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Unformatted text preview: Chapter 5/ Entropy and the Second and Third Laws of ThermodynamiCS ACE”, = 2x 127.6 J K“ moi“ — 219.2J K“ mol“ = 36.0 J K“ moi“
310.K
215° 310K =AS° 298.15K +AC. in
R( ) ’*( ) 1”” 298.15K
=175.0 J K“ moi“ +36.0 J K“ mol“ xln 310' K
298.15K = 176.4 J K“ moi“ AH; (310. K) = AH; (298.15 K)+ACP’mAT _ —74.1 M moi“ +36.0x10‘3 kJ K‘lmol‘l x 12.0 K = 437 kJ moi“ A50 _ 2111; (310. K) _ 73.7 k] moi“
surroundings _ *
T 310. K + A5; = 238 J K“ moi“ +1764 J K“ moi“ = 2238 J K‘1 mol—I
AS° =AS 414 JK“ mol‘l
P528) The amino acid glycine dimerizes to form the dipeptide gly 2Glycine (s) ——> Glycylglycine(s) H20 (1)
7 Calculate, AS ASSW, and ASwmz at T = 298 K. Useful thermodynamic data are: AS;e =2x103.5 I K‘I mol‘I +190. JK‘I m0
AH; = 2X537.2 kJ mol“ —746 kJ mol‘1 0 —l
: AHR: 42.6kJm61 =_143JK_1m01_,
T 298.0K —285.8 kJ mol”1 2 42.6 kJ mol‘1 surr ASW, = 442.95 J K" mol”1 +53.0 J K21 mol‘1 = —90.0 J K‘1 mol‘I ASWOMMW, and ASW, , verify that this transformation is given by CP(H20(Z)) = 75.3 J K“
AHﬁW = 6.008 kJ mol“ at 0.000 spontaneous at —2.25°C. The heat capacities are mol*1 and
CP(H20(S)): 37.7 J K“ moi“, and C. Assume that the surroundings
, —2.25°C) —> H20(s, —2.25°C) and , ~2.25°C). Because S is a state For pathway b) H200, —2.25°C) —> H20(l, 000°C) —4 H200, 000°C) —4
H200, —2.25°C) 108 cylglycine according to the reaction 270.90 K
273.15 K _ AHﬁzsioﬂ + "CPS," (.9)an
O 90 K 273.15 K _.
27 . 273 15 K 1 6008 Jmol
' 1—1 KIxm ' “1 “‘0 X 273.15 K
x753 1910 270.90 K I ‘1 270.90 K
9* +1 molx37.7 Jmot K X1“ 273.15 K 1
‘ K‘1 =—21.7 JK
sf_JK“ 422.0 JK 1—0.312 J lculate AHP : ‘1'
Cillate ASsurroundings’ we ﬁrSt ca ' 2; C ) Cp( q ‘ z —5928 J = q
__q 5928 J = 21.9 JK'1
” 270.90 K ""T. surroundings ' taneous.
‘1 — '1 > 0. The process 13 spon
~21.9 J K“ —21.7 JK — 0.2 JK total "' u r unive se
, I S .
. . . 1y
Cent1 rade, respective 04 d ees Fahrenheit correspond to 0, 21.1 and 40.6 degrees g 2.0, 70.0 and 1 egr ~1__ 03 J S~l
—77><power=3.15><1.25><103 JS —3.94><1
11"  103 J s“
.qhotzqmu 2+1 —5.19>< 1 61mm 394X103 J S“ 213.4 J K“ moi“ s T 294.3 K
old 3 _l
qhot JS Th 313.8K ()l 21 11 S4
_ —1 17] 5—1 J K m0
—1 1—1 51—134 JK m0
AS+AS =16.SJK m0 surr —1 =16.5 J K“ moi“ s _ — d T 2
P531) Using your result .t is constant between T ‘ 300. K an d‘t‘ s T = 310K. Assume the heat capa01 y
physiological con 1 10n , 310. K. . K and 310. K.
capacity is constant at its 300. K value between 300 "GT3 ~1 lel K 91602+0 77610T—0.0019511T2 +2.76403X10 21352 J K m0
310. _—__._______.______________________________._.__. S°(31(). K) = M T 109 Ch
apter 5/ Entropy and the Second and Third Laws of Th d.
ermO ynamics 7
Chapter 5/ Entropy and the Second and Third Laws of Thermodynamics S°(310. K) =S°(300. K) +CP mg
7,.
