, NJ: Wiley-Interscience, 1969.
J. W. Tester and M. Modell, Thermodynamics and Its Applications. Upper Saddle River,
NJ: Prentice-Hall, 1997.

Problems

69

PROBLEMS
Problems marked with an icon may be more conveniently
solved on a personal computer with a mathematical program.
2 .1 How high can a person (assume a weight of 70 kg) climb
on one ounce of chocolate, if the heat of combustion (628 kJ) can
be converted completely into work of vertical displacement?
2 .2 A mole of sodium metal is added to water. How much
work is done on the atmosphere by the subsequent reaction if
the temperature is 25 ⬚C?
2 .3 You want to heat 1 kg of water 10 ⬚C, and you have the
following four methods under consideration. The heat capacity
of water is 4.184 J K⫺1 g⫺1 .
(a) You can heat it with a mechanical eggbeater that is powered by a 1-kg mass on a rope over a pulley. How far does
the mass have to descend in the earth’s gravitational field
to supply enough work?
(b) You can send 1 A through a 100-⍀ resistor. How long will
it take?
(c) You can send the water through a solar collector that has
an area of 1 m2 . How long will it take if the sun’s intensity
on the collector is 4 J cm⫺2 min⫺1 ?
(d) You can make a charcoal fire. The heat of combustion of
graphite is ⫺393 kJ mol⫺1 . That is, 12 g of graphite will produce 393 kJ of heat when it is burned to CO2 (g) at constant
pressure. How much charcoal will have to burn?
2 .4

Show that the differential df is inexact.
df  dx ⫺

x
dy
y

Thus, the integral 冮 df depends on the path. However, we can
define a new function g by
dg 

1
df
y

which has the property that dg is exact. Show that dg is exact, so
that

冯 dg  0
2.5

Show that the function f (x, y ) defined by
df (x, y )  (x  2y ) dx ⫺ x dy

is inexact. Test to see whether the integrating factor 1/x 3 makes
it an exact differential.
2.6 Show that the function defined by

2 .8 (a ) The surface tension of water at 25C is 0.072 N m⫺1 .
How much work is required to form a surface 100 m by 100 m?
(b ) The force on a wire is due to a 75-kg person in the earth’s
gravitational field. If the wire stretches 1 m, how much work is
done on the wire? (c ) A gas expands 1 L against a constant external pressure of 1 bar. How much work is done on the gas?
2 .9 Over narrow ranges of temperature and pressure, the
differential expression for the volume of a fluid as a function
of temperature and pressure can be integrated to obtain
V  Ke␣T e⫺␬P
(␣ and ␬ are defined in Section 4.10). Show that V is a state
function.
2 .10 One mole of nitrogen at 25 ⬚C and 1 bar is expanded reversibly and isothermally to a pressure of 0.132 bar. (a) What
is the value of w ? (b) What is the value of w if the nitrogen is
expanded against a constant pressure of 0.132 bar?
2 .11 (a) Derive the equation for the work of reversible isothermal expansion of a van der Waals gas from V1 to V2 . (b)
A mole of CH4 expands reversibly from 1 to 50 L at 25 ⬚C.
Calculate the work in joules assuming (1) the gas is ideal and
(2) the gas obeys the van der Waals equation. For CH4 (g), a 
2 .283 L2 bar mol⫺2 and b  0.042 78 L mol⫺1 .
2 .12 Liquid water is vaporized at 100 ⬚C and 1.013 bar. The
heat of vaporization is 40.69 kJ mol⫺1 . What are the values of
(a ) wrev per mole, (b ) q per mole, (c ) ⌬U, and (d) ⌬H ?
2 .13 An ideal gas expands reversibly and isothermally from
10 bar to 1 bar at 298.15 K. What are the values of (a ) w per
mole, (b ) q per mole, (c ) ⌬U, and (d ) ⌬H ? (e ) The ideal gas
expands isothermally against a constant pressure of 1 bar. How
much work is done on the gas?
2 .14 Calculate H ⬚ (2000 K) ⫺ H ⬚ (0 K) for H(g).
2 .15 The heat capacities of a gas may be represented by
CP  ␣  ␤T  ␥T 2  ␦T 3
For N2 , ␣  28.883 J K⫺1 mol⫺1 , ␤  ⫺1.57 ⫻ 10⫺3 J K⫺2
mol⫺1 , ␥  0.808 ⫻ 10⫺5 J K⫺3 mol⫺1 , and ␦  ⫺2.871 ⫻ 10⫺9
J K⫺4 mol⫺1 . How much heat is required to heat a mole of N2
from 300 to 1000 K?
2 .16 (a) In a reversible adiabatic expansion of an ideal gas
with ␥  CP /CV independent of temperature, the
pressure and volume are related by
PV ␥  constant

df (x, y )  (y 2 ⫺ xy ) dx ⫺ x 2 dy
is inexact. Test the integrating factor 1/xy 2 to see whether it produces an exact differential.
2 .7 What are the partial derivatives (⭸z /⭸x )y and (⭸z /⭸y )x
of the following functions? (a ) z  xy , (b ) z  x /y , (c ) z 
ln(xy ), (d ) z  ln(x /y ), and (e ) z  exp(xy ).

Show that the work of adiabatic expansion from
P1 , V1 to P2 , V2 is
w  (P2 V2 ⫺ P1 V1 )/(␥ ⫺ 1)
(b)

Check this equation to be sure it gives the same
amount of work as Example 2 .7.

70

Chapter 2

First Law of Thermodynamics

2 .17 Calculate the temperature increase and final pressure of
helium if a mole is compressed adiabatically and reversibly from
44.8 L at 0 ⬚C to 22 .4 L.
2 .18 A mole of argon is allowed to expand adiabatically and
reversibly from a pressure of 10 bar and 298.15 K to 1 bar. What
is the final temperature, and how much work is done on the
argon?
2 .19 A tank contains 20 L of compressed nitrogen at 10 bar and
25 ⬚C. Calculate w when the gas is allowed to expand reversibly
to 1 bar pressure (a ) isothermally and (b ) adiabatically.
2 .20 An ideal monatomic gas at 298.15 K and 1 bar is expanded
in a reversible adiabatic process to a final pressure of 12 bar. Calculate q per mole, w per mole, and ⌬U.
2 .21 An ideal monatomic gas at 1 bar and 300 K is expanded
adiabatically against a constant pressure of 12 bar until the final
pressure is 12 bar. What are the values of q per mole, w per mole,
⌬U , and ⌬H ? Given: CV  32 R .
2 .22 Derive the equation for calculating the work involved in a
reversible, adiabatic pressure change of one mole of an ideal gas
so that the work can be calculated from the initial temperature
T1 , initial pressure P1 , and final pressure P2 .
2 .23 Calculate ⌬r H298
⬚ for
H2 (g)  F2 (g)  2HF(g)
H2 (g)  Cl2 (g)  2HCl(g)

The sum of the three reactions is
2C(s)  2H2 O(g)  CH4 (g)  CO2 (g)
What is ⌬r H ⬚ at 500 K for each of these reactions? Check that
the sum of the ⌬r H ⬚ values of the first three reactions is equal
to ⌬r H ⬚ for the fourth reaction. From the standpoint of heat
balance, would it be better to develop a process to carry out
the overall reactions in three separate reactors or in a single
reactor?
2 .27 What is the heat evolved in freezing water at ⫺10 ⬚C given
that
H2 O(l)  H2 O(s)

⌬ H ⬚ (273 K)  ⫺6004 J mol⫺1

CP (H2 O, l)  75.3 J K⫺1 mol⫺1
CP (H2 O, s)  36.8 J K⫺1 mol⫺1
2 .28 What is the enthalpy change for the vaporization of
water at 0 ⬚C? This value may be estimated from Table C.2
by assuming that the heat capacities of H2 O(l) and H2 O(g) are
independent of temperature from 0 to 25 ⬚C.
2 .29 Calculate the standard enthalpy of formation of methane
at 1000 K from the value at 298.15 K using the HT⬚ ⫺ H298
⬚ data
in Table C.3.
2 .30 For a diatomic molecule the bond energy is equal to the
change in internal energy for the reaction

H2 (g)  Br2 (g)  2HBr(g)

X2 (g)  2X(g)

H2 (g)  I2 (g)  2HI(g)

at 0 K. Of course, the change in internal energy and the change in
enthalpy are the same at 0 K. Calculate the enthalpy of dissociation of O2 (g) at 0 K. The enthalpy of formation of O(g) at 298.15
K is 249.173 kJ mol⫺1 . In the range of 0–298 K the average value
of the heat of capacity of O2 (g) is 29.1 J K⫺1 mol⫺1 and the average value of the heat capacity of O(g) is 22 .7 J K⫺1 mol⫺1 .
What is the value of the bond energy in electron volts? (When
the changes in heat capacities in the range of 0–298 K are
taken into account, the enthalpy of dissociation at 0 K is 493.58
kJ mol⫺1 .)
2 .31 One gram of liquid benzene is burned in a bomb
calorimeter. The temperature before ignition was 20.826 ⬚C,
and the temperature after the combustion was 25.000 ⬚C. This
was an adiabatic calorimeter. The heat capacity of the bomb,
the water around it, and the contents of the bomb before the
combustion was 10 000 J K⫺1 . Calculate ⌬f H ⬚ for C6 H6 (l) at
298.15 K from these data. Assume that the water produced in
the combustion is in the liquid state and the carbon dioxide
produced in the combustion is in the gas state.
2 .32 An aqueous solution of unoxygenated hemoglobin containing 5 g of protein (M  64 000 g mol⫺1 ) in 100 cm3 of
solution is placed in an insulated vessel. When enough molecular oxygen is added to the solution to completely saturate
the hemoglobin, the temperature rises 0.031 ⬚C. Each mole of
hemoglobin binds 4 mol of oxygen. What is the enthalpy of
reaction per mole of oxygen bound? The heat capacity of the
solution may be assumed to be 4.18 J K⫺1 cm⫺3 .

2 .24 The following reactions might be used to power rockets:
(1)

H2 (g)  12 O2 (g)  H2 O(g)

(2)

CH3 OH(l)  1 12 O2 (g)  CO2 (g)  2H2 O(g)

(3)

H2 (g)  F2 (g)  2HF(g)

(a ) Calculate the enthalpy changes at 25 ⬚C for each of these reactions per kilogram of reactants. (b ) Since the thrust is greater
when the molar mass of the exhaust gas is lower, divide the heat
per kilogram by the molar mass of the product (or the average
molar mass in the case of reaction 2) and arrange the above reactions in order of effectiveness on the basis of thrust.
2 .25 Calculate ⌬r H for the dissociation
O2 (g)  2O(g)
at 0, 298, and 3000 K. In Section 14.3 the enthalpy change for
dissociation at 0 K will be found to be equal to the spectroscopic
dissociation energy D0 .
2 .26 Methane may be produced from coal in a process represented by the following steps, where coal is approximated by
graphite:
2C(s)  2H2 O(g)  2CO(g)  2H2 (g)
CO(g)  H2 O(g)  CO2 (g)  H2 (g)
CO(g)  3H2 (g)  CH4 (g)  H2 O(g)

Problems
2 .33 Calculate the heat of hydration of Na2 SO4 (s) from the integral heats of solution of Na2 SO4 (s) and Na2 SO4 ⭈10H2 O(s) in
infinite amounts of H2 O, which are ⫺2 .34 and 78.87 kJ mol⫺1 ,
respectively. Enthalpies of hydration cannot be measured directly because of the slowness of the phase transition.
2 .34 We want to determine the enthalpy of hydration of CaCl2
to form CaCl2 ⭈6H2 O:
CaCl2 (s)  6H2 O(l)  CaCl2 ⭈6H2 O(s)
We cannot do this directly for a couple of reasons: (1) reactions in the solid state are slow, and (2) there is a series of
hydrates and so a mixture of different hydrates would probably be obtained. We can, however, determine the heats of
solution of CaCl2 (s) and CaCl2 ⭈6H2 O(s) in water at 298 K and
take the difference. The experimental heats of solution are as
follows:
CaCl2 (s)  Aq  CaCl2 (ai)

⌬r H  ⫺81.33 kJ mol⫺1

CaCl2 ⭈6H2 O(s)  Aq  CaCl2 (ai)  6H2 O(l)
⌬r H  15.79 kJ mol⫺1
where Aq represents a large amount of water. What is the enthalpy of hydration?
2.35 The change in internal energy ⌬c U ⬚ in the combustion
of C60 (s) is ⫺25 968 kJ mol⫺1 at 298.15 K [Kolesov et al., J.
Chem. Thermodyn. 28:1121 (1996)]. (a ) What is the enthalpy
of combustion ⌬c H ⬚ ? (b ) What is the enthalpy of formation
⌬f H ⬚ [C60 (s)]? (c ) What is the enthalpy of vaporization of
C60 (s) to C(g) per mole of C(g)? (d ) How does this compare
with the enthalpy of vaporization of graphite and diamond to
C(g)?
2 .36 How much work is done when a person weighing 75 kg
(165 lb) climbs the Washington Monument, 555 ft high? How
many kilojoules must be supplied to do this muscular work,
assuming that 25% of the energy produced by the oxidation
of food in the body can be converted to muscular mechanical
work?
2 .37 The average person generates about 2500 kcal of heat
a day. How many kilowatt-hours of energy is this? If walking
briskly dissipates energy at 500 W, what fraction of the day’s energy does walking one hour represent? How many kilograms
of water would have to be evaporated if this were the only
means of heat loss? (The heat of vaporization of water at 35 ⬚C
is 2400 J g⫺1 .)
2 .38 The surface tension of water is 71.97 ⫻ 10⫺3 N m⫺1 or
71.97 ⫻ 10⫺3 J m⫺2 at 25 ⬚C. Calculate the surface energy in
joules of 1 mol of water dispersed as a mist containing droplets
of 1 ␮m (10⫺4 cm) in radius. The density of water may be taken
as 1.00 g cm⫺3 .
2 .39 Are the following expressions exact differentials?
x
dx
(a) xy 2 dx ⫺ x 2 y dy
(b)
⫺ 2 dy
y
y

2 .40

71

Show that
dq  dU  P dV  CV dT  RT d ln V

is not an exact differential, but
dq
 CV d ln T  R d ln V
T
is an exact differential.
2 .41 Show that the differential dV of the molar volume of an
ideal gas is an exact differential.
2 .42 Calculate w for a reversible isothermal (298.15 K) expansion of a mole of N2 from 1 to 10 L assuming it is (a ) an ideal gas
and (b ) a van der Waals gas (see Table 1.3).
2 .43 An ideal gas at 25 ⬚C and 100 bar is allowed to expand
reversibly and isothermally to 5 bar. Calculate (a) w per mole,
(b) the heat absorbed per mole, (c) ⌬U, and (d) ⌬H.
2 .44 Ammonia gas is condensed at its boiling point at 1.013 25
bar at ⫺33.4 ⬚C by the application of a pressure infinitesimally
greater than 1 bar. To evaporate ammonia at its boiling point
requires the absorption of 23.30 kJ mol⫺1 . Calculate (a ) wrev
per mole, (b ) q per mole, (c ) ⌬U, and (d ) ⌬H.
2 .45 What is w per mole for a reversible isothermal expansion
of ethane from 5 to 10 L mol⫺1 at 298 K assuming (a ) ethane is
an ideal gas and (b ) it follows the van der Waals equation? (Van
der Waals constants are in Table 1.3.)
2 .46 According to Table C.3, how much heat is required to
raise the temperature of a mole of oxygen from 298 to 3000 K at
constant pressure?
2.47 From the following data calculate the value of
(H298
⬚ ⫺ H0⬚ ) for Al2 O3 (s).

T /K

C ⬚P /J K⫺1
mol⫺1

T

C ⬚P

T

C ⬚P

T

C ⬚P

10
20
30
40
50
60
70
80

0.009
0.076
0.263
0.691
1.492
2.779
4.582
6.895

90
100
110
120
130
140
150
160

9.69
12.84
16.32
20.06
23.96
27.96
31.98
35.99

170
180
190
200
210
220
230
240

39.94
43.79
47.53
51.14
54.60
57.92
61.10
64.13

250
260
270
280
290
298.16
273.16

67.01
69.76
72.37
74.84
77.19
79.01
73.16

2 .48 One mole of hydrogen at 25 ⬚C and 1 bar is compressed
adiabatically and reversibly into a volume of 5 L. Assuming ideal
gas behavior, calculate (a ) the final temperature, (b ) the final
pressure, and (c ) the work done on the gas.
2 .49 One mole of argon at 25 ⬚C and 1 bar pressure is allowed to expand reversibly to a volume of 50 L (a ) isothermally
and (b ) adiabatically. Assuming ideal gas behavior, calculate
the final pressure in each case and the work done on the
gas.

72

Chapter 2

First Law of Thermodynamics

2 .50 A mole of monatomic ideal gas at 1 bar and 273.15 K is
allowed to expand adiabatically against a constant pressure of
0.395 bar until equilibrium is reached. (a ) What is the final temperature? (b ) What is the final volume? (c ) How much work is
done on the gas in this process? (d ) What is the change in the
internal energy of the gas in this process?
2 .51 A tank contains 20 L of compressed nitrogen at 10 bar
and 25 ⬚C. Calculate the maximum work that can be obtained
when the gas is allowed to expand reversibly to 1 bar pressure
(a ) isothermally and (b ) adiabatically. The heat capacity of nitrogen at constant volume can be taken to be 20.8 J K⫺1 mol⫺1
independent of temperature.
2 .52 Compare the enthalpies of combustion of CH4 (g) to
CO2 (g) and H2 O(g) at 298 and 2000 K.
CH4 (g)  2O2 (g)  CO2 (g)  2H2 O(g)
2 .53 Compare the enthalpy of combustion of CH4 (g) to
CO2 (g) and H2 O(l) at 298 K with the sum of the enthalpies
of combustion of graphite and 2H2 (g), from which CH4 (g) can,
in principle, be produced.
2 .54 The enthalpy change for the combustion of toluene to
H2 O(l) and CO2 (g) is ⫺3910.0 kJ mol⫺1 at 25 ⬚C. Calculate the
enthalpy of formation of toluene.
2 .55 Calculate ⌬H (298 K) per gram of fuel (exclude oxygen)
for
(a)

H2 (g)  12 O2 (g)  H2 O(g)

(b)

CH4 (g)  2O2 (g)  CO2 (g)  2H2 O(g)

(c)

CH3 OH(l)  32 O2 (g)  CO2 (g)  2H2 O(g)

(d) C6 H14 (g)  9 12 O2 (g)  6CO2 (g)  7H2 O(g)
2 .56 A 1:3 mixture of CO and H2 is passed through a catalyst
to produce methane at 500 K.
CO(g)  3H2 (g)  CH4 (g)  H2 O(g)
How much heat is liberated in producing a mole of methane?
How does this compare with the heat obtained from burning
a mole of methane at this temperature? How does the heat of
combustion of CH4 compare with the heat of combustion of
CO  3H2 ?
2 .57 Calculate ⌬r H ⬚ for the dissociation
H2 (g)  2H(g)
at 0, 298, and 3000 K. The value at 0 K is equal to the spectroscopic dissociation energy D0 .
2 .58 In principle, methanol can be produced from methane in
two steps or one:
I.

CH4 (g)  H2 O(g)  CO(g)  3H2 (g)
CO(g)  2H2 (g)  CH3 OH(g)

II.

CH4 (g)  H2 O(g)  CH3 OH(g)  H2 (g)

What is ⌬r H ⬚ at 500 K for each of these reactions? From the
standpoint of heat balance, would it be better to develop a pro-

cess to carry out the overall reaction in two separate reactors or
in a single reactor?
2 .59 Calculate the heat of vaporization of water at 25 ⬚C. The
specific heat of water may be taken as 4.18 J K⫺1 g⫺1 . The heat
capacity of water vapor at constant pressure in this temperature
range is 33.5 J K⫺1 mol⫺1 , and the heat of vaporization of water
at 100 ⬚C is 2258 J g⫺1 .
2 .60 Calculate the enthalpy of dissociation of H2 (g) at
3000 K using ⌬f HT⬚ [H(g), 298.15 K]  217.999 kJ mol⫺1 and
HT⬚ ⫺ H298
⬚ values in Appendix C.3.
2 .61 Calculate the dissociation energy of CH4 into atoms at
298.15 K using ⌬H ⬚ for the dissociation reaction at 0 K and
HT⬚ ⫺ H298
⬚ values from Appendix C.3.
2 .62 The reaction of heated coal (approximated here by
graphite) with superheated steam absorbs heat. This heat is
usually provided by burning some of the coal. Calculate ⌬r H ⬚
(500 K) for both reactions.
2 .63 Ammonia is to be oxidized to NO2 (g) to make nitric acid.
What temperature will be reached if the only reaction is
NH3 (g)  74 O2 (g)  NO2 (g)  32 H2 O(g)
and a stoichiometric amount of oxygen is used?
2 .64 In an adiabatic bomb calorimeter, oxidation of 0.4362 g of
naphthalene (C10 H8 ) caused a temperature rise of 1.707 ⬚C. The
final temperature was 298 K. The heat capacity of the calorimeter and water was 10 290 J K⫺1 , and the heat capacity of the
products can be neglected. If corrections for the oxidation of the
wire and residual nitrogen are neglected, what is the molar internal energy of combustion of naphthalene? What is its enthalpy
of formation?
2 .65 The combustion of oxalic acid in a bomb calorimeter
yields 2816 J g⫺1 at 25 ⬚C. Calculate (a) ⌬c U ⬚ and (b) ⌬c H ⬚ for
the combustion of 1 mol of oxalic acid (M  90.0 g mol⫺1 ).
2 .66 A mole of liquid sulfuric acid (98 g) is added to a certain
quantity of water at 25 ⬚C, and it is found that the temperature
is 100 ⬚C! The NBS Tables of Chemical Thermodynamic Properties yield the following information:
⌬f H ⬚ /kJ mol⫺1
H2 SO4 (l)

⫺813.99

H2 SO4 (ai)

⫺909.27

The ai indicates this value is for sulfuric acid that is completely
ionized in water. What is the enthalpy of solution of sulfuric acid
in enough water to completely ionize it? Let’s make some approximations and estimate the mass of the “certain quantity of
water” in the first line. Assume that the solution has the same
heat capacity as water (CP  75.3 J K⫺1 mol⫺1 ), independent
of temperature, and ignore the mass of the sulfuric acid added.
How much water was used?

Problems
2 .67 When nitroglycerin explodes, the chemical reaction that
occurs can be assumed to be
C3 H5 N3 O9 (l)  3CO2 (g)  52 H2 O(g)  32 N2 (g)  14 O2 (g)
(a ) Calculate ⌬r H ⬚ and ⌬r U ⬚ for this reaction at 298 K,
given that the enthalpy of formation of liquid nitroglycerin is
⫺372 .4 kJ mol⫺1 . (b ) Consider 0.20 mol of nitroglycerin at 25 ⬚C
completely filling a constant-volume cell of 0.030 L. Calculate
the maximum temperature and pressure that would be generated by the explosion of the nitroglycerin if the constant-volume
cell did not burst (or vaporize). You may assume that (1) the explosion occurs so rapidly that the conditions are adiabatic, (2)
the pressure cell comes to immediate thermal equilibrium with
the products, (3) the products are ideal gases, and (4) the
total constant-volume heat capacity of products plus cell has the
temperature-independent value of 100 J K⫺1 .

Computer Problems
2.A In Section 2.8 we have seen that the knowledge of the
temperature dependence of CP⬚ , which can be represented by
a polynomial in T (see Table 2.2), makes it possible to calculate the change in molar enthalpy H ⬚ with temperature. In Section 2.13 we have seen that the standard enthalpy of reaction at
other temperatures can be calculated from ⌬r H ⬚(298.15 K) by
integration of ⌬r CP⬚ dT . It is difficult to make these calculations
with a hand-held calculator, but they can be conveniently made
using a mathematical application that can integrate. By putting
in empirical equations for CP⬚ , we can solve the following
problems:

73

(a) Print out the values of CP⬚ at 298.15 K and 1000 K for the
following gases: N2 , O2 , H2 , CO, CO2 , H2 O, NH3 , and CH4 .
Compare these values with those in Table C.3 of Appendix C.
(b) Plot CP⬚ for CO2 , H2 O, NH3 , and CH4 versus temperature
from 298.15 K to 1500 K.
(c) Calculate H ⬚ (1000 K) ⫺ H ⬚ (298.15 K) for N2 and compare it with the values in Table C.3.
(d) Calculate ⌬r H (1000 K) for the following gas reactions:
H2  12 O2  H2 O
1
2

N2  32 H2  NH3

CO2  4H2  CH4  2H2 O
CO  3H2  CH4  H2 O
2.B Plot the molar heat capacities in J K⫺1 mol⫺1 at constant
pressure for gaseous He, N2 , H2 O, CO2 , NH3 , and CH4 from
300 to 1800 K using the parameters in Table 2.2.
2.C Calculate the standard enthalpy of reaction at 298 K and
1000 K for the gas reaction CO2  4H2  CH4  2H2 O.
2.D Calculate the standard enthalpy of formation of CO2 at
1000 K from the standard enthalpy of formation at 298.15 K and
empirical equations for the heat capacities. The standard molar
heat capacities of graphite are given in Table C.3 as a function
of temperature. It is convenient to fit these data to a function of
temperature.
2.E (a) Calculate the work of reversible isothermal expansion of a mole of carbon dioxide at 298.15 K from an initial volume of 5 L to a final volume of 20 L on the assumption that
carbon dioxide is an ideal gas.
(b) Calculate the reversible work using the van der Waals constants for carbon dioxide.
(c) Explain the difference.

3

Second and Third Laws
of Thermodynamics

3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
3.9

Entropy as a State Function
The Second Law of Thermodynamics
Entropy Changes in Reversible Processes
Entropy Changes in Irreversible Processes
Entropy of Mixing Ideal Gases
Entropy and Statistical Probability
Calorimetric Determination of Entropies
The Third Law of Thermodynamics
Special Topic: Heat Engines

The first law of thermodynamics states that when one form of energy is converted
to another, the total energy is conserved. It does not indicate any other restriction
on this process. However, we know that many processes have a natural direction,
and it is with the question of direction that the second law is concerned. For example, a gas expands into a vacuum, but the reverse never occurs, although it would
not violate the first law. For a metal bar at uniform temperature to become hot
at one end and cold at the other would not be a violation of the first law, yet we
know this never occurs spontaneously.
The second law of thermodynamics is one of the most important generalizations in science. It is important in chemistry because it can tell us whether a process
or reaction will occur in the forward or backward direction. The quantity that tells
us whether a chemical reaction or a physical change can occur spontaneously in
an isolated system is the entropy S. Entropy is a function of the state of the system, as is the internal energy U. Thermodynamics does not deal with the rate of
approach to equilibrium, only with the equilibrium state. Some time is required
even for a gas to expand into another container, and for some chemical reactions
the rate of approach to equilibrium is very slow.
The third law of thermodynamics allows us to obtain the absolute value of the
entropy of a substance.

3.1 Entropy as a State Function

3.1

ENTROPY AS A STATE FUNCTION

Historically, the second law of thermodynamics and the concept of entropy arose
from considering the efficiencies of heat engines, but heat engines are quite different from chemical systems. Therefore, we are going to introduce the state function
entropy from the more mathematical viewpoint of the preceding two chapters.
To understand the need for a second law of thermodynamics, consider what
the first law of thermodynamics has not provided. According to the first law, energy is conserved in a process that takes a system from one state to another, but
this provides no information as to whether the process of chemical reaction can
proceed spontaneously or not. Yet we know that processes and chemical reactions proceed spontaneously in one direction and not in the opposite direction.
For example, gases always expand into a vacuum. Heat is always transferred from
a hot body to a cold body. In the presence of a catalyst, molecular hydrogen and
molecular oxygen react to form water. These are referred to as spontaneous processes. The reverse processes, nonspontaneous processes, can be accomplished
only by performing work on the system. A gas can be compressed to a smaller
volume by use of mechanical work. Heat can be transferred to a hot body from a
cold body by use of a refrigeration device. Water can be electrolyzed to molecular
hydrogen and molecular oxygen by use of electrical work from a battery.
When we see a movie run backward, we often laugh because we know the
event could not happen that way. The first law of thermodynamics does not tell
us the direction in which a process can occur spontaneously. The laws of classical mechanics and quantum mechanics also do not tell us the direction in which
time increases. When a movie of a collision between two particles is run backward
we don’t laugh, because it looks as reasonable as the movie run in the forward
direction. Indeed, the mechanical equations of motion are invariant under time
reversal. It is useful to be able to predict whether a physical change or a chemical
reaction will go spontaneously in the forward or backward direction, and so the
second law is very important.
To see how a state function can be introduced to identify a spontaneous process, let us consider the transfer of heat, which obviously involves the issue of the
temperature. In the preceding chapter we saw that heat is not a state property of a
system, even though it can be expressed in terms of state properties for a specified
process, i.e., along a particular path. For example, the first law shows that when
an ideal gas is heated reversibly,
dU  CV dT  dqrev  dw  dqrev ⫺ P dV  dqrev ⫺

nRT
dV
V

(3.1)

The reversible heat is not a state function because dqrev is an inexact differential
(Section 2.3). This is indicated by the symbol used here, but it can be confirmed
by applying the test of exactness, (⭸M /⭸y )x  (⭸N /⭸x )y , to equation 3.1 written
in the form
nRT
dqrev  CV dT 
dV
(3.2)
V
Taking the derivative of the coefficient of dT with respect to V at constant T
yields
⭸CV

冢 ⭸V 冣

T

0

(3.3)

75

76

Chapter 3

Second and Third Laws of Thermodynamics

since for an ideal gas CV is independent of the volume. Taking the derivative of
the coefficient of dV with respect to T at constant V yields



⭸(nRT /V )
⭸T





V

nR
V

(3.4)

Since the mixed partial derivatives are not equal, qrev is not a state function.
However, we saw in Section 2.3 that there may be an integrating factor that
will convert an inexact differential to an exact differential. Since the transfer of
heat depends on the temperature, let us try 1/T as an integrating factor. This is
accomplished by multiplying both sides of equation 3.2 by 1/T .
dqrev
CV
nR

dT 
dV
T
T
V

(3.5)

Applying the test for exactness to dqrev /T yields




⭸(CV /T )
⭸V
⭸(nR /V )
⭸T




0

(3.6)

0

(3.7)

T

V

Since both of these derivatives are equal to zero, dqrev /T is the exact differential
of a state function. This proof applies to an ideal gas, but the fact that this is a
general conclusion is shown in Section 3.9 on heat engines.
T1, P1, V1

T1, P2, V2
A

T

Show that 冮 dqrev /T is independent of path, by considering the following reversible changes
in the state of an ideal gas:

C

B

Example 3.1 Demonstration that the integral of the differential of the
reversible heat divided by T is a state function

T2, P3, V2

V
(a)

T3, P1, V2

(a )

ideal gas(T1 , P1 , V1 )  ideal gas(T1 , P2 , V2 )

reversible, isothermal

(b )

ideal gas(T1 , P1 , V1 )  ideal gas(T2 , P3 , V2 )

reversible, adiabatic

(c )

ideal gas(T2 , P3 , V2 )  ideal gas(T1 , P2 , V2 )

reversible, constant volume

(d )

ideal gas(T1 , P1 , V1 )  ideal gas(T3 , P1 , V2 )

reversible, constant pressure

(e )

ideal gas(T3 , P1 , V2 )  ideal gas(T1 , P2 , V2 )

reversible, constant volume

Our first law knowledge is sufficient to calculate 冮 dqrev /T for each of these changes, and
plots for an ideal gas are given in Fig. 3.1. The value of 冮 dqrev /T is calculated for each of
these changes in state as follows:
(a )

D
T

E

(b )

A
T1, P2, V2

T1, P1, V1

For an ideal gas, the internal energy depends only on the temperature, and so
dU  0. Therefore, dqA  ⫺dw  P dV  RT dV /V . Thus 冮 dqA /T  R ln(V2 /V1 ).
Since the change is adiabatic, dqB  0 and 冮 dqB /T  0, and from the expression
derived in Chapter 2,
R /CV

冢 冣

T1
V
 2
T2
V1
V
(b)

Figure 3.1 Reversible changes in
the state of an ideal gas.

(c )

Since the volume is constant, w  0, and from the first law dqC  dU  CV dT .
Thus 冮 dqC /T  冮 CV dT /T  CV ln(T1 /T2 ).

As indicated in Fig. 3.1a , the sum of B and C should equal A for state functions; that is,
冮 dqA /T  冮 dqB /T  冮 dqC /T or R ln(V2 /V1 )  CV ln(T1 /T2 ). This is in agreement with

3.2 The Second Law of Thermodynamics
the expression for a reversible adiabatic expansion. This confirms that 冮 dqrev /T is a state
function.
(d )
(e )

At constant pressure, 冮 dqD /T  冮 dHD /T , which is equal to 冮 CP dT /T  CP ln(T3 /T1 ).
Note that since the pressure is the same in the initial and final states, T3 /T1  V2 /V1 .
At constant V , dwE  0, and so from the first law, 冮 dqE /T  冮 dUE /T 
冮 CV dT /T  CV ln(T1 /T3 ).

As indicated in Fig. 3.1b , the sum of D and E should equal A for state functions; that is,
冮 dqD /T  冮 dqE /T  冮 dqA /T or CP ln(T3 /T1 )  CV ln(T1 /T3 )  R ln(T3 /T1 ) should be
equal to R ln(V2 /V1 ). This is true, as indicated in considering d .

3.2

THE SECOND LAW OF THERMODYNAMICS

We have seen that dqrev /T is the differential of a state function. Clausius named
this state function the entropy S, and so for an infinitesimal change in state
dS 

dqrev
T

(3.8)

and for a finite change in state,
⌬S 



dqrev
T

(3.9)

Summing around a closed cycle yields





qrev
0
T

dqrev
0
T

冱 ⌬S  0

冯 dS  0

(finite isothermal steps)

(3.10)

(infinitesimal steps)

(3.11)

(finite isothermal steps)

(3.12)

(infinitesimal steps)

(3.13)

We are indebted to Clausius for much more than a name, because in 1854 he
extended these equations to cycles containing irreversible steps by showing that



dq
ⱕ0
T

(3.14)

This Clausius theorem is the mathematical statement of the second law of thermodynamics. The cyclic integral is to be understood in the following way: (a) If
any part of the cyclic process is irreversible (spontaneous), the inequality applies
and the cyclic integral is negative. (b) If the cyclic process is reversible, the equality applies. (c) It is impossible for the cyclic integral to be greater than zero. Note
that the temperature that appears in the cyclic integral in equation 3.14 is that of
the heat reservoir or surroundings. When the process is reversible, the temperature of the system is equal to the temperature of the surroundings. Equation 3.14
leads to the following inequality for a noncyclic process:
dS ⱖ

dq
T

(3.15)

77

78

Chapter 3

Second and Third Laws of Thermodynamics

If the process is reversible, dS  dqrev /T , and if the process is irreversible, dS ⬎
dqirrev /T .
Thus there are two parts of the second law of thermodynamics:
1.
2.

There is a state function called the entropy S that can be calculated from
dS  dqrev /T .
The change in entropy in any process is given by dS ⱖ dq /T , where the
⬎ applies to a spontaneous process (irreversible process) and the equality
applies to a reversible process. This means that in order to calculate ⌬S for a
change in state, one must use a reversible process.

Example 3.2 Demonstration that dS is greater than the differential heat for
an irreversible process divided by T
Apply the Clausius theorem to the following isothermal irreversible cycle to obtain dS ⬎
dqirrev /T .
state 1

irreversible

state 2

reversible

state 1

Since this cycle is irreversible, equation 3.14 yields
2

冮1

dqirrev

T

1

冮2

dqrev
⬍0
T

Since the second step is reversible, dqrev /T can be replaced by dS , and the limits can be
interchanged, with a change in sign.
2

冮1

dqirrev

T

2

冮1 dS ⬍ 0

Thus,
⌬S  S2 ⫺ S1 

2

2

冮1 dS ⬎ 冮1

dqirrev
T

We can also write
dS ⬎

dqirrev
T

Thus, for an infinitesimal irreversible process the change in entropy is greater than the
differential of the heat divided by the temperature.

We can restate the second law by saying that the entropy increases in a spontaneous process in an isolated system because for an isolated system, dq  0, and
therefore ⌬S ⬎ 0 for a spontaneous process. The entropy of an isolated system (in
which the internal energy is constant) can continue to increase as long as spontaneous processes occur. When there are no more possible spontaneous processes,
the entropy is at a maximum; for any further infinitesimal process in the system,
dS  0. Thus the entropy change tells us whether a process or chemical reaction
can occur spontaneously in an isolated system.
This reasoning can be applied to a system that is not isolated by treating the
system plus its surroundings as an isolated system. When a spontaneous change
occurs in the system of interest, the entropy change dStotal in the system plus the
surroundings is given by
dStotal  dSsyst  dSsurr
(3.16)

3.2 The Second Law of Thermodynamics

79

First, consider a reversible process in which the system gains heat dqrev from the
surroundings. Since the surroundings gain heat equal to ⫺qrev , equation 3.16 becomes
dqrev
0  dSsyst ⫺
(3.17)
Tsurr
since dStotal  0 for a reversible process in an isolated system. Thus, for a reversible process in the system plus surroundings,
dSsyst 

dqrev
dqrev

Tsurr
T

(3.18)

since for a reversible process, the temperature of the system is equal to the temperature of the surroundings.
Second, consider an irreversible process in which the system gains heat dqirrev
from its surroundings. In this case, it is convenient to write equation 3.16 as
dSsyst  dStotal ⫺ dSsurr

(3.19)

If the transfer of heat occurs reversibly in the surroundings, the entropy change
in the surroundings is ⫺dqirrev /Tsurr and equation 3.19 becomes
dqirrev
(3.20)
Tsurr
since dStotal ⬎ 0 for a spontaneous change in the total (isolated) system. Equations
3.18 and 3.20 can be combined to obtain
dq
dSsyst ⱖ
(3.21)
Tsurr
dSsyst ⬎

where the inequality applies when the process is irreversible and the equality applies when the process is reversible. For a finite process,
⌬Ssyst ⱖ

冮T

dq

(3.22)

surr

As a spontaneous change occurs in an isolated system, the entropy increases
with time as shown in Fig. 3.2 and eventually levels off. When a process occurs reversibly in an isolated system, the entropy does not change. When a spontaneous
process occurs in an isolated system, the entropy increases. The direction of spontaneous change in any system is in the direction of increasing the entropy of the
universe, and thus the increase in entropy indicates the time sequence of a spontaneous process. The entropy is sometimes referred to as the “arrow of time.” The
second law of thermodynamics contrasts with the equations of classical mechanics
and quantum mechanics, which are reversible in time.
Equation 3.21 describes three types of processes:
dS ⬎ dq /T
dS  dq /T
dS ⬍ dq /T

spontaneous and irreversible process
reversible process
impossible process

(3.23)

S

0

t

Figure 3.2 Change in entropy of an
The simplest place to apply equation 3.21 is to an isolated system, because isolated system with time. The endq  0. For a finite change the three possibilities are
tropy of the system increases spontaneously until equilibrium is reached.
⌬S ⬎ 0
spontaneous and irreversible process
Thermodynamics does not deal with
⌬S  0
reversible process
the question of how long it will take
⌬S ⬍ 0
impossible process
(3.24) to reach equilibrium.

