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Unformatted text preview: Pal) Calculate and AG” for the reaction CalL5H) + 15.""‘Og{g) —} 6C03(g) + 3H30IU) at .mcn‘on .I'earri'on
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 AG:9PL'5£EJ!J‘DJ.‘ = AH:om5a::fax fascicmbimfari =6.S°(C‘(33.g)l 351 H30?) 5°(C‘5H5J) 15;“2.S"[03. g) = 6x 213.8 J Inol'l K'l — 3.x TOD J mol'1 t<:'1 173.4 J mol'1 K'l 15_.."2>< 205.2 J mol'1 K'l
= 219.6 J mol'1 K'l = 3263x103 kJ mor 298.15 Kx( 219.6 J inol'l K 1:) = 3203x103 kt mol'1 = T
:AH:ari_'5a';r‘mr AiPVJng.sr.;rm IAS‘:0m5u::for
= AGE...5......... ' Te9§.n.s.s.s.. Mle T
= AG:MIb_,m.m_I AMRT
where AM is the change in the number of moles of gas phase species in the reaction
aa;mmm_. = 3203x103 kJmol'l  l.5><S.314.Tmol'1K 1x298.15 K = 3199x103 kJmol'l The change in PV for the liquid can be neglected because liquids are essentially incompressible over the pressure range in this problem. P64) Consider the equilibrium NOQLQ) . * N0(g) + l."'203(g). One mole of NOgt'g) is placed in a vessel and allowed to come to equilibrium at a total pressure of 1 bar. An analysis of the contents of the vessel
gives the following results: I 600. K 900. K 3.0 3w; 0 . 224 5. 12 a. Calculate K: at 600. and 900. K. b. Calculate AG}: and for this reaction at 298.15 K. using only the data in the problem. Assume that is independent of temperature.
c. Calculate using the data tables and compare your answer with that obtained in part b. . , 1
a} 110313) 411013} I :Og(g) .. (Ex0 P3 P0 )1}: Kn = I
. Rm: PD
Rio 1
At 600. K. ' = 0.224 and P0 = —P._.O
Pm; " 3 '
‘eral' : 1250 I Joya I P0. = 1. bar 1bar=0.224P_,_..0: 1 Pm: 1 0.112 Pm:
1 bar: 1.336113%: PKG" = 0.7485 bar
[U.224x0.?485}x\/0.112x 0.?485
Kn = '  0.0649
‘ 0.?485
P... 1
At 900. K. ‘1'” = 5.12 and Pa = — Pk...)  2 '
Bram. = Pm I Pm: I 333
11301 = 5.12110 I Rm I 2 5.6 13.0 = 8 "'8 Rm
PE: 2 0.115311211‘
_ [S.12><0.11521><1f256><01152
go — ' = 2.?8
‘ 0.1152
b) Assmning that AHF’WWH is independent of temperature.
111 KP = AH:mrrfm_l / l 1 \
KP [600 K) R .. 900. K 600. K;
“K.[900. K)‘
R.th 3 ,K;[600.K], 1 1
AHrmm = 56.2x10' .T 11101'
‘ 1 1
\ 900. K 600. K,
' i / 1 1 \
In KP {298.15 K) = 1n KP[600. K} x‘ _
' ' R . 298.15 K 600.
éGfemm (298.15 K) = 112111 KP(298.151() = 8.314.111101'1K'1x298.15K><[ 14.15}
= 35.1x103 Jmol'1
em” =AH}(_\'O.g) 0H}[N03.g) =91.3x10‘ 1111011 33.2x103 Jmol'1=58.1x10‘ 1111011 14.15 P66) Consider the equilibrium C3H6(g) . ‘ C3H4t’g) — Big). At 1000. K and a constant total pressure of 2.00 bar. (5115(31) is introduced into a reaction vessel. The total pressure is held constant at 2.00 bar and
at equilibrium The composition of The liiixture in mole percent is Hg(g): 20.4%. C 3111(53): 20.4%. and
C;H5(§)i 59.2%. a. Calculate K; at 1000. K. b. If AHJQMOH = 13?.0 k.T111o1_1. calculate the value (3pr at 298.1531.
c. Calculate AGQWH for this reaction at 298.15 K.
. .1; .r a . '1 .
314100010: h” = —0"D4>_‘ 0‘04 = 0.0303
' ’ Kcﬁs 0 . a 92 1241000. K) = K1[1000. K)x = 2x 0.0703 2 0.141 . . 11H. *’ 1 1 ‘1
h) 111KF129815K1= KP11000.K1I " _ ’ ’ R . 298.131: 1000.11,. 13301031 11 ’ 1 1 \
= 1.9618TIx—11m1xl— —1
0314311101 K .290153; 1000K, KPI’zssis K): 2.00><1015
9) AG_.(298.151{)= 31011139129015 K)=1.01x10‘ Jmol" P6. 7') The pressure dependence of G is quite different for gases and condensed phases. Calculate AG?”
for the processes (C. 301ml. graphite. 1 bar. 298.15 K)—> (C. 303M. graphite. 250. bar. 298.15. K} and (He.