.7 J K n’lOlh1 In ' : 1 J _] ‘1 if" ' 1 ‘ external P5 32) O I 300. K . K mo] f z
 n6 mo e of an ideal ' RP
gas Wlth CV,” : 5R Und nRdetemal _ T n ﬁler/701
_ . , a . + f _ 1‘ "CV :11
followmg list from an initial state described b T ergoes the transﬁkmations described in 3 C Pr" 3 Pi
A Y = 325 K = ‘
U 3 AH , and A9 for each process. and P 300 bar' calculate qa ’ + Rgx’e’m/ 2 5x8 314 J mol'1 K" + 8314 J mOI—l K4 XO'SOObaI
a The gas under t . goes a reVersrble adlabam ex * =325 Kx —————————————————————’—
pansmn until the ﬁnal re . . . . RP”  , 8.314 J moi" K" x0.500bar
P ssure IS half its initial value, + "“PL‘L 2'5X8‘3 14 J m01 1 K 1 + 1_ 50 bar .1 1
wanVmATzlmolxsxgsmimd K x(279K—325 K)=965J I 1
nCPmAT=1molx7X8314Jm01 K x(279K325 K)=—1.35x103J q ~ 0 because the process IS adiabatic S Initial Value 2
1 n P, T,
3: E 7 1y P 17 T 7 H 11 _nR1n_L+nCPmln_L
7:  a PE, 2 ‘f = i T] 7 i ’ I i f Ti P ’F = T 4 1" K" 279 K
141 I I} Trnol><8.314lmol’l K“><1n1'50 bar +1 mol><7xg'31 Jmo xln————
5  3.00 bar 2 325 K 6 JK‘1—4.49 JK'1 = 1.28 JK‘1 Q_ 3.00 bar 7 i
I: 150 bar 5 =(2.00) 7 :0820
T f : 0820X325 K = 267 K onstant external pressure of zero bar until the ﬁnal gas undergoes an expansion against a c essure is equal to half of its initial value. AU = W = "Ci/MAT =1 molx 5X8314 .lmol’l K'1 .5 = —8.314 Jmol'I K“ xln—————— = 5.76 JK“
3.00 bar AH = nCP,I)2AT :1 mle 7X83 Jmol‘] 1(1 a
2 ><(267K—325 K)=—1.70x103 J AS = 0 because qrevmme = 0‘
C, what is the maximum heat leak (in watts) that ed at 200°C and the room temperature is 35° lerated? Assume that the coefﬁcient of performance is 50. % of the maximum theoretical value. ﬁnal pressure is half its initial value q = 0 because the process is adiabatic. _
ﬁcient of performance is
: T.