80

Chapter 3

Second and Third Laws of Thermodynamics

As an illustration of the use of the fact that dS ⱖ 0 for an isolated system,
consider two phases in a system of constant volume that is surrounded by insulation, as shown in Fig. 3.3. Suppose that one phase is at temperature T␣ and the
other is at temperature T␤ . We can imagine the transfer of an infinitesimal quantity of heat dq from phase ␣ to phase ␤ . The change in the entropy of the system
is given by




Figure 3.3 Two phases in an
isolated system of constant volume.
The container is surrounded by
insulation.

dS 



dq
dq
1
1
 dq


T␤
T␣
T␤
T␣



(3.25)

If this process occurs spontaneously, dS ⬎ 0, and so this equation shows that
T␣ ⬎ T␤ . In other words, heat flows spontaneously to the phase with the lower
temperature. If the two phases are at thermal equilibrium, dS  0 and therefore
T␣  T␤ . Thus the second law has led us to the conclusion that for two phases to
be in equilibrium, they must be at the same temperature.
Since S is a state function, we can integrate dS between two states of a system.
For the change in state
A(state 1)  A(state 2)



S2

S1

dS 



2

1

dqrev
 S2 ⫺ S1
T

(3.26)
(3.27)

Thus to determine the change in entropy in a process, we have to integrate along
the path of a reversible process connecting states 1 and 2. The ⌬S for an irreversible process can be calculated if we can devise a reversible path and use it
for the integration in equation 3.27. There is no change in entropy in a reversible
adiabatic process.

(a)

Example 3.3

The entropy of expansion of an ideal gas in an isolated system

Is the expansion of an ideal gas into a larger volume thermodynamically spontaneous
in an isolated system? More specifically, consider the expansion of an isolated ideal gas
initially at 298 K into a volume that is twice as large as its initial volume as shown in
Fig. 3.4a.
Remember that Joule found that there is no change in temperature when a dilute
gas is allowed to expand in an isolated system. To determine the change in entropy between the initial state and the final state, the reversible isothermal expansion described in
Fig. 3.4b can be used. As we have seen earlier, the work per mole done on the gas is
⫺RT ln 2, so that the heat absorbed by the gas is
qrev  RT ln 2
(b)

Figure 3.4 (a) Irreversible expansion of an ideal gas at 298 K into
twice the initial volume with no heat
or work. (b) Reversible isothermal
expansion of an ideal gas to twice
the initial volume at 298 K. In
(b) the piston must be surrounded
by a heat reservoir.

since ⌬U  0. Thus, the change in entropy of the gas is
⌬Sb  R ln 2  5.76 J K⫺1 mol⫺1
Since the change in state of the gas for the process described in Fig. 3.4a is the same
as that for Fig. 3.4b, ⌬Sa ⬎ 0. Thus, we can conclude that the expansion is spontaneous,
as we knew all along. The reverse of the process described in Fig. 3.4a, that is, the gas
flowing back spontaneously into the initial volume, is impossible since ⌬S ⬍ 0. It is important to note that this problem has nothing to do with minimizing the energy, which is
constant.

3.3 Entropy Changes in Reversible Processes

3.3

ENTROPY CHANGES IN REVERSIBLE PROCESSES

We will now consider some simple processes for which entropy changes are readily
calculated. It is especially easy to calculate entropy changes for reversible isothermal processes.
The transfer of heat from one body to another at an infinitesimally lower temperature is a reversible change, since the direction of heat flow can be reversed
by an infinitesimal change in the temperature of one of the bodies. For example,
consider the vaporization of a pure liquid into its vapor at the equilibrium vapor
pressure P :
liquid(T, P ) y vapor(T, P )

(3.28)

Since T is constant, the integration of equation 3.18 yields
S2 ⫺ S1  ⌬S 

qrev
T

(3.29)

where qrev represents the heat absorbed in the reversible change. Since the pressure is constant, the reversible heat is equal to the change in enthalpy ⌬H, so that
⌬S 

⌬H
T

(3.30)

This equation may also be used to calculate the entropy of sublimation, the entropy of melting, and the entropy change for a transition between two forms of a
solid.
Example 3.4 The molar entropy of vaporization of n-hexane
What is the change in the molar entropy of n -hexane when it is vaporized at its boiling
point?
n -Hexane boils at 68.7 ⬚C at 1.013 25 bar, and the molar enthalpy of vaporization is
28 850 J mol⫺1 at this temperature. If n -hexane is vaporized into the saturated vapor at
this temperature, the process is reversible and the molar entropy change is given by
⌬S 

⌬H
28 850 J mol⫺1

 84.41 J K⫺1 mol⫺1
T
341.8 K

The overbars indicate that we are dealing with molar changes.

The molar entropy of a vapor is always greater than that of the liquid with
which it is in equilibrium, and the molar entropy of the liquid is always greater
than that of the solid at the melting point. According to the statistical interpretation of entropy to be discussed in Section 3.6, in which the entropy is a measure
of the disorder of the system, the molecules of the gas are more disordered than
those of the liquid, and the molecules of the liquid are more disordered than those
of the solid.
Now let us apply the second law to the vaporization of a liquid into its saturated vapor. To apply equations 3.24, we must consider an isolated system. In this
case the liquid and vapor (the system) and the heat reservoir at T (the surroundings) form an isolated overall system. The total entropy change is given by
⌬S  ⌬Ssys  ⌬Ssurr

(3.31)

81

82

Chapter 3

Second and Third Laws of Thermodynamics

Since the heat gained by the system is equal to that lost by the surroundings, the
entropy change for the surroundings is the negative of the entropy change for
the system if the vaporization is carried out reversibly; for both the system and
surroundings taken together, the total change of entropy ⌬S is zero if the transfer
of heat is carried out reversibly. This is in agreement with the second form in
equations 3.24.
Heating and cooling a substance are other examples of processes that can be
carried out reversibly. The change in entropy when a substance is heated or cooled
can be calculated using

C/T

∆S

T1

T2
T

Figure 3.5 When a substance is
heated from T1 to T2 , without a
phase change, the increase in
entropy is given by the indicated
area. If the volume is constant, nCV
is used for C ; and if the pressure is
constant, nCP is used.

dS 

C dT
dqrev

T
T

(3.32)

where C is CP for a process at constant pressure and CV for a process at constant
volume. If the heat capacity is independent of temperature, and the temperature
is changed from T1 to T2 , then
⌬S 



T2

T1

C
T2
dT  C ln
T
T1

(3.33)

If the heat capacity is a function of temperature, this function can be substituted into the integral form of equation 3.33, or the entropy change can be obtained from a numerical integration, as shown in Fig. 3.5.

Example 3.5

The change in molar entropy on heating an ideal gas

Oxygen is heated from 300 to 500 K at a constant pressure of 1 bar. What is the increase in
molar entropy?
The coefficients in an empirical equation for the heat capacity at constant pressure as
a function of temperature are given in Table 2.2. Using equation 3.33, we find
⌬S 

T2

冮T

1

 ␣ ln

CP
dT 
T

T2



冮T 冢T  ␤  ␥T 冣 dT
1


T2
 ␤ (T2 ⫺ T1 )  (T 22 ⫺ T 12 )
2
T1

 25.503 ln

500
1
 (13.612 ⫻ 10⫺3 )(200) ⫺ (42.553 ⫻ 10⫺7 )(500 2 ⫺ 300 2 )
300
2

 15.41 J K⫺1 mol⫺1
The units have been omitted in this calculation, but you should check that they cancel
properly.

The entropy change is readily calculated for a reversible isothermal expansion
of an ideal gas. Since the internal energy of an ideal gas is independent of volume at constant temperature (Section 2.6), dq  ⫺dw  P dV. For a reversible
isothermal expansion from V1 to V2 ,

3.3 Entropy Changes in Reversible Processes

ideal gas(T ,V1 ,P1 ) y ideal gas(T, V2 , P2 )
⌬S 



V2

V1

P
dV  nR
T

 nR ln



V2

V1

(3.34)

1
dV
V

V2
V1

 ⫺nR ln

P2
P1

(3.35)

where the final form simply comes from the fact that the volume is inversely proportional to pressure at constant temperature.
Equation 3.35 can be applied to the isothermal expansion of a mole of an ideal
gas from its standard pressure P  to some other pressure P :
S  S  ⫺ R ln

冢P  冣
P

(3.36)

where S  is the molar entropy in the standard state.
Example 3.6 The change in entropy for a system and its surroundings in a
reversible process
Half a mole of an ideal gas expands isothermally and reversibly at 298.15 K from a volume
of 10 L to a volume of 20 L. (a) What is the change in the entropy of the gas? (b) How
much work is done on the gas? (c) What is qsurr ? (d) What is the change in the entropy of
the surroundings? (e) What is the change in the entropy of the system plus the
surroundings?
(a )
(b )
(c )
(d )

⌬S  nR ln(V2 /V1 )  (0.5 mol)(8.3145 J K⫺1 mol⫺1 ) ln 2  2.88 J K⫺1
wrev  ⫺nRT ln(V2 /V1 )
wrev  ⫺(0.5 mol)(8.3145 J K⫺1 mol⫺1 )(298.15 K) ln 2  ⫺859 J
Since the gas is ideal there is no change in its internal energy; ⌬U  q  w  0.
Thus qsys  859 J, and qsurr  ⫺859 J.
Since heat flows out of the surroundings, it has a decrease in entropy.
⌬Ssurr  ⫺859 J/298.15 K  ⫺2.88 J K⫺1

(e )

Since the entropy of the gas increases and the entropy of the surroundings decreases
by the same amount, there is no change in entropy for the system and its surroundings. The gas and the surroundings can be considered to be an isolated system. The
process is reversible, and so we expect ⌬S  0.

Example 3.7 The change in entropy for a system and its surroundings in an
irreversible process
Now consider that the expansion in the preceding example occurs irreversibly by simply
opening a stopcock and allowing the gas to rush into an evacuated bulb of 10-L volume.
(a) What is the change in the entropy of the gas? (b) How much work is done on the gas?
(c) What is qsurr ? (d ) What is the change in the entropy of the surroundings? (e) What is
the change in the entropy of the system plus the surroundings?

83

84

Chapter 3

Second and Third Laws of Thermodynamics
(a )
(b )
(c )
(d )
(e )

C

The change in entropy is the same as above because entropy is a state function.
No work is done in the expansion.
No heat is exchanged with the surroundings.
The entropy of the surroundings does not change.
The entropy of the system plus surroundings increases by 2.88 J K⫺1 . Since this is
an irreversible process we expect the entropy to increase.

ALCULATION OF ⌬S FOR VARIOUS CHANGES IN STATE

General
⌬S 



dqrev
T

One must find a reversible path to go from the initial state to the final state in order to
calculate the change in entropy; however, since entropy is a state function, the change in
entropy for the irreversible path between the same initial and final states is the same.

Specific Cases
(a)

Constant V heating of one mole of substance

substance(T1 , V )  substance(T2 , V )
T
冮T12

CV dT /T , where CV is the constant-volume heat capacity of the subThen, ⌬S 
stance. If CV is not a function of T , then ⌬S  CV ln(T2 /T1 ).
(b)

Constant P heating of one mole of substance

substance(T1 , P )  substance(T2 , P )
Then, ⌬S 
stance.
(c)

T
冮T12

CP dT /T , where CP is the constant-pressure heat capacity of the sub-

Phase change at constant T and P

H2 O(l, 373 K, 1 atm)  H2 O(g , 373 K, 1 atm)
Then ⌬S  ⌬H /T , where ⌬H is the heat of vaporization in this case.
(d)

Ideal gas changes in state at constant T

ideal gas(P, V, T )  ideal gas(P ⬘, V ⬘, T )
Then ⌬S  R ln(V ⬘/V )  R ln(P /P ⬘). Note the position of the P and P ⬘, and the
position of the V and V ⬘. Remember that ⌬S is positive if the volume increases at constant T .
(e)

Mixing of ideal systems at constant T and P

nA A(T, P )  nB B(T, P )  n mixture(T, P )
Here, n  nA  nB is the total number of moles in the mixture. Then,

⌬S  ⫺nR (yA ln yA  yB ln yB )
where the y terms are the mole fractions: yA  nA /(nA  nB ) and yB  nB /(nA  nB ).
Note that since the mole fractions are less than 1, the change in entropy is positive, as
expected for a spontaneous process.

3.4 Entropy Changes in Irreversible Processes

Example 3.8 The molar entropy of an ideal gas as a function of T and P
Calculate the change in entropy of an ideal monatomic gas B in changing from P1 , T1 to
P2 , T2 . What does this indicate about the form of the expression for the molar entropy of
the gas?
A reversible path is
B(T1 , P1 ) y B(T2 , P1 ) y B(T2 , P2 )
For the first step,
⌬S  CP

T2

冮T

1

T
T
dT
 CP ln 2  nCP ln 2
T
T1
T1

For the second step,
⌬S  ⫺nR ln

P2
P1

For the sum of the two steps,

冦 冤冢 冣 冥 冢 冣冧
T2
T1

⌬S  nR ln

5/2

⫺ ln

P2
P1

since CP  5/2 R .
This indicates that the expression for the molar entropy of an ideal monatomic gas is
of the form

冦 冤冢 冣 冥 冢 冣冧

S  R ln

T
T

5/2

⫺ ln

P
P

 const

where T ⬚ is the reference temperature, P ⬚ is the reference pressure, and the constant is a
characteristic of the particular gas. In Chapter 16 on statistical mechanics we will see that
for an ideal monatomic gas the constant term is equal to ⫺1.151 693R  32 R ln Ar when
P ⬚  1 bar, where Ar is the relative atomic mass of the gas. This equation, which will be
derived from statistical mechanics in Section 16.3, is referred to as the Sackur–Tetrode
equation.

Example 3.9 The change in entropy of a monatomic gas when both T and P
are changed
What is the change in molar entropy of helium in the following process?
1 He(298 K, 1 bar) y 1 He(100 K, 10 bar)
For an ideal monatomic gas, CP 
example,
⌬S  52 R ln

5
2R.

Using an equation from the preceding

100
10
⫺ R ln
298
1

 ⫺41.84 J K⫺1 mol⫺1

3.4

ENTROPY CHANGES IN IRREVERSIBLE PROCESSES

To obtain the change in entropy in an irreversible process we have to calculate
⌬S along a reversible path between the initial state and the final state. We have
already illustrated this in Section 3.2 by the expansion of an isolated ideal gas into
a vacuum.

85

86

Chapter 3

Second and Third Laws of Thermodynamics

As a second illustration we will consider the spontaneous (irreversible)
freezing of water below its freezing point (see Fig. 3.6). The freezing of a mole
of supercooled water at ⫺10 ⬚C is an irreversible change, but it can be carried
out reversibly by means of the three steps for which the entropy changes are
indicated:
H2O(l)

H2 O(l, 0 ⬚ C)

q



⌬S ⌬H /T

273 Cliq
⌬S  冮263 T dT

Heat reservoir
at –10 °C

H2 O(s, 0 ⬚ C)


Q

263 C ice
⌬S  冮273 T dT

H2 O(l, ⫺10 ⬚ C)

(3.37)

H2 O(s, ⫺10 ⬚ C)

For the crystallization of liquid water at 0 ⬚C, ⌬H  ⫺6004 J mol⫺1 . The heat
capacity of water may be taken to be 75.3 J K⫺1 mol⫺1 , and that of ice may be
taken to be 36.8 J K⫺1 mol⫺1 over this range. Then the total entropy change
of the water when 1 mol of liquid water at ⫺10 ⬚C changes to ice at ⫺10 ⬚C is
simply the sum of the three entropy changes:
H2O(s)
Heat reservoir
at –10 °C

Figure 3.6 When a mole of liquid
water freezes at ⫺10 ⬚C, its entropy
changes by ⫺20.54 J K⫺1 mol⫺1 ,
corresponding to the increase
in structural order. The entropy
of the reservoir increases by
21.37 J K⫺1 mol⫺1 because of the
heat transferred to it. The system
as a whole increases in entropy, as
expected for an irreversible process
in an isolated system.

⌬S  (75.3 J K⫺1 mol⫺1 ) ln

273
(⫺6004 J mol⫺1 )

263
273 K

 (36.8 J K⫺1 mol⫺1 ) ln

263
273

 ⫺20.54 J K⫺1 mol⫺1

(3.38)

The decrease in entropy corresponds to the increase in structural order when water freezes.
The statement that the entropy of an isolated system increases in a spontaneous process may be illustrated by considering supercooled water at ⫺10 ⬚C
in contact with a large heat reservoir at this temperature. The entropy change
for the isolated system upon freezing includes the entropy change of the reservoir as well as the entropy change of the water. Since the heat reservoir is large,
the heat evolved by the water on freezing is absorbed by the reservoir with
only an infinitesimal change in temperature. The transfer of heat to a reservoir at the same temperature is a reversible process. Since the heat of fusion of
water at ⫺10 ⬚C is ⌬H (263 K)  (75.3 J K⫺1 mol⫺1 )(10 K) ⫺ 6004 J mol⫺1 ⫺
(36.8 J K⫺1 mol⫺1 )(10 K)  ⫺5619 J mol⫺1 , the entropy change of the reservoir
in the transfer of heat is
⌬S 

(5619 J mol⫺1 )
 21.37 J K⫺1 mol⫺1
263 K

(3.39)

The entropy change of the water is ⫺20.54 J K⫺1 mol⫺1 , and the total entropy
change of the system water plus reservoir is
⌬S  (21.37 ⫺ 20.54) J K⫺1 mol⫺1  0.83 J K⫺1 mol⫺1

(3.40)

Thus, the total entropy of the isolated system, including water and reservoir, increases, as required by inequality 3.24.

3.5 Entropy of Mixing Ideal Gases

ENTROPY OF MIXING IDEAL GASES

Figure 3.7 shows that amount n1 of gas 1 at P and T is separated from amount n2
of gas 2 at the same pressure and temperature. When the partition is withdrawn
the gases will diffuse into each other at constant temperature and pressure if they
are ideal gases. To calculate the change in entropy in this irreversible process we
need to find a way to carry it out reversibly. This can be done in two steps. First,
we expand each gas isothermally and reversibly to the final volume V  V1  V2 .
These volumes are not molar volumes, but are actual volumes. When extensive
volumes are used, the ideal gas law is written PV  nRT .
Using equation 3.35, the entropy changes for the two gases are

P, T, V1

P, T, V2

n1

n2
Initial state

P, T, V1 + V2
n1 + n2

V1
n1
 ⫺n1 R ln
 ⫺n1 R ln y1
V
n1  n2

(3.41)

Final state

V2
n2
⌬S2  ⫺n2 R ln
 ⫺n2 R ln
 ⫺n2 R ln y2
V
n1  n2

(3.42)

Figure 3.7 Mixing of ideal gases.
The partition between n1 mol of gas
1 at P and T and n2 mol of gas 2 at
P and T is withdrawn so that the
gases can mix.

where yi is the mole fraction. The entropy of mixing ⌬mix S is the sum of the entropy changes for the two gases:
⌬mix S  ⫺n1 R ln y1 ⫺ n2 R ln y2

(3.43)

In the second step the expanded gases are mixed reversibly at constant volume.
To see how this might be done we have to imagine two semipermeable membranes,
arranged as shown in Fig. 3.8, one of which (represented by dashes) is permeable
only to gas 1, and the other (represented by dots) is permeable only to gas 2. The
membrane that is permeable to gas 1 and an impermeable membrane, which is separated from it by volume V, are moved at the same infinitesimally slow rate to the
left. As shown by the diagram for the intermediate stage in the reversible mixing
process, the 兩 membrane combination has to be moved to the left against a pressure of P1 (to the left of 兩 ) plus P2 (to the left of 兩 ). But the pressure on the
right-hand side of is also P1  P2 . Therefore, no work is required to move the
membrane in this frictionless device. The internal energies of the two ideal gases
are functions only of T (Section 2.6), and so according to the first law no heat is absorbed by the gas in this step. Consequently, there is no entropy change associated
with the second step, and so equation 3.43 gives the total entropy change for the
isothermal mixing of the two ideal gases.
Equation 3.43 can be generalized to
⌬mix S  ⫺R 冱 ni ln yi  ⫺nt R 冱 yi ln yi

Example 3.10 Entropy of mixing ideal gases
What is the entropy of mixing of 1 mol of oxygen with 1 mol of nitrogen at 25 ⬚C, assuming
that they are ideal gases?
Equation 3.44 becomes
 11.526 J K⫺1

You might try other proportions and see what happens.

Gas 2

V

Gas 1

Gas 2

V

(3.44)

Since yi ⬍ 1, ln yi ⬍ 0, and ⌬mix S is always positive. A plot of the entropy of
mixing two ideal gases is shown in Fig. 3.9.

⌬mix S  ⫺(2 mol)(8.3145 J K⫺1 mol⫺1 ) 冢 12 ln 12  12 ln 12 冣

Gas 1

Vacuum

⌬S1  ⫺n1 R ln

Gases
1 and 2

3.5

87

Gases
1 and 2

Vacuum

V

Figure 3.8 Reversible isothermal
mixing of ideal gases 1 and 2 using a
membrane permeable only to gas 1
(shown by dashes), a membrane
permeable to gas 2 (shown by dots),
and an impermeable membrane
(shown by the solid line).

Chapter 3

Second and Third Laws of Thermodynamics
0.7
0.6
0.5

∆Smix /R

88

0.4
0.3
0.2
0.1
0

0

0.2

0.4

0.6

0.8

1

y2

Figure 3.9 Plot of the entropy of mixing of two ideal gases to produce one mole of an
ideal mixture. (See Computer Problem 3.C.)

Example 3.11 The Gibbs paradox
For the mixing of two gases that have the same temperature and pressure, equation 3.43
can be written in the form
⌬mix S  n1 R ln

Vf
V
 n2 R ln f
V1
V2

where V1 and V2 are the volumes of the two gases and Vf  V1  V2 . Suppose that two
equal volumes of the same gas are mixed. What is the change in entropy?
If we apply this equation to the mixing of two equal volumes of the same gas, we obtain
⌬mix S  (n1  n2 )R ln 2
This answer is wrong because there is no change of state in this process, and so ⌬S  0.
This is known as the Gibbs paradox. The answer to this paradox was not properly understood until the development of quantum mechanics. According to quantum mechanics, the
molecules of a single species are indistinguishable; therefore, equation 3.44 does not apply.
If two different species have properties that are very nearly the same, they are still different
from a quantum mechanical point of view, and equation 3.44 does apply.

(a)

3.6

(b)

Figure 3.10 (a) Equilibrium
gaseous system after a hole is
punched in the diaphragm.
(b) Highly improbable state of
the gaseous system after the
hole has been punched.

ENTROPY AND STATISTICAL PROBABILITY

Using the macroscopic approach of thermodynamics, we have found that the
equilibrium state of an isolated system is that state in which the entropy has its
maximum value. From a microscopic point of view, we might expect that the
equilibrium state of an isolated system would be the state with the maximum statistical probability. For simple systems we can use the molecular point of view to
calculate the statistical probabilities of different final states. For example, assume
that one mole of an ideal gas is in a container that is connected to a container of
equal volume through a stopcock; this expansion has already been discussed in
Example 3.3. Actually, it is a little easier to think about the reverse process, and so
we will do that. As shown in Fig. 3.10, we will start with an opening between the
two chambers and ask, what is the statistical probability that all of the molecules
will be in the original chamber? The probability that a particular molecule will be
in the original chamber is 1/2. The probability that two particular molecules will
be in the original chamber is (1/2)2 , and the probability that all the molecules

3.6 Entropy and Statistical Probability

will be in the original chamber is (1/2)N , where N is the number of molecules in
the system. If the system contains a mole of gas molecules, the statistical probability that all the molecules will be in the original chamber is
(1/2)6.022⫻10  e⫺4.174⫻10
23

23

(3.45)

Boltzmann postulated that
S  k ln ⍀

(3.46)

where k is Boltzmann’s constant (R /NA ) and ⍀ is the number of equally probable
microscopic arrangements for the system. This relation can be used to calculate
⌬S for the transformation from an initial state with entropy S and ⍀ equally
probable arrangements to a final state with entropy S ⬘ and ⍀ ⬘ equally probable
arrangements:
⌬S  S ⬘ ⫺ S  k ln(⍀ ⬘/⍀)

(3.47)

It is often difficult to count the number of equally probable arrangements in the
final state and in the initial state, but the ratio ⍀ ⬘/⍀ in this case is equal to the ratio
of the probability that all the molecules are in one chamber to the probability that
they are all in one chamber or the other. The probability that all the molecules
are in the original chamber or the other chamber is, of course, unity. Thus, for
the system in the preceding paragraph, we have already calculated the ratio
of the probabilities that is equal to ⍀ ⬘/⍀ . The change in entropy in going from
the state with the gas distributed between the two chambers to the state with
all the molecules in the original chamber is
⌬S  k ln e⫺4.174⫻10

23

 (1.381 ⫻ 10⫺23 J K⫺1 )(⫺4.174 ⫻ 1023 )
 ⫺5.76 J K⫺1

(3.48)

Since we are considering one mole of an ideal gas, the change in entropy for the
expansion process in Example 3.3 is ⌬S  5.76 J K⫺1 mol⫺1 , in agreement with
the result using Boltzmann’s hypothesis. This confirms that the Boltzmann constant is indeed given by k  R /NA .
If the gas molecules were all to be found in one chamber after having been
distributed between the two chambers, we would say that the second law had been
violated. We have just seen that the probability that such a thing might happen is
not zero. It is, however, so small that we could never expect to be able to observe
all the molecules in one chamber, even for systems containing much, much less
than one mole of gas. If, however, we considered a system of only two molecules,
then we could find both molecules in one chamber with reasonable probability.
This shows that the laws of thermodynamics are based on the fact that macroscopic systems contain very large numbers of molecules.
The equation S  k ln ⍀ embodies an important concept, but it is not used
very often because it is difficult to calculate ⍀ . In Chapter 16 on statistical mechanics we will use other equations to calculate the entropy.
The collision of two gas molecules is reversible in the sense that the reverse
process can also happen. If, after the molecules are moving away from each other,
we could simply reverse the direction of the velocity vectors, the molecules would
move along the same trajectories in the reverse direction. In short, the movie of
the reverse process is just as reasonable as the movie of the forward process. This is
true for both classical mechanics and quantum mechanics. If molecular collisions

89

90

Chapter 3

Second and Third Laws of Thermodynamics

are reversible, then why is the expansion of a gas into a vacuum, or the mixing of
two gases, irreversible? If we could take a movie of the expansion of a gas into a
vacuum that would show the locations of all the molecules, we could tell whether
the movie was being run forward or backward. However, if we were to look at
each molecular collision we would find that each followed the laws of mechanics,
and was reversible. If we were to look at the movie being run backward, we would
feel that it was depicting something that could not happen. But why couldn’t it
happen? As a matter of fact, it could, but only if we could give all the molecules
the positions they have at the end of the movie but then reverse their velocity
vectors. If we then looked at the individual collisions, we would find that they
would take all the molecules to the region from which they had expanded. The
reason this does not happen in real life is that it takes an extraordinarily special
set of molecular coordinates and velocities. This set of coordinates and velocity
vectors is so unlikely that thermodynamics says that the reverse process can never
happen.
Since dS ⱖ dq /T , the entropy is a measure of the flow of heat between a system and its environment. When heat is absorbed by the system from its surroundings, q is positive and the entropy of the system increases. The energy flowing into
the sysem is “dispersed” in the sense that it goes into increasing the energy of various molecular motions in the system. This concept of the dispersal of energy also
applies to the expansion of an ideal gas into a vacuum. In this case q is zero, but
the total energy of the gas is dispersed over a larger volume. Thus entropy is a
measure of the dispersal of energy among the possible microstates of molecules
in a system.
Sometimes entropy is referred to as “disorder,” and a messy desk is referred
to as a state of high entropy. Or shuffling a deck of playing cards is said to result
in an increase in entropy of the cards. But this is misleading from a scientific viewpoint because moving macroscopic objects around does not involve an increase in
entropy.* Another source of confusion about entropy comes from the use of this
term in information theory, which was introduced by Shannon in 1948. The quantity entropy in information theory is not the entropy of thermodynamics because
it does not deal with the transfer of heat and the dispersal of energy among the
microstates of a system.
The concept of temperature is necessarily involved in understanding thermodynamic entropy because it indicates the thermal environment of the particles in a system. These particles are involved in the ever-present thermal
motion that makes spontaneous change possible because it is the mechanism
by which molecules can occupy new microstates when the external conditions are
altered.

Comment:
The entropy of mixing of ideal gases brings up a very important idea, namely, that
some processes happen spontaneously even though they do not reduce the energy
of a system. Our experience with mechanics leads us to expect that if something
happens spontaneously, there is necessarily a decrease in energy. Now we know
that is not true, and we can expect to find chemical reactions that occur because of
the contribution of a positive ⌬S .
*F. L. Lambert, J. Chem. Educ. 79:187–192 (2002).

91

3.7 Calorimetric Determination of Entropies

3.7

CALORIMETRIC DETERMINATION OF ENTROPIES

The entropy of a substance at any desired temperature relative to its entropy
at absolute zero may be obtained by integrating dqrev /T from absolute zero to
the desired temperature. This requires heat capacity measurements down to the
neighborhood of 0 K as well as enthalpy of transition measurements for all transitions in this temperature range. Since measurements of CP cannot be carried
to 0 K, the Debye function (equation 16.104) is used to represent CP below the
temperature of the lowest measurements.
If data on the enthalpy of fusion at the melting point Tm and the enthalpy of
vaporization at the boiling point Tb are available, the entropy at a temperature T
above the boiling point Tb relative to that at 0 K may be calculated from

0

⌬fus H ⬚
CP⬚ (s)

dT 
T
Tm



Tb

Tm

⌬vap H ⬚
CP⬚ (l)

dT 
T
Tb



T

Tb

CP⬚ (g)
dT
T
(3.49)

If there are various solid forms with enthalpies of transition between the forms,
the corresponding entropies of transition would have to be included in this sum.
Heat capacity measurements down to these very low temperatures are made
with special calorimeters in which the substance is heated electrically in a carefully insulated system and the input of electrical energy and the temperature are
measured accurately.
The attainment of very low temperatures in the laboratory involves successive application of different methods. Vaporization of liquid helium (b.p. 4.2 K
at 1 bar) at low pressures produces temperatures down to about 0.3 K. Lower
temperatures may be reached by use of adiabatic demagnetization. A paramagnetic (Section 22.6) salt such as gadolinium sulfate is cooled with liquid helium
in the presence of a strong magnetic field. The salt is thermally isolated from its
surroundings, and the magnetic field is removed. The salt undergoes a reversible
adiabatic process in which the atomic spins become disordered. Since the energy must come from the crystal lattice, the salt is cooled. Temperatures of about
0.001 K may be reached in this way. Adiabatic demagnetization of nuclear spins
can then be used to obtain temperatures of the order of 10⫺6 K.
As an illustration of the determination of the entropy of a substance relative to its entropy at 0 K, the measured heat capacities for SO2 are shown as
a function of T and of log T in Fig. 3.11. Solid SO2 melts at 197.64 K, and the
heat of fusion is 7402 J mol⫺1 . Liquid SO2 vaporizes at 263.08 K at 1.013 25 bar,
and the heat of vaporization is 24 937 J mol⫺1 . The calculation is summarized in
Table 3.1.
Table 3.1

Entropy of Sulfur Dioxide

100

CP/J K–1 mol–1



Tm

80

Liquid
B.P.

60

Solid
M.P.

40

Gas

20
0

0

50 100 150 200 250 300
T/K
(a)

100

CP/J K–1 mol–1

ST⬚ ⫺ S0 ⬚ 

80

Liquid

40

Solid

Method of Calculation

⌬S ⬚/J K⫺1 mol⫺1

0–15
15–197.64
197.64
197.64–263.08
263.08
263.08–298.15

Debye function (CP  constant T 3 )
Graphical, solid
Fusion, 7402/197.64
Graphical, liquid
Vaporization, 24 937/263.08
From CP of gas

1.26
84.18
37.45
24.94
94.79
5.23

S ⬚ (298.15 K) ⫺ S ⬚(0 K)  247.85

M.P.
Gas

20

T /K

B.P.

60

0
1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4 2.6
log (T/K)
(b)

Figure 3.11 Heat capacity of sulfur dioxide at a constant pressure
of 1 bar at different temperatures.
[Graph a redrawn from W. F.
Giauque and C. C. Stephenson,
J. Am. Chem. Soc. 60:1389 (1938).]

92

Chapter 3

Second and Third Laws of Thermodynamics

3.8


ST – S0

∆ tr S
Monoclinic

Rhombic

0

368.5
T/K

Figure 3.12 Molar entropies of
monoclinic and rhombic sulfur from
absolute zero to the transition point
(368.5 K).

THE THIRD LAW OF THERMODYNAMICS

In the early years of the twentieth century T. W. Richards and W. Nernst independently studied the entropy changes of certain isothermal chemical reactions and found that the change in entropy approached zero as the temperature
was reduced. The entropy change for a chemical reaction cannot be determined
calorimetrically using ⌬S  qrev /T because reactions do not occur reversibly.
We will see in Chapter 5 how the entropy change for a chemical reaction may
be determined. In the meantime we can discuss the measurement of the change
in entropy for a phase change for a single substance. As an example we will
consider the phase change
S(rhombic)  S(monoclinic)

(3.50)

As for chemical reactions, the change in entropy for a phase change approaches
zero as the temperature is reduced to absolute zero.
Figure 3.12 shows the molar entropies of monoclinic and rhombic sulfur
down to absolute zero as determined from
ST ⫺ S0 



T

0

CP
dT
T

(3.51)

Rhombic sulfur is the stable form below the transition temperature of 368.5 K.
However, monoclinic sulfur may be supercooled below this temperature, and its
heat capacity may be measured down to the neighborhood of absolute zero.
The entropy change for the phase change (equation 3.50) at the transition
temperature calculated on the assumption that S 0 is zero for both forms is
1.09 J K⫺1 mol⫺1 . This is in agreement with the entropy change at 368.5 K calculated from the difference in enthalpy of 401 J mol⫺1 between the two forms
at 368.5 K:
⌬tr S 

401 J mol⫺1
 1.09 J K⫺1 mol⫺1
368.5 K

where ⌬tr S is the entropy change for the transition per mole. The difference in
enthalpy between the two forms is simply equal to the difference between their
heats of combustion at 368.5 K.
It can be seen from Fig. 3.12 that as T y 0, ⌬tr S y 0. It has been found that
⌬tr S y 0 as T y 0 for many other isothermal phase transitions and chemical
reactions. In 1905 these observations led Nernst to the conclusion that as temperature approaches 0 K, ⌬r S for all reactions approaches zero.
lim ⌬r S  0

T y0

(3.52)

In 1913 Max Planck took this idea one step further, and we will take his statement as the third law of thermodynamics: The entropy of each pure element or
substance in a perfect crystalline form is zero at absolute zero.
We will see later (Chapter 16) that statistical mechanics gives a reason for
picking this value. As the derivation in Section 3.6 suggests, this corresponds
to a single quantum state (⍀  1) for a perfect crystal at absolute zero. Thus,
according to the third law, S0 ⬚ of the preceding section can be taken as zero if
the substance has a perfect crystalline form in the neighborhood of absolute zero.
Heat capacity measurements down to temperatures of nearly 0 K are therefore

3.8 The Third Law of Thermodynamics

often said to yield “third-law entropies.” Thus, the calculation of the preceding
section yielded the third-law entropy of SO2 (g) at 298.15 K.
Third-law entropies can be tested against what we would expect from two
other types of measurements, namely measurements of equilibrium constants
and of spectroscopic data. As we will see in Chapter 5, if the equilibrium constant
for a reaction is measured over a range of temperatures, then both ⌬H ⬚ and ⌬S ⬚
can be calculated. This ⌬S ⬚ can be compared with the value expected from the
third law if the heat capacities of all reactants and products have been measured
down to the neighborhood of absolute zero.
In Chapter 16, on statistical mechanics, we will see that the entropies of relatively simple gases at any desired temperature may be calculated from molar
masses and certain spectroscopic information. The molar entropy of a gas at a
certain temperature calculated using statistical mechanics may be compared with
the molar entropy obtained from calorimetric measurements, assuming that the
entropy of the pure crystalline substance is zero at absolute zero.
In general the tests of the third law described in the preceding two paragraphs confirm the third law, but there are some apparent violations. For example, the entropy of N2 O(g) at 298.15 K determined from heat capacity measurements is 5.8 J K⫺1 mol⫺1 smaller than that calculated from spectroscopic data.
This indicates an entropy of 5.8 J K⫺1 mol⫺1 at absolute zero for crystals of N2 O.
This is an asymmetric linear molecule, NNO. The residual entropy of the crystal is due to disorder in the arrangement of N2 O molecules. In solid N2 O, the
molecules are arranged with random head–tail alignments (such as NNO, ONN,
NNO, NNO, ONN) instead of being perfectly ordered (NNO, NNO, NNO, NNO,
NNO). If the orientation were perfectly random, the crystal might be regarded
as a mixed crystal with equal mole fractions of NNO and ONN. The entropy of
the mixed crystal would then be the entropy change of mixing. Using equation
3.44, which applies to ideal crystals as well as ideal gases, we see that
⌬mix S  ⫺R

冢 12 ln 12  12 ln 12 冣  5.76 J K⫺1 mol⫺1

(3.53)

Therefore, the statistical mechanical value is taken as the correct entropy, and
this is the value that will be found in tables. Thus the apparent violation of the
third law is understood and can be calculated for ideal crystals.
Another example of an imperfect crystal from the standpoint of the third law
is H2 O. Crystals of H2 O have a residual entropy of 3.35 J K⫺1 mol⫺1 at 0 K. In
ice, the hydrogen atoms are arranged around each oxygen atom in a tetrahedral
manner. Two of the four atoms are covalently bonded to that oxygen atom, and
two are hydrogen bonded (Section 11.10) to that oxygen atom. Since the arrangements of the two types of hydrogen atoms around the oxygen atoms in
the crystal are random, it may be shown that the entropy of the crystal should
approach R ln 32  3.37 J K⫺1 mol⫺1 at absolute zero.
There are two types of randomness in crystals at absolute zero that are not
considered in calculating entropies at absolute zero if these entropies are to be
used only for chemical purposes:
1.

2.

Most crystals are made up of a mixture of isotopic species, but the entropy of
mixing isotopes is ignored because the reactants and products in a reaction
or phase change contain the same mixtures of isotopes.
There is a nuclear spin degeneracy at absolute zero that is ignored because
it exists in both reactants and products.

93

94

Chapter 3

Second and Third Laws of Thermodynamics

There is a corollary to the third law that is very much in the spirit of the
Clausius statement of the second law, since it states an impossibility. According
to this corollary, it is impossible to reduce the temperature of a system to 0 K in
a finite number of steps. This conclusion that absolute zero is unattainable may
be derived from the third law.
The third law is important because it makes possible the calculation of the
equilibrium constant for a chemical reaction purely from calorimetric measurements on the reactants and products. The entropies of a number of substances at
298.15 K are given in Table C.2. The entropies of a smaller number of substances
are given in Table C.3 for 0, 298, 500, 1000, 2000, and 3000 K. These values and
others in the much larger tables from which they have been taken come from
four sources:
1.
2.
3.
4.

Heat capacities and enthalpies as a function of temperature
Statistical mechanical calculations using molecular structure and energy
levels
Temperature variation of equilibrium constants (Section 5.5)
Calculations from enthalpies and Gibbs energies from other sources such as
electromotive force measurements (Section 7.6)

Where the entropies have been determined calorimetrically, corrections have
been made for imperfections of crystals encountered in the neighborhood of absolute zero in the few cases where this occurs and for gas nonideality at 1 bar.
The standard states are, of course, the same as discussed in Section 2.11. The
entropy of H(ao) is arbitrarily assigned the value of zero, and this makes it
possible to calculate entropies of other aqueous ions. Some ions have negative
entropies because of this arbitrary convention.
In comparing standard entropy values from various sources it is important
to be aware of the standard pressure used. The adjustment of standard entropies
from a standard state pressure of 1 atm to 1 bar is discussed in Problem 3.35.

3.9

SPECIAL TOPIC: HEAT ENGINES

A heat engine is an engine that uses heat to generate mechanical work by carrying a “working substance” through a cyclic process. The arrangements for a
Carnot heat engine are shown in Fig. 3.13. The arrows indicate that in one cycle
this engine receives heat 兩q1 兩 from the high-temperature reservoir, rejects heat
兩q2 兩 to the low-temperature reservoir, and does work 兩w 兩 on its surroundings. The
absolute value signs are used because the signs of these algebraic quantities are
set by conventions that require in this case that q1 is positive, q2 is negative, and
w is negative. In the operation of the Carnot heat engine a working fluid, which
we will refer to as a gas, is taken through a sequence of four steps that return it
to its initial state. The engine itself consists of an idealized cylinder with a piston
that can slide without friction and can do work on the surroundings or have the
surroundings do work on the gas. The temperatures of the two reservoirs are
represented by T1 and T2 .
The cycle for the Carnot heat engine consists of the following four steps,
which are represented in Fig. 3.14:

95

3.9 Special Topic: Heat Engines

Heat reservoir at
high temperature T1
 q1
Engine
containing
working
fluid

 w

Surroundings

 q2

Heat reservoir at
low temperature T2

Figure 3.13 Carnot heat engine.

1.

Reversible isothermal expansion of the gas from state A to state B. During this
expansion step, the piston does work 兩w1 兩 on the surroundings, and heat 兩q1 兩 is
absorbed by the gas from the high-temperature reservoir. Note that according
to the convention that w is work done on the gas, w1 is negative. To ensure
that the heat transfer is reversible, the temperature of the high-temperature
reservoir is only infinitesimally higher than the temperature T1 of the gas.
Reversible adiabatic expansion of the gas from state B to state C. For this
step we assume that the piston and cylinder are thermally insulated so that
no heat is gained or lost. The expansion continues until the temperature of
the gas has dropped to T2 . During this expansion step, the piston does work
兩w2 兩 on the surroundings.
Reversible isothermal compression of the gas from state C to state D. During
this compression step the surroundings do work 兩w3 兩 on the gas, and heat 兩q2 兩
flows out of the gas to the low-temperature heat reservoir. Note that according to the convention that q is heat absorbed by the gas, q2 is negative. The
temperature of the low-temperature reservoir is only infinitesimally lower
than the temperature T2 of the gas so that the heat transfer is reversible.
Reversible adiabatic compression of the gas from state D to state A. This
step completes the cycle by bringing the gas back to its initial state at temperature T1 . During this compression step, the surroundings do work 兩w4 兩
on the gas, but no heat is gained or lost.

2.

3.

4.

| q1|

A

B

T1

P

A first-law analysis of a Carnot cycle can be summarized as follows:
Step
Step
Step
Step

1.
2.
3.
4.

D

Isothermal expansion, ⌬U1  q1  w1
Adiabatic expansion, ⌬U2  w2
Isothermal compression, ⌬U3  q2  w3
Adiabatic compression, ⌬U4  w4

| q2|
V

The change in the internal energy of the gas for the cycle ⌬Uc y is the sum of
the changes in the four steps, and it is equal to zero because the gas is returned
to its initial state. Thus,
⌬Ucy  0  (q1  q2 )  (w1  w2  w3  w4 )
 qcy  wcy

T2
C

(3.54)

Figure 3.14 Plot of P versus V for
the working fluid in a Carnot engine. Heat 兩q1 兩 is absorbed in the
isothermal expansion compression
at T1 , and heat 兩q2 兩 is evolved in the
isothermal compression at T2 . The
other two steps are adiabatic.

96

Chapter 3

Second and Third Laws of Thermodynamics

Thus, the work 兩wcy 兩 done by the engine in a cycle is equal to the difference
between the “heat in” 兩q1 兩 and the “heat out” 兩q2 兩:
⫺wcy  qcy  q1  q2
兩wcy 兩  兩q1 兩 ⫺ 兩q2 兩

(3.55)
(3.56)

The engine absorbs heat q1 from the high-temperature reservoir and does work
on the surroundings so that w is negative, but the first law does not tell us the
relative amounts of work and of heat rejected at T2 .
In practice, the low-temperature reservoir of a heat engine is often the atmosphere, so that the economic cost of producing work in the surroundings is
mainly that of supplying q1 . The efficiency ⑀ of a heat engine is defined as the
ratio of the work done on the surroundings to the heat input at the higher temperature. Thus, the efficiency is defined by

⑀

⫺wcy
q1  q2

q1
q1

(3.57)

where the negative sign is required by the fact that wcy is negative and the efficiency is, of course, positive. It is convenient to express the efficiency in terms of
the magnitudes of the heats,

⑀ 1⫺

兩q2 兩
兩q1 兩

(3.58)

because this reminds us that 0 ⬍ ⑀ ⬍ 1.
From equation 3.58 it is clear that to improve the efficiency of a Carnot heat
engine one would like to reduce the magnitude of the heat rejected to the cold
reservoir, 兩q2 兩. But it has been found impossible to reduce 兩q2 兩 to zero.
Kelvin showed that a temperature scale could be set up by taking the temperature of a heat reservoir in a reversible Carnot engine to be proportional to
the heat transferred in a cycle. Thus
qh
Th

qc
Tc

(3.59)

where h indicates the hot reservoir and c indicates the cold reservoir, and the
heats are absolute values. This equation can be derived by writing the equation
for the cyclic integral of the entropy for the Carnot cycle. Figure 3.14 shows that
this cyclic integral is given by
qh
qc
0

Th
Tc

(3.60)

since there are no changes in entropy in the adiabatic steps. The temperatures in
Fig. 3.14 are based on the ideal gas scale, but since equations 3.59 and 3.60 are
the same, these two scales agree for an ideal gas. The advantage of the Kelvin
definition of the temperature scale, however, is that a reversible Carnot cycle
can be carried out with any fluid and so the definition is not dependent on the
special properties of an ideal gas. Because of his contributions in establishing
the absolute temperature scale, the unit of temperature is named the kelvin. By
taking T  273.16 K at the triple point of water, the Kelvin scale becomes iden-

3.9 Special Topic: Heat Engines

tical with the ideal gas scale. However, the ideal gas scale is more practical for laboratory use and is used in establishing secondary standards for
temperature.
Note that when equation 3.59 is used in equation 3.58, we find that the efficiency ⑀ of a heat engine is given by

⑀ 1⫺

T2
T1

(3.61)

Thus, high efficiencies are obtained if the ratio T2 /T1 is small; in practice T2 is
generally close to room temperature and so T1 is made as high as possible.


1.

2.

3.

4.
5.

6.

7.
8.

Eight Key Ideas in Chapter 3
Since the first law provides no information as to whether a process or chemical reaction can proceed spontaneously or not, thermodynamics needs a
state function that can be used for this purpose. We know that when there
is a temperature difference in a system, heat is transferred, but heat is not
a state function. However, the differential heat divided by temperature is
an exact differential, and so dq /T is the differential of a state function. This
state function is the entropy S, and so 冱 ⌬S  0 around a closed cycle,
where ⌬S is the entropy change in a step.
According to the second part of the second law, the change in entropy in
any process is given by dS ⱖ dq /T , where the inequality applies to a spontaneous process and the equality applies to a reversible process. This means
that the entropy increases in a spontaneous process in an isolated system.
Entropy changes can be determined using reversible processes, like the vaporization of a liquid, the expansion of a gas, or the heating of a gas, liquid,
or solid. To obtain the change in entropy in an irreversible process, we have
to calculate ⌬S along a reversible path.
When ideal gases are mixed, there is an entropy of mixing that is always
positive.
The entropy of a system is a measure of the dispersal of energy among the
molecules in microstates. The relation between the entropy of a system and
its microstates will be discussed in Chapter 16, on statistical mechanics.
The entropy of a substance at any desired temperature relative to its entropy
at absolute zero can be obtained by integrating dqrev /T from absolute zero
to the desired temperature.
According to the third law of thermodynamics, the entropy of each pure
element or substance in a perfect crystalline form is zero at absolute zero.
Historically the second law was first discovered in considerations of the efficiencies of heat engines. The maximum efficiency of a heat engine in converting heat to work is given by 1 ⫺ T2 /T1 , where T1 is the temperature
of the high-temperature reservoir and T2 is the temperature of the lowtemperature reservoir. Kelvin showed that a temperature scale can be set
up by taking the temperature of a heat reservoir in a reversible Carnot engine to be proportional to the heat transferred in a cycle. This has the advantage over the ideal gas temperature scale introduced in Chapter 1 that
it is independent of the working substance.

97

98

Chapter 3

Second and Third Laws of Thermodynamics

REFERENCES
M. Bailyn, A Survey of Thermodynamics. New York: American Institute of Physics, 1994.
K. E. Bett, J. S. Rowlinson, and G. Saville, Thermodynamics for Chemical Engineers.
Cambridge, MA: MIT Press, 1975.
H. B. Callen, Thermodynamics and an Introduction to Thermostatistics, 2nd ed. Hoboken,
NJ: Wiley, 1985.
J. W. Gibbs, The Collected Works of J. Willard Gibbs. New Haven, CT: Yale University
Press, 1948.
K. S. Pitzer, Thermodynamics, 3rd ed. New York: McGraw-Hill, 1995.
J. M. Smith, H. C. Van Ness, and M. M. Abbott, Introduction to Chemical Engineering
Thermodynamics, 5th ed. New York: McGraw-Hill, 1996.
J. W. Tester and M. Modell, Thermodynamics and Its Applications. Upper Saddle River,
NJ: Prentice-Hall, 1997.

PROBLEMS
Problems marked with an icon may be more conveniently solved on a personal computer with a mathematical program.
3.1 Show that (⭸CV /⭸V )T  0 for an ideal gas, a gas following P  nRT /(V ⫺ nb ), and a van der Waals gas.
3.2 Show that qrev is not a state function for a gas obeying
the equation of state P (V ⫺ nb )  nRT , but that qrev /T is.
3.3 Show that qrev is not an exact differential for a gas obeying the van der Waals equation, but that qrev /T is.
3.4 An ideal gas initially at P1 , V1 , T1 undergoes a reversible
isothermal expansion to P2 , V2 , T1 . The same change in state
of the gas can be accomplished by allowing it to expand adiabatically to P3 , V2 , T2 and then heating it at constant volume
to P2 , V2 , T1 . Show that the entropy change for the reversible
isothermal expansion is the same as the sum of the entropy
changes in the reversible adiabatic expansion and the reversible heating to P2 , V2 , T1 . This shows that ⌬S is independent of path and is therefore a state function.
3.5 Water is vaporized reversibly at 100 ⬚C and 1.013 25 bar.
The heat of vaporization is 40.69 kJ mol⫺1 . (a) What is the
value of ⌬S for the water? (b) What is the value of ⌬S for
the water plus the heat reservoir at 100 ⬚C?
3.6 Assuming that CO2 is an ideal gas, calculate ⌬H ⬚ and
⌬S ⬚ for the following process:
1 CO2 (g, 298.15 K, 1 bar) y 1 CO2 (g, 1000 K, 1 bar)
Given: CP⬚  26.648  42.262 ⫻ 10⫺3 T ⫺ 142.4 ⫻ 10⫺7 T 2 in
J K⫺1 mol⫺1 .
3.7 The temperature of an ideal monatomic gas is increased
from 300 to 500 K. What is the change in molar entropy of the
gas (a) if the volume is held constant and (b) if the pressure is
held constant?

3.8 Ammonia (considered to be an ideal gas) initially at 25 ⬚C
and 1 bar pressure is heated at constant pressure until the volume has trebled. Calculate (a) q per mole, (b) w per mole,
(c) ⌬H, (d) ⌬U , and (e) ⌬S . Given: CP  25.895  32.999 ⫻
10⫺3 T ⫺ 30.46 ⫻ 10⫺7 T 2 in J K⫺1 mol⫺1 .
3.9 Two blocks of the same metal are the same size but are at
different temperatures, T1 and T2 . These blocks of metal are
brought together and allowed to come to the same temperature. Show that the entropy change is given by
⌬S  CP ln

(T1  T2 )2
4T1 T2

if CP is constant. How does this equation show that the change
is spontaneous?
3.10 In the reversible isothermal expansion of an ideal gas at
300 K from 1 to 10 L, where the gas has an initial pressure
of 20.27 bar, calculate (a) ⌬S for the gas and (b) ⌬S for all
systems involved in the expansion.
3.11 A mole of oxygen is expanded reversibly from 1 to 0.1
bar at 298 K. What is the change in entropy of the gas, and
what is the change in entropy for the gas plus the heat reservoir with which it is in contact?
3.12 Three moles of an ideal gas expand isothermally and reversibly from 90 to 300 L at 300 K. (a) Calculate ⌬U, ⌬S, w,
and q for this system. (b) Calculate ⌬U, ⌬S, w per mole, and
q per mole. (c) If the expansion is carried out irreversibly by
allowing the gas to expand into an evacuated container, what
are the values of ⌬U, ⌬S, w per mole, and q per mole?
3.13 (a) A system consists of a mole of ideal gas that undergoes the following change in state:
1X(g, 298 K, 10 bar) y 1X(g, 298 K, 1 bar)

Problems
What is the value of ⌬S if the expansion is reversible? What is
the value of ⌬S if the gas expands into a larger evacuated container so that the final pressure is 1 bar? (b) The same change
in state takes place, but we now consider the gas plus the heat
reservoir at 298 K to be our system. What is the value of ⌬S
if the expansion is reversible? What is the value of ⌬S if the
gas expands into a larger container so that the final pressure is
1 bar?
3.14 An ideal gas at 298 K expands isothermally from a pressure of 10 bar to 1 bar. What are the values of w per mole, q
per mole, ⌬U, ⌬H, and ⌬S in the following cases? (a) The expansion is reversible. (b) The expansion is free. (c) The gas and
its surroundings form an isolated system, and the expansion is
reversible. (d) The gas and its surroundings form an isolated
system, and the expansion is free.
3.15 An ideal monatomic gas is heated from 300 to 1000 K
and the pressure is allowed to rise from 1 to 2 bar. What is the
change in molar entropy?
3.16 The purest acetic acid is often called glacial acetic acid
because it is purified by fractional freezing at its melting point
of 16.6 ⬚C. A flask containing several moles of acetic acid
at 16.6 ⬚C is lowered into an ice–water bath briefly. When
it is removed, it is found that exactly 1 mol of acetic acid
has frozen. Given: ⌬fus H (CH3 CO2 H)  11.45 kJ mol⫺1 and
⌬fus H (H2 O)  5.98 kJ mol⫺1 . (a) What is the change in entropy of the acetic acid? (b) What is the change in entropy
of the water bath? (c) Now consider that the water bath and
acetic acid are in the same system. What is the entropy change
for the combined system? Is the process reversible or irreversible? Why?
3.17 In Problem 2.21 one mole of an ideal monatomic gas at
1 bar and 300 K was expanded adiabatically against a constant
pressure of 12 bar until the final pressure was 12 bar; a temperature of 240 K was reached. What is the value of ⌬S for this
process?
3.18 Ten moles of H2 and two moles of D2 are mixed at 25 ⬚C
and 1 bar. What is the value of ⌬S ⬚ ? Assume ideal gases.
3.19 (a ) Write the expression for the entropy of a mixture
of ideal gases A, B, and C at T and P using yi for the mole
fraction of gas i . (b ) Now let us carry out the combination of
terms contributing to the entropy of the system in two steps.
First, imagine that gases A and C are mixed to form a mixture
with mole fractions rA and rC within the A–C mixture, but
that B remains unmixed. Write the equation for the entropy of
the system with two terms, one for the nI  nA  nC moles
of the A plus C mixture, which contributes pressure PI  PA
PC , and the other nB SB . (c ) Second, imagine that B is mixed
with mixture I, considered as one species, and show that this
equation is the same as that obtained in (a ).
3.20 One mole of A at 1 bar and one mole of B at 2 bar are
separated by a partition and surrounded by a heat reservoir.
When the partition is withdrawn, how much does the entropy
change?

99

3.21 Use the microscopic point of view of Section 3.6 to
show that for the expansion of amount n of an ideal gas by a
factor of 2, ⌬S  nR ln 2. In this expression S is an extensive
property.
3.22 Calculate the change in molar entropy of aluminum that
is heated from 600 to 700 ⬚C. The melting point of aluminum
is 660 ⬚C, the heat of fusion is 393 J g⫺1 , and the heat capacities of the solid and the liquid may be taken as 31.8 and
34.4 J K⫺1 mol⫺1 , respectively.
3.23 Steam is condensed at 100 ⬚C, and the water is cooled
to 0 ⬚C and frozen to ice. What is the molar entropy change
of the water? Consider that the average specific heat of liquid
water is 4.2 J K⫺1 g⫺1 . The enthalpy of vaporization at the
boiling point and the enthalpy of fusion at the freezing point
are 2258.1 and 333.5 J g⫺1 , respectively.
3.24 Calculate the molar entropy of carbon disulfide at
25 ⬚C from the following heat capacity data and the heat of
fusion, 4389 J mol⫺1 , at the melting point (161.11 K):
T /K
CP /J K⫺1 mol⫺1

15.05
6.90

20.15
12.01

29.76
20.75

42.22
29.16

T /K
CP /J K⫺1 mol⫺1

57.52
35.56

75.54
40.04

89.37
43.14

99.00
45.94

T /K
CP /J K⫺1 mol⫺1

108.93
48.49

119.91
50.50

131.54
52.63

156.83
56.62

161–298
75.48

3.25 Ten grams of molecular hydrogen at 1 bar expand to
triple the volume (a) isothermally and reversibly and (b)
adiabatically and reversibly. In each case what are ⌬S(H2 ),
⌬S(surr), and ⌬S(H2  surr)?
3.26 Theoretically, how high could a gallon of gasoline lift an
automobile weighing 2800 lb against the force of gravity, if it
is assumed that the cylinder temperature is 2200 K and the
exit temperature 1200 K? (Density of gasoline  0.80 g cm⫺3 ;
1 lb  453.6 g; 1 ft  30.48 cm; 1 L  0.2642 gal; heat of
combustion of gasoline  46.9 kJ g⫺1 .)
3.27 (a) What is the maximum work that can be obtained
from 1000 J of heat supplied to a steam engine with a hightemperature reservoir at 100 ⬚C if the condenser is at 20 ⬚C?
(b) If the boiler temperature is raised to 150 ⬚C by the use of
superheated steam under pressure, how much more work can
be obtained?
3.28 The term adiabatic lapse rate used by meteorologists
is the decrease in temperature with height that results from
the adiabatic expansion of an air mass as it is pushed up a
mountain by the wind. Similarly, the wind coming down
the mountain slope warms up. This adiabatic expansion is represented by
P0 , V0 , T0 y P, V, T

⌬S3  0

where P0 , V0 , T0 represents the sea level conditions. The calculation of the entropy change can be carried out in two steps

100

Chapter 3

Second and Third Laws of Thermodynamics

(the first at constant pressure, the second at constant temperature):
P 0 , V0 , T 0 y P 0 , V ⴱ , T
P0 , V ⴱ , T y P, V, T

T
T0
P
⌬S2  nR ln 0
P
⌬S1  nCP ln

We have seen in Section 1.11 that the pressure of the atmosphere drops off exponentially if temperature is independent
of height h . (a ) Since ⌬S1 ⌬S2  ⌬S3 , what is the expression
for ⌬T /⌬h ? (b ) If the temperature at the foot of a 14,000-foot
mountain is 25 ⬚ C, what temperature would you expect at the
summit from this adiabatic lapse rate? For this calculation
the molar mass of air can be taken to be M  29 g mol⫺1 and
its heat capacity can be taken as 29.1 J K⫺1 mol⫺1 .
3.29 When an ideal gas is allowed to expand isothermally in
a piston, ⌬U  q  w  0. Thus, the work done by the system on the surroundings is equal to the heat transferred from
the reservoir to the gas, and the efficiency of turning heat into
work is 100%. Explain why this is not a violation of the second
law.
3.30 In Problem 2.50 we found that when an ideal monatomic
gas at 1 bar and 273.15 K is allowed to expand adiabatically against a constant pressure of 0.395 bar until it reaches
equilibrium, the final temperature is 207.04 K. What is the
value of ⌬S for this process? Given: CV  32 R and therefore
CP  52 R .
3.31 Compare the entropy difference between 1 mol of liquid water at 25 ⬚C and 1 mol of vapor at 100 ⬚C and 1.013 25
bar. The average specific heat of liquid water may be taken as
4.2 J K⫺1 g⫺1 , and the heat of vaporization is 2259 J g⫺1 .
3.32 Calculate the increase in the molar entropy of nitrogen
when it is heated from 25 to 1000 ⬚C (a) at constant pressure
and (b) at constant volume. Given: CP  26.9835 
5.9622 ⫻ 10⫺3 T ⫺ 3.377 ⫻ 10⫺7 T 2 in J K⫺1 mol⫺1 .
3.33 Assuming that the heat capacity of water is independent of temperature, calculate the net change in entropy
when 1 mol of water at 0 ⬚C is mixed with 1 mol of water at 100 ⬚C. Assume that the heat capacity of water is
(4.184 J K⫺1 g⫺1 )(18 g mol⫺1 )  75.3 J K⫺1 mol⫺1 and that
the heat capacity of the calorimeter is negligible.
3.34 In Section 3.3 we derived the expression
S2 ⫺ S1  nR ln

V2
V1

for the reversible isothermal expansion of an ideal gas. We
ought to derive this same expression by another reversible
path. Do this by imagining that the gas is first expanded adiabatically and reversibly from V1 to V2 . Since the temperature falls from T to T ⬘, heat must be added reversibly and at
constant volume in a second step to restore the temperature
to T .

3.35 Show that the standard molar entropy S ⬚ (T ) for an
ideal gas at 1 bar can be calculated from the standard molar
entropy S ⴱ (T ) at 1 atm using
S ⬚ (T ) ⫺ S ⴱ (T )  R ln

Pⴱ
P⬚

 0.109 J K⫺1 mol⫺1
The conversion for liquids and solids is negligible. Why?
3.36 Argon undergoes the following change in state:
P1  20 bar, T1  300 K y P2  1 bar, T2  200 K
What is the change in molar entropy, assuming that argon is
an ideal gas?
3.37 A flask containing several moles of liquid benzene at
its freezing temperature (5.5 ⬚C) is placed in thermal contact with an ice–water bath. When the flask is removed
from the ice–water bath, it is found that 1 mol of benzene has frozen. Given: ⌬fus H (C6 H6 )  9.87 kJ mol⫺1 ,
⌬fus H (H2 O)  5.98 kJ mol⫺1 . (a) What is the change in
entropy of benzene? (b) What is the change in entropy of the
water bath? (c) Considering the flask of benzene and the water
bath as an isolated system, what is the change in entropy of
the isolated system? Is the process reversible or irreversible?
3.38 One mole each of H2 (g), N2 (g), and O2 (g) are mixed at
25 ⬚C. What is ⌬mix S ?
3.39 According to the Debye equation (Section 16.12) the
heat capacity of a solid is proportional to the temperature
cubed at low temperatures:
CP (T )  (const)T 3
Show that the entropy at a temperature T ⬘ is given by
S (T ⬘) 

CP (T ⬘)
3

when the Debye equation holds.
3.40 Calculate the molar entropy of liquid chlorine at its
melting point, 172.12 K, from the following data obtained by
W. F. Giauque and T. M. Powell:
T /K
CP /J K⫺1 mol⫺1

15
3.72

20
7.74

25
12.09

30
16.69

35
20.79

T /K
CP /J K⫺1 mol⫺1

40
23.97

50
29.25

60
33.47

70
36.32

90
40.63

T /K
CP /J K⫺1 mol⫺1

110
43.81

130
47.24

150
51.04

170
55.10

172.12
m.p.

The heat of fusion is 6406 J mol⫺1 . Below 15 K it may be assumed that CP is proportional to T 3 .
3.41 In running a Carnot cycle backward to produce refrigeration, the objective is to remove as much heat 兩q2 兩 from the
cold reservoir as possible for a given amount of work, and so

Problems

101

the coefficient of performance ␤ is defined as
兩q 兩
兩q2 兩
T2
␤ 2 

兩w 兩
兩q1 兩 ⫺ 兩q2 兩
T1 ⫺ T2
A household refrigerator operates between 35 and ⫺10 ⬚C.
How many joules of heat can in principle be removed per
kilowatt-hour of work?
3.42 A heat pump is used to heat a home in the winter when
the temperature in the ground is 0 ⬚C and the temperature of
the radiator in the house is 35 ⬚C. When a Carnot cycle is run
backward for this purpose, the objective is to obtain as much
heat in the radiator as possible for a given amount of electrical
work. The coefficient of performance ␤ ⬘ for a heat pump is
defined by

␤⬘ 

兩q1 兩
兩q1 兩
T1


兩w 兩
兩q1 兩 ⫺ 兩q2 兩
T1 ⫺ T2

What is the minimum amount of electrical work needed to produce a kilowatt-hour of heat?

Computer Problems
3.A Calculate the standard molar entropies at 1000 K for
the eight gases in Computer Problem 2.A using the empirical equations given there for the molar heat capacities as a
function of temperature and the values of the standard molar
entropies at 298.15 K in Table C.2. Compare these calculated
values with values in Table C.3 in Appendix C.
3.B Plot the standard molar entropy of water vapor versus
temperature from 300 to 1000 K by use of the empirical equation for the molar heat capacity of water vapor.
3.C Plot the entropy of mixing of two ideal gases to form a
mole of mixture versus the mole fraction of one of the species.
Investigate the slope of this plot as y2 y 1 and as y2 y 0.
3.D Plot the molar entropy in J K⫺1 mol⫺1 of a monatomic
ideal gas versus pressure P from 1 bar to 100 bar and T from
298.15 K to 500 K, assuming that its molar entropy is zero at
298.15 K and 1 bar.

4

Fundamental Equations
of Thermodynamics

4.1
4.2
4.3
4.4
4.5
4.6
4.7
4.8
4.9

Fundamental Equation for the Internal Energy
Definitions of Additional Thermodynamic Potentials
Using Legendre Transforms
Effect of Temperature on the Gibbs Energy
Effect of Pressure on the Gibbs Energy
Fugacity and Activity
The Significance of the Chemical Potential
Additivity of Partial Molar Properties
with Applications to Ideal Gases
Gibbs–Duhem Equation
Special Topic: Additional Applications
of Maxwell Relations

With the definitions of T, U, and S we have completed the set of necessary thermodynamic properties. Although the entropy provides a criterion of whether a
change in an isolated system is spontaneous, it does not provide a convenient criterion at constant T and V or constant T and P, the usual conditions in the laboratory. We need two additional thermodynamic properties (functions) to make
calculations at constant T and V or constant T and P more convenient than they
are with entropy. Fortunately, there is a general way to do this using Legendre
transforms. These two new thermodynamic properties are the Helmholtz energy
A and the Gibbs energy G. At constant T and V, spontaneous processes occur
with a decrease in A, and at constant T and P, spontaneous processes occur with
a decrease in G.
In order to discuss equilibria in systems with more than two phases and chemical equibrium, we need to introduce the chemical potential ␮i of a species. The
chemical potential of a species is the same in all of the phases of a system at equilibrium. The chemical potential is also the property that determines whether a
species will undergo chemical reaction. There are many relations between the
thermodynamic properties of a system, and the fundamental equations for the

4.1 Fundamental Equation for the Internal Energy

various thermodynamic potentials (U , H , A, and G ) are the source of these very
useful relations. In the next four chapters, these fundamental equations will lead
us to the quantitative treatment of chemical equilibrium, phase equilibrium, electrochemical equilibrium, and biochemical equilibrium.

4.1

FUNDAMENTAL EQUATION FOR THE INTERNAL ENERGY

The first law is given in equation 2.9 as
dU  d q  d w

(4.1)

and the second law is given by equation 3.15, which is
dS ⱖ dq /T

(4.2)

where the inequality applies when dq is for an irreversible process and the equal
sign applies when dq is for a reversible process. If we restrict consideration to
closed systems in which only reversible PV work is involved,
dw  ⫺P dV

(4.3)

dS  dq /T

(4.4)

and equation 4.2 becomes

so that the combined first and second law is
dU  T dS ⫺ P dV

(4.5)

Since equation 4.5 involves only state functions, it applies to both reversible and
irreversible processes. We saw in Table 2.1 that ⫺P and V are conjugate variables.
Now we can see that T and S are also conjugate variables.
In 1876 Gibbs made a very important addition to this equation in order to
discuss phase equilibrium and reaction equilibrium. He introduced the concept
of the chemical potential ␮i of a species and wrote the fundamental equation for
U as
dU  T dS ⫺ P dV  ␮1 dn1  ␮2 dn2  ⭈⭈⭈

(4.6)

where the additional terms are “chemical work” terms for each species and ni is
the amount of species i . The chemical potential is a measure of the potential a
species has to move from one phase to another or undergo a chemical reaction.
Equation 4.6 shows that if more of species i is added to a system at constant S
and V , there is a contribution ␮i dni to the internal energy. Note that for a system
in which there are no chemical reactions, ␮i and ni are conjugate variables. If a
system contains Ns different species, equation 4.6 can be written
Ns

dU  T dS ⫺ P dV  冱 ␮i dni

(4.7)

i 1

This equation shows that U is a function of S , V , and 兵ni 其, where 兵ni 其 is the set
of amounts of species i ; this can be represented by U (S, V, 兵ni 其). In view of the
importance of the variables S , V , and 兵ni 其 in equation 4.7, they are called the
natural variables of U . Note that the natural variables for the interal energy are

103

104

Chapter 4

Fundamental Equations of Thermodynamics

all extensive. The total differential of U can be written in terms of the differentials
of these natural variables.
dU 

⭸U
⭸S

冢 冣

V,兵ni 其

dS 

⭸U
⭸V

Ns

冢 冣

dV  冱

S,兵ni 其

i 1

⭸U

冢⭸n 冣

dni

(4.8)

i S,V,nj 苷ni

Comparison of equation 4.7 with equation 4.8 shows that
T 

⭸U

冢 ⭸S 冣

P ⫺

␮i 

(4.9)

V,兵ni 其

⭸U

冢⭸V 冣

(4.10)

S,兵ni 其

⭸U

冢⭸n 冣

(4.11)

i S,V,nj 苷ni

where nj 苷 ni means that species other than i are held constant. These three
equations are often referred to as equations of state because they give relations
between state properties. Note that the derivatives of extensive properties with
respect to extensive properties are intensive properties. Equations 4.9 to 4.11 are
very important because they show that if U of a system can be determined as
a function of its natural variables (S , V , and 兵ni 其), all the other thermodynamic
properties of the system—including the chemical potentials of all the species—can
be calculated. In Chapters 1 and 2 we saw that the extensive state of a system can
be described by specifying T , P , and 兵ni 其 or T , V , and 兵ni 其, but now we have to
understand that determination of U as a function of these variables is not enough
to allow us to calculate all the other thermodynamic properties of the system.
A fundamental equation such as 4.7 leads to relations of another type between
thermodynamic properties, which are referred to as Maxwell relations. They are
obtained by equating second cross-partial derivatives. We have already used equations like this in the test for exactness in Section 2.3. The Maxwell equations for
equation 4.7 are
⭸T

冢⭸V 冣

S,兵ni 其

⭸T

冢⭸n 冣

⭸P

冢⭸ S 冣

⭸␮i

冢 ⭸S 冣

冢⭸n 冣



冢 ⭸V 冣

冢 冣



冢 ⭸n 冣

⭸P

i S,V,nj 苷ni

⭸ ␮i
⭸ nj

S,V,ni 苷nj

(4.12)

V,兵ni 其



i S,V,nj 苷ni



⫺

(4.13)

V,兵ni 其

⭸ ␮i

(4.14)

S,兵ni 其

⭸ ␮j
i

(4.15)

S,V,nj 苷ni

This shows that these properties are related in interesting ways.
The internal energy provides a criterion for whether a process can occur spontaneously at constant S , V , and 兵ni 其. If we substitute the second law in the form
dS ⱖ dq /T and dw  ⫺Pext dV  冱 ␮i dni in equation 4.1, we obtain
Ns

dU ⱕ T dS ⫺ Pext dV  冱 ␮i dni
i 1

(4.16)

4.2 Definitions of Additional Thermodynamic Potentials Using Legendre Transforms

Thus if an infinitesimal change takes place in a system of constant entropy, volume,
and 兵ni 其,
(dU )S,V,兵ni 其 ⱕ 0

(4.17)

This is the criterion for spontaneous change and equilibrium in the system involving PV work and specified amounts of Ns species. At equilibrium, U at constant
S , V , and 兵ni 其 must be at a minimum.
At constant values of the intensive properties T, P, and ␮i , equation 4.7 can
be integrated to obtain
Ns

U  TS ⫺ PV  冱 ␮i ni

(4.18)

i 1

This equation can also be considered as a consequence of Euler’s theorem (see
Section 1.10).
The equations in this section are not very useful in the laboratory because the
entropy and volume are not easily controlled. Fortunately, however, more useful
thermodynamic properties can be defined, based on the internal energy.

4.2

DEFINITIONS OF ADDITIONAL THERMODYNAMIC
POTENTIALS USING LEGENDRE TRANSFORMS

The internal energy and other thermodynamic properties defined starting with
the internal energy are referred to as thermodynamic potentials. To define new
thermodynamic potentials, we use the method of Legendre transforms. We have
already seen an example of this with the enthalpy H , defined by H  U  PV
(see Section 2.7). We did not emphasize that this is a Legendre transform, but it
is, and now we are going to use two more Legendre transforms to define two thermodynamic potentials. A Legendre transform is a linear change in variables that
starts with a mathematical function and defines a new function by subtracting one
or more products of conjugate variables. This is different from the usual change
in variables in that a partial derivative of a thermodynamic potential becomes
an independent variable in the new thermodynamic potential. As explained in
Appendix D.5, no information is lost in this process. Thus a new thermodynamic
potential defined in terms of the internal energy contains all of the information
that is in U (S, V, 兵ni 其).*
Now we can make a more complete treatment of the enthalpy H than in Section 2.7. The total differential of the enthalpy is
dH  dU  P dV  V dP

(4.19)

Substituting equation 4.7 for dU yields the fundamental equation for the enthalpy
of a system involving pressure–volume work and Ns different species:
Ns

dH  T dS  V dP  冱 ␮i dni

(4.20)

i 1

*H. B. Callen, Thermodynamics, 2nd ed. Hoboken, NJ: Wiley, 1985; R. A. Alberty, J. M. G. Barthel,
E. R. Cohen, R. N. Goldberg, and E. Wilhelm, Use of Legendre Transforms in Chemical Thermodynamics, (an IUPAC Technical Report), Pure Appl. Chem. 73, 8 (2001).

105

106

Chapter 4

Fundamental Equations of Thermodynamics

Thus we see that
⭸H

冢 ⭸S 冣
⭸H
V 冢
⭸P 冣
⭸H
␮ 冢
⭸n 冣
T 

(4.21)

P,兵ni 其

(4.22)

S,兵ni 其

(4.23)

i

i S,P,兵nj 苷i 其

If H can be determined as a function of S, P, and the amounts of all the species,
then T, V, and ␮i can be determined by taking partial derivatives of H. Thus
all the thermodynamic properties of a system can in principle be obtained using H, just as all the thermodynamic properties of the system were determined
using U in the preceding section. This shows that no information is lost in
making a Legendre transform. Equation 4.20 provides a number of Maxwell
equations.
The enthalpy provides a criterion for whether a process can occur spontaneously for a system at constant S, P, and ni . Following the same reasoning we
used to obtain equation 4.17 we obtain
(dH )S,P,兵ni 其 ⱕ 0

(P  Pext )

(4.24)

This shows that a change can take place spontaneously at constant entropy, pressure, and amounts of species if the enthalpy decreases.
At constant values of T, P, and ␮i , equation 4.20 can be integrated to obtain
N

H  TS  冱 ␮i ni

(4.25)

i 1

Note that using H  U  PV , we can obtain equation 4.18.
The internal energy and the enthalpy do not provide very useful criteria for
spontaneous change because the entropy has to be held constant. This problem is
avoided by the use of Legendre transforms in which the product TS of conjugate
variables is subtracted from the internal energy and the enthalpy. This introduces
the intensive variable T as a natural variable in place of the extensive variable S.
The two Legendre transforms that define the Helmholtz energy A and the
Gibbs energy G are*
A  U ⫺ TS

(4.26)

G  U  PV ⫺ TS  H ⫺ TS

(4.27)

The total differential of the Helmholtz energy is given by
dA  dU ⫺ T dS ⫺ S dT

(4.28)

*The Helmholtz energy A is named in honor of Hermann L. F. von Helmholtz. The Gibbs energy
G is named in honor of J. Willard Gibbs of Yale University, whose many important generalizations
in thermodynamics have given him the position of one of the great geniuses of science. The Gibbs
energy is sometimes referred to as the Gibbs free energy or the free energy. See J. W. Gibbs, Trans.
Conn. Acad. Arts Sci., 1876–1878; The Collected Works of J. Willard Gibbs, Vol. 1. New Haven, CT:
Yale University Press, reprinted 1948.