‘QOXA— : 0.500x 275 K = 4.17
‘ Th0, Tc,,,d 308 K  275 K 110 f Thermodynamics Chapter 5/ Entropy and the Second and Third Laws 0
Chapter 5/ Entropy and the Second and Third Laws of Thermodynamics .1 ~1 lenﬂ: —1.4l><104 J
Input power x coefﬁcient of performance ., = —1_75 m01x8.314 J mol K x 4.00 L
=0.25hp x—7%W~X4.17=778W=778Js“ a
P  1041
3 4 ~‘ 375K~920K)—119><
_ .314Jm01 K ><(
TLLTb)=1.75 mol><2><8 nCV ,m ( 15.4 L 3 _ _._._ = 5.76><10 J
a RT 1n V" — —175 molx8.314 Jmol‘1 K 1 X375 Km 44.2 L +
,n c "'— — '
5 34) Consider the reversible Carnot cycle shown in Figure 5.2 with 1.75 mol of an ideal gas with V C )=+1.19x10“ J 7 3 ‘1 '1 920 K—375 K
7‘ 4 m — .314Jmol K ><(
0t I‘CSBI‘VQ‘ nCVm (Ta Td) 1.75 01x2><8
temperature 0f 3km a 920. K from an initial volume of 4.00 1, (Va ) to a volume of 11.50 L (V1,), SYStem then undergoes an adiabatic expansion until the temperature falls to T cold = 375 K. The syst 375K w_1.00><103J:1.69X1031
then undergoes an isothermal compression and a subsequent adiabatic compression until the initial st Tcold L _ _L._ _—_ 0592 q = __ n 6 . T1101
a. Calculate Vc and Vd. re 1 69 kJ of heat must be extracted from the s urroundings to do 1.00 U of work in the b. Calculate w for each step in the cycle and for the total cycle. 0. Calculate a and the amount of heat that is extracted from the hot reservoir to do 1.00 U of work 0°C and 100°C, the heat capacity of Hg(l) is given by
the surroundings, T
Cdeg’ I) = 30.093 4 4.944 x1073 E J K‘1 mol“1
a) We ﬁrst calculate VC and Vd. l erature from 000° to 750°C at
V. 1‘7 _ V r E te‘AH and AS if 1.75 moles of Hg(l) are raised in temp Vb 9V1, Tb
' l 3
V __5 w [T/K]
7%ﬂ 1—3 = 172% 2 23.84; Vc=44.2L 0_3(348152_273‘152)]Jm011
b 920K 920 [30093434815—273.15)—2.472x1 
1
T. _ V. V. T. W J
7:, V; ’Va 7;
_ —l
4 3 T~ —3 5—27315) JK1 H101
V 5 74,944x10 (348.1
—= 373513: 3—75—K— 2:3.84; V;=15.4L
Va 920K 920K [T/K]=1.75 molx 30.093 in}— I b) We next calculate w for each step in the cycle, and for the total cycle. r, . " ",5,
g a n P  d?" I L? T1 r ‘ ' determine Tm. You can
atron occurs reversibly. Hint: Determine AHden graphically, then ' ' d a
rveﬁttmg routine an
ntegration numerically using either a Spread sheet program or a cu culator (see Example Problem 5.9) ulti 1y by the
e area under the curve into rectangles of width 1.0 K, add the areas and m P 112 Chapter 5/ Entropy and the Second and Third Laws of Thermodyna .
mics
Chapter 5/ Entropy and the Second and Third Laws of Thermodynamics molecular mass of 14000. mol'l Th ' ‘ '
g . e vertical ax1s IS C 1.)," / T and the units for the areas in the ﬁgure below are J K‘1 g'l. The result is 190 kJ mol'l. T o Pf f
5(200. “C, 0.500 bar) = S (298.15 K)—ann—‘—+nCP 111—.
P T l I 0.500 bar :1 molx188.8 J K“ mol‘1—1 molx8.314 J K—1 mol“ xln +1 molx33.58 J K“ mol’l 1 bar
498.15 K
x1n———————
298.15 K
= 212 J K'1
P5.3 9) The heat capacity of a —quartz is given by
C m, (a—quartz,s) T 5 T2 _1 _, =46.94+34.31x103——11.30x10—7
JK mol K K The coefﬁcient of thermal expansion is given by ﬂ = 0.3530 x 104 K‘1 and Vm = 22.6 cm3 mol—1. Calculate AS,“ for the transformation a —quartz (380°C, 1 atm) —> a —quartz (315°C, 850. atm). From equations 5.23 and 5.24 7’ dT 333 "as ass TemperatureSK m g AS,” 2 ijPJn ? ‘ VIB (P f ‘ Pi )
P5.37) Calculate AS I
mrroundin s and 0,0 for art f P 2
State of th d g r l P c o roblem P5 .6. Is the process a spontaneous The 588 15 46.94 + 34.31 x 10'3 $241.3 ><10‘5 e surrou ' ‘ = _ 
C The gas undergoes a reversible i 3”” I
sothe 1 ' .
I‘ma expansron at 300. K until the pressure is half of its in't‘ 1 588.15 K _ 2 . _
Value. “31 = 46.94x1n—————+34.31><103x(498—298)—5.65x105x(588.152#311.152) JKl mol]
311.15 K
V. 3 5
w=~ z RTI __f_=_ _ _ _ 1m _ 1.0125X10 Pa
q n n Vi L95 molx8.314 Jmoli K 1X290. K M 33.2mm J —22.6 cm3 mol1 x 06 cm3 x03530x1041 K ‘ x849 atmxT ‘ d) :1: —3.26><103 J _1 = 29.9 J K‘1 mor1 +0.0218 J K" mol“ —14.1 J K—1 moiI —0.0686 J K“ mol'l = 25.2 J K" moi"
mgs T 290K ——11.2JK P540)
V. .
= _f _
AS ann V —1 95 molx8.314 J mOI I K ‘ xln 2 :11 2 JK 1 a. Calculate AS if 1 mol of liquid water is heated from 000° to 100°C under constant pressure if CE,”
= 75.3 J K‘1 mol—1. [oral : AS + A"Sisy'zzrroundmgs :1 JI<.1 —1 JI<—I : O
b. The melting point of water at the pressure of interest is 000°C and the enthalpy of fusion is 6.010 M There is no natural di t' ' 
rec 1011 of change in this process because it is reversible. mol’]. The boiling point is 100°C and the enthalpy of vaporization is 40.65 M mol—1. Calculate AS P538 C l
) a culate the entropy of one mole of water vapor at 200. for the transformation H20(s, 000°C) —> H20(g, 100°C). 0C . I .
in the data tables. and 0.500 bar usmg the Informatlon a) The heat input is the same for a reversible and an irreversible process. 114
115 Chapter 5/ Entropy and the Second and Third Laws of Thermodynamics dq : dqreversible 2 "Cam
C
_ ,m I] T.
AS _ n ;‘ T 2 CP m In; AS AHvapmVaImn C 1 Tf
’ ‘ Z —n g + n ’ m n l
1 l l Tvapm'ization I , vaporization
= m 1 v.
0 x75 3 JmOl K ‘1 2731511: ##185 g steam X40650 Jmol’] + 18.5 g steam X75 3 JK] m01_1)‘<1n344 K
=2.71 J K 1 ' 7 18.02 g rnol'1 373 K 18.02 g steam mol” ' 373 K
= —l 18 J K—1 b) AS is calculated for the water. It consists of heating the water from 310. K to 344 K.
fusion __ 1 molx J mOl—l _~R— nusion K — J I<.I fusion '— 7; 305 g x753 JK'I moi" X111 344 K = 134 JK“ AS = nC. ln = #——————
1 18.02 g mol" 298 K “vaporization : w 2 1 mol X J mOI—l fusion m I’aporization 15 K = J Kl
AS __ 7101a] '_ AS fusion + =154JK" ASt0m1=l34 J K'1 —1 18 J K‘} = 16.4 J K‘l. The process is spontaneous. P5.42) The following heat capacity data have been reported for Lalanine: 40. 60. 80. 100. 140. 180. 220. 260. 300.
l7.45 30.99 42.59 52.50 68.93 83.l4 96.14 109.6 122.7 By a graphical treatment, obtain the molar entropy of Lalanine at T = 300. K. You can perform the + 118,..an = (22.00 + 108.95 + 23.49) J K" greater than 373 K.  , calculated tern 
perature W1” be integration numerically using either a spread sheet program or a curveﬁtting routine and a graphing
n H20
steam Va oriza [on '+ , H20 __ Seam ‘ I
p , nmamcpmz (Tf 7; z )+ "Hzoc (Tf ‘TH20) 2 0 calculator (see Example Problem 5 9)
T ._ nsteam szozsleam + n CHZOTH2O 3 1 ‘— «S
nywamcgza + nH OCIISQU ‘ 1011 2 ..