4.2 Definitions of Additional Thermodynamic Potentials Using Legendre Transforms

107

Substituting equation 4.7 for dU yields the fundamental equation for the Helmholtz
energy:
N

dA  ⫺S dT ⫺ P dV  冱 ␮i dni

(4.29)

i 1

Thus the natural variables for A are T, V, and ni . Thus we see that
⭸A

冢⭸T 冣
⭸A
P  ⫺冢
⭸V 冣
⭸A
␮ 冢
⭸n 冣
S ⫺

(4.30)

V,兵ni 其

(4.31)

T,兵ni 其

(4.32)

i

i T,V,兵nj 苷i 其

Equation 4.29 also provides a number of Maxwell relations.
The Helmholtz energy A provides the criterion for spontaneous change at
specified T, V, and 兵ni 其:
(dA)T,V,兵ni 其 ⱕ 0

(4.33)

Thus a change can take place spontaneously at constant temperature, volume, and
amounts of species if the Helmholtz energy decreases (see Table 4.1). Integration
of the fundamental equation for A at constant values of the properties T, P, and
兵␮i 其 yields
Ns

A  ⫺PV  冱 ␮i ni

(4.34)

i 1

The Helmholtz energy is less useful in chemistry than the Gibbs energy because
processes and reactions are more often carried out at constant pressure rather
than constant volume.
The total differential of the Gibbs energy is
dG  dU  P dV  V dP ⫺ T dS ⫺ S dT

(4.35)

Substituting equation 4.7 for dU yields the fundamental equation for the Gibbs
energy:
Ns

dG  ⫺S dT  V dP  冱 ␮i dni

(4.36)

i 1

Thus, we see that
⭸G
S ⫺
⭸T

冢 冣
⭸G
V 冢
⭸P 冣
⭸G
␮ 冢
⭸n 冣

(4.37)

P,兵ni 其

(4.38)

T,兵ni 其

i

i T,P,兵nj 苷i 其

(4.39)

Table 4.1

Criteria for Irreversibility and Reversibility
for Processes Involving
Only Pressure–Volume
Work

For Irreversible
Processes

For Reversible
Processes

(dS )V,U,兵ni 其 ⬎ 0
(dU )V,S,兵ni 其 ⬍ 0
(dH )P,S,兵ni 其⬍ 0
(dA)T,V,兵ni 其 ⬍ 0
(dG )T,P,兵ni 其⬍ 0

(dS )V,U,兵ni 其  0
(dU )V,S,兵ni 其  0
(dH )P,S,兵ni 其 0
(dA)T,V,兵ni 其  0
(dG )T,P,兵ni 其 0

108

Chapter 4

Fundamental Equations of Thermodynamics

If G can be determined as a function of its natural variables T, P, and 兵ni 其, then
S, V, and ␮i can be calculated by taking partial derivatives of G . Note that the
chemical potential of species i is equal to the partial molar Gibbs energy. Thus
equation 4.36 can be written with ␮i  Gi , where Gi is the partial molar Gibbs
energy of species i . The partial molar volume Vi of a species was introduced in
Section 1.10.
Equations 4.37 and 4.38 tell us something interesting about the dependence
of G on T and P . Since the entropy of a system is always positive, G decreases
with increasing temperature at constant pressure. Since S is greater for a gas than
for the corresponding solid, the temperature coefficient of the Gibbs energy is
more negative for a gas than for the corresponding solid. Since the volume of a
system is always positive, G increases with increasing P at constant T. Since V is
greater for a gas than for the corresponding solid, the pressure coefficient of G is
larger for a gas than for the corresponding solid.
Now we are in a position to illustrate the fact that if a thermodynamic potential is known as a function of its natural variables, we can calculate all of the
thermodynamic properties of the system. Suppose that G for a system containing
a single species has been determined as a function of temperature and pressure.
The entropy and volume of the system can be calculated from
S ⫺

⭸G

冢 ⭸T 冣

V 

and

P

⭸G

冢 ⭸P 冣

(4.40)

T

Then U, H, and A can be calculated using the equations
⭸G

冢 ⭸P 冣

U  G ⫺ PV  TS  G ⫺ P

T

H  G  TS  G ⫺ T

⫺T

⭸G

冢 ⭸T 冣

⭸G

冢 ⭸T 冣

(4.41)

P

(4.42)

P

A  G ⫺ PV  G ⫺ P

⭸G

冢 ⭸P 冣

(4.43)

T

This is not possible when G is known as a function of V and T or P and V.
The Gibbs energy provides a criterion for whether a process can occur spontaneously at constant T, P, and 兵ni 其. Following the same reasoning we used to obtain
equation 4.17, we obtain
(dG )T,P,兵ni 其 ⱕ 0

G

t

Figure 4.1 When a system undergoes spontaneous change at
constant T and P, the Gibbs energy decreases until equilibrium is
reached.

(T  Tsurr , P  Pext )

(4.44)

where the subscript 兵ni 其 indicates that the amounts of all species are held constant. A change can take place spontaneously at constant temperature, pressure,
and amounts of species if the Gibbs energy decreases. This equation provides the
means for discussing phase equilibrium and chemical equilibrium. Figure 4.1 illustrates the way G changes when a system spontaneously goes from an initial state
to the equilibrium state.
Since the Gibbs energy decreases in an irreversible process at constant T and
P, it becomes a minimum in the final equilibrium state. We can, however, imagine
a reversible process occurring at equilibrium; for example, we may imagine the
evaporation of an infinitesimal amount of water from the liquid into the vapor

4.2 Definitions of Additional Thermodynamic Potentials Using Legendre Transforms

phase that is saturated with water vapor at constant temperature and pressure.
For such a process, dG  0.
These same relations may be applied to finite changes as well as infinitesimal
changes, replacing the d’s by ⌬’s. It must be remembered, however, that spontaneous changes always go to the minimum (as in the case of the Gibbs energy at
constant T and P ) or to the maximum (as in the case of the entropy of an isolated
system).
Integration of the fundamental equation for G at constant values of the intensive properties T, P, and ␮i yields
Ns

G  冱 ␮i ni

(4.45)

i 1

This is a very interesting and important equation because it shows that the Gibbs
energy is made up of a sum of terms with one for each of the N species when
the intensive properties for G are constant. The corresponding equations for U,
H, and A when their intensive properties are held constant are given in equations 4.18, 4.25, and 4.34. The importance of equation 4.45 will be discussed in
Section 4.7.
Equation 4.36 yields the following Maxwell relations:



⭸S

冢⭸P 冣

T,兵nj 其

⭸S

冢⭸n 冣

i T,P,兵nj 其

⭸V

冢⭸n 冣

i T,P,兵nj 苷i 其



⭸V

冢⭸T 冣

(4.46)

P,兵ni 其

⭸␮i



冢 ⭸T 冣



冢 ⭸P 冣

P,兵ni 其

⭸␮i

T,兵ni 其

 ⫺Si

(4.47)

 Vi

(4.48)

where Si is the partial molar entropy of species i , and Vi is the partial molar
volume of species i. There is also a Maxwell equation for each pair of species,
which will be discussed later.
The Gibbs energy has been defined so that we will have a criterion for spontaneous change at constant T and P that will not require us to specifically consider
what is happening in the surroundings, as we did in the derivation in Section 3.2.
For an isolated system consisting of the system of interest and the surroundings
at constant T and P, the entropy criterion is that ⌬Ssys  ⌬Ssurr must increase for
a spontaneous process. Since the temperature is constant, ⌬Ssurr  ⫺⌬Hsys /T, so
that ⌬Ssys ⫺ ⌬Hsys /T  ⫺⌬Gsys /T must increase in a spontaneous process at constant T and P, or ⌬Gsys must decrease. Thus, the Gibbs energy G simply provides
a more convenient thermodynamic property than the entropy for the application
of the second law at constant T and P.
Although these criteria show whether a certain change is irreversible, it does
not necessarily follow that the change will take place with an appreciable speed.
Thus, a mixture of 1 mol of carbon and 1 mol of oxygen at 1 bar pressure and 25 ⬚C
has a Gibbs energy greater than that of 1 mol of carbon dioxide at 1 bar and 25 ⬚C,
and so it is possible for the carbon and the oxygen to combine to form carbon
dioxide at this constant temperature and pressure. Although carbon may exist for
a very long time in contact with oxygen, the reaction is theoretically possible. The
reverse of a thermodynamically spontaneous change is, of course, a nonsponta-

109

110

Chapter 4

Fundamental Equations of Thermodynamics

neous change. Thus, the decomposition of carbon dioxide to carbon and oxygen
at room temperature, which involves an increase in Gibbs energy, is nonspontaneous. It can occur only with the aid of an outside agency.
When work other than PV work occurs in a system, it contributes a term to
the fundamental equation for the internal energy, as indicated in equation 2.43.
These terms carry forward into the fundamental equation for the Gibbs energy,
so that if, for example, extension work and surface work are involved,
Ns

dG  ⫺S dT  V dP  冱 ␮i dni  f dL  ␥ dAs

(4.49)

i 1

where f is the force of extension, L is the length, ␥ is the surface tension, and As
is the surface area. Thus we see that
⭸G

冢 ⭸L 冣
⭸G
␥ 冢
⭸A 冣

f ⫺

(4.50)

T,P,兵ni 其,As

(4.51)

s T,P,兵ni 其,L

This opens up the possibility of many more Legendre transforms and Maxwell
equations.
Comment:
Legendre transforms look pretty mathematical, so we should stop and think
about why they are used. The first law introduces the internal energy U, which
is related to heat and work, and the second law introduces the entropy S, which
increases when a spontaneous change occurs in an isolated system. That is really
all we know about thermodynamic systems. Any thermodynamic question can
be answered by using U and S, but these explanations become very complicated
when we want to use P and T as independent variables. The natural variables
of U are S and V, and S and V may be impossible or very difficult to control
in the laboratory. And so we use Legendre transforms to introduce intensive
variables that are more convenient. The Gibbs energy G is especially important
for chemistry because its natural variables are T and P.
If various kinds of work are involved, the first law is
dU  dq  dw

(4.52)

⫺ dU  T dS ⱖ ⫺ dw

(4.53)

Using equation 3.21 yields

At constant temperature, this becomes
⫺d(U ⫺ TS ) ⱖ ⫺dw

(4.54)

(dA)T ⱕ dw

(4.55)

or

where A is the Helmholtz energy. The symbol A actually comes from arbeit, the
German word for work. Thus, in a reversible process at constant temperature,

4.3 Effect of Temperature on the Gibbs Energy

the work done on the system is equal to the increase in the Helmholtz energy. In
general, equation 4.54 shows that the decrease in A is an upper bound on the total
work done in the surroundings. When the system does work on the surroundings
in a real process, the work done on the surroundings (remember that it is negative)
is less than the decrease in the Helmholtz energy of the system.
The Gibbs energy is especially useful when non-PV work is involved. In this
case the first law can be written
dU  dq ⫺ Pext dV  dwnonpv

(4.56)

so that the inequality T dS ⱖ dq can be written
⫺dU ⫺ Pext dV  T dS ⱖ ⫺dwnonpv

(4.57)

The external pressure is represented by Pext . At constant temperature and P 
Pext  constant, this can be written
⫺d(U  PV ⫺ TS ) ⱖ ⫺dwnonpv
(dG )T,P ⱕ dwnonpv

(4.58)
(4.59)

For a reversible process at constant T and P the change in Gibbs energy is equal
to the non-PV work done on the system by the surroundings. Thus, when work is
done on the system, the Gibbs energy increases, and when the system does work
on the surroundings, the Gibbs energy decreases. In general, equation 4.58 shows
that the decrease in G is an upper bound on the non-PV work done on the surroundings. When the system does work on the surroundings, the work done (remember that it is negative) is less than the decrease in Gibbs energy.
Let us consider inequality 4.59 in more detail by applying it to the charging
and discharging of an electrochemical cell at constant temperature and pressure.
The electrochemical cell is charged by an electrical generator, and we will imagine
a perfect direct-current generator that consumes a known amount of mechanical
work. When the electrochemical cell is charged, the increase in the Gibbs energy
of the cell is less than the electrical work done by the generator on the system
in a real process and is equal to the electrical work in the theoretical limit of a
reversible process.
When the electrochemical cell is discharged by operating an idealized electrical motor that does mechanical work, the Gibbs energy of the cell decreases
and the work done is negative. According to inequality 4.59, the work done by
the electrical motor is more positive than the decrease in G in a real process.
Thus, the amount of work done in the surroundings is less than the decrease in
the Gibbs energy of the electrochemical cell, except in the theoretical limit of a
reversible process. This provides a simple interpretation for the change in Gibbs
energy for a system. For a reversible process at constant temperature and pressure, the change in the Gibbs energy for the system is equal to the non-PV work
done on the system by the surroundings.

4.3

EFFECT OF TEMPERATURE ON THE GIBBS ENERGY

For an open system, equation 4.37 shows that (⭸G /⭸T )P,兵ni 其  ⫺S . Since S is a
positive quantity, the Gibbs energy G necessarily decreases as the temperature in-

111

112

Chapter 4

Fundamental Equations of Thermodynamics

creases at constant pressure and constant amounts of species. An important equation is obtained by using this equation to eliminate S from G  H ⫺ TS :
G  H T

⭸G

冢 ⭸T 冣

(4.60)

P,兵ni 其

Since this equation involves both the Gibbs energy and its temperature derivative,
it is more convenient to rearrange it so that only a temperature derivative appears.
This can be accomplished by first differentiating G /T with respect to temperature
at constant pressure and constant amounts of species:



⭸ G /T
⭸T



P,兵ni 其

⫺

G
1 ⭸G

T2
T ⭸T

冢 冣

(4.61)

P,兵ni 其

Eliminating G from the right-hand side by use of equation 4.60 yields
H  ⫺T 2



⭸G /T
⭸T



(4.62)

P,兵ni 其

This is referred to as the Gibbs–Helmholtz equation. For a change between state
1 and state 2, this equation can be written
⌬H  ⫺T 2



⭸(⌬G /T )
⭸T



(4.63)

P,兵ni 其

This equation is very useful because if we can determine ⌬G for a process or a
reaction as a function of temperature, the enthalpy change for the process or reaction can be calculated without using a calorimeter. It is also useful because if ⌬H
and ⌬G are known at one temperature, equation 4.63 can be integrated to calculate ⌬G at another temperature assuming that ⌬H is independent of temperature
(see Section 5.5).

4.4

EFFECT OF PRESSURE ON THE GIBBS ENERGY

The equation (⭸G /⭸P )T,兵ni 其  V (equation 4.38) may be integrated to obtain the
value of G at another pressure, provided that its value is known at one pressure
and V is known as a function of pressure at constant temperature:



G2

dG 

G1

G2  G1 



P2



P2

V dP

(4.64)

V dP

(4.65)

P1

P1

This equation always applies, and, as we have noted, the Gibbs energy of a single
substance always increases with the pressure. There are two special cases where
equation 4.65 leads to simple relationships. If the volume is nearly independent
of pressure, as it is for a liquid or solid, then
G2  G1  V (P2 ⫺ P1 )

(4.66)

The Gibbs energy of a gas is more dependent on pressure than that of a liquid. For an ideal gas the dependence of Gibbs energy on pressure is obtained

4.4 Effect of Pressure on the Gibbs Energy

by substituting V  nRT /P in equation 4.64:

冮⬚

dG  nRT

G

6

P

冮 ⬚ d ln P

4

(4.67)

P

G  G ⬚  nRT ln

P
P⬚

(4.68)

(∆fG –∆fG°)/kJ mol–1

G

2

0

2
4
6
8
where P ⬚ is the standard state pressure. The Gibbs energy is like the internal
P/bar
energy and the enthalpy in that its absolute value is not known, but the Gibbs
–2
energy of a species can be determined with respect to the elements it contains.
Thus equation 4.68 can be written ⌬f Gi  ⌬f Gi⬚  RT ln(Pi /P ⬚ ). The standard
–4
Gibbs energy G ⬚ has different values at different temperatures. The logarithmic
dependence of the molar Gibbs energy of an ideal gas on the pressure of the gas
–6
is illustrated in Fig. 4.2.
Alternatively, equation 4.65 may be integrated between any two pressures to Figure 4.2 Dependence of the
Gibbs energy of formation of an
obtain

P2
⌬G  G2 ⫺ G1  nRT ln
P1

(4.69)

Example 4.1 Calculation of molar thermodynamic properties for an ideal gas
Since the molar Gibbs energy of an ideal gas is given by G  G ⬚  RT ln(P /P ⬚ ), derive
the corresponding expressions for V, U, H, S, and A.
Using equations 4.38, 4.41, 4.42, 4.37, and 4.43, respectively,
V 

RT
P

U  U ⬚  H ⬚ ⫺ RT
H  H ⬚  G ⬚ T S ⬚
P
P⬚
P
A  A ⬚  RT ln
P⬚
S  S ⬚ ⫺ R ln

where S ⬚  ⫺(⭸G ⬚ /⭸T )P and U ⬚  G ⬚  T S ⬚ ⫺ RT. Note that the internal energy U
and enthalpy H of an ideal gas are independent of pressure and volume.

Example 4.2 Calculation of changes in thermodynamic properties in the
reversible isothermal expansion of an ideal gas
An ideal gas at 27 ⬚C expands isothermally and reversibly from 10 to 1 bar against a pressure
that is gradually reduced. Calculate q per mole and w per mole and each of the thermodynamic quantities ⌬U, ⌬H, ⌬G, ⌬A, and ⌬S.
Since the process is carried out isothermally and reversibly,
w  ⫺RT ln

10
V2
P
 ⫺RT ln 1  ⫺(8.3145 J K⫺1 mol⫺1 )(300.15 K) ln
1
P2
V1

 ⫺5746 J mol⫺1
⌬A  w  ⫺5746 J mol⫺1

113

ideal gas on the pressure of the gas.
(See Computer Problem 4.D.)

10

114

Chapter 4

Fundamental Equations of Thermodynamics
Since the internal energy of an ideal gas is not affected by a change in volume,
⌬U  0
q  ⌬U ⫺ w  0  5746  5746 J mol⫺1
⌬H  ⌬U  ⌬(P V )  0  0  0
and since P V is constant for an ideal gas at constant temperature,
⌬G 

1

冮10 V dP

 RT ln

1
 (8.3145 J K⫺1 mol⫺1 )(300.15 K)(⫺2.3026)
10

 ⫺5746 J mol⫺1
⌬S 

qrev
5746 J mol⫺1

 19.14 J K⫺1 mol⫺1
T
300.15 K

Also,
⌬S 

⌬H ⫺ ⌬G
0  (5746 J mol⫺1 )

 19.14 J K⫺1 mol⫺1
300.15 K
T

Example 4.3 Calculation of changes in thermodynamic properties in the
irreversible isothermal expansion of an ideal gas
An ideal gas expands isothermally at 27 ⬚C into an evacuated vessel so that the pressure drops from 10 to 1 bar; that is, it expands from a vessel of 2.463 L into a connecting
vessel such that the total volume is 24.63 L. Calculate the change in thermodynamic
quantities.
This process is isothermal, but is not reversible.
w  0 because the system as a whole is closed and no external work can be done.
⌬U  0 because the gas is an ideal gas.
q  ⌬U ⫺ w  0  0  0
⌬U, ⌬H, ⌬G, ⌬A, and ⌬S are the same as in Example 4.2 because the initial and final
states are the same.

Example 4.4 Calculation of the Gibbs energy of formation of gaseous and
liquid methanol as a function of pressure
The effect on the Gibbs energy of a gas due to changing the pressure is much greater
than the effect on the Gibbs energy of the corresponding liquid because the molar volume
of the gas is much larger. To see how this works for a specific substance, consider the liquid
and gas forms of methanol, for which the standard Gibbs energy of formation and standard
enthalpy of formation at 298.15 K are given in Table C.2. The standard Gibbs energy of formation for liquid CH3 OH at 298.15 K is ⫺166.27 kJ mol⫺1 , and that for gaseous CH3 OH
is ⫺161.96 kJ mol⫺1 . The density of liquid methanol at 298.15 K is 0.7914 g/cm3 . (a) Calculate ⌬f G(CH3 OH , g) at 10 bar at 298.15 K assuming methanol vapor is an ideal gas. (Note
that the superscript ⬚ has been deleted because the pressure is not the standard pressure
of 1 bar.) (b) Calculate ⌬f G(CH3 OH , l) at 10 bar at 298.15 K.

4.5 Fugacity and Activity
–156
Gas
–158

∆fG/kJ mol–1

–160
–162
–164
Liquid

–166
–168
–170

0

2

4

6

8

10

P/bar

Figure 4.3 Plots of the Gibbs energies of formation of gaseous and liquid methanol at
various pressures at 298.15 K. Note that the Gibbs energy of formation of the gas increases
with pressure much more rapidly than the Gibbs energy of formation of the liquid. The
curves cross at the vapor pressure of liquid methanol at this temperature. (See Computer
Problem 4.D.)
(a) The effect of pressure on ⌬G(g) is given by equation 4.68:
⌬f G  ⌬f G ⬚  RT ln(P /P ⬚ )
 ⫺161.96 kJ mol⫺1  (8.3145 ⫻ 10⫺3 kJ K⫺1 mol⫺1 )(298.15 K) ln 10
 ⫺156.25 kJ mol⫺1
This point is plotted in Fig. 4.3 along with points at lower pressures.
(b) The effect of pressure on ⌬G(l) is given by equation 4.66:
⌬f G  ⌬f G ⬚  V (P ⫺ P ⬚ )
The molar volume of methanol is
V  (32.04 g mol⫺1 )/(0.7914 g cm⫺3 )  40.49 cm3 mol⫺1
 (40.49 cm3 mol⫺1 )(10⫺2 m/cm)3  40.49 ⫻ 10⫺6 m3 mol⫺1
and
⌬f G  ⫺166.27 kJ mol⫺1  (40.49 ⫻ 10⫺6 m3 mol⫺1 )(9 ⫻ 10 5 Pa)/(10 3 J kJ⫺1 )
 ⫺166.23 kJ mol⫺1
This point is plotted in Fig. 4.3 along with points at lower pressures. Note that the liquid
and gas curves cross at a pressure less than 1 bar. The intersection is at the vapor pressure of methanol at 298.15 K; at this pressure the liquid and gas have the same molar
Gibbs energy of formation ⌬f G. Note that ⌬f G(g) of an ideal gas becomes ⫺⬁ at zero
pressure.

4.5

FUGACITY AND ACTIVITY

The Gibbs energy of a real gas is not given by equation 4.68, which we derived
for an ideal gas. However, G. N. Lewis recognized that it would be convenient to
keep the same form of equation for a real gas. He accomplished this by defining

115

116

Chapter 4

Fundamental Equations of Thermodynamics

the fugacity f , which is a function of T and P, using
G  G ⬚  RT ln

f
P⬚

(4.70)

f
1
P

lim

Py0

(4.71)

The fugacity has the units of pressure. As the pressure approaches zero, the gas
approaches ideal behavior and the fugacity approaches the pressure. The fugacity
is simply a measure of the molar Gibbs energy of a real gas, but it has the following
advantage over the molar Gibbs energy: f goes from 0 to ⬁, while G goes from
⫺⬁ to ⬁ (see Fig. 4.2).
The fugacity of a real gas at a particular temperature and pressure can be calculated if the equation of state (Section 1.5) of the gas is known. As we will see in
the following derivation, it is convenient to use the virial equation of state written
in terms of pressure (equation 1.12). Since (⭸G /⭸P )T  V, the differential Gibbs
energy at constant temperature is dG  V dP for a real gas and dG id  V id dP
for an ideal gas. The difference in Gibbs energy between a real gas and an ideal gas
can be integrated from some low pressure P ⴱ to the pressure at which we would
like to know the fugacity:



P

Pⴱ

d(G ⫺ G id ) 

(G ⫺ G id )P ⫺ (G ⴱ ⫺ G ⴱid )P ⴱ 



P



P

Pⴱ

Pⴱ

(V ⫺ V id ) dP

(4.72)

(V ⫺ V id ) dP

(4.73)

Now, if we let P ⴱ y 0, then G ⴱ y G ⴱid , and
(G ⫺ G id )P 



P

(V ⫺ V id ) dP

(4.74)

0

Introducing equation 4.70 for G and equation 4.68 for G id yields

冢 冣

f
1

P
RT

ln



P



P

(V ⫺ V id ) dP

(4.75)

0

or



f
1
 exp
P
RT

(V ⫺ V id ) dP

0



(4.76)

The ratio of the fugacity to the pressure is called the fugacity coefficient ␾ ; ␾ 
f /P. The fugacity coefficient is frequently used as a measure of the nonideality of
a gas in connection with its phase equilibrium or chemical equilibrium properties.
When PVT data are available on a gas, a plot may be prepared of the difference
between its molar volume and the molar volume of an ideal gas versus pressure at
the temperature of interest. The integral of this plot up to the pressure of interest
is then used in equation 4.76 to calculate ␾  f /P. Equation 4.76 may be written
in terms of the compressibility factor Z . Since V  RTZ /P,



f
1
 exp
P
RT

冮 冢
P

0

冣 冥

冤冮

RTZ
RT

dP  exp
P
P

P

0

Z ⫺1
dP
P



(4.77)

4.5 Fugacity and Activity

Thus, the fugacity of a gas is readily calculated at some pressure P if Z is known
as a function of pressure up to that particular pressure.

Example 4.5 Expression of the fugacity in terms of virial coefficients
Given the expression for the compressibility factor Z as a power series in P (equation
1.12), what is the expression for the fugacity in terms of the virial coefficients?
Using equation 4.77, we find
ln

f

P

P

冮0 (B ⬘  C ⬘P  ⭈⭈⭈) dP  B ⬘P 

C ⬘P 2
 ⭈⭈⭈
2

Example 4.6 The fugacity of a van der Waals gas
Using the expression for the compressibility factor Z of a van der Waals gas given in equation 1.26, what is the expression for fugacity of a van der Waals gas?
As an approximation, terms in P 2 and higher in the series expansion are omitted.

冤 冢 冣冥

Z  1 b ⫺

ln

f

P


a
RT

P

Z ⫺1
dP
P

P

冤 冢 冣冥

冮0 冢
冮0



b⫺

a
RT

冤 冢 冣冥

 b⫺

P
RT

1
dP
RT

a
RT

P
RT

冤冢

a
P
RT RT

f  P exp b ⫺

冣 冥

Example 4.7 Estimating the fugacity of nitrogen gas at 50 bar and 298 K
Given that the van der Waals constants of nitrogen are a  1.408 L2 bar mol⫺2 and b 
0.03913 L mol⫺1 , estimate the fugacity of nitrogen gas at 50 bar and 298 K.

冤冢

f  P exp b ⫺

 (50 bar) exp

冣 冥

a
P
RT RT

冦 冋0.03913 ⫺ (0.083 145)(298) 册 (0.083 145)(298) 冧
1.408

50

 48.2 bar

Now we are in a better position to understand the standard state of a gas that
is used for thermodynamic tables, such as Tables C.2 and C.3. The standard state is

117

118

Chapter 4

Fundamental Equations of Thermodynamics

the pure substance at a pressure of 1 bar in a hypothetical state in which it exhibits
ideal gas behavior as shown in Fig. 4.4. The solid line gives the behavior of a real
gas. As the pressure is reduced, the real gas approaches ideal behavior. This ideal
gas is then compressed to 1 bar along the dashed line as a hypothetical ideal gas.
G. N. Lewis introduced the activity a as a means of dealing with real substances in the gas, liquid, and solid state. In analogy to equation 4.70, the activity
of a pure substance, or its activity in a mixture, is defined by

f /bar

Standard
state
1

0
0

1
P/bar

Figure 4.4 Plot of fugacity versus
pressure for a real gas. The dashed
line is for an ideal gas. The standard
state is the pure substance at a
pressure of 1 bar in a hypothetical
state in which it exhibits ideal gas
behavior.

␮i  ␮i⬚  RT ln ai

(4.78)

Thus, the activity ai is simply a means for expressing the chemical potential of a
species in a mixture. The activity is dimensionless, and ai  1 in the reference
state for which ␮i  ␮i⬚ . For a real gas ai  fi /P ⬚ , where fi is the fugacity. For
an ideal gas, ai  Pi /P ⬚ . We will see later in dealing with solutions that it is convenient to write the activity ai as the product of an activity coefficient ␥i and a
concentration.
The activity of a pure solid or liquid can be taken as unity if the pressure
is close enough to the standard state pressure so that the effect of pressure on
the chemical potential is negligible. If the effect of pressure is not negligible, the
activity of a solid or liquid can be readily calculated because the molar volume
V can be assumed to be constant at all reasonable pressures. For a pure solid or
liquid, equation 4.66 can be written

␮ (T, P )  ␮ ⬚ (T )  V (P ⫺ P ⬚ )

(4.79)

Comparison with equation 4.78 shows that RT ln a  V (P ⫺ P ⬚ ) or
a  e V (P ⫺P ⬚ )/RT

(4.80)

Small changes in pressure do not have a significant effect on the activity of a solid
or liquid because of the smallness of the exponent.

Example 4.8 Calculating the activity of liquid water at 10 and 100 bar
What is the activity of liquid water at 1, 10, and 100 bar at 25 ⬚C, assuming that V is constant?
At P  1 bar,
a1
At P  10 bar,
a  exp
 exp

V (P ⫺ P ⬚ )
RT
(0.018 kg mol⫺1 )(9 bar)
(0.083 145 L bar K⫺1 mol⫺1 )(298 K)

 1.007
At P  100 bar, a  1.075.

4.6

THE SIGNIFICANCE OF THE CHEMICAL POTENTIAL

The chemical potential ␮i of a species was introduced in equation 4.6, and we saw
in equation 4.11 that it is equal to the partial derivative of the internal energy of

4.6 The Significance of the Chemical Potential

a homogeneous mixture with respect to the amount of species i in the system at
constant S, V, and 兵nj 其. Later we found in equations 4.23, 4.32, and 4.39 that this
property of a species can be obtained in three other ways:

␮i 

⭸U

冢⭸n 冣

i S,V,兵nj 苷i 其



⭸H

冢⭸n 冣

i S,P,兵nj 苷i 其



⭸A

冢⭸n 冣

i T,V,兵nj 苷i 其



⭸G

冢⭸n 冣

(4.81)

(dG )T,P  ⫺␮i (␣ ) dni  ␮i (␤ ) dni  dni [␮i (␤ ) ⫺ ␮i (␣ )]

(4.82)

For this transfer to occur spontaneously, dG ⬍ 0 and therefore ␮i (␣ ) ⬎ ␮i (␤ );
in other words, a species diffuses spontaneously from the phase where its chemical potential is higher to the phase where its chemical potential is lower. In this
way the chemical potential is analogous to the electric potential and mechanical
potential. If the phases are in equilibrium (i.e., no spontaneous change can take
place), then there is no change in Gibbs energy in the transfer, dG  0, and
(4.83)

Thus, at equilibrium, the chemical potential of a species is the same in all of the
phases of a system.
We will later see that at equilibrium the chemical potential of a species is the
same in all of the phases in a system, even if the phases are at different pressures,
as in a small liquid droplet in equilibrium with its vapor (Section 6.9) or in an
osmotic pressure experiment (Section 6.7). In the next chapter, we will see that it
is the chemical potentials of the species involved that determine whether they will
undergo a chemical reaction. In Chapter 7 we will see that the chemical potential
of an ion in a multiphase system with phases at different electric potentials is the
same in each phase at equilibrium.
The partial molar entropy and partial molar volume of a species can be calculated from measurements of the chemical potential of the species as a function
of temperature and pressure by use of two Maxwell relations for the fundamental
equation for G (see equations 4.47 and 4.48):
⫺Si 
Vi 

⭸␮i

冢 ⭸T 冣

(4.84)

P,兵ni 其

⭸␮i

冢 ⭸P 冣

T,兵ni 其

µi(β)

µi(α)

i T,P,兵nj 苷i 其

Since it is impossible to hold S constant (except in reversible adiabatic processes)
and since experiments are not often carried out at constant volume, it is this last
definition of the chemical potential that is most often used, and at constant T and
P the chemical potential can be referred to as the partial molar Gibbs energy;
␮i  Gi .
The concept of the chemical potential ␮i of a species is extremely important,
and we can see that in considering two phases that are at constant T and P, as
shown in Fig. 4.5. We have already seen that at equilibrium these phases have
the same temperature (Section 3.2) and the same pressure. The two phases may
contain many species, and we will consider transferring an infinitesimal amount
dni of species i from phase ␣ to phase ␤ at constant T and P. The total differential
of the Gibbs energy for this two-phase system is given by equation 4.36, which
becomes

␮i (␣ )  ␮i (␤ )

119

(4.85)

Figure 4.5 Two phases at the same
temperature and pressure. Many
species may be present, but we will
focus on species i.

120

Chapter 4

Fundamental Equations of Thermodynamics

As a simple example, consider a mixture of ideal gases. Since the partial molar
volume for a species in an ideal gas is equal to the molar volume of the mixture
(i.e., Vi  V ),

RT

µ i – µ i°

0

1

2

V 

Pi / P °

–RT

RT
⭸␮i

P
⭸P

冢 冣

T,兵ni 其

冋冢 冣冢 冣册
⭸␮i
⭸Pi

⭸Pi
⭸P

T,兵ni 其

 xi

⭸␮i

冢⭸P 冣

(4.86)

i T,兵ni 其

Dividing both sides by xi yields
⭸␮i

冢⭸P 冣

–2RT

Figure 4.6 Chemical potential ␮i
of an ideal gas as a function of
pressure relative to the chemical potential ␮i⬚ of the gas at the standard
pressure of 1 bar. This plot is closely
related to Figure 4.2, but it is given
because of the importance of the
chemical potential.



i T,兵ni 其



RT
Pi

(4.87)

Thus,
␮i

冮⬚
␮i

d␮i  RT

Pi

冮⬚
P

d Pi
Pi

(4.88)

Pi
P⬚

(4.89)

Integration yields

␮i  ␮i⬚  RT ln

where P ⬚ is the standard state pressure. The standard state pressure is taken as
10 5 Pa  1 bar. A similar equation was given earlier (cf. equation 4.68) for a pure
gas. See Fig. 4.6.
We can derive the expression for the partial molar entropy of an ideal gas in
a mixture by using equations 4.84 and 4.89. Substituting equation 4.89 in equation 4.84 yields
Si  S i⬚ ⫺ R ln

Pi
P⬚

(4.90)

The equation for the molar entropy of a pure ideal gas was derived in Example 4.1.
Comment:
The chemical potential is one of the most important concepts in chemical
thermodynamics. In both chemical reactions and phase changes, the chemical
potential of a species times its differential amount (reacted or transferred)
determines the change in U, H, A, or G, depending on the variables that are
held constant during the change. Although the chemical potential is a very general
concept, we will use it most frequently in discussing systems at specified
T and P. The chemical potential of a species is determined by its partial pressure
in an ideal gas mixture and by its concentration in an ideal solution. For real
mixtures, whether gaseous or liquid, the chemical potential of a species is a
much more complicated function of the composition, but we can always think
of it as a simple function of the activity a, which was introduced in Section 4.5.

4.7

ADDITIVITY OF PARTIAL MOLAR PROPERTIES
WITH APPLICATIONS TO IDEAL GASES

Since G  冱 ni ␮i  冱 ni Gi (equation 4.45), all the other extensive properties
of a one-phase system are also additive. One way of looking at this is that if the

4.7 Additivity of Partial Molar Properties with Applications to Ideal Gases

Gibbs energy of a system is known as a function of T, P, and amounts of species,
the entropy of the system can be calculated by taking the negative temperature
derivative of the Gibbs energy (see equation 4.37):
Ns

S  ⫺冱
i 1

⭸␮i
⭸T

Ns

冢 冣

P,兵ni 其

ni  冱 Si ni

(4.91)

i 1

where Si is the partial molar entropy of species i .
Equation 4.38 shows that the volume of a system is equal to the derivative of
the Gibbs energy with respect to pressure. Substituting equation 4.48 in equation
4.38 yields
Ns

V  ⫺冱
i 1

⭸␮i
⭸P

Ns

冢 冣

P,兵ni 其

ni  冱 Vi ni

(4.92)

i 1

where Vi , the partial molar volume of species i, is defined in equation 4.38. The
additivity of the partial molar volumes of a mixture of ideal gases was discussed
in Chapter 1.
Equation 4.61 shows the relation between the Gibbs energy and the enthalpy
for a system. Substituting equation 4.45 in equation 4.62 yields
Ns

H  ⫺T 2 冱
i 1



⭸(␮i /T )
⭸T



Ns

P,兵ni 其

ni  冱 Hi ni

(4.93)

i 1

where the partial molar enthalpy of species i is defined by
Hi 

⭸H

冢⭸n 冣

(4.94)

i T,P,兵nj 苷i 其

Derivations of relations between partial molar properties

Example 4.9

Take the derivatives of G  H ⫺ TS and ⫺S  (⭸G /⭸T )P,兵ni 其 with respect to ni to obtain
the corresponding equations for the partial molar properties.
⭸G

冢⭸ n 冣

i T,P,兵nj 苷i 其



⭸H

冢⭸ n 冣

⫺T

i T,P,兵nj 苷i 其

⭸S

冢⭸n 冣

i T,P,兵nj 苷i 其

␮ i ⬅ G i  H i ⫺ T Si


⭸S

冢⭸n 冣

i T,P,兵nj 苷i 其



冤 冢 冣冥
⭸ ⭸G
⭸ ni ⭸ T

⫺ Si 

P,n T,P,兵n 其
j 苷i

冢 冣
⭸Gi
⭸T



P,兵nj 苷i 其



冤 冢 冣
⭸ ⭸G
⭸T ⭸ni



T,P,兵nj 苷i 其 P,兵n 其
j 苷i

⭸␮i

冢 ⭸T 冣

P,兵nj 苷i 其

This last equation is actually a Maxwell relation from equation 4.36. Note that whereas V
and S for a system are always positive, Vi and Si may be negative.

There are similar equations for U and A, which can be obtained from Legendre transforms. These additivity equations are general, but they are most easily
applied to mixtures of ideal gases.
Thermodynamics alone cannot lead to the conclusion that a mixture of ideal
gases will behave as an ideal gas. Nevertheless, it is found that at low pressures

121

122

Chapter 4

Fundamental Equations of Thermodynamics

mixtures of real gases do behave as ideal gases; that is, they behave as if there are
no interactions between the molecules of the gas. Such mixtures are referred to
as ideal mixtures to indicate the additional assumption involved.
The expression for the chemical potential of a species in an ideal gas mixture
is given by equation 4.89, but we will find it convenient to write it as

␮i  ␮i⬚  RT ln

yi P
P⬚

(4.95)

since the partial pressure Pi of any species i in an ideal gas mixture is defined by
Pi ⬅ yi P

(4.96)

where yi is the mole fraction and P is the total pressure. We will use y to represent
the mole fraction in the gas phase, and x to represent the mole fraction in the
liquid phase.
When we substitute equation 4.95 in G  冱 ni␮i , we obtain the expression
for the Gibbs energy of an ideal mixture of ideal gases:
G  冱 ni␮i⬚  RT
 nt

冋冱

冱 ni ln yi  冱 ni RT ln(P /P ⬚ )

yi␮i⬚  RT



冱 yi ln yi  RT ln(P /P ⬚ )

 nt G

(4.97)

where nt  冱 ni is the total amount of gas in the system and G  G /nt is the
molar Gibbs energy of the mixture. This expression for the Gibbs energy of an
ideal gas mixture is written in terms of the natural variables T and P. Therefore,
we can use equations 4.40–4.43 to derive the expressions for S, V, U, H, and A for
an ideal gas mixture.
S ⫺
 nt

⭸G

冢 ⭸T 冣

冋冱

P,兵ni 其

 冱 ni S ⬚i ⫺ R 冱 ni ln yi ⫺ 冱 ni R ln(P /P ⬚ )



yi S ⬚i ⫺ R 冱 yi ln yi ⫺ R ln(P /P ⬚ )

 nt S

(4.98)

where S  S /nt is the molar entropy of the mixture.
The enthalpy of the ideal gas mixture can be calculated from H  G  TS
since we have expressions for both G and S:
H  冱 ni (␮i⬚  T S ⬚i )  冱 ni H ⬚i
 nt

冋冱 ⬚册

y i H i  nt H ⬚

(4.99)

since H ⬚i  ␮i⬚  T S ⬚i . Note that the enthalpy of a mixture of ideal gases is independent of the pressure. This is a result of the fact that the molecules in a mixture
of ideal gases do not interact with each other.
The volume of an ideal gas mixture is given by
V 

⭸G

冢 ⭸P 冣

T,兵ni 其

 冱 ni RT /P  冱 ni Vi

(4.100)

4.7 Additivity of Partial Molar Properties with Applications to Ideal Gases

123

where the last form simply comes from the definition of the partial molar volume.
The partial molar volume of each species in the mixture is the same: Vi  RT /P,
as we saw in Section 1.10. Gibbs commented that “every gas is as a vacuum to
every other gas.” These statements apply only to ideal gas mixtures.