18.5 g steam ’m ’ /
R _ 100 e ”
18.02 g mol‘1X75'3 JK‘ moi1x373 K + M 1 ' //
18.02 gHzo moi] X753JK‘ mol‘x310 K : . //
80 ~
+ 18.5 g steam ' // g Steam g morl J mol'1 : //
x753 JK" 1 305 gH 0(1 60 —
18.02 g mOl—l mol + #x75 3 JKJ _I  /
1} =344 K 18‘” gH20 mol" ‘ mol Z “V
40 4
AS IS calculated for 7
th ~ ‘
e steam It con51sts of condensmg the steam t 3 _
water to 344 K, a 73 K and COOIing the resulting 20 7
— gig/1 1  1 1 1 4 I 1 1 1 1 l 4 1 1 1 L 1 1 1 1 I 1 1  L l
t 50 100 150 200 250 300 The above graph shows C p, m as a function of T. The line is the best ﬁt to the data of a polynomial of the
form a + bx +cx2 +dx3.given by 9.1602 +0.77610 x0.0019511x2+ 2.76403 x 10'6 x3.
117 116 Chapter 5/ Entropy and the Second and Third Laws of Thermodynamics 50600 K): —9.1602+ 0.77610T—0.0019511T2 +2.76403x104’T3 3x8.314 Jmor‘K‘l
T T : 135.2 J K“ mol‘l ——————————— 2
5 x8314 J mol‘lK“ 3
———————————>< 189 K—300 K):—2.31><10 J
AH = nCPJnAT =1 “101x 2 ( 3
J 189 K—300 K)=—1.38x10 J
’ ~ AT =1 molx x(
U:W—~I’ZCVSm because the contribution to S below 10, K is very small '— e = U AS = 0 because = H AS = L ' . total
AS 0 becaus greversible  surroundings q b) w =2 0 because AV: 0. 3x8.3l4 Jmol'lK'l
——'—_'—2"'_‘___
5x8.314 Jmor‘K'1
———"—"‘—2—‘_—"‘—_ 3
K—189 K)=1.38x10 J
 =1molx X(300
AU 5: q _' "CVJHAT 3
C AT 1mol>< x(300K—189 K)=2.3lx10 J
AH=n Pym = The area required for the solar panel is o . 2 746 J S.1
efﬁciencyxﬂux 0.168 x 2.00 J cm‘2 min'] TV 300 K _ 0 b
= 2 2 __L=0.314 barx —0.50 at
><lmin/60s><104cm2/m2 13'3m IT, 189K
P5 44) An ideal gas sample containing one mole for Wthh CV,” =3/2R undergoes the following In Pf" C In ff
2 —— R ; + n m ’—
revers1ble cyclical process from an initial state characterized by T = 300. K and P = 1.00 bar' AS n Pr P, T’ 8 314 J 14K" 300 K
. 5X . m0 1 a It 13 ex anded rev 5 bl d ' .  W l MX n p er 1 y an adiabatically until the volume doubles : _1 m01x8.314 J mol 1K I X In 0314 bar +1 m0 X 2 189 K = —3.84 JK" + 9.60 JK'1 = 5.76 JK" 3
AS ___“‘I.__=“_1_3£>fl’_l=_4.61 JK'1 surroundings‘ T K surroundings 
AS =5.76JK'1—4.6‘l JK'1=l.15 JK1 total :AS+AS surroundings 0) AH: AU: 0 because AT: 0. P
7}. = 0.630x300. K = 189 K V i
w 2 —q = Z “HRTll’l— V, Pf
The initial and ﬁnal volume and the ﬁnal pressure are calculated. l _ 3
= ~1mo1x8.314Jmol"K"><300 leng “1‘73X10 J
nRT. l.00mol><8.3l4Jmol‘lK‘Ix300 K 1.00 bar 1 V324: ‘ _ 2 3 l '1 1 EM ‘ . .____ 1? 105Pa 315x“) m As=—nR1n—’— 1m01X8314Jm01 K ><1n0.500 bar V,. =2V, =6.32x10'2m3 ' ’ 173x103J 576JK.1 V. T. 1 189K ——_q—*="”.: ‘ P' barx—x 20.315 b ASsurroundm .s _ ’ Vf 1'; 2 300K ar g T surroundings q = 0 because the expansion is adiabatic. AS +AS=—5.76JK‘1+5.76JK“ :0 total =AS surroundings 118 119 Chapter 5/ Entropy and the Second and Third Laws of Thermodynamics Chapter 5/ Entropy and the Second and Third Laws of Thermodynamics Chapter 5/ Entropy and the Second and Third Laws of Thermodynamics . ' L _ [rs Fo1 the ent1re cycle, ,p 5C dT : 1 , _,._..
T l T y 4
1 0062 JK‘1 g" 0.083 JK1 g1 0.125 JK" g" 0.167 JK1 g w,omg=—l.38x103J + 0 +1.73 x103] = 344 J . 1 ‘ + + qW=0+1.38x103J1.73x103J=—344J W4. 292K + 293 K 294K 295 K 4 ‘1 3 3 ‘ ~ ~ 29 ~1 4 0 334 J K“ g1 0.480 J K‘1 g“ 0.627 I K‘1 g“ + 0836 K g
AU,o,al=—1.38x10J+1.38><10J+0—0 +‘WJ. ' 297K + 298K + 299K 300K ‘ 296 K _ _ .1 1 AHWLZIM X “PM 2‘31X103J + 0 2 0 104 J K“ g1 1.23 J K‘lg'1+1.38 J K" g" +1.40 J K1 g1+1.33 J K g 1 —1 4"!“4’ 302 K 303 K 304 K 305 K
48m]: 0 —5.76 J K + 5.76 J K = 0 301K _ 4 _1 76 J K _1 .1 0 940 jK'I g] 0731 J K" g‘ + 0.522 J K g +0.3 g
ASsurWdzngFAS mm; = 0 — 4.61 J K'1+ 5.76 J K'1 =1.14 J K'1 307 K + 308 K 309 K 310 K
— l
’5 50“ 0.167 J K" g" + 0.084 J K1 g
P5.45 For rotein denaturation, the excess entro of denaturation is deﬁned as AS = P dT , 313 K
) p py den 312 K
T T1 where 5C? is the transition excess heat capacity. The way in which 5C2” can be extracted from differential scanning calorimetry (DSC) data is discussed in section 4.10 and shown in Figure 4.7. The DSC data below are for a protein mutant which denatures between T1: 288 K and T 2 = 318 K. Using the equation for AS,“ given above, calculate the excess entropy of denaturation. In your calculations, use the dashed curve as the heat capacity base line which deﬁnes 5 1’.” as shown in Figure 4.7. Assume the molecular weight of the protein is 14000. g mol'l. You can perform the integration numerically using either a spread sheet program or a curveﬁtting routine and a graphing calculator (see Example Problem 5.10) We break up the area under the curve into rectangles and evaluate the area which is C p x AT. We then divide by T for each rectangle and sum the contributions and ﬁnally multiply by the molar mass. This estimate approaches the value of the integral as the AT value approaches zero. The result is 429 J mol'1 Kl @mpsratars; tr: Th6 graph above shows C p as a function of T. 121 ChaPter 5/ Entro
py and the Second and Th'
1rd Laws of Therm . .