Example 4.10 Calculation of changes in thermodynamic properties on mixing
ideal gases
In Section 3.5 we considered the mixing of two ideal gases that are initially at the same
temperature and pressure but are separated from each other by a partition. Use equations
4.97–4.100 to calculate the changes in Gibbs energy, entropy, enthalpy, and volume when
the partition is withdrawn. Note that the final total pressure is the same as the initial pressure of each gas.
The initial values of these quantities are
G  n1 ␮1⬚  n1 RT ln(P /P ⬚ )  n2 ␮2⬚  n2 RT ln(P /P ⬚ )
 n1 ␮1⬚  n2 ␮2⬚  (n1  n2 )RT ln(P /P ⬚ )
S  n1 S 1⬚ ⫺ n1 R ln(P /P ⬚ )  n2 S 2⬚ ⫺ n2 R ln(P /P ⬚ )
 n1 S 1⬚  n2 S 2⬚ ⫺ (n1  n2 )R ln(P /P ⬚ )
H  n1 H 1⬚  n2 H 2⬚
V  n1 RT /P  n2 RT /P  (n1  n2 )RT /P
+1

The values after mixing are given by equations 4.97–4.100. Therefore the changes upon
pulling out the partition are
⌬mix G  RT (n1 ln y1  n2 ln y2 )

∆ mix G
nt RT

0

⌬mix S  ⫺R (n1 ln y1  n2 ln y2 )
⌬mix H  0

–1

⌬mix V  0

0

1
y2
(a)

Since the mole fractions in a mixture are less than unity, the logarithmic terms
are negative, and ⌬mix G ⬍ 0. This corresponds to the fact that the mixing of gases
is a spontaneous process at constant temperature and pressure. In other words, if
two gases at the same pressure and temperature are brought into contact, they
will spontaneously diffuse into each other until the gas phase is macroscopically
homogeneous.
The Gibbs energy change for mixing two ideal gases is plotted versus the mole
fraction of one of the gases in Fig. 4.7a. The greatest Gibbs energy change on
mixing is obtained for y1  y2  12 . The dependence of the entropy of mixing on
the mole fraction of one of the two gases is shown in Fig. 4.7b.
When ideal gases are mixed at constant temperature and pressure, no heat is
produced or consumed. This corresponds to the fact that molecules of ideal gases
do not attract or repel each other. Thus, from an energy standpoint, it makes no
difference whether the gases are separated or mixed. The driving force for mixing
arises exclusively from the change in entropy. From the viewpoint of statistical

+1

∆ mix S
nt R

0

–1
0

1
y2
(b)

Figure 4.7 Thermodynamic quantities for the mixing of two ideal gases
to form an ideal mixture. The total
amount of gas is represented by nt .

124

Chapter 4

Fundamental Equations of Thermodynamics

mechanics (Chapter 16), the mixed state is found at equilibrium because it is more
probable, as discussed in Section 3.6 for the expansion of a gas.

4.8

GIBBS–DUHEM EQUATION

In making Legendre transforms of the internal energy, we introduced the intensive variables P and T as natural variables. This process can be continued to introduce the intensive variables ␮1 , ␮2 , . . . , ␮N by making the complete Legendre
transform
Ns

U ⬘  U  PV ⫺ TS ⫺ 冱 ni␮i  0

(4.101)

i 1

It is evident from equation 4.18 that this transformed internal energy U ⬘ is equal
to zero. The differential of U ⬘ is
Ns

Ns

i 1

i 1

dU  P dV  V dP ⫺ T dS ⫺ S dT ⫺ 冱 ni d␮i ⫺ 冱 ␮i dni  0 (4.102)
Subtracting this from the fundamental equation (4.17) for U yields
Ns

V dP ⫺ S dT ⫺ 冱 ni d␮i  0

(4.103)

i 1

which is known as the Gibbs–Duhem equation. Note that it deals with changes
only of the intensive variables for the system. Because of this relation, the intensive variables for a system are not independent. This is in agreement with the conclusion discussed in Section 1.9 that there are Ns  1 independent intensive variables for a one-phase system. In a multiphase system there is a separate Gibbs–
Duhem equation for each phase.
For a system with two species at constant temperature and pressure that contains 1 mol of material,
x1 d␮1  x2 d␮2  0

(4.104)

x1 d␮1  (1 ⫺ x1 ) d␮2  0

(4.105)

where x1 is the mole fraction of component 1 and (1 ⫺ x1 ) is the mole fraction
of component 2. Thus, when the composition is changed at constant T and P, the
change in the chemical potential of species 2 is not independent of the change
in the chemical potential of species 1. Later, in Section 6.6, we will use this form
of the Gibbs–Duhem equation to show that if Henry’s law holds for the solute
(species 2), Raoult’s law holds for the solvent (species 1).
Comment:
We can generalize on what we have here by pointing out that the Gibbs–Duhem
equation is the complete Legendre transform for a system with a certain set of
intensive variables. You may have thought of the intensive variables of a system
as being independent, as the extensive variables are, but they are not. Suppose
that the intensive variables for a system are T, P, ␮1 , ␮2 , and ␮3 . According to
the Gibbs–Duhem equation, if you specify the values of T, P, ␮1 , and ␮2 , then ␮3
can have only a particular value. This will be even more evident when we discuss
equilibrium constants in the next chapter.

4.9 Special Topic: Additional Applications of Maxwell Relations

4.9

SPECIAL TOPIC: ADDITIONAL APPLICATIONS
OF MAXWELL RELATIONS

A number of applications have already been made of Maxwell relations, but some
others are of special interest. For the purpose of making calculations about a mole
of a substance, the Maxwell relations from the fundamental equations for U, H,
A, and G can be written
⭸T

冢⭸V 冣

⫺

S

⭸P

冢⭸ S 冣

(4.106)

V

冢⭸P 冣  冢⭸S 冣

(4.107)

冢 冣 冢⭸T 冣

(4.108)

冢 冣 冢 冣

(4.109)

⭸T

⭸V

S

⭸S
⭸V



⭸S
⭸P

P



⭸P

V

T



T

⭸V
⭸T

P

In Section 2.6, we found that (⭸U /⭸V )T  0 for an ideal gas. Now we can calculate
this partial derivative for any gas. Dividing the combined first and second law for
a mole of a substance by dV and imposing constant temperature yields

冢 冣
⭸U
⭸V

T

T

冢 冣
⭸S
⭸V

⫺P

(4.110)

T

Using one of the Maxwell equations (4.108) yields

冢 冣
⭸U
⭸V

T

T

⭸P

冢⭸T 冣

⫺P

(4.111)

V

As an example, this equation can be applied to a van der Waals gas. Differentiation of P  RT /(V ⫺ b ) ⫺ a /V 2 with respect to T at constant V yields
(⭸P /⭸T ) V  R /(V ⫺ b ). Substituting this relation in equation 4.111 yields

冢 冣
⭸U
⭸V

T





RT
RT
RT
a
⫺P 

⫺ 2
V ⫺b
V ⫺b
V ⫺b
V



(4.112)

a
 2
V
Thus, for a van der Waals gas the internal pressure is inversely proportional to the
square of the molar volume.

Example 4.11 Calculation of the change in molar internal energy in the
expansion of propane gas, assuming it is a van der Waals gas
Propane gas is allowed to expand isothermally from 10 to 30 L. What is the change in molar
internal energy?

125

126

Chapter 4

Fundamental Equations of Thermodynamics
The change in internal energy for a given change in volume at constant temperature
is given by
U2

冮U

dU 

1

V2

冮V

⌬U  a

V2

冢 冣

1

a
1
dV  a ⫺
V2
V





1
1

V1
V2

V1

According to Table 1.3, a  8.779 L2 bar mol⫺2 , but we need to convert this to SI base
units to calculate ⌬U in J mol⫺1 :
a  (8.779 L2 bar mol⫺2 )(10 5 Pa bar⫺1 )(10⫺3 m3 L⫺1 )2
 0.8779 Pa m6 mol⫺2
⌬U  a



1
1

V1
V2


冢10 ⫻ 10

 (0.8779 Pa m6 mol⫺2 )
⫺1

1
⫺3

m3 mol⫺1



1
30 ⫻ 10⫺3 m3 mol⫺1



 58.5 J mol

Example 4.12 Calculation of the molar entropy of the isothermal expansion
of a van der Waals gas
Derive the equation for the molar entropy of isothermal expansion of a van der Waals gas.
P 

RT
a

V ⫺b
V2

冢 冣 冢 冣
⭸S
⭸V
S2

冮S



T

⭸P
⭸T

dS  R

1


V

V2

冮V

⌬S  R ln

1

R
V ⫺b

1
dV
V ⫺b

V2 ⫺ b
V1 ⫺ b

In dealing with derivatives of the volume of a fluid with respect to T and P ,
it is convenient to use ␣ for the cubic expansion coefficient,
1 ⭸V
V ⭸T

冢 冣

␣

P



冢 冣

1 ⭸V
V ⭸T

(4.113)

P

and ␬ for the isothermal compressibility,

␬⫺

1 ⭸V
V ⭸P

冢 冣

T

⫺

冢 冣

1 ⭸V
V ⭸P

(4.114)

T

For an ideal gas these quantities are ␣  1/T and ␬  1/P . The derivative of U
with respect to V at constant T can be written in terms of ␣ and ␬ . According to
the cyclic rule (Appendix D.3),
⭸P

冢⭸T 冣

V

⫺


(⭸V /⭸T )P


(⭸V /⭸P )T

(4.115)

4.9 Special Topic: Additional Applications of Maxwell Relations

Substituting this in equation 4.111 yields

冢 冣
⭸U
⭸V



T

␣T ⫺ ␬P


(4.116)

The dependence of the enthalpy on the pressure can be obtained by dividing
dH  T dS  V dP by the differential of the pressure to obtain

冢 冣
⭸H
⭸P

T

T

冢 冣
⭸S
⭸P

V

(4.117)

T

Substituting equation 4.109 yields

冢 冣
⭸H
⭸P

 ⫺T

T

冢 冣
⭸V
⭸T

V

(4.118)

P

We found in Section 2.8 that
CP ⫺ CV  P 

冢 冣 冢 冣
⭸U
⭸V

T

⭸V
⭸T

(4.119)

P

Inserting (⭸V /⭸T )P  V␣ and using equation 4.116 we obtain
CP ⫺ CV 

T V␣ 2


(4.120)

Since it is difficult to measure CV , its value is generally calculated from measurements of CP , the molar volume V, the cubic expansion coefficient ␣ , and the
isothermal compressibility ␬ .


1.

2.

3.

4.

Nine Key Ideas in Chapter 4
The combined first and second law plus the introduction of chemical work
terms by Gibbs yields the fundamental equation for the internal energy. This
equation provides equations of state for T, P, and 兵␮i 其, Maxwell equations,
and an integrated equation for U.
The internal energy provides a criterion for spontaneous change and equilibrium when its natural variables S, V, and 兵ni 其 are held constant. However,
S and V are usually not convenient independent variables in the laboratory.
Legendre transforms can be used to define more useful thermodynamic potentials for this purpose. A Legendre transform is the definition of a new
thermodynamic potential by subtracting the product of conjugate variables
from another thermodynamic potential.
The Helmholtz energy is defined by A  U ⫺ TS and the Gibbs energy is
defined by G  H ⫺ TS. The Helmholtz energy provides the criterion for
spontaneous change at specified T, V, and amounts of species. The Gibbs
energy provides the criterion for spontaneous change at specified T, P, and
amounts of species. Since T and P are convenient independent variables, the
Gibbs energy is widely used as a criterion for phase equilibrium and chemical
equilibrium.
The Gibbs energy decreases in a spontaneous process at constant T and P,
and for a reversible process, the change in Gibbs energy is equal to the nonPV work that can be done on the surroundings by the system.

127

128

Chapter 4

Fundamental Equations of Thermodynamics

5.

6.

7.
8.

9.

The determination of ⌬G for a process or chemical reaction at a series of temperatures makes it possible to calculate ⌬H by use of the Gibbs–Helmholtz
equation.
The molar Gibbs energy for an ideal gas depends on its partial pressure, and
the molar Gibbs energy for a nonideal gas depends on its fugacity. The activity
of a real gas is equal to the ratio of its fugacity to the standard pressure.
At equilibrium the chemical potential of a species is the same in all of the
phases of the system.
Since the Gibbs energy of a phase is equal to the sum of the products of
amounts times molar Gibbs energies of species, all the other extensive thermodynamic properties of a one-phase system are also additive.
The complete Legendre transform for a system provides a relation between
the differentials of all of the intensive variables for a system, and so the intensive variables for a system are not independent. This remarkable equation is
known as the Gibbs–Duhem equation.

REFERENCES
M. Bailyn, A Survey of Thermodynamics. New York: American Institute of Physics, 1994.
J. A. Beattie and I. Oppenheim, Principles of Thermodynamics. New York: Elsevier Scientific, 1979.
K. E. Bett, J. S. Rowlinson, and G. Saville, Thermodynamics for Chemical Engineers. Cambridge, MA: MIT Press, 1975.
H. B. Callen, Thermodynamics and an Introduction to Thermostatistics. Hoboken, NJ: Wiley, 1985.
J. W. Gibbs, The Collected Works of J. Willard Gibbs. New Haven, CT: Yale University
Press, 1948.
W. Greiner, L. Neise, and H. Stöcker, Thermodynamics and Statistical Mechanics. New
York: Springer-Verlag, 1995.
K. S. Pitzer, Thermodynamics, 3rd ed. New York: McGraw-Hill, 1995.
S. I. Sandler, Chemical and Engineering Thermodynamics. Hoboken, NJ: Wiley, 1999.
J. M. Smith, H. C. Van Ness, and M. M. Abbott, Introduction to Chemical Engineering
Thermodynamics, 5th ed. New York: McGraw-Hill, 1996.
J. W. Tester and M. Modell, Thermodynamics and Its Applications. Upper Saddle River,
NJ: Prentice-Hall, 1997.

PROBLEMS
Problems marked with an icon may be more conveniently
solved on a personal computer with a mathematical program.
4.1 One mole of nitrogen gas is allowed to expand from 0.5 to
10 L. Calculate the change in molar entropy using (a) the ideal
gas law and (b) the van der Waals equation.
4.2 Derive the relation for CP ⫺ CV for a gas that follows the
van der Waals equation.
4.3 Earlier we derived the expression for the entropy of an
ideal gas as a function of T and P. Now that we have the Maxwell
relations, derive the expression for dS for any fluid.

4.4 What is the change in molar entropy of liquid benzene
at 25 ⬚C when the pressure is raised to 1000 bar? The coefficient of thermal expansion ␣ is 1.237 ⫻ 10⫺3 K⫺1 , the density
is 0.879 g cm⫺3 , and the molar mass is 78.11 g mol⫺1 .
4.5 Derive the expression for CP ⫺ CV for a gas with the following equation of state:
(P  a /V 2 )V  RT
4.6 What is the difference between the molar heat capacity of
iron at constant pressure and constant volume at 25 ⬚C? Given:

Problems

␣  35.1 ⫻ 10⫺6 K⫺1 , ␬  0.52 ⫻ 10⫺6 bar⫺1 , and the density
is 7.86 g cm⫺3 .
4.7 In equation 1.26 we saw that the compressibility factor of
a van der Waals gas can be written as
Z  1





1
a
b⫺
P  ⭈⭈⭈
RT
RT

(a) To this degree of approximation, derive the expression
for (⭸H /⭸P )T for a van der Waals gas. (b) Calculate
(⭸H /⭸P )T for CO2 (g) in J bar⫺1 mol⫺1 at 298 K. Given: a 
3.640 L2 bar mol⫺2 and b  0.042 67 L mol⫺1 .
4.8 Derive the expression for (⭸U /⭸V )T (the internal
pressure) for a gas following the virial equation with Z 
1  B /V.
4.9 In Section 3.4 we calculated that the enthalpy of freezing
water at ⫺10 ⬚C is ⫺5619 J mol⫺1 , and we calculated that the entropy of freezing water is ⫺20.54 J K⫺1 mol⫺1 at ⫺10 ⬚C. What
is the Gibbs energy of freezing water at ⫺10 ⬚C?
4.10 (a) Integrate the Gibbs–Helmholtz equation to obtain an
expression for ⌬G2 at temperature T2 in terms of ⌬G1 at T1 ,
assuming that ⌬H is independent of temperature. (b) Obtain an
expression for ⌬G2 using the more accurate approximation that
⌬H  ⌬H1  (T ⫺ T1 )⌬CP , where T1 is an arbitrary reference
temperature.
4.11 When a liquid is compressed its Gibbs energy is increased.
To a first approximation the increase in molar Gibbs energy can
be calculated using (⭸G /⭸P )T  V, assuming a constant molar
volume. What is the change in molar Gibbs energy for liquid
water when it is compressed to 1000 bar?
4.12 An ideal gas is allowed to expand reversibly and isothermally (25 ⬚C) from a pressure of 1 bar to a pressure of 0.1 bar.
(a) What is the change in molar Gibbs energy? (b) What would
be the change in molar Gibbs energy if the process occurred irreversibly?
4.13 The standard entropy of O2 (g) at 298.15 K and 1 bar is
listed in Table C.2 as 205.138 J K⫺1 mol⫺1 , and the standard
Gibbs energy of formation is listed as 0 kJ mol⫺1 . Assuming that
O2 is an ideal gas, what will be the molar entropy and molar
Gibbs energy of formation at 100 bar?
4.14 Helium is compressed isothermally and reversibly at
100 ⬚C from a pressure of 2 to 10 bar. Calculate (a) q per mole,
(b) w per mole, (c) ⌬G, (d) ⌬A, (e) ⌬H, ( f ) ⌬U, and (g) ⌬S,
assuming that helium is an ideal gas.
4.15 Toluene is vaporized at its boiling point, 111 ⬚C. The heat
of vaporization at this temperature is 361.9 J g⫺1 . For the vaporization of toluene, calculate (a) w per mole, (b) q per mole, (c)
⌬H, (d) ⌬U, (e) ⌬G, and ( f ) ⌬S.
4.16 If the Gibbs energy varies with temperature according to
G /T  a  b /T  c /T 2
how will the enthalpy and entropy vary with temperature?
Check that these three equations are consistent.

129

4.17 Calculate the change in molar Gibbs energy G when
supercooled water at ⫺3 ⬚C freezes at constant T and P. The
density of ice at ⫺3 ⬚C is 0.917 ⫻ 10 3 kg m⫺3 , and its vapor
pressure is 475 Pa. The density of supercooled water at ⫺3 ⬚C
is 0.9996 ⫻ 103 kg m⫺3 , and its vapor pressure is 489 Pa.
4.18 Calculate the molar Gibbs energy G of fusion when supercooled water at ⫺3 ⬚C freezes at constant T and P. The enthalpy of fusion of ice is 6000 J mol⫺1 at 0 ⬚C. The heat capacities
of water and ice in the vicinity of the freezing point are 75.3 and
38 J K⫺1 mol⫺1 , respectively.
4.19 At 298.15 K and a particular pressure, a real gas has
a fugacity coefficient ␾ of 2.00. At this pressure, what is the
difference in the chemical potential of this real gas and an
ideal gas?
4.20 As shown in Example 4.6, the fugacity of a van der Waals
gas is given by a fairly simple expression if only the second virial
coefficient is used. To this degree of approximation, derive the
expressions G, S, A, U, H, and V.
4.21 A mole of a van der Waals gas is expanded isothermally
from V1 to V2 . Derive the expressions for the changes in Helmoltz energy and internal energy.
4.22 A one-component system has three natural variables, as
is evident from
dU  T dS ⫺ P dV  ␮ dn
We have seen how three additional potentials can be defined
by making Legendre transforms and how a complete Legendre
transform yields the Gibbs–Duhem equation, which in a certain
sense is like a thermodynamic potential but has the value zero.
The total number of thermodynamic potentials for a system with
D natural variables is 2D . This is the number of Legendre transforms that can be made taking all possible pairs of conjugate
variables, two pairs, three pairs, . . . Since 23  8, there are three
more thermodynamic potentials for this system that can be defined by Legendre transforms. Write their fundamental equations; let’s call them X , Y , and Z .
4.23 Using the relation derived in Example 4.6, calculate the
fugacity of H2 (g) at 100 bar at 298 K.
4.24 Show that if the compressibility factor is given by Z 
1  BP /RT the fugacity is given by f  P eZ ⫺1 . If Z is not
very different from unity, eZ ⫺1  1  (Z ⫺ 1)  ⭈⭈⭈ ⬵ Z so that
f  PZ . Using this approximation, what is the fugacity of H2 (g)
at 50 bar and 298 K using its van der Waals constants?
4.25 Calculate the partial molar volume of zinc chloride
in 1 molal ZnCl2 solution using the following data:
% by weight
of ZnCl2
Density/g cm⫺3

2
1.0167

6
1.0532

10
1.0891

14
1.1275

18
1.1665

4.26 Calculate ⌬mix G and ⌬mix S for the formation of a quantity of air containing 1 mol of gas by mixing nitrogen and oxygen at 298.15 K. Air may be taken to be 80% nitrogen and 20%
oxygen.

130

Chapter 4

Fundamental Equations of Thermodynamics

4.27 A mole of gas A is mixed with a mole of gas B at 1 bar
and 298 K. How much work is required to separate these gases
to produce a container of each at 1 bar and 298 K?
4.28 The fundamental equation for the enthalpy is given by
equation 4.20. Show that the fundamental equation for the internal energy can be obtained by using the inverse Legendre
transform U  H ⫺ PV . This is an example of what is meant by
saying that there is no loss of information in making a Legendre
transform.
4.29 Derive

冢 冣

⭸2 G
CP  ⫺ T
⭸T 2

4.30 In studying statistical mechanics we will find (see Table
16.1) that for a monatomic ideal gas, the molar Gibbs energy is
given by
G  ⫺T ln

G  ⫺ 52 T ln T  T ln P
where the numerical factors have been omitted so that only
the functional dependence on the natural variables, T and
P , is shown. If we want to treat the thermodynamics of an
ideal monatomic gas at specified T and V without losing any
information, we cannot simply replace P with T /V and use
T
5
G  ⫺ T ln T  T ln
2
V
even though this relation is correct. If we want to treat the
thermodynamics of an ideal monatomic gas at specified T
and V without losing any information, we have to use the
following Legendre transform to define the molar Helmholtz
energy A:
A  G ⫺ PV
Use the expression for A obtained in this way to calculate S , V ,
H , and U for an ideal monatomic gas as a function of T and V .
Show that these expressions agree with the expressions obtained
in the preceding problem.
4.32 We already know enough about the thermodynamics of
a monatomic ideal gas to express V, U, and S in terms of the
natural variables of G, namely T, P, and n .
V  nRT /P
U  32 nRT



S⬚
 ln
R

冤冢 冣 冢 ⬚ 冣冥 冧
T
T⬚

5/2

The fundamental equation for the Gibbs energy is
dG  ⫺S dT  V dP  ␮ dn
Show that the correct expressions for S, V, and ␮ are obtained by
using the partial derivatives of G indicated by this fundamental
equation.
4.33 Show that
⭸U

⭸H

冢 ⭸S 冣  冢 ⭸S 冣
V

T 5/2
P

where numerical constants have been omitted so that only the
functional dependence on the natural variables of G , that is, T
and P , is shown. Derive the corresponding equations for S , H ,
V , U , and A.
4.31 Statistical mechanics shows that for a monatomic ideal
gas, the molar Gibbs energy is given by

S  nR

The last equation is the Sackur–Tetrode equation, where S ⬚ is
the molar entropy at the standard temperature T ⬚ (298.15 K)
and standard pressure P ⬚ (1 bar). The Gibbs energy G (T, P, n )
of the ideal monatomic gas can be calculated by using the Legendre transform
G  U  PV ⫺ TS

P
P

P

⭸H

⭸G

冢 ⭸P 冣  冢 ⭸P 冣
S

T

4.34 Earlier we derived the expression for the entropy of an
ideal gas as a function of T and V. Now that we have the Maxwell
relations, derive the expression for dS for any fluid.
4.35 The coefficient of thermal expansion ␣ of Fe(s) at 25 ⬚C
is 355 ⫻ 10⫺7 K⫺1 . What is the change in molar entropy of iron
when the pressure is raised to 1000 bar? (The density of iron at
25 ⬚C is 7.86 g cm⫺3 .)
4.36 Show that CP and CV for an ideal gas are independent of
volume and pressure.
4.37 Derive the thermodynamic equation of state

冢 冣
⭸U
⭸P

 V (␬P ⫺ ␣T )

T

4.38 Derive the expression for (⭸H /⭸P )T for a gas following
the virial equation
P V  RT  B (T )P
4.39 Assuming that the density of water is independent of pressure in the range 1 to 50 bar, what is the change in molar Gibbs
energy of water when the pressure is raised this amount?
4.40 An ideal gas is compressed isothermally from 1 to 5 bar
at 100 ⬚C. (a) What is the molar Gibbs energy change? (b) What
would have been the change in molar Gibbs energy if the compression had been carried out at 0 ⬚C?
4.41 At 298 K, for H2 (g), S ⬚  130.684 J K⫺1 , ⌬f H ⬚ 
0 kJ mol⫺1 , and ⌬f G ⬚  0 kJ mol⫺1 . What are the values of the
molar entropy, enthalpy of formation, and Gibbs energy of formation at 10⫺2 bar, assuming that H2 is an ideal gas?
4.42 The heat of vaporization of liquid oxygen at 1.013 25 bar
is 6820 J mol⫺1 at its boiling point, ⫺183 ⬚C, at that pressure. For
the reversible vaporization of liquid oxygen, calculate (a) q per
mole, (b) ⌬U, (c) ⌬S, and (d) ⌬G.
4.43 An ideal gas is expanded isothermally and reversibly at
1
0 ⬚C from 1 to 10
bar. Calculate (a) w per mole, (b) q per mole,
(c) ⌬H, (d) ⌬G, and (e) ⌬S for the gas. One mole of an ideal
gas in 22.71 L is allowed to expand irreversibly into an evacu-

Problems
ated vessel such that the final total volume is 227.1 L. Calculate
( f ) w per mole, (g) q per mole, (h) ⌬H, (i) ⌬G, and ( j ) ⌬S
for the gas. Calculate (k) ⌬S for the system and its surroundings involved in the reversible isothermal expansion and calculate (l) ⌬S for the system and its surroundings involved in an irreversible isothermal expansion in which the gas expands into
an evacuated vessel such that the final total volume is 227.1 L.
4.44 Steam is compressed reversibly to liquid water at the boiling point 100 ⬚C. The heat of vaporization of water at 100 ⬚C and
1.013 25 bar is 2258 J g⫺1 . Calculate w per mole and q per mole
and each of the thermodynamic quantities ⌬H, ⌬U, ⌬G, ⌬A,
and ⌬S.
4.45 An ideal gas at 300 K has an initial pressure of 15 bar and
is allowed to expand isothermally to a pressure of 1 bar. Calculate (a) the maximum work that can be obtained from the expansion, (b) ⌬U, (c) ⌬H, (d) ⌬G, and (e) ⌬S.
4.46 Calculate the molar entropy changes for the gas plus
reservoir in Examples 4.2 and 4.3.
4.47 An ideal gas is compressed reversibly at 100 ⬚C from 2 to
10 bar. Calculate ⌬H, ⌬S, and ⌬G. Show that you can get the
same value of ⌬G in two ways.
4.48 Toluene is vaporized at its boiling point (111 ⬚C). The
heat of vaporization at this temperature is 361.9 J g⫺1 .
Calculate ⌬H, ⌬S, and ⌬G.
4.49 What is the change in molar Gibbs energy for the freezing of water at ⫺10 ⬚C? The vapor pressure of H2 O(l, ⫺10 ⬚C)
is 286.5 Pa, and the vapor pressure of H2 O(s, ⫺10 ⬚C) is 260
Pa. At ⫺10 ⬚C the molar volume of supercooled water is
1.80 ⫻ 10⫺5 m3 mol⫺1 and the molar volume of ice is 2.00 ⫻
10⫺5 m3 mol⫺1 .
4.50 At low pressures the compressibility factor for a van der
Waals gas is given by
Z 





PV
a
P
 1 b ⫺
RT
RT RT

Derive the expression for ⌬G for a change in pressure from P1
to P2 .
4.51 For a gas that follows the equation of state P (V ⫺ b ) 
RT, show that the fugacity is given by
f  P ebP /RT
4.52 The apparent specific volume v of a solute (i.e., volume
contributed to the solution by one kilogram of solute) is equal to
the volume V of the solution minus the volume of pure solvent
it contains divided by the mass of solute:
V ⫺ (V␳s ⫺ m )/␳0
v
m
where m is the mass of solute, ␳s is the density of the solution,
and ␳0 is the density of the solvent. The density of a solution of

131

serum albumin containing 15.4 g of protein per liter is 1.0004 ⫻
103 kg m⫺3 at 25 ⬚C (␳0  0.977 07 ⫻ 103 kg m⫺3 ). Calculate the
apparent specific volume.
4.53 A solution of magnesium chloride, MgCl2 , in water containing 41.24 g L⫺1 has a density of 1.0311 ⫻ 103 kg m⫺3 at 20 ⬚C.
The density of water at this temperature is 0.998 ⫻ 103 kg m⫺3 .
Calculate (a) the apparent specific volume (see the equation in
Problem 4.52) and (b) the apparent molar volume of MgCl2 in
this solution.
4.54 Calculate ⌬G and ⌬S for mixing 2 mol of H2 with 1 mol of
O2 at 25 ⬚C under conditions where no chemical reaction occurs.
4.55 Liquid water can be superheated to 120 ⬚C at 1.01325 bar.
Calculate the changes in entropy, enthalpy, and Gibbs energy
for the process of superheated water at 120 ⬚C and 1.01325 bar
changing to steam at the same temperature and pressure. The
enthalpy of vaporization is 40.58 kJ mol⫺1 at 100 ⬚C and 1.01325
bar. Given: CP (H2 O, l)  75.3 J K⫺1 mol⫺1 and CP (H2 O, g) 
(36  0.013T ) J K⫺1 mol⫺1 , where T is in kelvins.

Computer Problems
4.A We know a good deal about the thermodynamics of
monatomic ideal gases, as we have already seen. Plot (a ) ␮ , (b )
S , and (c ) V versus temperature from 100 to 500 K and pressure
from 0.1 to 20 bar and discuss the slopes of each of the plots of
␮ in the two directions. The necessary equations are
(a) ␮  ⫺T ln(T 5/2 /P )
(b) S  ln(T 5/2 /P )
(c) V  T /P
4.B Given the virial equation for N2 in terms of pressure at
298.15 K, plot the compressibility factor and the fugacity from 0
to 1000 bar versus the pressure.
4.C Given the virial coefficients for O2 at 298.15 K in Table
1.1, plot the compressibility factor and the fugacity of the gas
versus its pressure up to 1000 bar.
4.D Plot the Gibbs energy of formation of an ideal gas relative
to its standard Gibbs energy at 298.15 K from 0 to 10 bar.
4.E The 3D plots of molar entropy and molar volume for a
monatomic ideal gas made in Computer Problem 4.A can be
made in another way by using
Sm  ⫺(⭸␮ /⭸T )P
Vm  (⭸␮ /⭸P )T
(a ) Plot Sm , (b ) plot Vm , (c ) plot ⫺(⭸␮ /⭸T )P , and (d ) plot
(⭸␮ /⭸P )T .
4.F (a ) Calculate the fugacity of molecular hydrogen at 100
bar and 25 ⬚ C using the virial coefficients in Table 1.1. (b ) Plot
the fugacity of molecular hydrogen from P  0 to 1000 bar at
25 ⬚ C.

5

Chemical Equilibrium

5.1
5.2
5.3
5.4
5.5
5.6
5.7
5.8
5.9
5.10
5.11

Derivation of the General Equilibrium Expression
Equilibrium Constant Expressions for Gas Reactions
Determination of Equilibrium Constants
Use of Standard Gibbs Energies of Formation
to Calculate Equilibrium Constants
Effect of Temperature on the Equilibrium Constant
Effect of Pressure, Initial Composition, and Inert Gases
on the Equilibrium Composition
Equilibrium Constants for Gas Reactions Written
in Terms of Concentrations
Heterogeneous Reactions
Degrees of Freedom and the Phase Rule
Special Topic: Isomer Group Thermodynamics
Special Topic: Chemical Equations as Matrix Equations

In 1864 Guldberg and Waage showed experimentally that in chemical reactions an
equilibrium is reached that can be approached from either direction. They were
apparently the first to realize that there is a mathematical relation between the
concentrations of reactants and products at equilibrium. In 1877 van’t Hoff suggested that in the equilibrium expression for the hydrolysis of ethyl acetate, the
concentration of each reactant should appear to the first power, corresponding to
the stoichiometric numbers in the balanced chemical equation.
The fundamental equation provides the basis for understanding chemical
equilibrium. The basic equations in terms of the chemical potential are completely general, but we will emphasize ideal gas reactions because of the simple
relation between the chemical potential of a species and its partial pressure. There
are brief discussions of multireaction equilibria and of gas–solid reactions. Chemical equilibrium in the liquid phase is discussed in later chapters. In discussing
systems at chemical equilibrium, it is important to know how many independent
variables there are; this question is answered by the phase rule, which is derived
in this chapter. The choice of independent variables is somewhat arbitrary, but

5.1 Derivation of the General Equilibrium Expression

the number is specified by the phase rule. The measurement of an equilibrium
constant yields the standard reaction Gibbs energy, and the measurement of the
temperature coefficient yields the standard reaction enthalpy.

5.1

DERIVATION OF THE GENERAL EQUILIBRIUM EXPRESSION

When discussing chemical equilibrium at a specified temperature and pressure,
we must add terms to the fundamental equation for dU (equation 4.5) similar to
those we added in Section 4.1 to get equation 4.16. These terms allow for changes
in the number of moles of all the species in the chemical reaction. This leads to
the fundamental equation for the Gibbs energy for a closed system when chemical
reactions are considered:
Ns

dG  ⫺S dT  V dP  冱 ␮i dni

(5.1)

i 1

where Ns is the number of species. When chemical reactions are involved, we
have already seen in Section 2.11 that the various dni in equation 5.1 are not
independent variables. If there is a single reaction, the amounts ni of the various
species at any time are given by
ni  ni 0  ␯i ␰

(5.2)

where ni 0 is the initial amount of species i , ␯i is the stoichiometric number of
the species in the reaction, and ␰ is the extent of reaction. Note that the extent
of reaction for a system is expressed in moles. Since dni  ␯i d␰, substituting this
relation into equation 5.1 yields
dG  ⫺ S dT  V d P 



Ns

冱 ␯i ␮i
i 1



d␰

(5.3)

Thus at specified T and P ,
⭸G
⭸␰

冢 冣

T,P

Ns

 冱 ␯i ␮i

(5.4)

i 1

At chemical equilibrium and constant T and P G has its minimum value; thus,
the derivative in equation 5.4 is zero:
Ns

冱 ␯i ␮i,eq  0

(5.5)

i 1

This equilibrium condition applies to all chemical equilibria, whether they involve
gases, liquids, solids, or solutions.
It is convenient to refer to the derivative in equation 5.4 as the reaction Gibbs
energy and represent it with ⌬r G:
⌬r G 

⭸G

冢 ⭸␰ 冣

T,P

(5.6)

133

134

Chapter 5

Chemical Equilibrium

Therefore, we will use equation 5.4 in the form

Completion

Ns

G

ξ

0

ξe

(a)

Completion

G

⌬r G  冱 ␯i ␮i

(5.7)

i 1

The reaction Gibbs energy is the change in Gibbs energy when the extent of reaction ␰ changes by 1 mol, as specified by a balanced chemical equation, at specified
partial pressures or concentrations of the species involved.
When a reaction takes place at constant temperature and pressure, the Gibbs
energy decreases, and the reaction continues until the Gibbs energy has reached
a minimum value, as shown by Fig. 5.1. It is of interest to note that the thermodynamic condition for chemical equilibrium (equation 5.5) has the same form as
the reaction to which it applies (equation 2.88), except that molecular formulas
are replaced by the corresponding chemical potentials of the reactant and product species. Substituting the expression for the chemical potential of a species at
equilibrium,

␮i,eq  ␮i⬚  RT ln ai,eq

(5.8)

into equation 5.5 yields
Ns

Ns

i 1

i 1

冱 ␯i ␮i⬚  ⫺RT 冱 ␯i ln ai,eq
0

ξe

ξ

(5.9)

The stoichiometric numbers can be put in the exponential position to obtain
(b)

Figure 5.1 (a) Gibbs energy as a
function of extent of reaction ␰ at
constant T and P for a reaction in
a single phase that goes nearly to
completion. (b) Gibbs energy as a
function of ␰ for a reaction that does
not go very far toward completion.

Ns

Ns

冱 ␯i ␮i⬚  ⫺RT 冱 ln(ai,eq ) ␯
i 1

i

(5.10)

i 1

and the sum of logarithms can be replaced by the logarithm of a product:
Ns

Ns

i 1

i 1

冱 ␯i ␮i⬚  ⫺RT ln 冲 ai,␯eqi

(5.11)

Example 5.1 Changing a summation of logarithmic terms to a product of
exponential terms
Show how equation 5.10 is converted to equation 5.11. For a simple example, the sum is
given by
Sum  ln(a1␯1 )  ln(a2␯2 )
The product is given by
ln(a1␯1 a2␯2 )  ln(a1␯1 )  ln(a2␯2 )
This shows that the sum of logarithmic terms is equal to the logarithm of the product given
in equation 5.11.

Since the product of activities at equilibrium in equation 5.11 is so useful, it
is defined as the equilibrium constant K of the reaction:
K 

Ns

冲 ai,␯eq
i

i 1

(5.12)

5.1 Derivation of the General Equilibrium Expression

The equilibrium constant is a dimensionless quantity, but its magnitude depends
on the way the chemical equation is written because of the stoichiometric numbers. To interpret an equilibrium constant it is necessary to know the balanced
chemical equation to which it applies and the standard states of the species on
which the activities are based (see Section 4.5).
The quantity 冱 ␯i ␮i⬚ in equation 5.11 is equal to ⌬r G ⬚ , the standard reaction
Gibbs energy, and so equation 5.11 can be written
⌬r G ⬚  ⫺RT ln K

(5.13)

This is a very important equation because the equilibrium constant, which can be
determined experimentally, tells us the change in the standard Gibbs energy for
the reaction. Conversely, since ⌬r G ⬚ can be evaluated by other methods, K can
be calculated. Note that since ⌬r G ⬚ is a function only of T, K is also a function
only of T.
Now we go back to equation 5.7 to see how to calculate the reaction Gibbs energy ⌬r G under a particular set of conditions when we know the equilibrium constant. If we substitute ␮i  ␮i⬚  RT ln ai for the chemical potential of a species
in equation 5.7, we obtain
Ns

⌬r G  冱 ␯i ␮i⬚  RT
i 1

Ns

Ns

冱 ␯i ln ai  ⌬r G ⬚  RT ln 冲 ai␯
i 1

i

(5.14)

i 1

As we can see from this equation, ⌬r G ⬚ is the change in Gibbs energy for the reaction when the activities of reactants and products are all unity. In other words,
⌬r G ⬚ is the change in Gibbs energy when separate reactants in their standard
states are converted to the separate products in their standard states. ⌬r G is the
change in the Gibbs energy in a specified reaction when separated reactants at
specified activities are converted to separated products at specified activities. The
product in the last term is much like an equilibrium constant, except that the activities of reactants and products can have any values we want. This product of
activities is called the reaction quotient and is represented by Q :
Ns

冲 ai␯

Q 

(5.15)

i

i 1

so that
⌬r G  ⌬r G ⬚  RT ln Q

(5.16)

This equation gives the change in Gibbs energy for a specified chemical reaction
when the reactants and products have activities ai , so it can be used to test for
spontaneity in the forward direction (⌬r G ⬍ 0) or backward direction (⌬r G ⬎ 0).
We can substitute the definition of the Gibbs energy (G  H ⫺ TS ) in equation 5.6 to obtain
⌬r G 

⭸H

冢 ⭸␰ 冣

⫺T

T,P

 ⌬r H ⫺ T ⌬r S

⭸S

冢 ⭸␰ 冣

T,P

(5.17)

135

136

Chapter 5

Chemical Equilibrium

which provides a logical introduction of the reaction enthalpy ⌬r H, in agreement
with equation 2.93, and the reaction entropy ⌬r S that is given later, in equation
5.42. This equation also applies when the reactants and products are each in their
standard states, so that ⌬r G ⬚  ⌬r H ⬚ ⫺ T ⌬r S ⬚ .

Example 5.2 The general expression for an equilibrium constant
What is the most general equilibrium constant expression for the following reaction?
3C(graphite)  2H2 O(g)  CH4 (g)  2CO(g)
K 



2
aCH4 aCO
3 a2
aC
H2 O



eq

If the pressure is not too high, the graphite can be considered to be in its standard state so
that aC  1. The activities of the gases can be replaced by fi /P ⬚ or, if the pressure is low
enough, by Pi /P ⬚ .

5.2

EQUILIBRIUM CONSTANT EXPRESSIONS
FOR GAS REACTIONS

For real gases the activity is given by ai  fi /P ⬚ , where fi is the fugacity of the i th
species and P ⬚ is the standard state pressure.
In Section 4.5 we saw that the partial molar Gibbs energy, which we can now
refer to as the chemical potential, of a gas is given by
fi
P⬚

␮i  ␮i⬚  RT ln

(5.18)

This relation can be substituted into equation 5.5 and the same operations carried
out to obtain
K 

Ns

冲冢

i 1

fi,eq
P⬚

␯i



(5.19)

This equation is not used very often because of the difficulty in evaluating fi in a
mixture of gases, but it is the most general expression for the equilibrium constant
of a reaction involving real gases.
For ideal gases, we have seen earlier that
Pi
P⬚

␮i  ␮i⬚  RT ln

(5.20)

Substituting this relation in equation 5.5 and carrying out the operations in Section
5.1 yields
K 

Ns

␯i

冲冢 P⬚ 冣

i 1

Pi,eq

(5.21)

5.2 Equilibrium Constant Expressions for Gas Reactions

This equilibrium constant is also a function only of temperature. For real gases the
right-hand side of equation 5.21 will depend on pressure since in general for real
gases fi 苷 Pi . The term thermodynamic equilibrium constant is often used for the
equilibrium constant obtained by use of equation 5.19 or by using equation 5.21 at
low pressure. Calculations of K using tables of Gibbs energies of formation yield
thermodynamic equilibrium constants.
The value of an equilibrium constant cannot be interpreted unless it is accompanied by a balanced chemical equation and a specification of the standard
state of each reactant and product. The values of the stoichiometric numbers are
arbitrary to the extent that a chemical equation may be multiplied or divided by a
positive or negative integer. In this section we are considering gas reactions, and
so the standard state of each reactant and product is the pure gas at 1 bar in the
ideal gas state. Later, in Section 7.6, we will discuss the standard states of substances in liquid solution in more detail. In using equations 5.12, 5.19, 5.21, and
similar equations, we will omit the subscript eq.
The equilibrium extent of a chemical reaction in ideal gas mixtures depends
on just three independent variables: (1) pressure, (2) initial composition, and (3)
temperature. We will examine the effects of each of these variables and also the
effect of adding inert gases to the reaction mixture.

Example 5.3 How partial pressures determine whether a reaction goes
forward or backward
(a) A mixture of CO(g), H2 (g), and CH3 OH(g) at 500 K with PCO  10 bar, PH2  1 bar,
and PCH3 OH  0.1 bar is passed over a catalyst. Can more methanol be formed? Given:
⌬r G ⬚  21.21 kJ mol⫺1 .
CO(g , 10 bar)  2H2 (g , 1 bar)  CH3 OH(g , 0.1 bar)
⌬r G  ⌬r G ⬚  RT ln Q
 21.21 kJ mol⫺1  (0.008 314 5 kJ K⫺1 mol⫺1 )(500 K) ln
 2.07 kJ mol⫺1

(0.1)
(10)(1)2

Thus, the reaction as written is not spontaneous. (b) Can the following conversion occur at
500 K?
CO(g , 1 bar)  2H2 (g , 10 bar)  CH3 OH(g , 0.1 bar)
⌬r G  21.21 kJ mol⫺1  (0.008 314 5 kJ K⫺1 mol⫺1 )(500 K) ln

(0.1)
(1)(10)2

 ⫺7.51 kJ mol⫺1
Under these conditions the reaction is thermodynamically spontaneous.

To illustrate why gas reactions never go to completion, let us consider a simple
isomerization of ideal gas A to ideal gas B at constant pressure:
A(g)  B(g)

(5.22)

According to equation 4.47, the Gibbs energy of the reaction mixture at any extent
of reaction is
G  nA ␮A  nB ␮B

(5.23)

137

138

Chapter 5

Chemical Equilibrium

where nA is the amount of A and nB is the amount of B. If the reaction is started
with 1 mol of A, the amounts of A and B at a later time are given in terms of
extent of reaction ␰ by
nA 1 ⫺ ␰

(5.24)

nB ␰

(5.25)

Thus,
G  (1 ⫺ ␰)␮A  ␰␮B

(5.26)

From equation 5.20, the chemical potentials of A and B in the mixture of ideal
gases are given by

␮A  ␮A⬚  RT ln(PA /P ⬚ )
 ␮A
⬚  RT ln yA  RT ln(P /P ⬚ )
 ␮A
⬚  RT ln(1 ⫺ ␰)  RT ln(P /P ⬚ )
␮B  ␮B⬚  RT ln yB  RT ln(P /P ⬚ )
 ␮B⬚  RT ln ␰  RT ln(P /P ⬚ )

(5.27)

where P is the total pressure at equilibrium. Substituting these equations into
equation 5.26,
G  (1 ⫺ ␰)␮A
⬚  ␰␮B⬚  RT ln(P /P ⬚ )  RT [(1 ⫺ ␰) ln(1 ⫺ ␰)  ␰ ln ␰]
 ␮A
(5.28)
⬚ ⫺ (␮A⬚ ⫺ ␮B⬚ )␰  RT ln(P /P ⬚ )  ⌬mix G ⬚
where ⌬mix G ⬚ is the Gibbs energy of mixing (1 ⫺ ␰) mol of A with ␰ mol of B. Figure 5.2 gives a plot of G versus ␰. The first three terms give the linear function. It
is the mixing term that causes the minimum in the plot of G versus extent of reaction ␰. At constant temperature and pressure, the criterion of equilibrium is that
the Gibbs energy is a minimum (see Section 4.2). Thus, starting with A, the Gibbs
energy can decrease along the curve until ␰eq mol of B have been formed. Starting
with B, the Gibbs energy can decrease until (1 ⫺ ␰eq ) mol of A have formed.
Even though B has the lower value of the standard molar Gibbs energy
(␮A
⬚ ⬎ ␮B⬚ ), the system can achieve a lower Gibbs energy by having some A
present at equilibrium with the resulting Gibbs energy of mixing. Generalizing
from this example, we can say that no chemical reaction of gases goes to completion; nevertheless, it may be very difficult to detect reactants at equilibrium if the
products have a very much lower Gibbs energy.
A surprising feature of Fig. 5.2 is that the slope dG /d␰ approaches ⫺⬁ as
␰ y 0 and ⬁ as ␰ y 1. This can be shown by differentiating equation 5.28 with
respect to ␰ and looking at the limits.

5.3

DETERMINATION OF EQUILIBRIUM CONSTANTS

If the initial concentrations of the reactants are known and only one reaction occurs, it is necessary to determine the concentration of only one reactant or product at equilibrium to be able to calculate the concentrations or pressures of the
others by means of the balanced chemical equation. Chemical methods based on
chemical reaction with one of the reactants or products can be used for such analyses only when the reaction being studied can be stopped at equilibrium, as by a

40
39.5
39
38.5
38
37.5
37

15
10
5

dG/dG
ξ

G

5.3 Determination of Equilibrium Constants

0

0.2

0.4

0.6

0.8

1

–5
–10

36.5
0.2

0.4

0.6

0.8

–15

1

(b)

ξ
(a)
100

d2G/dG
ξ2

80
60
40
20
0.2

0.4

0.6

0.8

1

ξ
(c)

Figure 5.2 (a) Gibbs energy of the reaction system A(g)  B(g) versus the extent of
reaction ␰ at constant temperature. (b) Derivative of the Gibbs energy with respect to
extent of reaction, which is zero at the minimum Gibbs energy. (c) Second derivative of
the Gibbs energy with respect to the extent of reaction, which is positive over the whole
range. Note that the second derivative has to be positive for the equilibrium to be stable.
(See Computer Problem 5.N.)

very sudden chilling to a temperature where the rate of further chemical change is
negligible, or by destruction of a catalyst. Otherwise, the concentrations will shift
during the chemical analysis.
Measurements of physical properties, such as density, pressure, light absorption, refractive index, electromotive force, and electrical conductivity, are especially useful for determining the concentrations of reactants at equilibrium since,
for these methods, it is unnecessary to “stop” the reaction.
It is essential to know that equilibrium has been reached before the analysis
of the mixture can be used for calculating the equilibrium constant. The following
criteria for the attainment of equilibrium at constant temperature are useful:
1.
2.

The same value of the equilibrium constant should be obtained when the
equilibrium is approached from either side.
The same value of the equilibrium constant should be obtained when the
initial concentrations of reacting material are varied over a wide range.

The determination of the density of a partially dissociated gas provides one
of the simplest methods for measuring the extent to which the gas is dissociated.
When a gas dissociates, more molecules are produced, and at constant temperature and pressure the volume increases.
If the equilibrium extent of reaction ␰ has been determined, we need the expression for the equilibrium constant in terms of the equilibrium extent of reaction and the pressure. We will assume that the gases are ideal. The first step is
to express the mole fractions of reactants and products, in terms of the extent of
reaction ␰.

139

140

Chapter 5

Chemical Equilibrium

For example, consider the dissociation of an initial amount n0 (N2 O4 ) of N2 O4
to NO2 at a given temperature and pressure. If the equilibrium extent of reaction is ␰, the equilibrium amount of N2 O4 is n0 (N2 O4 ) ⫺ ␰ and the equilibrium
amount of NO2 is 2␰. However, since the equilibrium composition at a given temperature and pressure is independent of the size of the system, we might as well
divide these expressions by n0 (N2 O4 ) and represent the equilibrium amounts by
the dimensionless quantities 1 ⫺ ␰⬘ for N2 O4 and 2␰⬘ for NO2 . The quantity ␰⬘ is a
dimensionless extent of reaction defined by ␰⬘  ␰/n0 (N2 O4 ). In working equilibrium problems we will leave off the prime to simplify the notation and write 1 ⫺ ␰
rather than 1 mol ⫺␰. The following format is recommended:
N2 O4 (g)  2NO2 (g )
Initial amount 1
0
Equilibrium amount 1 ⫺ ␰
2␰ Total amount  1  ␰
1⫺␰
2␰
Equilibrium mole fraction
1␰
1␰
(PNO2 /P ⬚ )2
PN2 O4 /P ⬚
兵[2␰/(1  ␰)](P /P ⬚ )其2

[(1 ⫺ ␰)/(1  ␰)](P /P ⬚ )
4␰ 2 P /P ⬚

1⫺␰2

(5.29)

K 

(5.30)

At equilibrium the amount of N2 O4 (g) is 1 ⫺ ␰ and the amount of NO2 (g) is 2␰,
so that the total amount is 1  ␰. The partial pressures of the reactants at equilibrium are obtained by multiplying their equilibrium mole fractions by the total
pressure P. Equation 5.30 gives the relation between the equilibrium constant K,
the equilibrium extent of reaction ␰, and the total pressure P. This may be solved
for the equilibrium extent of reaction to obtain

␰

1
[1  (4/K )(P /P ⬚ )]1/2

(5.31)

This is the dimensionless extent of reaction obtained by dividing the extent of
reaction ␰ by n0 (N2 O4 ). Equations 5.30 and 5.31 apply to any dissociation reaction
of the type A(g)  2B(g) (see Fig. 5.3), but not to dissociation reactions of the
type A(g)  B(g)  C(g). For reactions of the latter type, the factor 4 in equation
5.31 is replaced by 1.
1.0
K = 10

0.8
0.6
ξ

0.4

K=1

0.2

K = 0.1

0.0

0

1

2

3
P/bar

4

5

Figure 5.3 Extent of reaction
␰ at equilibrium for the reaction
A(g)  2B(g) as a function of
P /P ⬚ , for various values of K, using equation 5.31. (See Computer
Problem 5.H.)

5.3 Determination of Equilibrium Constants

The equilibrium extent of reaction 5.29 is readily determined by measuring
the density of the partially dissociated gas. For example, assume that we start with
a mass m of N2 O4 (g). The initial volume is V1  mRT /M1 P , where M1 is the
molar mass of N2 O4 (92.01 g mol⫺1 ). If the gas is held at a constant pressure and
temperature, the equilibrium volume is given by V2  mRT /M2 P , where M2 is
the average molar mass of the partially dissociated gas, which is defined by M2 
MN2 O4 yN2 O4  MNO2 yNO2 . Thus , V1 /V2  M2 /M1 . This ratio is equal to 1/(1  ␰),
and so

␰

M1 ⫺ M2
M2

(5.32)

Example 5.4 Calculation of the equilibrium constant for a gas reaction from
the equilibrium density
When nitrogen tetroxide is held in a container at constant T and P near room temperature
or higher, it rather quickly reaches an equilibrium degree of dissociation. If 1.588 g of nitrogen tetroxide gives a total pressure of 1.0133 bar when partially dissociated in a 500⫺cm3
glass vessel at 25 ⬚ C, what is the extent of reaction? What is the value of K? What is the
extent of reaction at a total pressure of 0.5 bar?
The average molar mass of the gas at equilibrium is given by
M2 

RT m
(0.083 145 L bar K⫺1 mol⫺1 )(298.15 K)(1.588 ⫻ 10⫺3 kg)

P V
(1.0133 bar)(0.5 L)

 77.70 g mol⫺1

␰
K 

92.01 ⫺ 77.70
 0.1842
77.70
4␰ 2 (P /P ⬚ )
(4)(0.1842)2 (1.0133)

 0.143
2
1⫺␰
1 ⫺ (0.1842)2

The extent of reaction at 0.5 bar is calculated using equation 5.31:

␰



0.143
0.143  4(0.5)



1/2

 0.258

As an example of a more complicated gas reaction, consider that ammonia is
produced by holding an initial mixture containing equal amounts of nitrogen and
hydrogen at constant temperature and high pressure in contact with a catalyst.
Note that reactants are not always added in stoichiometric proportions. In the actual production of ammonia, hydrogen is, of course, the more expensive reactant.
Again we use the dimensionless extent of reaction:
N2 (g)  3H2 (g) 
Initial amount 1
Equilibrium amount 1 ⫺ ␰
1⫺␰
Equilibrium mole fraction
2 ⫺ 2␰

1
1 ⫺ 3␰
1 ⫺ 3␰
2 ⫺ 2␰

2NH3 (g)

(5.33)

0
2␰ Total amount  2 ⫺ 2␰
2␰
2 ⫺ 2␰

141

142

Chapter 5

Chemical Equilibrium

Note that when ␰ moles of N2 have been used up, 3␰ moles of H2 will have reacted,
and 2␰ moles of NH3 will have been formed. Thus, the total amount of reactants
and products at equilibrium is 2 ⫺ 2␰. The equilibrium constant is given by
K 

16␰ 2 (1 ⫺ ␰)
(1 ⫺ 3␰)3 (P /P ⬚ )2

(5.34)

If the balanced chemical equation is divided by 2, the equilibrium constant for
it will be the square root of K as expressed by equation 5.34. If the balanced
chemical equation is reversed, the equilibrium constant for it will be the reciprocal
of K in equation 5.34.

Example 5.5
to ammonia

The total pressure to obtain 10% conversion of nitrogen

What total pressure must be used to obtain a 10% conversion of nitrogen to ammonia
at 400 ⬚C, assuming an initially equimolar mixture of nitrogen and hydrogen and ideal gas
behavior? The equilibrium constant for the formation of NH3 (g) according to reaction 5.33
with a standard state pressure of 1 bar is 1.60 ⫻ 10⫺4 at 400 ⬚C.
We use equation 5.34 to obtain
1.60 ⫻ 10⫺4 

16(0.10)2 (0.90)
(0.70)3 (P /P ⬚ )2

P
 51.2
P⬚
P  51.2 bar

Example 5.6 Calculation of the equilibrium constant for the forward reaction
and the reverse reaction
(a) What is the value of the standard Gibbs energy for reaction 5.33 at 400 ⬚C? (b) What
is the value of the equilibrium constant and the standard reaction Gibbs energy when
reaction 5.33 is divided by 2? (c) What is the value of the equilibrium constant and the
standard reaction Gibbs energy when reaction 5.33 is reversed?
(a )

⌬r G ⬚  ⫺RT ln K
 ⫺(8.315 J K⫺1 mol⫺1 )(673 K) ln(1.60 ⫻ 10⫺4 )
 48.91 kJ mol⫺1

(b )

K  (1.60 ⫻ 10⫺4 )1/2  0.012 65
⌬r G ⬚  ⫺RT ln 0.012 65
 24.46 kJ mol⫺1

(c )

K 

1
 6250
(1.60 ⫻ 10⫺4 )

⌬r G ⬚  ⫺RT ln 6250
 ⫺48.91 kJ mol⫺1

5.4 Use of Standard Gibbs Energies of Formation to Calculate Equilibrium Constants

The calculation of the equilibrium composition of a reaction system involving
two or more reactions is more complicated, because the equilibrium composition
is the one that satisfies a set of independent equilibrium constant expressions for
the system and a set of independent conservation equations for components (see
Sections 5.9 and 5.11). This solution of simultaneous nonlinear equations cannot
be obtained analytically, but requires an iterative process in which some kind of
initial guess is improved using the Newton–Raphson process. Computer programs
such as equcalc (see Computer Problem 5.K) have been written to do this, and
Mathematica also provides Solve, which can do this.

5.4

USE OF STANDARD GIBBS ENERGIES OF FORMATION TO
CALCULATE EQUILIBRIUM CONSTANTS

There are three ways in which ⌬r G ⬚ for a reaction may be obtained: (1) ⌬r G ⬚
may be calculated from a measured equilibrium constant using equation 5.13, (2)
⌬r G ⬚ may be calculated from
⌬r G ⬚  ⌬r H ⬚ ⫺ T ⌬r S ⬚

(5.35)

using ⌬r H ⬚ obtained calorimetrically and ⌬r S ⬚ obtained from the third law
entropies, and (3) for gas reactions ⌬r G ⬚ may be calculated using statistical
mechanics (Chapter 16) and certain information about molecules obtained from
spectroscopic data. Methods 2 and 3 make it possible to calculate equilibrium
constants of reactions that have never been studied in the laboratory. In method
2 the necessary data are obtained solely from thermal measurements, including
heat capacity measurements down to the neighborhood of absolute zero. The
calculation of equilibrium constants for gas reactions using statistical mechanics
is even more remarkable in that only properties of the individual molecules are
used to calculate equilibrium constants for reactions of ideal gases. For the simplest reactions ⌬r H ⬚ may be calculated from spectroscopic data; however, for
more complicated reactions, the calculation of ⌬r H ⬚ requires calorimetric data.
Rather than tabulating values of equilibrium constants of reactions, or of
⌬r G ⬚ values calculated using equation 5.13, it is more convenient to tabulate values of the standard Gibbs energy of formation ⌬f Gi⬚ , which is the standard Gibbs
energy for the formation of a mole of i from its elements. The standard Gibbs
energy of formation of i is related to the standard enthalpy of formation of i and
the standard entropy of i by
⌬f Gi⬚  ⌬f Hi⬚ ⫺ T 冢S ⬚i ⫺ 冱 ␯e S ⬚e 冣

(5.36)

where 冱 ␯e S ⬚e is the sum of the standard entropies of the elements in the formation reaction for species i . Since values of ⌬f Gi⬚ are tabulated, the equation for
⌬r G ⬚ used in equation 5.14 (⌬r G ⬚  冱 ␯i ␮i⬚ ) can be written as
Ns

⌬r G ⬚  冱 ␯i ⌬f Gi⬚

(5.37)

i 1

where the ␯i ’s are the stoichiometric numbers from the balanced chemical equation. The standard Gibbs energies of formation of elements in their reference
states are zero at all temperatures.

143

144

Chapter 5

Chemical Equilibrium

Standard Gibbs energies of formation at 298.15 K and 1 bar are provided for
about 15 000 species in the NBS Tables of Chemical Thermodynamic Properties.*
Some values for those tables are given in Table C.2. Standard Gibbs energies of
formation at temperatures up to 6000 K are given in the JANAF tables.† Some
values from those tables are given in Table C.3. Gibbs energies of formation of
several hundred organic compounds up to 1000 K are given in Stull et al.‡ The
values in the latter table need to be converted to joules, and to a standard state
pressure of 1 bar, before being used with values in the other tables.§
Most chemical reactions that occur are exothermic; that is, ⌬r H ⬚ is negative. However, some endothermic reactions do occur. Endothermic reactions have
equilibrium constants greater than unity only if T ⌬r S ⬚ is sufficiently positive to
give a negative ⌬r G ⬚ . In the case of gaseous reactions, this may happen if there
are more molecules on the right-hand side of the chemical equation than the left.
Note that in the limit of high temperature, reactions with positive ⌬r S ⬚ have equilibrium constants greater than unity, independent of ⌬r H ⬚ .

Example 5.7
H2O (g)

Calculation of the standard Gibbs energy of formation for

Given the following calorimetric information, calculate the standard Gibbs energy of formation of H2 O(g) at 298.15 K.
⌬f H ⬚ /kJ mol⫺1

S ⬚ /J K⫺1 mol⫺1

⫺241.818
0
0

188.825
130.684
205.138

H2 O(g)
H2 (g)
O2 (g)

The standard Gibbs energy of formation is the standard reaction Gibbs energy for the
following reaction.
H2 (g)  12 O2 (g)  H2 O(g)
⌬f G ⬚ (H2 O, g)  ⌬f H ⬚ (H2 O, g) ⫺ T ⌬f S ⬚ (H2 O, g)

 ⫺241.818 ⫺ (298.15)[188.825 ⫺ 130.684 ⫺ (0.5)(205.138)](10⫺3 )
 ⫺228.572 kJ mol⫺1

Example 5.8 Use of tables to calculate equilibrium constants
Calculate the equilibrium constants for the following reactions at the indicated
temperature.
at 25 ⬚C

(a )

3O2 (g)  2O3 (g)

(b )

CO(g)  2H2 (g)  CH3 OH(g)

at 500 K

*D. D. Wagman et al., The NBS Tables of Chemical Thermodynamic Properties, J. Phys. Chem. Ref.
Data 11 (suppl. 2) (1982).
†M. W. Chase et al., JANAF Thermochemical Tables, J. Phys. Chem. Ref. Data 14 (suppl. 1) (1985).
‡D. R. Stull, E. F. Westrum, and G. C. Sinke, The Chemical Thermodynamics of Organic Compounds.
New York: Wiley, 1969.
§ R. D. Freeman, J. Chem. Educ. 62:681 (1985).

5.5 Effect of Temperature on the Equilibrium Constant
(a)

Use Table C.2 to obtain
⌬r G ⬚  2⌬f G ⬚ (O3 , g) ⫺ 3⌬f G ⬚ (O2 , g)
 2(163.2)  326.4 kJ mol⫺1




K  exp ⫺
 exp ⫺

⌬r G ⬚
RT



326 400
(8.3145)(298.15)



 6.62 ⫻ 10⫺58
(b)

Use Table C.3 to obtain
⌬r G ⬚  ⫺134.27 ⫺ (⫺155.41)  21.14 kJ mol⫺1



K  exp ⫺

21 140
(8.3145)(500)



 6.19 ⫻ 10⫺3

Comment:
There are two ways to store information on equilibrium constants and use it
to calculate equilibrium constants of reactions that have not been studied. One
way is to tabulate reactions and equilibrium constants and then calculate new
equilibrium constants by adding and subtracting reactions from the tabulation.
When two reactions are added, the equilibrium constant of the new reaction
is equal to the product of the two equilibrium constants. When one reaction is
subtracted from the other, the equilibrium constant of the new reaction is equal
to the ratio of the two equilibrium constants. The other way, which is used in
thermodynamic tables, makes use of equation 5.37 to calculate standard Gibbs
energies of formation for species and tabulate them. This has the advantage that
although a species can be involved in a large number of reactions, it requires only
one entry in a table. Note that equilibria between different phases can be handled
in the same way.

5.5

EFFECT OF TEMPERATURE ON THE EQUILIBRIUM CONSTANT

The effect of temperature on chemical equilibrium is determined by ⌬r H ⬚ , as
shown by the Gibbs–Helmholtz equation 4.63. If ⌬r G ⬚  ⫺RT ln K is substituted into this equation, we obtain
⌬r H ⬚  ⫺T 2





d(⌬r G ⬚ /T )
d ln K
 RT 2
dT
dT





(5.38)

or
⌬r H ⬚

冢 dT 冣  RT
d ln K

2

(5.39)

This equation is often called the van’t Hoff equation. Note that since K is independent of P for an ideal gas, the left-hand side need not be written as a partial
derivative.

145

146

Chapter 5

Chemical Equilibrium

Thus, for an endothermic reaction the equilibrium constant increases as the
temperature is increased, and for an exothermic reaction the equilibrium constant
decreases as the temperature is increased.
According to Le Châtelier’s principle, when an equilibrium system is perturbed, the equilibrium will always be displaced in such a way as to oppose the
applied change. When the temperature of an equilibrium system is raised, this
change cannot be prevented by the system, but what happens is that the equilibrium shifts in such a way that more heat is required to heat the reaction mixture
to the higher temperature than would have been required if the mixture were inert. In other words, when the temperature is raised, the equilibrium shifts in the
direction that causes an absorption of heat.
If ⌬r H ⬚ is independent of temperature, the integral of equation 5.39 from T1
to T2 yields
ln

⌬r H ⬚ (T2 ⫺ T1 )
K2

K1
RT1 T2

(5.40)

If ⌬r H ⬚ is independent of temperature, then ⌬r CP⬚ is zero. If ⌬r CP⬚ is zero,
then ⌬r S ⬚ is also independent of temperature, as shown later in equation 5.46.
Thus, when ⌬r CP⬚  0, the temperature dependence of the equilibrium constant
is given by ⌬r G ⬚  ⫺RT ln K  ⌬r H ⬚ ⫺ T ⌬r S ⬚ , or
ln K  ⫺

⌬r H ⬚
⌬r S ⬚

RT
R

(5.41)

According to this equation a plot of ln K versus 1/T is linear over a temperature
range in which ⌬r H ⬚ and ⌬r S ⬚ for the reaction are constant.

Example 5.9 Calculation of the standard enthalpy of reaction and standard
entropy of reaction from the equilibrium as a function of temperature
Calculate ⌬r H ⬚ and ⌬r S ⬚ for the reaction
N2 (g)  O2 (g)  2NO(g)
from the following values of K :
T /K
K /10⫺4

1900
2.31

2000
4.08

2100
6.86

2200
11.0

2300
16.9

2400
25.1

2500
36.0

2600
50.3

These data are plotted in Fig. 5.4.
Since the plot is linear, we can use equation 5.41. The slope of the plot is ⫺2.19 ⫻ 104 K
and so
⌬r H ⬚  ⫺slope ⫻ R  ⫺(⫺2.19 ⫻ 10 4 K)(8.3145 J K⫺1 mol⫺1 )
 182 kJ mol⫺1
The intercept of Fig. 5.4 at 1/T  0 can be calculated from the experimental value of K
at some temperature and the slope. The intercept may be used to calculate the standard
entropy change ⌬r S ⬚ according to equation 5.41.
⌬r S ⬚
 3.13
R
⌬r S ⬚  (3.13)(8.3145 J K⫺1 mol⫺1 )  26.0 J K⫺1 mol⫺1

5.5 Effect of Temperature on the Equilibrium Constant
–5

ln K

–6

–7

–8

–9
0.003 75 0.0004 0.000 425 0.000 45 0.000 475 0.0005 0.000 525 0.000 55
K/T

Figure 5.4 Plot of ln K against reciprocal absolute temperature for the reaction N2 (g) 
O2 (g)  2NO(g). The standard enthalpy of reaction is calculated from the slope of the
straight line. (See Computer Problem 5.B.)

The standard reaction entropy ⌬r S ⬚ for a reaction 冱 ␯i Bi  0 is equal to
the change in entropy when the separated reactants, each in its standard state,
are completely converted to separated products, each in its standard state, at the
specified temperature. ⌬r S ⬚ can be calculated by taking the partial derivative of
⌬r G ⬚ with respect to temperature T , as indicated by equation 4.39. The standard
reaction entropy ⌬r S ⬚ at a particular temperature is given by
Ns

⌬r S ⬚  冱 ␯i Si ⬚
i 1

where Ns is the number of species involved in the reaction.
Example 5.10 Calculation of standard reaction entropies using tables
Calculate the standard reaction entropies for the following reactions at 298 K:
(a )

H2 (g)  12 O2 (g)  H2 O(l)

(b )

N2 (g)  3H2 (g)  2NH3 (g)

(c )

CaCO3 (s, calcite)  CaO(s)  CO2 (g)

(d )

N2 O4 (g)  2NO2 (g)
The following reaction entropies are calculated using Table C.2:

(a )

⌬r S ⬚  69.91 ⫺ 130.68 ⫺ 12 (205.14)
 ⫺163.34 J K⫺1 mol⫺1

(b )

⌬r S ⬚  2(192.45) ⫺ 191.61 ⫺ 3(130.68)
 ⫺198.75 J K⫺1 mol⫺1

(c )

⌬r S ⬚  39.75  213.74 ⫺ 92.9
 160.59 J K⫺1 mol⫺1

(d )

⌬r S ⬚  2(240.06) ⫺ 304.29
 175.83 J K⫺1 mol⫺1

(5.42)

147

148

Chapter 5

Chemical Equilibrium
Note that since the molar entropies of gases are greater than those of liquids and solids,
the entropy always increases when the reaction produces more moles of gaseous products
than reactants.

In general, ⌬r H ⬚ and ⌬r S ⬚ depend on the temperature because the heat capacities of the reactants depend on temperature, and we have seen earlier that
⌬f Hi⬚ (T )  ⌬f Hi⬚ (298.15 K) 
Si⬚ (T )  Si⬚ (298.15 K) 





T

298.15

T

298.15

CPi⬚ dT ⬘

(5.43)

CPi⬚
dT ⬘
T⬘

(5.44)

where CPi⬚ can be represented by a power series in T (see Table 2.2). Since
⌬r H ⬚  冱 ␯i ⌬f Hi⬚ and ⌬r S ⬚  冱 ␯i Si⬚ ,
⌬r H ⬚ (T )  ⌬r H ⬚ (298.15 K) 
⌬r S ⬚ (T )  ⌬r S ⬚ (298.15 K) 



T

⌬r CP⬚ dT ⬘

(5.45)

⌬r CP⬚
dT ⬘
T⬘

(5.46)

298.15



T

298.15

Substituting these relations in ⌬r G ⬚ (T )  ⌬r H ⬚ (T ) ⫺ T ⌬r S ⬚ (T ) yields
⌬r G ⬚ (T )  ⌬r G ⬚ (298.15) 



T

298.15

⌬r CP⬚ dT ⬘ ⫺ T



T

298.15

⌬r CP⬚
dT ⬘ (5.47)
T⬘

Since ln K  ⫺⌬r G ⬚ /RT ,
ln K (T ) 

(298.15)
1
ln K (298.15 K) ⫺
T
RT



T

298.15

⌬r CP⬚ dT ⬘ 

1
R



T

298.15

⌬r CP⬚
dT ⬘
T⬘
(5.48)

The calculation of ln K (T ) in this way is pretty tedious without a computer, but
it can easily be made using a mathematical program that can integrate.
Another way to do this calculation is to use the Gibbs–Helmholtz equation
(see equation 4.62):





⌬r H ⬚
d(⌬r G ⬚ /T )
⫺
T2
dT

(5.49)

The effect of temperature on an endothermic reaction A(g)  2B(g) is shown
in Fig. 5.5 for three total pressures. As the temperature is increased, the equilibrium extent of reaction increases. At a given temperature the extent of reaction
is greater at a lower pressure.
Example 5.11 The dependence of the equilibrium constant on temperature
when ⌬r CP⬚ is constant
When ⌬r H ⬚ and ⌬r S ⬚ depend only slightly on T, it may be sufficient to assume that ⌬r CP⬚
is constant. Derive the expression for ln K(T ) when ⌬r CP⬚ is constant.

5.5 Effect of Temperature on the Equilibrium Constant
1
0.1 bar
0.8
1 bar
0.6
ξ

0.4

10 bar

0.2
0
300

320

340

360

380

400

T/ K

Figure 5.5 Extent of reaction ␰ versus temperature for an endothermic reaction A(g) 
2B (g ), ⌬r H ⬚  50 kJ mol⫺1 , at three total pressures. (See Computer Problem 5.I.)
The temperature dependencies of the standard enthalpy of reaction and standard entropy of reaction are given by
⌬r HT⬚  ⌬r H298
⬚  ⌬r CP⬚ (T ⫺ 298.15 K)
⌬r ST⬚  ⌬r S298
⬚  ⌬r CP⬚ ln

T
298.15 K

Substituting these relations in ⫺RT ln K  ⌬r HT⬚ ⫺ T ⌬r ST⬚ yields
ln K  ⫺

⌬r H298
⌬r S298
⌬r CP⬚
T
298.15 K


1⫺

⫺ ln

RT
R
R
T
298.15 K





Example 5.12 Dependence of the standard Gibbs energy of reaction
on temperature when the standard enthalpy of reaction is independent of
temperature
At any temperature,
⌬r G ⬚ (T )  ⌬r H ⬚ (T ) ⫺ T ⌬rS ⬚ (T )
When the standard enthalpy of reaction is independent of temperature because ⌬r CP⬚ is
independent of temperature, the standard entropy of reaction is also independent of temperature (see equations 5.43 and 5.44). Thus the standard Gibbs energy of a reaction at a
temperature other than 298.15 K can be calculated using
⌬r G ⬚ (T )  ⌬r H ⬚ (298.15 K) ⫺ T ⌬r S ⬚ (298.15 K)
When ⌬r G ⬚ (298.15) K and ⌬r H ⬚ (298.15) K are available, the standard entropy of reaction
can be calculated using
⌬r S ⬚ (298.15 K) 

⌬r H ⬚ (298.15 K) ⫺ ⌬r G ⬚ (298.15 K)
298.15 K

Substituting this into the expression for ⌬G ⬚ (T ) yields
⌬r G ⬚ (T )  (T /298.15 K)⌬r G ⬚ (298.15 K)  (1 ⫺ T /298.15 K) ⌬r H ⬚ (298.15 K)

149

150

Chapter 5

Chemical Equilibrium

5.6

EFFECT OF PRESSURE, INITIAL COMPOSITION, AND INERT
GASES ON THE EQUILIBRIUM COMPOSITION

For an ideal mixture of ideal gases the equilibrium partial pressures of the reactants and products can be expressed in terms of their equilibrium mole fractions
yi and the total pressure P of reactants and products:
K 冲
i

␯i

冢 冣
yi P
P⬚

 冲 yi␯ i
i

冲冢
i

P
P⬚

␯i







冢 冣K
P
P⬚

y

(5.50)

In this equation ␯  冱 i ␯i , and Ky is the equilibrium constant written in terms
of mole fractions at a particular total pressure.
The value of Ky is a function only of temperature at a constant total pressure
P:
Ky 

Ns

冲 yi␯ i 

i 1

⫺␯

冢 冣
P
P⬚

K

(5.51)

The equilibrium constant written in terms of mole fractions depends on the pressure as well as the temperature, but, as we will see, it is very useful in calculating the equilibrium extent of reaction because it is written in terms of amounts.
If the amount of gaseous products is equal to the amount of gaseous reactants,
then ␯  冱 i ␯i  0, Ky  K, and changing the total pressure of reactants does
not affect the equilibrium mole fractions of the reactants and products. If a reaction causes an increase in the number of molecules, then ␯ ⬎ 0, and Ky decreases
as the pressure is increased at constant temperature. Thus, raising the pressure
decreases the equilibrium mole fractions of the products and increases the equilibrium mole fractions of the reactants; in short, raising the pressure pushes the
reaction backward.
Le Châtelier’s principle provides a quick way to check conclusions like this
about the effects of changes in independent variables on chemical equilibrium.
According to Le Châtelier, when an independent variable of a system at equilibrium is changed, the equilibrium shifts in the direction that tends to reduce the
effect of the change. When pressure is increased, the equilibrium shifts in the direction to reduce the number of molecules.
If a reaction involves only solids and liquids, the effect of pressure on the
equilibrium is small.