Odynamics
Chapter 6: Chemical Equilibrium P5.46
) The standard entropy of Pb(s) at 298.15 K is 64 80 J K" Pb(S) is given b mOITI. Assu .
y me that the heat capacny of Solutions Problem numbers in italics indicate that the solution is included in the Student’s Manual. CP,m b9 S) Questions on Concepts Jmol_1K_1=22.13+0.01172£+1.00x10“55 _ (26.1) It is found that KP is independent of T for a particular chemical reaction. What does this tell you 1
. Assu ' ‘7
me that _ about the reaction. CP,m T
J 161111014 =32.51—0.00301E danI,
dT is proportional to AH. This is true if AHreaction = 0 because a. b. quilibrium for 5 = 0.1. What does this tell you about the Calculate the standard entropy of Pb(l) at 650 °C
. . (26.2) The reaction A + B —> C + D is at e C l
a culate AH for the transformation Pb(s, 25 06C) _) Pba 650 ° . g _ .  . 9
a) S”, (PbJ, 923.15 K) : S: (Pb33329815 K) variation of Gpm With cf . +60055 C19,”; AH , _ 92315
J ﬂmon + I CP,m °C). It tells you that AG; for A + B is less than that for C + D. If the AG; of reactants and products were 5 T / K
298.] [ l 71%,” 60055 [T / K] dIT/K]
= 64.80 J mol“ K“ nearly the same, 5.“, would be near 0.5. If the AG}. for the products were much less than those for the
600.55 + I reactants, 58., would be near 1. 298.15 [T/K] [T/K]
+ 4770 J mol‘1 Q63) Under what condition is Kp = Kx? 600.55 K
9231.15 3251—0_00301[T/K] This is the case for a reaction involving gases if An = 0 because as shown in Equation (6.78),
\K‘d 600.55 [T/K] [T/K] P ~Av = 64.80 J mol‘1 K" + 20 Kx = KP .  J 111011 K'1 _ _ P
=10611m0rl K—I JmOll JmOll K1 . .
Q6.4) What is the relationship between the KP for the two reactions 3/2H2(g) + l/2N2(g) —> NH3(g) and 3H2(g) + N2(g) “> 2NH3(8)? 600.55 b) AH,” = Csolid 773.15
I 29:15 PM d[T/K]+AHfusion+ I Cfff’Z’ddH / Kl 600.55 = 8918 J mor‘ + 4770 J moiI + 9748 J 11101"
= 23.4><103 1 mol'1 Kp for the second reaction is the square of KP for the ﬁrst reaction as can be seen in the reactions quotients
* 2
PNH3 PNH3
P° a d P°
————————————— n ——~—————.
PHZ 3/2 PNZ 1/2 PHZ 3 PNZ
P° P° P° P° Q65) Under what conditions is dA S 0‘ a condition that defines the spontaneity of a process? This is the case at constant T and V if no nonexpansion work is possible. 122
123 Chapter 6/ Chemical Equilibrium
Chapter 6/ Chemical Equilibrium , show that if a valve separating vesse
l of pure A from a vessel containing a mixture of A and B '
is op ened, mixing will occur. Bot fuel cell? ‘ The Chemic '  . a
al potential of A in the mixture is less than in the pure gas b S dwmmpmo"
ecause ' “AlTaP)=#°(T +RT1£¢ o
A \ l n Po , and PA < P . Because mass ﬂows fmm . . ‘ _ ‘ u . _
reglons 0f high chemical 34.18 refer to the reaction system C0(g) + 1/202(g) —>C02(g) at equilibrium for —283.0 kJ moi“. let the change in the partial pressure of C02 as the temperature is increased at constant total ase, following Le Chatelier’s principle because the reaction is exothermic. dictkthe change in the partial pressure of C02 as the pressure is increased at constant ase, following Le Chatelier’s principle because An < 0. edict the change in the partial pressure of C02 as Xe gas is introduced into the reaction vessel pressure and temperature. C Otto oppose the decrease in volume. __i_ct the change in the partial pressure of C02 as Xe gas is introduced into the reaction vessel Q6.11) The reaction A + B —> C + D is at equilibrium for ._
é: Vessel at constant volume and temperature. . = 0. . 