Example 5.13 The effect of pressure on the ammonia synthesis reaction
In Example 5.5, we saw that when an initially equimolar mixture of nitrogen and hydrogen
is placed in contact with an ammonia catalyst at 400 ⬚C and 51.2 bar, there is a 10% conversion to ammonia at equilibrium. What pressure is required to obtain a 15% conversion?
Using the equation for the equilibrium constant, we obtain
1.60 ⫻ 10⫺4 

16(0.15)2 (0.85)
(0.65)(P /P ⬚ )2

P
 83.45
P⬚
P  83.45 bar

5.6 Effect of Pressure, Initial Composition, and Inert Gases on the Equilibrium Composition
According to Le Châtelier’s principle, raising the total pressure will cause the reaction
to shift in the direction of the product NH3 because there is a decrease in moles of gas in
the forward reaction.

To discuss the effect of initial composition on the equilibrium composition
for a reaction, we will use the equilibrium constant expressed in terms of mole
fractions. At any time during a reaction the amounts of each reactant and product
may be expressed in terms of the initial amount ni 0 and the extent of reaction ␰:
Ky  冲
i

ni 0  ␯i ␰
n0  ␯␰



␯i







1
n0  ␯␰



冣 冲(n

i 0  ␯i ␰ )

␯i

(5.52)

i

In this equilibrium expression the amount of gas is represented by

冱 ni  冱(ni 0  ␯i ␰)  冱 ni 0  ␰ 冱 ␯i  n0  ␰␯

(5.53)

i

where the initial amount of gaseous reactants and products is represented by n0
and ␯  冱 i ␯i . Thus, the calculation of the amounts of reactants and products at
equilibrium from the initial composition and the value of Ky simply comes down
to the solution of a polynomial in ␰. The polynomials arising here always have one
positive real root. Quadratic equations are readily solved, and higher-order polynomials can be solved by iterative methods or by using a mathematical application
in a computer.
If an inert gas is added to an equilibrium mixture of gases at constant temperature and volume, there is no effect on the equilibrium. But adding an inert
gas at constant temperature and pressure has the same effect as lowering the
pressure. When inert gases are present, equation 5.52 has to be modified to include the number of moles of inert gas in the denominator of each mole fraction;
thus n0  ␯␰ becomes n0  ␯␰  ninerts . Substituting the modified form of equation
5.52 in equation 5.50 yields



P /P ⬚
K 
n0  ␯␰  ninerts

␯ N
s

冣冲

(ni 0  ␯i ␰) ␯ i

(5.54)

i 1

If ␯ ⬍ 0, the addition of an inert gas at constant pressure reduces the sum of the
partial pressures of the reactants and products, and the reaction shifts to the left
to compensate for this. Equation 5.54 applies generally to ideal mixtures of ideal
gases, but if the partial pressure of the inert gases is known, this partial pressure
may be subtracted from the total pressure, and equation 5.50 may be used with P
equal to the sum of the partial pressures of the reactants and products.

Example 5.14 The effect of initial composition, pressure, and inert gases on
equilibrium
As an illustration of the effect of initial composition, pressure, and the addition of an inert
gas, consider the equilibrium for the production of methanol from CO and H2 :
CO(g)  2H2 (g)  CH3 OH(g)
The value of K at 500 K is 6.23 ⫻ 10⫺3 . (a) A gas stream containing equimolar amounts of
CO and H2 is passed over a catalyst at 1 bar. What is the extent of reaction at equilibrium?

151

152

Chapter 5

Chemical Equilibrium

Initial amount
Equilibrium amount
Mole fraction

CO

H2

CH3 OH

1
1⫺␰
1
2

1
1 ⫺ 2␰
1 ⫺ 2␰
2(1 ⫺ ␰)

0


2(1 ⫺ ␰)

K 

Total amount  2(1 ⫺ ␰)

4␰(1 ⫺ ␰)
 6.23 ⫻ 10⫺3
(1 ⫺ 2␰)2 12

Solving this quadratic equation indicates that ␰  0.001 55, so that yCO  0.5000,
yH2  0.4992, and yCH3 OH  0.0008.
(b) To attain a more complete reaction the pressure is raised to 100 bar and 2 mol of
hydrogen is used per mole of CO. What is the equilibrium extent of reaction?

Initial amount
Equilibrium amount
Mole fraction

K 

CO

H2

CH3 OH

1
1⫺␰
1⫺␰
3 ⫺ 2␰

2
2 ⫺ 2␰
2 ⫺ 2␰
3 ⫺ 2␰

0


3 ⫺ 2␰

Total amount  3 ⫺ 2␰

␰(3 ⫺ 2␰)2
 6.23 ⫻ 10⫺3
(1 ⫺ ␰)(2 ⫺ 2␰)2 (100)2

Solution of this equation for the extent of reaction by successive approximation yields ␰ 
0.817, so that, yCO  0.134, yH2  0.268, and yCH3 OH  0.598. (You might check that this
gives the right value for the equilibrium constant.)
(c) If the reactant gases contain a mole of nitrogen in addition to 1 mol of CO and 2
mol of hydrogen, what is the equilibrium extent of reaction at 100 bar?
The first two lines of the table in (b) are unchanged. In the third line, the total number
of moles is now 4 ⫺ 2␰.
K 

␰(4 ⫺ 2␰)2
 6.23 ⫻ 10⫺3
(1 ⫺ ␰)(2 ⫺ 2␰)2 (100)2

Solution of this equation yields ␰  0.735, so that yCO  0.105, yH2  0.210, yCH3 OH 
0.291, and yN2  0.395. Here the presence of an inert gas reduces the equilibrium conversion to product, but for a reaction for which ␯  冱 ␯i is positive, addition of an inert gas
will cause the reaction to go further to the right.

5.7

EQUILIBRIUM CONSTANTS FOR GAS REACTIONS WRITTEN
IN TERMS OF CONCENTRATIONS

Since thermodynamic tables for gases are based on a standard state pressure
of 1 bar for the ideal gas, the ⌬f Gi⬚ values lead directly to equilibrium constants in terms of pressure (or fugacities). However, in connection with chemical kinetics it is useful to express equilibrium constants of gas reactions in terms
of concentrations, since rate equations are written in terms of concentrations
(Section 18.2). These two types of equilibrium constants will be represented by
KP and Kc . To obtain a general expression for the equilibrium constant Kc

5.7 Equilibrium Constants for Gas Reactions Written in Terms of Concentrations

in terms of concentrations for ideal gases, we replace Pi in equation 5.21 with
Pi  ni RT /V  ci RT :
KP 

Ns

冲冢

i 1

Pi
P⬚

␯i



冲
i



ci RT
P⬚

␯i



(5.55)

To define a dimensionless equilibrium constant in terms of concentration, we
introduce the standard concentration c ⬚ , which represents one mole per liter.
Introducing this standard concentration into each term of equation 5.55 yields
KP 

冲 冤冢
Ns

i 1

ci
c⬚

冣冢

c ⬚ RT
P⬚

冱 ␯i

冣冥 冢 冣
c ⬚ RT

P⬚

␯i

冲 冢c i⬚ 冣
c

i

冱 ␯i

冢 冣

c ⬚ RT

P⬚

␯i

(5.56)

Kc

where the equilibrium constant expressed in terms of concentration,
Kc  冲
i

␯i

冢 冣
ci
c⬚

(5.57)

is a function only of temperature for a mixture of ideal gases. If c ⬚  1 mol L⫺1
and P ⬚  1 bar, then c ⬚ RT /P ⬚  24.79 at 298.15 K.

Example 5.15 A gas equilibrium constant expressed in concentrations
What is the value of the equilibrium constant Kc for the dissociation of ethane into
methyl radicals at 1000 K?
C2 H6 (g)  2CH3 (g)
⌬r G ⬚  2⌬f G ⬚ (CH3 ) ⫺ ⌬f G ⬚ (C2 H6 )
 2(159.82) ⫺ 109.55  210.09 kJ mol⫺1



KP  exp ⫺
 exp

⌬r G ⬚
RT



(⫺210.09)
(8.3145 ⫻ 10⫺3 )(1000)

 1.062 ⫻ 10⫺11
Kc 

P⬚
([CH3 ]/c ⬚ )2
 KP
c ⬚ RT
[C2 H6 ]/c ⬚

 (1.062 ⫻ 10⫺11 )

(1 bar)
(1 mol L⫺1 )(0.083145)(1000 K)

 1.278 ⫻ 10⫺13
Thus, at equilibrium [CH3 ]2 /[C2 H6 ]  1.278 ⫻ 10⫺13 mol L⫺1 , where the brackets indicate concentrations in moles per liter.

153

154

Chapter 5

Chemical Equilibrium

5.8

HETEROGENEOUS REACTIONS

A reaction involving more than one phase that does not involve equilibria of a
species between phases is referred to as a heterogeneous reaction. Examples are
CaCO3 (s)  CaO(s)  CO2 (g)

(5.58)

CH4 (g)  C(s)  2H2 (g)

(5.59)

Depending on the initial conditions, reactions of this type can go to completion,
whereas reactions in a single phase do not go to completion because of the entropy
of mixing (Section 3.5). The equilibrium constant for reaction 5.58 is equal to the
partial pressure of CO2 gas that is measured at equilibrium when all three phases
are present. If the pressure applied to the system is less than K , reaction 5.58 will
go to completion to the right. If the pressure applied to the system is greater than
K , reaction 5.58 will go to completion to the left.
Equilibrium constants for reactions like 5.58 and 5.59 can be written without
terms for the pure solids, provided they are present. The reason for doing this is
that the activities of the pure solid phases are very nearly equal to unity for moderate pressures. If the gases are ideal, then the equilibrium constant expressions
for reactions 5.58 and 5.59 are
PCO2
K 
(5.60)
P⬚
PH2 2
K 
(5.61)
PCH4 P ⬚
These equilibrium constants are independent of the amount of pure solid phase
present. As long as graphite is present, the second reaction behaves like a homogeneous reaction in that the reaction does not go to completion because of the
entropy of mixing in the gas phase.
Table 5.1 gives the pressure of CO2 (g) in equilibrium with CaCO3 (s) and
CaO(s) at a series of temperatures. The natural logarithm of the equilibrium pressure is plotted versus 1/T in Fig. 5.6. The three phases are at equilibrium only along
the line.
Example 5.16 Calculation of reaction properties at high temperatures
assuming ⌬r CP⬚ is constant
Calculate ⌬r G ⬚ , ⌬r H ⬚ , and ⌬r S ⬚ for reaction 5.58 at 1000 K using data in (a) Table 5.1
and (b) Table C.2 with the assumption that ⌬r CP⬚ is independent of temperature.
(a ) ⌬r G ⬚  ⫺RT ln K  ⫺(8.3145 J K⫺1 mol⫺1 )(1000 K)(⫺3.00)  24.9 kJ mol⫺1
⌬r H ⬚  ⫺R(slope)  ⫺(8.3145 J K⫺1 mol⫺1 )(⫺2.055 ⫻ 10 4 K)
 171 kJ mol⫺1
Table 5.1
t / ⬚C
PCO2 /P ⬚
t / ⬚C
PCO2 /P ⬚

Pressures of CO2 (g) in Equilibrium with CaCO3 (s) and
CaO(s)
500
9.2 ⫻ 10⫺5
897
0.987

600
2.39 ⫻ 10⫺3
1000
3.820

700
2.88 ⫻ 10⫺2
1100
11.35

800
0.2217
1200
28.31

5.9 Degrees of Freedom and the Phase Rule

4
2

ln K

0
–2
–4
–6
–8

0.0006

0.0007

0.0008

0.0009

0.0010
K/T

0.0011

0.0012

0.0013

Figure 5.6 Logarithm of the equilibrium constant for reaction 5.58 as a function of reciprocal temperature. The equilibrium partial pressuer of CO2 (g) is essentially atmospheric
pressure when ln K  0. Above the line CaCO3 (s)+CO2 (g) are stable, and below the line
CaO(s)+CO2 (g) are stable. (See Computer Problem 5.L.)
⌬r H ⬚ ⫺ ⌬r G ⬚
T
(171 ⫺ 24.9) ⫻ 10 3 J mol⫺1

1000 K

⌬r S ⬚ 

 146 J K⫺1 mol⫺1
(b) According to Table C.2 the values at 298 K are ⌬r G ⬚  130.40 kJ mol⫺1 , ⌬r H ⬚ 
78.32 kJ mol⫺1 , and ⌬r S ⬚  160.59 J K⫺1 mol⫺1 when calcite is the reactant.
⌬r CP  C ⬚P (CaO)  C ⬚P (CO2 ) ⫺ C ⬚P (CaCO3 )
 42.80  37.11 ⫺ 81.88  ⫺1.97 J K⫺1 mol⫺1
Using the equations in Example 5.11,
⌬r H1000
⬚  178.32 kJ mol⫺1 ⫺ (1.97 ⫻ 10⫺3 kJ K⫺1 mol⫺1 )(701.85 K)
 176.94 kJ mol⫺1
⌬r S1000
⬚  160.59 J K⫺1 mol⫺1 ⫺ (1.97 J K⫺1 mol⫺1 ) ln(1000/298.15)
 158.21 J K⫺1 mol⫺1
⌬r G1000
⬚  ⌬H1000
⬚ ⫺ T ⌬S1000

 176.94 kJ mol⫺1 ⫺ (1000 K)(158.21 ⫻ 10⫺3 kJ K⫺1 mol⫺1 )
 18.73 kJ mol⫺1

5.9

DEGREES OF FREEDOM AND THE PHASE RULE

In Section 1.9 we discussed the fact that for a one-phase system without chemical reactions, F  Ns  1 variables have to be specified in order to describe the

155

156

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Chemical Equilibrium

intensive state of the system at equilibrium, where F is the number of degrees of
freedom and Ns is the number of different kinds of species. For a one-phase system involving a single species, F  2 so that it is sufficient to specify T and P , but
other properties could be used. For a two-phase system containing a single species,
F  1, so only T or P has to be specified. For a three-phase system containing a
single species, F  0, so that no variables have to be specified. We also discussed
the fact that for these systems containing a single species, D  F  p  2  p
variables are required to describe the extensive state of the system, where p is the
number of different phases. Thus the descriptions for the one-phase, two-phase,
and three-phase systems containing a single species each require the specification
of D  3 variables since amounts of phases have to be specified. In Section 2.2
these ideas were confirmed in our discussion of the internal energy U of a system in a certain intensive state or a certain extensive state. In Sections 4.1 and
4.2, we noted that the D  Ns  2 variables in the fundamental equations are
also involved in the criteria, such as (dG )T,P,兵ni 其 ⱕ 0, for spontaneous change and
equilibrium. In Section 4.8 we saw that the Gibbs–Duhem equation for a nonreaction system, which gives a relation between the intensive variables for a phase in
a system, is in agreement with F  Ns  1 of these variables being independent.
However, when a chemical reaction occurs and is at equilibrium, this provides a relationship between the chemical potentials of the species involved, and
so the intensive state of the system is described by specifying one fewer intensive variable. For example, to describe the intensive state of a system containing N2 , H2 , and NH3 at chemical equilibrium, it is necessary to specify only T, P,
and y (H2 )/y (N2 ) or T, P, and n (NH3 )/n (N2 ). The choice of independent intensive variables is optional, but the number is not. We can generalize this discussion
by stating that the number of intensive variables that have to be specified to describe the intensive state of a one-phase system is Ns ⫺ R  1, where R is the
number of independent chemical reactions that are at equilibrium. The reactions
in a set are independent if no reaction in the set can be obtained by adding and
subtracting other reactions in the set. (We will discuss this in more detail in Section 5.11.)
In using Ns ⫺ R  1 to count the number of intensive variables for a reaction
system, it is convenient to introduce a new concept, which is component. Components are the things that are conserved in a chemical reaction system. An obvious
choice of components are the amounts of atoms of elements, but we will see later,
in Section 5.11, that this is not the only choice; molecules or groups of atoms can
be used as components. We will use C to represent the number of components
in a reaction system, and for each of these C components there is a conservation
equation. But only an independent set of conservation equations can be used in an
equilibrium calculation, just as only an independent set of R chemical reactions
can be used. Thus C is the number of independent components that is given by
C  Ns ⫺ R . The number of intensive variables that have to be specified to describe the intensive state of a one-phase reaction system is C  1. We can look at
it this way: When there are reactions at equilibrium, we do not have to specify the
concentrations of all species because they can be calculated using the equilibrium
constant expressions. The number of intensive variables that have to be specified
to describe the intensive state of a one-phase system is Ns  1 if there are no reactions and C  1 if there are chemical reactions at equilibrium. The number of
variables required to describe the extensive state of a one-phase reaction system
is D  C  2.

5.9 Degrees of Freedom and the Phase Rule

Example 5.17 The choices of independent intensive properties
(a) Oxygen is in equilibrium with ozone: 3O2 (g)  2O3 (g). How many intensive variables
have to be specified to describe the intensive state of the system? (b) The water gas shift
reaction, H2 O(g)CO(g)  H2 (g)CO2 (g), is at equilibrium. How many intensive variables have to be specified to describe the state of the system? Give three possible choices
of intensive variables. (c) The equilibrium behavior of a system involving the following
two reactions is to be investigated:
H2 O(g)  CO(g)  H2 (g)  CO2 (g)
3H2 (g)  CO(g)  CH4 (g)  H2 O(g)
How many intensive variables have to be specified to describe the state of the system?
Give a possible choice. Someone suggests that the equilibrium calculations be made with
the following two reactions:
H2 O(g)  CO(g)  H2 (g)  CO2 (g)
2H2 (g)  2CO(g)  CH4 (g)  CO2 (g)
Is that all right?
(a) There is one component; C  Ns ⫺ R  2 ⫺ 1  1. Therefore, two intensive
variables have to be specified.
(b) There are three components; C  Ns ⫺ R  4 ⫺ 1  3. Therefore, four intensive variables have to be specified. Three possible choices of intensive variables are T, P,
n (H2 O)/n (CO), n (H2 )/n (CO); T, P, n (CO)/n (CO2 ), n (H2 )/n (CO2 ); T , P, n (H2 O)/n (CO),
P (H2 ).
(c) There are three components; C  Ns ⫺ R  5 ⫺ 2  3. Thus four intensive variables have to be specified. Note that this is the same as in the preceding example, and so
any of the choices in (b) can be used. Since the suggested reactions are independent, they
are just as good as the original set for equilibrium calculations.

In 1876, Gibbs derived a general expression for the number of intensive variables (the number of degrees of freedom) that have to be specified for a multiphase system at equilibrium. It is important to emphasize that we are only interested in describing the intensive state of the system; thus we are not concerned
with the relative amounts of the various phases. The number of degrees of freedom F of a system is the number of intensive variables that must be specified to
describe the state of the system completely.
To derive an expression for F, we will consider a system that consists of p
phases. If a phase contains C components, its composition may be specified by
stating (C ⫺ 1) mole fractions—one less than the number of components because
the mole fraction of one component can be obtained from 冱 xi  1, where xi
represents the mole fraction of component i . Thus, the total number of concentrations to be specified for the whole system is (C ⫺ 1) for each of the p phases or
p (C ⫺ 1) concentrations. In general, there are two more intensive variables that
have to be considered, temperature and pressure. Thus,
Number of intensive variables  p (C ⫺ 1)  2

(5.62)

Next we consider the number of relationships that must be satisfied for phase
equilibrium. The chemical potential ␮ for each component is the same in each
phase ␣ , ␤ , ␥ , and so on, and so ␮i,␣  ␮i,␤  ␮i,␥  ⭈⭈⭈ for component i . There
are p phases but only ( p ⫺ 1) equilibrium relationships of the type ␮i,␣  ␮i,␤ for
each component. For example, if there are two phases, there is only one equilibrium relationship for each component that gives its distribution between the two

157

158

Chapter 5

Chemical Equilibrium

phases. Altogether there are C components, each one of which can be involved
in an equilibrium between phases. Thus,
Number of independent equations  C ( p ⫺ 1)

(5.63)

The difference between the number of variables and the number of independent equations is the number of independent variables, which is referred to as the
number of degrees of freedom:
F  [ p (C ⫺ 1)  2] ⫺ C ( p ⫺ 1)
or
F  C ⫺ p 2

(5.64)

This is the important phase rule of Gibbs. The number of degrees of freedom F
is equal to the number of intensive variables that can be set arbitrarily. For example, for a one-component system F  3 ⫺ p . Under conditions where a single
phase is present, F  2 and the pressure and temperature can both be set arbitrarily. Under conditions where two phases are in equilibrium, F  1 and either
the temperature or pressure may be set arbitrarily. Under conditions where three
phases are in equilibrium, F  0 and the pressure and temperature are fixed by
the equilibrium. Such a system is said to be invariant and is represented by a point
in a plot of pressure versus temperature.
It can be seen from the phase rule that the greater the number of components
in a system, the greater the number of degrees of freedom. On the other hand,
the greater the number of phases, the smaller is the number of variables such as
temperature, pressure, and mole fraction that must be specified to describe the
system completely.
Equations 5.62 and 5.63 are based on the assumption that pressure and temperature are both variables. If the pressure, for example, is fixed, the phase rule
becomes F  C ⫺ p  1. On the other hand, if the system is affected by both
temperature and pressure and another variable, such as magnetic field strength,
the phase rule becomes F  C ⫺ p  3.
When there are special conditions involved in the specification of the system,
the number of these special constraints s must be included in the phase rule to give
F  C ⫺ p  2 ⫺ s . An example of a special constraint would be taking the initial
amounts of two reactants in the ratio of their stoichiometric numbers in the reaction.
The number D of independent properties required to describe the extensive
state of a system can be readily counted in the fundamental equation for a thermodynamic potential: D is the number of differential terms on the right-hand side.
The number F of properties required to describe the intensive state of a system
can be readily counted in the Gibbs–Duhem equation for the system: F is the
number of differential terms in the Gibbs–Duhem equation. This is illustrated
first by a one-phase system containing one species at a specified T and P . The
appropriate thermodynamic potential is the Gibbs energy, and the fundamental
equation is
dG  ⫺S dT  V dP  ␮ dn
(5.65)
The differential terms on the right-hand side indicate that the natural variables
are T , P , and n , and so D  3. The corresponding Gibbs–Duhem equation is
0  ⫺S dT  V dP ⫺ n d␮

(5.66)

The intensive variables T , P , and ␮ for the system are indicated by the differential
terms, but only two of them are independent according to equation 5.66. Any

5.9 Degrees of Freedom and the Phase Rule

two of the intensive variables can be chosen as degrees of freedom, and F 
2. This is in agreement with the phase rule, which gives F  1 ⫺ 1  2  2. The
relation D  F  p  2  1  3 is satisfied. The criterion for equilibrium is
(dG )T,P,n ⱕ 0.
Now consider a one-phase system at specified T and P involving the chemical
reaction A  B  C. The appropriate thermodynamic potential is the Gibbs energy, and the fundamental equation for the system before equilibrium is reached
is
dG  ⫺S dT  V dP  ␮A dnA  ␮B dnB  ␮C dnC
(5.67)
We now substitute the equilibrium relation ␮A  ␮B  ␮C into equation 5.67 to
eliminate ␮C , yielding
dG  ⫺S dT  V dP  ␮A dncA  ␮B dncB

(5.68)

where ncA  nA  nC and ncB  nB  nC are the amounts of the A and B components. The differential terms on the right-hand side of equation 5.68 indicate
that when the reaction is at equilibrium the natural variables are T , P , ncA , and
ncB , so that D  4 for this one-phase, three-species system involving one reaction. Note that the amounts of components A and B are independent variables
for the system at equilibrium, but the amounts of species nA , nB , and nC are not.
The corresponding Gibbs–Duhem equation is
0  ⫺S dT  V dP ⫺ ncA d␮A ⫺ ncB d␮B

(5.69)

The intensive variables T , P , ␮A , and ␮B for the system are indicated by the differential terms, but because they are related by equation 5.69 only three of them are
independent, so that F  3. This is in agreement with the phase rule, which gives
F  N ⫺ R ⫺ p  2  3 ⫺ 1 ⫺ 1  2  3. The relation D  F  p  3  1  4
is satisfied. The criterion for equilibrium is (dG )T,P,ncA ,ncB ⱕ 0.
The relationship between D and the fundamental equation and the relationship between F and the Gibbs–Duhem equations are illustrated by a two-phase
system containing one species at specified T and P. The appropriate thermodynamic potential is the Gibbs energy, and the fundamental equation is
dG  ⫺S dT  V dP  ␮␣ dn␣  ␮␤ dn␤

(5.70)

where the two phases are labeled ␣ and ␤ . The equilibrium relation ␮␣  ␮␤ (see
Section 4.6) can be substituted in equation 5.70 to obtain
dG  ⫺S dT  V dP  ␮ dn

(5.71)

where ␮  ␮␣  ␮␤ and n  n␣  n␤ . The differential terms on the right-hand
side indicate that the natural variables are T, P, and n , and so D  3. There is a
separate Gibbs–Duhem equation for each phase:
0  ⫺S␣ dT  V␣ dP ⫺ n␣ d␮␣

(5.72)

0  ⫺S␤ dT  V␤ dP ⫺ n␤ d␮␤

(5.73)

Since the equilibrium relation indicates that d␮␣  d␮␤ , there are only three
intensive variables. Equations 5.72 and 5.73 provide two relationships between
these three intensive variables, and so only one is independent. The independent
variable can be taken to be T, P, or ␮ ; thus F  1, as expected from the phase
rule. The relation D  F p  12  3 is also satisfied. The criterion for equilibrium is (dG )T,n␣ ,n␤ ⱕ 0.

159

160

Chapter 5

Chemical Equilibrium

Example 5.18 Identifying properties to describe the intensive state of a system
and the extensive state of a system
Identify sets of properties to specify the extensive and intensive states of the following
three systems by examining their fundamental equations for the Gibbs energy. In each
case, show that the number of properties required to specify the intensive state is given
by F  Ns ⫺ R ⫺ p  2  C ⫺ p  2 and the number of properties required to specify
the extensive state is given by D  F  p . Also state the criterion for equilibrium and
spontaneous change for each system.
(a )

One-phase system containing two species

(b )

One-phase system containing A and B, which are in equilibrium with each other

(c )

Two-phase system containing two species
(a ) The fundamental equation for G is
dG  ⫺S dT  V dP  ␮1 dn1  ␮2 dn2

The differential terms on the right-hand side indicate that the natural variables are T , P ,
n1 , and n2 , and so D  4. The corresponding Gibbs–Duhem equation is
0  ⫺S dT  V dP ⫺ n1 d␮1 ⫺ n2 d␮2
The intensive variables T , P , ␮1 , and ␮2 for the system are indicated by the differential
terms, but only three of them are independent. Any three of them can be chosen, and
F  3. This agrees with the phase rule, which gives F  2 ⫺ 1  2  3. The relation
D  F  p  3  1  4 is satisfied. The criterion for equilibrium is (dG )T,P,n1 ,n2 ⱕ 0.
(b ) The fundamental equation for G is
dG  ⫺S dT  V dP  ␮A dnA  ␮B dnB
This form of the fundamental equation can be used to derive the equilibrium relation
␮A  ␮B . Substituting this relation into the fundamental equation for G yields
dG  ⫺S dT  V dP  ␮c dnc
where ␮c  ␮A  ␮B and nc  nA  nB . The differential terms on the right-hand side
indicate that the natural variables are T , P , and nc , and so D  3. Note that the amount
of the component (nA  nB ) is an independent variable for the system, but nA and nB are
not. The corresponding Gibbs–Duhem equation is
0  ⫺S dT  V dP ⫺ n d␮
The intensive variables T , P , and ␮ for the system are indicated by the differential terms,
but only two of them are independent so that F  2. This is in agreement with the phase
rule, which gives F  2 ⫺ 1 ⫺ 1  2  2. The relation D  F  p  2  1  3 is satisfied.
The criterion for equilibrium is (dG )T,P,nc ⱕ 0, where nc is the amount of the component.
(c ) The fundamental equation for G is
dG  ⫺S dT  V dP  ␮1␣ dn1␣  ␮2␣ dn2␣  ␮1␤ dn1␤  ␮2␤ dn2␤
This fundamental equation can be used to derive the equilibrium relations ␮1␣  ␮1␤ and
␮2␣  ␮2␤ . Substituting these relations into the fundamental equation yields
dG  ⫺S dT  V dP  ␮1␣ dn1  ␮2␣ dn2
where n1  n1␣  n1␤ and n2  n2␣  n2␤ . The differential terms on the right-hand side
indicate that the natural variables can be taken as T , P , n1 , and n2 , and so D  4. There
is a Gibbs–Duhem equation for each phase:
0  ⫺S␣ dT  V␣ dP ⫺ n1␣ d␮1␣ ⫺ n2␣ d␮2␣
0  ⫺S␤ dT  V␤ dP ⫺ n1␤ d␮1␤ ⫺ n2␤ d␮2␤

5.10 Special Topic: Isomer Group Thermodynamics
But the equilibrium relations indicate that d␮1␣  d␮1␤ and d␮2␣  d␮2␤ , so that there
are only four intensive variables remaining. The two Gibbs–Duhem equations provide two
relationships between these four intensive variables, and so only two are independent.
They can be taken to be T and P . This is in agreement with the phase rule F  2 ⫺ 2 
2  2. The relation D  F  p  2  2  4 is satisfied. The criterion for equilibrium is
(dG )T,P,n1 ,n2 ⱕ 0 or (dG )T,P,n␣ ,n␤ ⱕ 0, where n␣  n1␣  n2␣ and n␤  n1␤  n2␤ .

Example 5.19 Calculating degrees of freedom
The reaction CaCO3 (s)  CaO(s)  CO2 (g) is at equilibrium. (a) How many degrees of
freedom are there when all three phases are present at equilibrium? (b) How many degrees
of freedom are there when only CaCO3 (s) and CO2 (g) are present? (See the introduction
to components early in this section.)
(a )

C  Ns ⫺ R  3 ⫺ 1  2
F  C ⫺ p 2  2 ⫺ 32  1

Therefore, only the temperature or pressure may be varied independently.
(b )

F  2 ⫺ 22  2

Therefore, both the temperature and pressure may be varied without destroying a phase.

5.10

SPECIAL TOPIC: ISOMER GROUP THERMODYNAMICS

There is a special case of chemical equilibrium that is of sufficient importance to
be discussed separately, and that is the chemical equilibrium between isomers.
We have already had an example of isomerism A  B in Section 5.2. In Section
5.9, we saw that the number of components is given by C  Ns ⫺ R , where Ns
is the number of species in the system and R is the number of independent reactions. If Ns isomers are in equilibrium, there are Ns ⫺ 1 independent reactions,
and C  Ns ⫺ (Ns ⫺ 1)  1. Thus when a group of isomers is in equilibrium,
they form a single component, and we can calculate the standard thermodynamic
properties of the isomer group as a whole if we know the standard thermodynamic properties of the individual isomers. When a system involving isomers is
in chemical equilibrium, the isomer group can be treated as a pseudospecies in a
larger system because the distribution of isomers in an isomer group is a function
of temperature only and is not affected by the presence of other reactants. Thus
an isomer group has a standard Gibbs energy of formation, a standard enthalpy
of formation, a standard entropy, and a standard heat capacity at constant pressure. This greatly simplifies equilibrium calculations on organic mixtures, where
the number of isomers in a homologous series increases approximately exponentially with carbon number.
As an example, consider the three isomers of pentane (n -pentane, isopentane,
and neopentane) in equilibrium with each other. For ideal gases, the equilibrium
constants of the three formation reactions are of the form Ki  (Pi /P ⬚ )/(PH2 /P ⬚ )6
 exp(⫺⌬f Gi⬚ /RT ). The equilibrium mole fractions ri of the three isomers within
the isomer group are given by
ri 

Pi
exp(⫺⌬f Gi⬚ /RT )

冱 Pi
冱 exp(⫺⌬f Gi⬚ /RT )

(5.74)

161

162

Chapter 5

Chemical Equilibrium

since the partial pressure of molecular hydrogen cancels. We can represent the
denominator by exp[⫺⌬f G ⬚ (iso)/RT ], where ⌬f G ⬚ (iso) is the standard Gibbs energy of formation of the isomer group, and write
ri 

exp(⫺⌬f Gi⬚ /RT )
exp[⫺⌬f G ⬚ (iso)/RT ]

 exp兵[⌬f G ⬚ (iso) ⫺ ⌬f Gi⬚ ]/RT 其

(5.75)

where
NI

⌬f G ⬚ (iso)  ⫺RT ln 冱 exp(⫺⌬f Gi⬚ /RT )

(5.76)

i 1

The number of isomers in the isomer group is represented by NI .
Note that the standard Gibbs energy of formation of the isomer group is not
the mole fraction average of ⌬f Gi⬚ , but is actually more negative than the standard Gibbs energy of formation of the most stable isomer. This is to be expected
because in any equilibrium calculation on a larger system, the equilibrium mole
fraction of the isomer group has to be larger than that of the most stable isomer.
The value of ⌬f G ⬚ (iso) can also be calculated from the mole fraction average
Gibbs energy of formation plus the entropy of mixing.
The standard enthalpy of formation ⌬f H ⬚ (iso) of the isomer group is the mole
fraction weighted average of the ⌬f Hi⬚ , and it can be calculated using
NI

⌬f H ⬚ (iso)  冱 ri ⌬f Hi⬚

(5.77)

i 1

The expressions for ⌬f S ⬚ (iso) and ⌬f CP⬚ (iso) are readily derived.*
The fact that species in equilibrium can be treated as a single species in an
equilibrium calculation is illustrated by the fact that H2 O(l) and H(aq) are
shown as single species in thermodynamic tables. In these cases we do not know
the thermodynamic properties of the various species, which are rapidly interconverted, but we can treat the sum of species as a single species. Another example
is glucose(aq), which is made up of ␣ and ␤ forms. Later, in treating biochemical reactions, we will see that thermodynamic properties can be calculated for
adenosine triphosphate at a specified pH, although it is made up of several species.

Example 5.20

The standard Gibbs energy of an isomer group at equilibrium

The standard Gibbs energies of formation of the gaseous pentanes at 298.15 K are
as follows: n -pentane, ⫺8.33 kJ mol⫺1 ; isopentane, ⫺13.27 kJ mol⫺1 ; and neopentane,
⫺17.37 kJ mol⫺1 . Calculate the standard Gibbs energy of formation of the isomer group
at chemical equilibrium. What are the equilibrium mole fractions of the isomers at this
temperature?
⌬f G ⬚ (iso)  ⫺RT ln[exp(8.33/RT )  exp(13.27/RT )  exp(17.37/RT )]
 ⫺17.86 kJ mol⫺1
where RT is in kJ mol⫺1 .
*R. A. Alberty, Ind. and Eng. Chem. Fund. 22:318 (1983).

5.11 Special Topic: Chemical Equations as Matrix Equations
The equilibrium mole fractions are given by
rn  exp[(⫺17.86  8.33)/RT ]  0.021
riso  exp[(⫺17.86  13.27)/RT ]  0.157
rneo  exp[(⫺17.86  17.37)/RT ]  0.821

Comment:
When isomers are in equilibrium with one another, the mole fraction of an
isomer within the isomer group is dependent only on the temperature, for ideal
gases. Thus we can calculate a standard Gibbs energy of formation of the isomer
group if we know the standard Gibbs energies of formation of the individual
isomers. For ideal gases, the standard enthalpy of formation of the isomer group
is simply a weighted average of the various isomers. If isomers are interconverted
sufficiently rapidly, we may not be aware of the existence of isomers, so that
the substance is treated as a single species with standard formation properties
determined in the usual way.

5.11

SPECIAL TOPIC: CHEMICAL EQUATIONS
AS MATRIX EQUATIONS

So far, we have considered systems in which there is a single chemical reaction.
However, in many practical applications of chemical thermodynamics a number
of chemical reactions occur simultaneously. The equilibrium composition can be
calculated by using a set of independent chemical reactions that will represent all
possible chemical changes in the system. A set of reactions is independent if no
member of the set may be obtained by adding and subtracting other members of
the set. Different sets of reactions may be chosen to describe a given system, but
the number R of independent reactions is always the same.
For a simple system, a set of independent reactions can be found by inspection, but for a more complex system, matrix operations are required. As we saw
in Section 5.9, the number R of independent reactions is needed to calculate
the number C of components in a system at chemical equilibrium because C 
Ns ⫺ R , where Ns is the number of species. The determination of the number of
independent reactions and the number of components depends on the fact that
atoms of each element have to be conserved in a chemical reaction.
Chemical equations are really matrix equations, and the recognition of their
mathematical character is especially useful in considering multireaction systems.*
As a first step, consider the following chemical equation:
CO  12 O2  CO2

冋册 冋册 冋册

C 1
O 1

0
2

1
2

(5.78)

Molecular formulas can be interpreted as column matrices, as indicated below
the chemical equation. In this case, the top integer of the column matrix gives
the number of carbon atoms, and the bottom integer gives the number of oxygen
*R. A. Alberty, J. Chem. Educ. 68:984 (1991); R. A. Alberty, J. Chem. Educ. 69:493 (1992).

163

164

Chapter 5

Chemical Equilibrium

atoms. We have been representing a general chemical reaction as 0  冱 ␯i Bi ,
and so equation 5.78 can be written as
0  ⫺1CO ⫺ 12 O2  1CO2

(5.79)

When the molecular formulas are replaced by the corresponding column matrices,
we obtain
⫺1

冋册 冋册 冋册 冋册
1

1

0
1
0
1

2
2
0

1
2

(5.80)

This is one way to write the conservation equations for carbon and oxygen. Textbooks on linear algebra* show that this equation can be written in the form of a
matrix product.



1
1

0
2

⫺1
1
0
⫺ 12 
2
0
1



冋册

(5.81)

Notice that the stoichiometric numbers in equation 5.79 form a column matrix.
When you multiply a 2 ⫻ 3 matrix times a 3 ⫻ 1 matrix, you obtain a 2 ⫻ 1 matrix
(see Appendix D.8). This equation is a form of the conservation equations for the
two elements involved in this reaction.
We can generalize equation 5.81 by writing it as
A␯  0

(5.82)

where A is the conservation matrix, ␯ is the stoichiometric number matrix, and 0
is a zero matrix. This equation can be used to calculate the ␯ matrix from the A
matrix. The ␯ matrix is called the null space of the A matrix. For small matrices
the null space can be calculated by hand, and for larger matrices a computer with
the operations of linear algebra can be used (e.g., Mathematica† ). The first step
in calculating the null space by hand is to make a Gaussian reduction of the A
matrix to get it into row echelon form, that is, a matrix with an identity matrix
(see Appendix D.8) on the left. Rows in the A matrix can be multiplied by integers
and added or subtracted from other rows to obtain this row echelon form. If we
subtract the first row of the A matrix in equation 5.81 from the second row, we
obtain
A



1
0

0
2



1
1

(5.83)

If we then divide the second row by 2, we obtain
A



1

0

1

0

1

1
2



(5.84)

The rule for calculating the null space is to change the sign of the entries in the
*G. Strang, Linear Algebra and Its Applications. San Diego: Harcourt Brace Jovanovich, 1988.
† Wolfram Research, Inc., 100 Trade Center Drive, Champaign, IL 61820-7237.

5.11 Special Topic: Chemical Equations as Matrix Equations

column(s) to the right of the identity matrix and append an appropriate-size unit
matrix below it. This yields
⫺1

␯  ⫺ 12
1

(5.85)

The rows here are in the order of the columns in the A matrix, and so this yields
chemical equation 5.79.
The use of row reduction is important for another reason, namely, that the
rank of the A matrix is equal to the number C of components (see Section 5.9).
C  rank A

(5.86)

Equation 5.84 shows that the rank of the A matrix is 2. The number of components is often equal to the number of elements, as it is in this case, but there are
important exceptions. The rank of the ␯ matrix is equal to the number of independent columns, and so this gives us the number R of independent reactions for
a system:
R  rank ␯

(5.87)

The rank of the ␯ matrix in equation 5.85 is unity, which corresponds to the fact
that there is a single reaction. We have seen earlier (Section 5.9) that
Ns  C  R

(5.88)

Ns  rank A  rank ␯

(5.89)

and so

We can see from our simple example that the Gaussian reduction divides the
species in a system into C components and Ns ⫺ C noncomponents, where Ns
is the number of species, that is, the number of columns in the A matrix. Thus
the conservation matrix A is C ⫻ Ns , and the stoichiometric number matrix ␯ is
Ns ⫻ R .
To see how these operations work out for a larger system, let us consider a
reaction system containing CO, H2 , CO2 , H2 O, and CH4 . The conservation matrix
is
CO H2
C
AH
O

1
0
1

0
2
0

CO2

H2 O

CH4

1
0
2

0
2
1

1
4
0

(5.90)

Row reduction yields the following row echelon form:
CO
A  H2
CO2

CO H2
1
0
0
1
0
0

CO2
0
0
1

H2 O
⫺1
1
1

CH4
2
2
⫺1

(5.91)

165

166

Chapter 5

Chemical Equilibrium

Note that the components are now taken to be CO, H2 , and CO2 . Thus the ␯
matrix is
R1 R2
1 ⫺2
⫺1 ⫺2
⫺1
1
1
0
0
1

CO
H2
␯  CO2
H2 O
CH4

(5.92)

This corresponds to the chemical equations
H2  CO2  H2 O  CO

(5.93)

2CO  2H2  CH4  CO2

(5.94)

Equation 5.91 shows that CO, H2 , and CO2 become components, and H2 O
and CH4 become noncomponents. The fact that the A matrix came out this way
indicates that the composition of the system can be expressed in terms of CO,
H2 , and CO2 . If we had put other species first, they might have been chosen as
components, but of course the components must contain all of the elements. The
choice of components is somewhat arbitrary, but the number is not.
Equation 5.82 shows that the conservation matrix and the stoichiometric
number matrix are equivalent because the stoichiometric number matrix can be
calculated from the conservation matrix, as we have seen. Equation 5.82 can also
be written as

␯ T AT  0

(5.95)

where the superscript T indicates the transpose (see Appendix D.8). Thus AT is
the null space of ␯T so that we can start with a chemical equation, or a system
of independent chemical equations, and calculate the conservation matrix for the
system. Because of the nature of conservation equations and chemical equations,
neither the A matrix nor the ␯ matrix is unique. Conservation equations can be
written in various ways that are all equivalent, and a set of independent reactions
for a system can be written in different ways that are all equivalent, even though
we may prefer to see chemical reactions written in a particular way. Any set of
independent reactions for a system can be used to calculate the equilibrium composition. However, if we have two A matrices, or two ␯ matrices, and want to know
whether they are equivalent, we can make a Gaussian reduction to see if they give
the same row echelon form.
Usually the number C of components is equal to the number of elements in
the system. However, there are two types of situations where this is not true. If two
elements are always in the same ratio, they can be replaced by a pseudoelement,
and the number of components is smaller than the number of elements. Consider
the isomerization reaction
C4 H10 (n -butane)  C4 H10 (isobutane)
The A matrix is
A



4
10



4
10

(5.96)

(5.97)

5.11 Special Topic: Chemical Equations as Matrix Equations

which reduces to
A

冋 册
1 1
0 0

(5.98)

which has a rank of 1. Thus there is a single component. The intensive state of the
system at equilibrium is specified by giving T and P. In systems involving ions the
charge balance may be redundant.
The rank of A is larger than the number of elements when there are additional
constraints in a reaction system. This does not happen very often in chemistry, but
it is illustrated by the following example.
Example 5.21 The conservation matrix and the stoichiometric number matrix
for a reaction system
Consider a system in which only the following reaction occurs because of the catalyst used:
C6 H6 (g)  CH4 (g)  C6 H5 CH3 (g)  H2 (g)
Write out the A matrix and the ␯ matrix, and show that they are compatible. How many
degrees of freedom does this system have, and what is a possible choice of degrees of
freedom?
Note that aromatic rings are conserved as well as C and H. Thus the A matrix has
three rows, where the third row is for the aromatic component.
C
AH
ar

C 6 H6
6
6
1

CH4
1
4
0

C6 H5 CH3
7
8
1

H2
0
2
0

This matrix has a rank of 3, and so C  3. Thus the number of components is one more
than the number of elements. This matrix can be used to calculate the stoichiometric numbers in the chemical equation that occurs. Row reduction yields

C6 H6
A  CH4
C6 H5 CH3

C6 H6
1
0
0

CH4
0
1
0

C6 H5 CH3
0
0
1

H2
1
1
⫺1

The last column indicates the stoichiometric numbers of the four reactants in the order
listed, ␯ T [⫺1, ⫺1, 1, 1], in agreement with the balanced chemical equation. The number
of intensive variables that have to be specified to describe the intensive state of this system
at equilibrium is F  C ⫺ p 2  3 ⫺ 12  4, which can be taken to be T, P, n (H)/n (C),
and n (ar)/n (C).

Example 5.22 Calculation of the number of independent reactions
A system contains CaO(s), CO2 (g), and CaCO3 (s). How many independent reactions are
there?
The system formula matrix is
1
A 2
0

0
1
1

1
3
1

167

168

Chapter 5

Chemical Equilibrium
if the elements (rows) are in the order C, O, Ca and the species (columns) are in the order
CO2 , CaO, CaCO3 . Gaussian elimination yields
1
A 0
0

0
1
0

1
1
0

This corresponds to the stoichiometric number matrix
⫺1
␯  ⫺1
1
which corresponds to the equation
CaO(s)  CO2 (g)  CaCO3 (s)
The number of independent reactions is given by
R  Ns ⫺ rank A  3 ⫺ 2  1


1.

2.

3.

4.

5.

6.

7.

Ten Key Ideas in Chapter 5
The fundamental equation for the Gibbs energy can be used to derive the
expression for the equilibrium constant for a chemical reaction in terms of
the activities of the reactants and shows that ⫺RT ln K is equal to the standard Gibbs energy of reaction ⌬r G ⬚ . The value of an equilibrium constant
cannot be interpreted unless it is accompanied by a balanced chemical equation and a specification of the standard state of each reactant and product.
For gas reactions the activity can be expressed in terms of the fugacity (ai 
fi /P ⬚ ), but most of our calculations have been made for reactions in mixtures
of ideal gases. Because of the entropy of mixing, a plot of the Gibbs energy
versus extent of reaction always has a minimum at the equilibrium extent of
reaction.
For a mixture of ideal gases in which there is a single reaction, the expression
for the equilibrium constant can always be written in terms of the extent of
reaction and the total pressure. Thus if the equilibrium constant is known,
the equilibrium composition can always be calculated by solving a polynomial equation.
The standard Gibbs energy of reaction is made up of contributions from the
standard enthalpy of reaction ⌬r H ⬚ and the standard entropy of reaction:
⌬r G ⬚  ⌬r H ⬚ ⫺ T ⌬r S ⬚ . Therefore, the equilibrium constant for a reaction
can be calculated from calorimetric measurements that can be used to obtain
⌬r H ⬚ and third law calorimetric measurements that can be used to obtain
⌬r S ⬚ .
Equilibrium constants can be calculated for many reactions for which ⌬f Hi⬚
and Si⬚ have been tabulated for all the species since ⌬r H ⬚  冱 ␯i ⌬f Hi⬚ and
⌬r S ⬚  冱 ␯i Si⬚ , where the ␯i are stoichiometric numbers.
For an endothermic reaction, the equilibrium constant increases as the temperature is increased, and for an exothermic reaction the equilibrium constant decreases as the temperature is increased.
According to Le Châtelier’s principle, when an independent variable of a
system at equilibrium is changed, the equilibrium shifts in the direction that
tends to reduce the effect of the change.

Problems

8.

9.

10.

169

According to the phase rule, the number F of degrees of freedom for a system at equilibrium is given by Ns ⫺ R ⫺ p  2, where Ns is the number of
species, R is the number of independent reactions, and p is the number of
different phases. This is in agreement with the Gibbs–Duhem equation.
When a group of isomers is in equilibrium, they form a single component,
and we can calculate the standard thermodynamic properties of the isomer
group as a whole if we know the standard thermodynamic properties of the
individual isomers.
The conservation of atoms in a system of reactions can be represented in two
ways, by the conservation equations for each element and by an independent set of chemical equations. A Gaussian reduction of the conservation
equation yields a set of independent chemical equations.

REFERENCES
K. E. Bett, J. S. Rowlinson, and G. Saville, Thermodynamics for Chemical Engineers. Cambridge, MA: MIT Press, 1975.
K. S. Pitzer, Thermodynamics, 3rd ed. New York: McGraw-Hill, 1995.
S. I. Sandler, Chemical and Engineering Thermodynamics. Hoboken, NJ: Wiley, 1999.
J. M. Smith, H. C. Van Ness, and M. M. Abbott, Introduction to Chemical Engineering
Thermodynamics, 5th ed. New York: McGraw-Hill, 1996.
W. R. Smith and R. W. Missen, Chemical Reaction Equilibrium Analysis. Hoboken, NJ:
Wiley, 1982.
J. W. Tester and M. Modell, Thermodynamics and Its Applications. Upper Saddle River,
NJ: Prentice-Hall, 1997.

PROBLEMS
Problems marked with an icon may be more conveniently solved on a personal computer with a mathematical
program.
5.1 For the reaction N2 (g)3H2 (g)  2NH3 (g), K  1.60 ⫻
10⫺4 at 400 ⬚C. Calculate (a) ⌬r G ⬚ and (b) ⌬r G when the pressures of N2 and H2 are maintained at 10 and 30 bar, respectively, and NH3 is removed at a partial pressure of 3 bar. (c) Is
the reaction spontaneous under the latter conditions?
5.2 A 1:3 mixture of nitrogen and hydrogen was passed
over a catalyst at 450 ⬚C. It was found that 2.04% by volume
of ammonia gas was formed when the total pressure was maintained at 10.13 bar. Calculate the value of K for 32 H2 (g) 
1
2 N2 (g)  NH3 (g) at this temperature.
5.3 At 55 ⬚C and 1 bar the average molar mass of partially
dissociated N2 O4 is 61.2 g mol⫺1 . Calculate (a) ␰ and (b) K for
the reaction N2 O4 (g)  2NO2 (g). (c) Calculate ␰ at 55 ⬚C if the
total pressure is reduced to 0.1 bar.
5.4 A 1-liter reaction vessel containing 0.233 mol of N2 and
0.341 mol of PCl5 is heated to 250 ⬚C. The total pressure at equi-

librium is 29.33 bar. Assuming that all gases are ideal, calculate
K for the only reaction that occurs:
PCl5 (g)  PCl3 (g)  Cl2 (g)
5.5 An evacuated tube containing 5.96 ⫻ 10⫺3 mol L⫺1 of
solid iodine is heated to 973 K. The experimentally determined
pressure is 0.496 bar. Assuming ideal gas behavior, calculate K
for I2 (g)  2I(g).
5.6 Nitrogen trioxide dissociates according to the reaction
N2 O3 (g)  NO2 (g)  NO(g)
When one mole of N2 O3 (g) is held at 25 ⬚C and 1 bar total pressure until equilibrium is reached, the extent of reaction is 0.30.
What is ⌬r G ⬚ for this reaction at 25 ⬚C?
2HI(g)  H2 (g)  I2 (g)
at 698.6 K, K  1.83 ⫻ 10⫺2 . (a) How many grams of hydrogen
iodide will be formed when 10 g of iodine and 0.2 g of hydrogen
are heated to this temperature in a 3-L vessel? (b) What will be
the partial pressures of H2 , I2 , and HI?

170

Chapter 5

Chemical Equilibrium

5.8 Express K for the reaction

yiso 

CO(g)  3H2 (g)  CH4 (g)  H2 O(g)
in terms of the equilibrium extent of reaction ␰ when one mole
of CO is mixed with one mole of hydrogen.
5.9 What are the percentage dissociations of H2 (g), O2 (g),
and I2 (g) at 2000 K and a total pressure of 1 bar?
5.10 To produce more hydrogen from “synthesis gas” (CO 
H2 ) the water gas shift reaction is used:
CO(g)  H2 O(g)  CO2 (g)  H2 (g)
Calculate K at 1000 K and the equilibrium extent of reaction
starting with an equimolar mixture of CO and H2 O.
5.11 Calculate the extent of reaction ␰ of 1 mol of H2 O(g) to
form H2 (g) and O2 (g) at 2000 K and 1 bar. (Since the extent of
reaction is small, the calculation may be simplified by assuming
that PH2 O  1 bar.)
5.12 At 500 K CH3 OH, CH4 , and other hydrocarbons can be
formed from CO and H2 . Until recently the main source of the
CO mixture for the synthesis of CH3 OH was methane:
CH4 (g)  H2 O(g)  CO(g)  3H2 (g)
When coal is used as the source, the “synthesis gas” has a different composition:
C(graphite)  H2 O(g)  CO(g)  H2 (g)
Suppose we have a catalyst that catalyzes only the formation of
CH3 OH. (a) What pressure is required to convert 25% of the
CO to CH3 OH at 500 K if the synthesis gas comes from CH4 ?
(b) If the synthesis gas comes from coal?
5.13 Many equilibrium constants in the literature were calculated with a standard state pressure of 1 atm (1.013 25 bar).
Show that the corresponding equilibrium constant with a standard pressure of 1 bar can be calculated using
K (bar)  K (atm)(1.013 25)冱 ␯i
where the ␯i are the stoichiometric numbers of the gaseous
reactants.
5.14 Older tables of chemical thermodynamic properties are
based on a standard state pressure of 1 atm. Show that the corresponding ⌬f Gj⬚ with a standard state pressure of 1 bar can be
calculated using
⌬f Gj⬚ (bar)

 ⌬f Gj⬚ (atm) ⫺ (0.1094 ⫻ 10⫺3 kJ K⫺1 mol⫺1 )T

冱 ␯i

where the ␯i are the stoichiometric numbers of the gaseous reactants and products in the formation reaction.
5.15 Show that the equilibrium mole fractions of n -butane and
isobutane are given by
yn 

e⫺⌬f Gn⬚ /RT
⬚ /RT )
(e⫺⌬f Gn⬚ /RT  e⫺⌬f Giso

⬚ /RT
e⫺⌬f Giso

(e⫺⌬f Gn⬚ /RT

⬚ /RT )
 e⫺⌬f Giso

5.16 Calculate the molar Gibbs energy of butane isomers
for extents of reaction of 0.2, 0.4, 0.6, and 0.8 for the reaction
n -butane  isobutane
at 1000 K and 1 bar. At 1000 K
⌬f G ⬚ (n -butane)  270 kJ mol⫺1
⌬f G ⬚ (isobutane)  276.6 kJ mol⫺1
Make a plot and show that the minimum corresponds to the
equilibrium extent of reaction.
5.17 In the synthesis of methanol by CO(g)  2H2 (g) 
CH3 OH(g) at 500 K, calculate the total pressure required for
a 90% conversion to methanol if CO and H2 are initially in a
1:2 ratio. Given: K  6.09 ⫻ 10⫺3 .
5.18 At 1273 K and at a total pressure of 30.4 bar the equilibrium in the reaction CO2 (g)  C(s)  2CO(g) is such that 17
mol % of the gas is CO2 . (a) What percentage would be CO2 if
the total pressure were 20.3 bar? (b) What would be the effect
on the equilibrium of adding N2 to the reaction mixture in a
closed vessel until the partial pressure of N2 is 10 bar? (c) At
what pressure of the reactants will 25% of the gas be CO2 ?
5.19 When alkanes are heated up, they lose hydrogen and
alkenes are produced. For example,
C2 H6 (g)  C2 H4 (g)  H2 (g)
K  0.36 at 1000 K
If this is the only reaction that occurs when ethane is heated to
1000 K, at what total pressure will ethane be (a) 10% dissociated
and (b) 90% dissociated to ethylene and hydrogen?
5.20 At 2000 ⬚C water is 2% dissociated into oxygen and
hydrogen at a total pressure of 1 bar. (a) Calculate K for
H2 O(g)  H2 (g)  12 O2 (g). (b) Will the extent of reaction increase or decrease if the pressure is reduced? (c) Will the extent
of reaction increase or decrease if argon gas is added, when the
total pressure is held equal to 1 bar? (d) Will the extent of reaction change if the pressure is raised by the addition of argon at
constant volume to the closed system containing partially dissociated water vapor? (e) Will the extent of reaction increase or
decrease if oxygen gas is added while holding the total pressure
constant at 1 bar?
5.21 At 250 ⬚C, PCl5 is 80% dissociated at a pressure of 1.013
bar, and so K  1.80. What is the extent of reaction at equilibrium after sufficient nitrogen has been added at constant pressure to produce a nitrogen partial pressure of 0.9 bar? The total
pressure is maintained at 1 bar.
5.22 The following exothermic reaction is at equilibrium at
500 K and 10 bar:
CO(g)  2H2 (g)  CH3 OH(g)
Assuming that the gases are ideal, what will happen to the
amount of methanol at equilibrium when (a) the temperature is

Problems
raised, (b) the pressure is increased, (c) an inert gas is pumped
in at constant volume, (d) an inert gas is pumped in at constant
pressure, and (e) hydrogen gas is added at constant pressure?
5.23 The following reaction is nonspontaneous at room temperature and endothermic:
3C(graphite)  2H2 O(g)  CH4 (g)  2CO(g)
As the temperature is raised, the equilibrium constant will become equal to unity at some point. Estimate this temperature
using data from Table C.3.
5.24 The measured density of an equilibrium mixture of N2 O4
and NO2 at 15 ⬚C and 1.103 bar is 3.62 g L⫺1 , and the density at
75 ⬚C and 1.013 bar is 1.84 g L⫺1 . What is the enthalpy change
of the reaction N2 O4 (g)  2NO2 (g)?
5.25 Calculate Kc for the reaction in Problem 5.19 at 1000 K
and describe what it is equal to.
5.26

The equilibrium constant for the reaction
N2 (g)  3H2 (g)  2NH3 (g)

is 35.0 at 400 K when partial pressures are expressed in bars. Assume that the gases are ideal. (a) What is the equilibrium composition and equilibrium volume when 0.25 mol N2 is mixed with
0.75 mol H2 at a temperature of 400 K and a pressure of 1 bar? (b)
What is the equilibrium composition and equilibrium pressure if
this mixture is held at a constant volume of 33.26 L at 400 K?
5.27 Show that to a first approximation the equation of state
of a gas that dimerizes to a small extent is given by
PV
Kc
1⫺
RT
V
5.28 Water vapor is passed over coal (assumed to be pure
graphite in this problem) at 1000 K. Assuming that the only
reaction occurring is the water gas reaction
C(graphite)  H2 O(g)  CO(g)  H2 (g)

K  2.52

calculate the equilibrium pressures of H2 O, CO, and H2 at a
total pressure of 1 bar. [Actually, the water gas shift reaction
CO(g)  H2 O(g)  CO2 (g)  H2 (g)
also occurs, but it is considerably more complicated to take this
additional reaction into account.]
5.29 What is the equilibrium partial pressure of NO in air at
1000 K at a pressure of 1 bar?
1
1
2 N2 (g)  2 O2 (g)

5.30

 NO(g)

Starting with the fundamental equation for G in the form
dG  ⫺S dT  V dP  ⌬r G d␰

derive equations for calculating ⌬r S , ⌬r V , and ⌬r H from experimental data on ⌬r G for a chemical reaction as a function
of T and P .

5.31

171

What is ⌬r S ⬚ (298 K) for the following reaction?
H2 O(l)  H(ao)  OH⫺ (ao)

Why is this change negative and not positive?
5.32 Mercuric oxide dissociates according to the reaction
2HgO(s)  2Hg(g)O2 (g). At 420 ⬚C the dissociation pressure
is 5.16 ⫻ 10 4 Pa, and at 450 ⬚C it is 10.8 ⫻ 10 4 Pa. Calculate (a)
the equilibrium constants and (b) the enthalpy of dissociation
per mole of HgO.
5.33 The decomposition of silver oxide is represented by
2Ag2 O(s)  4Ag(s)  O2 (g)
Using data from Table C.2 and assuming ⌬r CP⬚  0, calculate
the temperature at which the equilibrium pressure of O2 is 0.2
bar. This temperature is of interest because Ag2 O will decompose to yield Ag at temperatures above this value if it is in contact with air.
5.34 The dissociation of ammonium carbamate takes place according to the reaction
(NH2 )CO(ONH4 )(s)  2NH3 (g)  CO2 (g)
When an excess of ammonium carbamate is placed in a previously evacuated vessel, the partial pressure generated by NH3
is twice the partial pressure of the CO2 , and the partial pressure
of (NH2 )CO(ONH4 ) is negligible in comparison. Show that
K 



PNH3
P⬚

2



3

冢 冣

PCO2
4 P

27 P ⬚
P⬚

where P is the total pressure.
5.35 At 1000 K methane at 1 bar is in the presence of hydrogen. In the presence of a sufficiently high partial pressure of
hydrogen, methane does not decompose to form graphite and
hydrogen. What is this partial pressure?
5.36 For the reaction
Fe2 O3 (s)  3CO(g)  2Fe(s)  3CO2 (g)
the following values of K are known.
t / ⬚C
K

250
100

1000
0.0721

At 1120 ⬚C for the reaction 2CO2 (g)  2CO(g)  O2 (g), K 
1.4 ⫻ 10⫺12 bar. What equilibrium partial pressure of O2 would
have to be supplied to a vessel at 1120 ⬚C containing solid Fe2 O3
just to prevent the formation of Fe?
5.37 When a reaction is carried out at constant pressure,
the entropy change can be used as a criterion of equilibrium
by including a heat reservoir as part of an isolated system
containing the reaction chamber. Show that ⫺⌬r G /T is the
global increase in entropy for the reaction system plus heat
reservoir.

172

Chapter 5

Chemical Equilibrium

5.38 The effect of temperature on KP is given by equation
5.51, and the effect of temperature on Kc is given by
⌬U ⬚  RT 2

⭸ ln Kc
⭸T





V

Is it possible for a gas reaction to have KP increase with increasing temperature, but Kc decrease with increasing T ? If so, what
has to be true?
5.39 Calculate the partial pressure of CO2 (g) over
CaCO3 (calcite)—CaO(s) at 500 ⬚C using the equation in Example 5.11 and data in Table C.2.
5.40 The NBS tables contain the following data at 298 K:

CuSO4 (s)
CuSO4 ⭈ H2 O(s)
CuSO4 ⭈ 3H2 O(s)
H2 O(g)

⌬f H ⬚ /kJ mol⫺1

⌬f G ⬚ /kJ mol⫺1

⫺771.36
⫺1085.83
⫺1684.31
⫺241.818

⫺661.8
⫺918.11
⫺1399.96
⫺228.572

(a) What is the equilibrium partial pressure of H2 O over a
mixture of CuSO4 (s) and CuSO4 ⭈ H2 O(s) at 25 ⬚C? (b) What
is the equilibrium partial pressure of H2 O over a mixture of
CuSO4 ⭈ H2 O(s) and CuSO4 ⭈ 3H2 O(s) at 25 ⬚C? (c) What are
the answers to (a) and (b) if the temperature is 100 ⬚C and ⌬CP⬚
is assumed to be zero?
5.41 One micromole of CuO(s) and 0.1 ␮ mol of Cu(s) are
placed in a 1-L container at 1000 K. Determine the identity and
quantity of each phase present at equilibrium if ⌬f G ⬚ of CuO(s)
is ⫺66.66 kJ mol⫺1 and that of Cu2 O(s) is ⫺77.94 kJ mol⫺1 at
1000 K. [From H. F. Franzen, J. Chem. Educ. 65:146 (1988).]
5.42 For the heterogeneous reaction
CH4 (g)  C(s)  2H2 (g)
derive the expression for the extent of reaction in terms of the
equilibrium constant and the applied pressure, when graphite
is in equilibrium with the gas mixture. Is this the same expression (equation 5.31) that was obtained for the reaction
N2 O4 (g)  2NO2 (g)?
5.43 Calculate the equilibrium extent of the reaction
N2 O4 (g)  2NO2 (g) at 298.15 K and a total pressure of 1 bar if
the N2 O4 (g) is mixed with an equal volume of N2 (g) before the
reaction occurs. As shown by Example 5.4, K  0.143. Do you
expect the same equilibrium extent of reaction as in Example
5.4 If not, do you expect a larger or smaller equilibrium extent
of reaction?
5.44 (a) A system contains CO(g), CO2 (g), H2 (g), and
H2 O(g). How many chemical reactions are required to describe chemical changes in this system? Give an example. (b)
If solid carbon is present in the system in addition, how many
independent chemical reactions are there? Give a suitable set.
5.45 For a closed system containing C2 H2 , H2 , C6 H6 ,
and C10 H8 , use Gaussian elimination to obtain a set of inde-

pendent chemical reactions. Perform the matrix multiplication
to verify A␯  0 .
5.46 The reaction A  B  C is at equilibrium at a specified
T and P. Derive the fundamental equation for G in terms of
components by eliminating ␮C .
5.47 The article by C. A. L. Figueiras in J. Chem. Educ.
69:276 (1992) illustrates an interesting problem you can get into
in trying to balance a chemical equation. Consider the following
reaction without stoichiometric numbers:
ClO3 ⫺  Cl⫺  H  ClO2  Cl2  H2 O
There are actually an infinite number of ways to balance this
equation. The following steps in unraveling this puzzle can be
carried out using a personal computer with a program such as
Mathematica, which can do matrix operations. Write the conservation matrix A and determine the number of components.
How many independent reactions are there for this system of
six species? What are the stoichiometric numbers for a set of
independent reactions? These steps show that chemical change
in this system is represented by two chemical reactions, not one.
5.48 A chemical reaction system contains three species: C2 H4
(ethylene), C3 H6 (propene), and C4 H8 (butene). (a) Write the
A matrix. (b) Row-reduce the A matrix. (c) How many components are there? (d) Derive a set of independent reactions from
the A matrix.
5.49 How many degrees of freedom are there for the following
systems, and how might they be chosen?
(a) CuSO4 ⭈ 5H2 O(cr) in equilibrium with CuSO4 (cr) and
H2 O(g)
(b) N2 O4 in equilibrium with NO2 in the gas phase
(c) CO2 , CO, H2 O, and H2 in chemical equilibrium in the gas
phase
(d) The system described in (c ) made up with stoichiometric
amounts of CO and H2
5.50 Graphite is in equilibrium with gaseous H2 O, CO, CO2 ,
H2 , and CH4 . How many degrees of freedom are there?
What degrees of freedom might be chosen for an equilibrium
calculation?
5.51 A gaseous system contains CO, CO2 , H2 , H2 O, and C6 H6
in chemical equilibrium. (a ) How many components are there?
(b ) How many independent reactions? (c ) How many degrees
of freedom are there?
5.52 At 500 ⬚C, K  5.5 for the reaction
CO(g)  H2 O(g)  CO2 (g)  H2 (g)
If a mixture of 1 mol of CO and 5 mol of H2 O is passed over a
catalyst at this temperature, what will be the equilibrium mole
fraction of H2 O?
5.53 At 400 ⬚C, K  79.1 for the reaction
NH3 (g)  12 N2 (g)  32 H2 (g)

Problems

173

Show that the fraction ␣ of NH3 dissociated at a total pressure as follows:
P is given by
⌬CP⬚ (bar)  ⌬CP⬚ (atm)
1
␣
⌬H ⬚ (bar)  ⌬H ⬚ (atm)
冪1  kP
⌬S ⬚ (bar)  ⌬S (atm)  (0.109 J K⫺1 mol⫺1 ) 冱 ␯i
and calculate the value of k in this equation.
5.54 For the reaction N2 O4 (g)  2NO2 (g), K at 25 ⬚C is 0.143. ⌬G ⬚ (bar)  ⌬G ⬚ (atm) ⫺ (0.109 ⫻ 10⫺3 kJ K⫺1 mol⫺1 )T 冱 ␯i
What pressure would be expected if 1 g of liquid N2 O4 were
K (bar)  K (atm)(1.013 25)冱 ␯i
allowed to evaporate into a 1-liter vessel at this temperature?
Assume that N2 O4 and NO2 are ideal gases.
where 冱 ␯i is the difference between the stoichiometric num5.55 The dissociation of N2 O4 is represented by N2 O4 (g) 
bers of gaseous products and gaseous reactants in the balanced
2NO2 (g). If the density of the equilibrium gas mixture is
chemical equation.
⫺1
3.174 g L at a total pressure of 1.013 bar at 24 ⬚C, what minimum pressure would be required to keep the degree of disso- 5.64 The reaction
ciation of N2 O4 below 0.1 at this temperature?
2NOCl(g)  2NO(g)  Cl2 (g)
5.56 At 250 ⬚C, 1 L of partially dissociated phosphorus pen- comes to equilibrium at 1 bar total pressure and 227 C when

tachloride gas at 1.013 bar weighs 2.690 g. Calculate the extent the partial pressure of the nitrosyl chloride, NOCl, is 0.64 bar.
of reaction ␰ and the equilibrium constant.
Only NOCl was present initially. (a) Calculate ⌬r G ⬚ for this
5.57 Derive the analogue of equation 5.34 for the reaction reaction. (b) At what total pressure will the partial pressure of
N2 (g)3H2 (g)  2NH3 (g) when stoichiometric amounts of ni- Cl2 be 0.1 bar?
trogen and hydrogen are used.
5.65 Acetic acid is produced on a large scale by the carbony5.58 Hydrogen is produced on a large scale from methane. lation of methanol at about 500 K and 25 bar using a rhodium
Calculate the equilibrium constant K for the production of H2 catalyst. What is Ky under these conditions? (⌬f G ⬚ for acetic
from CH4 at 1000 K using the reaction CH4 (g)  H2 O(g)  acid gas at 500 K is ⫺335.28 kJ mol⫺1 .)
CO(g)  3H2 (g).
5.66 Calculate the total pressure that must be applied to a mix5.59 (a) What is the equilibrium constant for the formation of ture of three parts of hydrogen and one part nitrogen to give a
CH4 from CO and H2 at 500 and 1000 K?
mixture containing 10% ammonia at equilibrium at 400 ⬚C. At
400 ⬚C, K  1.60 ⫻ 10⫺4 for the reaction N2 (g)  3H2 (g) 
CO(g)  3H2 (g)  CH4 (g)  H2 O(g)
2NH3 (g).
(b) What is the equilibrium constant for the formation of CH4
5.67 A mixture of one mole of nitrogen and one mole of hyfrom graphite and H2 O at 500 and 1000 K?
drogen is equilibrated over a catalyst for the ammonia reaction
at 500 K and 1 bar. (a) What are the equilibrium mole fractions
2C(graphite)  2H2 O(g)  CH4 (g)  CO2 (g)
of N2 , H2 , and NH3 ? (b) The experiment is repeated with 1.2
5.60 (a) Calculate the extent of dissociation of H2 (g) at 3000 mol of nitrogen and 1 mol of hydrogen. What is the equilibrium
K and 1 bar. A value of 0.072 was obtained experimentally by partial pressure of ammonia? (c) How do you explain this result
Langmuir. (b) Calculate the extent of dissociation of O2 (g) at in terms of Le Châtelier’s principle?
3000 K at 1 bar.
5.68 Assume that the following reaction is in equilibrium at
1000 K:
5.61 What is K for
I2 (g)  2I(g)

3C(graphite)  2H2 O(g)  CH4 (g)  2CO(g)

at 1000 K, and what is the degree of dissociation at 1 bar? At
0.1 bar?
5.62 Propene and cyclopropane are isomers. Their standard
thermodynamic properties in the gas phase are given in Table
C.2. If they were in equilibrium, what would be the standard
Gibbs energy of formation and the standard enthalpy of the
isomer group at 25 ⬚C? Show that the same value of the standard Gibbs energy of formation is obtained by calculating the
mole fraction average Gibbs energy and adding the entropy of
mixing.
5.63 Show that in going from a standard state pressure of 1 atm
to 1 bar, thermodynamic quantities for reactions are corrected

⌬r H ⬚ (1000 K)  182 kJ mol⫺1 . (a) What will be the effect on
the equilibrium composition of raising the temperature at a total pressure of 1 bar? (b) What will be the effect of raising the
pressure to 5 bar? (c) What will be the effect of adding nitrogen
at a constant pressure of 1 bar?
5.69 When N2 O4 is allowed to dissociate into NO2 at 25 ⬚C at
a total pressure of 1 bar, it is 18.5% dissociated at equilibrium,
and so K  0.141. (a) If N2 is added to the system at constant
volume, will the equilibrium shift? (b) If the system is allowed
to expand as N2 is added at a constant total pressure of 1 bar,
what will be the equilibrium degree of dissociation when the N2
partial pressure is 0.6 bar?

174

Chapter 5

Chemical Equilibrium

5.70 For the formation of nitric oxide

5.76

N2 (g)  O2 (g)  2NO(g)

CH4 (g)  2O2 (g)  CO2 (g)  2H2 O(g)

2.5 ⫻ 10⫺3 .

K at 2126.9 ⬚C is
(a) In an equilibrium mixture containing 0.1 bar partial pressure of N2 and 0.1 bar partial pressure
of O2 , what is the partial pressure of NO? (b) In an equilibrium
mixture of N2 , O2 , NO, CO2 , and other inert gases at 2126.9 ⬚C
and 1 bar total pressure, 80% by volume of the gas is N2 and
16% O2 . What is the percentage by volume of NO? (c) What is
the total partial pressure of inert gases?
5.71 Prove that for 2C2  C4 in the presence of other gases
in the gas phase, the equilibrium constant expression
K 

y4
y22 (P /P ⬚ )

5.72 The following data apply to the reaction Br2 (g) 
2Br(g):
1123
0.408

1172
1.42

1223
3.32

1273
7.2

Determine by graphical means the reaction enthalpy at 1200 K.
5.73 The average molar mass M of equilibrium mixtures
of NO2 and N2 O4 at 1.013 bar total pressure is given in the
following table at three temperatures:
t / ⬚C
M /g mol⫺1

25
77.64

45
66.80

at 298 and 1000 K.
5.77 What is ⌬r S ⬚ for H2 (g)  2H(g) at 298, 1000, and
3000 K?
5.78 Calculate the entropy changes for the following reactions
at 25 ⬚C:
(a)
(b)

Ag(ao)  Cl⫺ (ao)  AgCl(s)
HS⫺ (ao)  H(ao)  S2⫺ (ao)

5.79 From electromotive force measurements it has been
found that ⌬r S ⬚ for the reaction
1
2 H2 (g)  AgCl(s)

may be interpreted in two ways:
(i) The mole fractions are expressed only in terms of C2 and
C4 and P  P2  P4 .
(ii) The mole fractions are expressed in terms of all gases
present and P is the total pressure.

T /K
K /10⫺3

Calculate ⌬r S ⬚ for the reaction

65
56.51

(a) Calculate the degree of dissociation of N2 O4 and the equilibrium constant at each of these temperatures. (b) Plot log K
against 1/T and calculate ⌬r H ⬚ for the dissociation of N2 O4 .
(c) Calculate the equilibrium constant at 35 ⬚C. (d) Calculate
the degree of dissociation for NO2 at 35 ⬚C when the total pressure is 0.5 bar.
5.74 For a chemical reaction, ln K  a  b /T  c /T 2 . Derive
the corresponding expressions to calculate ⌬r G ⬚ , ⌬r H ⬚ , ⌬r S ⬚ ,
and ⌬r CP⬚ .
5.75 One mole of carbon monoxide is mixed with one mole of
hydrogen and passed over a catalyst for the following reaction:

 HCl(aq)  Ag(s)

is ⫺62.4 J K⫺1 mol⫺1 at 298.15 K. What is the value of
S ⬚ [Cl⫺ (aq)]?
5.80 The solubility of hydrogen in a molten iron alloy is found
to be proportional to the square root of the partial pressure of
hydrogen. How can this be explained?
5.81 Is magnetite (Fe3 O4 ) or hematite (Fe2 O3 ) the more stable ore thermodynamically at 25 ⬚C in contact with air?
5.82 An equimolar mixture of CO(g), H2 (g), and H2 O(g) at
1000 K is compressed. At what total pressure will solid carbon
start to precipitate out if there is chemical equilibrium? (See
Problem 5.28.)
5.83 At 50 C the partial pressure of H2 O(g) over a mixture of
CuSO4 ⭈ 3H2 O(s) and CuSO4 ⭈ H2 O(s) is 4.0 ⫻ 10 3 Pa and that
over a mixture of CuSO4 ⭈ 3H2 O(s) and CuSO4 ⭈ 5H2 O(s) is 6.3 ⫻
10 3 Pa. Calculate the change in Gibbs energy for the reaction
CuSO4 ⭈ 5H2 O(s)  CuSO4 ⭈ H2 O(s)  4H2 O(g)
5.84 Calculate (a) K and (b) ⌬r G ⬚ for the following reaction
at 20 ⬚C:
CuSO4 ⭈ 4NH3 (s)  CuSO4 ⭈ 2NH3 (s)  2NH3 (g)
The equilibrium pressure of NH3 is 8.26 kPa.
5.85 The vapor pressure of water above mixtures of CuCl2 ⭈
H2 O(s) and CuCl2 ⭈ 2H2 O(s) is given as a function of temperature in the following table:
t / ⬚C
P /bar

17.9
0.0049

39.8
0.0250

60.0
0.122

80.0
0.327

(a) Calculate ⌬r H ⬚ for the reaction

CO(g)  2H2 (g)  CH3 OH(g)

CuCl2 ⭈ 2H2 O(s)  CuCl2 ⭈ H2 O(s)  H2 O(g)

At 500 K and a total pressure of 100 bar, 0.40 mol of CH3 OH
is found at equilibrium. What is the value of the equilibrium
constant expressed in terms of P /P ⬚ , where P ⬚ is the reference pressure of 1 bar? What is the equilibrium constant expressed in terms of c /c ⬚ , where c ⬚ is the reference concentration of 1 mol L⫺1 ? Assume ideal gases.

(b) Calculate ⌬r G ⬚ for the reaction at 60 ⬚C. (c) Calculate ⌬r S ⬚
for the reaction at 60 ⬚C.
5.86 The equilibrium constant for the association of benzoic
acid to a dimer in dilute benzene solutions is as follows at 43.9 ⬚C:
2C6 H5 COOH  (C6 H5 COOH)2

Kc  2.7 ⫻ 102

Problems
Molar concentrations are used in expressing the equilibrium
constant. Calculate ⌬r G ⬚ , and state its meaning.
5.87 Superheated steam is passed ...


Anonymous
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