variation of Gpure with 9:? 5 What does this tell you about the I yst will have no effect because the system is already at equilibrium. A catalyst can increase the 'h'ch equilibrium is achieved, but cannot change the position of equilibrium. redict the change in the partial pressure of C02 as 02 is removed from the reaction vessel at t pressure and temperature. ecrease to reestablish equilibrium as the reactant oxygen is removed. 125 Chapter 6/ Chemical Equilibrium
Chapter 6/ Chemical Equilibrium ’AiGomction __ AH:eactmn 1 ‘/ f~ .
_R><298.15K R r, 29815K  " 58.1x103 Jmo1" _
6.3x10‘ Jmol x
: WWsm J K"moi" 525 K 298.15 K Problems P6.1) Calculate AAfeaam and Awaam for the reaction C6H6(l) + 15/202(g) —> 6C02(g) + 3H20(l) at
Because 298 K from the combustion enthalpy of benzene and the entropies of the reactants and products. All gaseous reactants and products are treated as ideal gases =AH° —TAST combustion combustion AG" combustion AS" 6S°(CO2,g)+3S° (H20,l)—S°(C6H6,l)—15/2S°(02,g) combustion _ = 6x213.8 J moi" K" + 3x70.0 J moi" K" —173.4 J moi" K" —15/2><205.2 J moi" K" 2 is placed in a reaction vessel and
= J mOIl K1 35 0 bar and 750. K in the reaction AGjambumn = ~3268x103 kJ moi" —298.15 Kx(—219.6 J moi" K"): —3203x103 kJ moi"
Motiombustion : Aljdmnbzution — ASCOombustion
: AHzom us ion —A ~ ' —TAS:om ZlSlOI’l  ' o b t ( o )wmmm b t 0 equation relating KP and the extent of reaction as in Example Problem 6
= AGCWHSW + T common * AUDV) — TASCOMMW” L ation solver calculate the number of moles of each specres present at
= AG" — AnR T ’ combustion Where An is the change in the number of moles of gas phase species in the reaction
AA° Muslim 2 —3203 X103 kJmol'1 +1.5 x 8.314 Jmol‘1K_I x 298.15 K C0 —3 199 ><103 kJ mol'1 (NH3,g) = —45.9><l()3 Jmol‘l Th . . . . . . . o ): _16.5X103 Jmol“
e change in PV for the liquid can be neglected because liquids are essentially incompresmble over th _ f 398’ pressure range in this problem. AG° (298.15 K) # AHZWW L 1 reaction — R T. _ 298.15 K
P6.2) Calculate KP at 525 K for the reaction N0(g) + 1/202(g) ——> N02(g) assuming that AH; is szggl 5 K j 3 ‘1 1
16.5X103Jmol" r 45.9x10 Jinol X 1 constant over the interval 298—600. K. Do you expect KP to increase or decrease as the temperature is + 8.314 Jmol" K" 750. K  298.15 K increased to 600. K? m Mumps—mama) __
K)“ 8.314 Jmoi" K" X298.15 K reaction _ =33.2x103 Jmoi" ~—91.3x103 Jmoi" =—58.1x103 Jmoi" K)=l.llx10_2 AG" 2A0; (N02 , g) — AG; (No, g) reaction _ =51.3x103 Jmoi" —87.6><103 Jmoi" =—36.3x103 Jmoi" 126 127 ...
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