3:. 1 bar. 298.15 K) —} (He. g. 250. bar. 298.15 K). By what factor is AG... greater for He than for
graphite'? For a solid or liquid. we can assume that the volume is indepent of pressure over a limited range in P. .3 AG = ’de? = V03. 12.)
R . . sf. .
AGmt'CJJSOhar}=AGm(C13.1bar)  13:413. 1?): GFE(C.3.1har)  —[P. 13)
. ., p . .
1' '3 '0 g .
=0 l wxmmxm' Pa: 133 Jinol'
22:10kg111" Treating He as an ideal gas.
Pr Gm [ He. g.250. bar) = (He. g.1 bar) I [WP P 250.13. ‘ . 
=0 I RTh1—'r=111101ex8.314.1ntol'1 K'1x298.15[{x hl—ru=l3.7><10‘ Jinol' P 1 bar I This result is a factor of 103. greater Than that for graphite. P6.11 Consider the equilibrium C‘O(g} + HgO(g} . ~ C‘Ogt'g — H3(g). At 1000. K. the composition ofthe reaction mixture is
Substance C(Dﬁg} Hﬁg} (0(3) HgOIEg)
131101693 27.1 211 33.9 33.9 :1. Calculate Kp and AG? at 1000. K. reaction same” b. Given the answer to part (a). use the of the reaction species to calculate AG; at 298.15 K. Asstune that AHommM is independent of temperature.
20 Mg} + H1014} £01131 + Hag) K (II1‘0: P3 ) x P3 ) P _ . = [Pm P) x [Rep For each component. P.. = .1'_. 13mm because AH = 0. K = K}. NCO: NH: _ 0.271X 0.3711 a; — —1.40
163 1323 0.22911 0.229 = Rfln K1: 331411111011K'1x1000.K><h1[1.40)
= 2.80x103 .1 mol'1
b} =AH}[CO_..g}I 53:11:03} 11H}(H30_g} = 393.5 111111011 1 110.5 111111011 1 241.8 11.1 111011
= 41.2 111Jmol'1 \Ve use the GibbsHehnholtz equation and the relationship between AG; and 111 KP. AH” . 1' \
11116999815K]=111KD[1000K1 ; 1
' ‘ ' a .298.15K 1000K;
41.2 to" .1 1'1 1' 1 1 \ —, 7 =12.0
8.314.T111o1 11 12981511 100011,. ao_;= RThiKF(29S.15K)= 29.? 11.11111111 P6213) C‘at’HC‘Ogjﬁsj decomposes at elevated temperatures according to the stoichiometric equation Ca(HC03)_«.(3) —> CaCO3(.s) — H20(g) + C'Oﬂg).
a. prure Ca(HC03)3(.5') is put into a sealed vessel. the air is pumped out. and the vessel and its contents are heated. the total pressure is 0.23.5. bar. Determine KP under these conditions. b. Ifthe vessel initially also contains 0.105 bar HgOﬁg}. what is the partial pressure of (‘03.(g) at
equilibritun‘?
3) C'a(HC'O_.)3(.s) < } CaCO3(.s)  HEOLQ) — (Dig) Partial pressure at Equilibrium. P. — NIP .fP cf? 1 The total pressure is made up of equal partial pressures ofH;O(g) and C‘Og(g). P {'32 ‘. I; 2%_\.IJ
a}.— H‘0 ‘Q _ 53 l _ 0' “1 —0.0138
P” P” ._ P” . i, 2 , / b) If one of the products is originally present C‘a(HC‘O_:)3(.s) < >CaCO3(.s) —H;O(g) — ('03(3)
Partial pressure at
Equilibrium. E. — NIP .fP + P.. a:
P R". r'.33I.i‘:’.\"r'1'5’\" I'1” \""P\"
is..— H‘0 ‘O _ = l _ 10.105 =0.0138
P3 P3 P3 ,. P3 , \ P3 y. P3 ,
P P» .
3 = ‘f— = 0.0?62; Pm =0.0T62 bar
P P ' 196.18) Calculate the maximum nonexpansion work that can be gained from the combustion of benzener and of Hg(g) on a per gram and a per mole basis under standard conditions. Is it apparent from this calculation why fuel cells based on H; oxidation are under development for mobile applications?
C5H5{}) + 15.:"20;(g} < > 6C0ﬁg} — 31—1200?) _ 15 = so; = 33G} [HJOJM sacs; (C‘()3.g) TAGHOE. 0) 3G}(C‘6H6.i') C! _ max I 1 .1 ' 1 '_ _ 1
1t}, _ _ r, = 3x[ 2:.'.lkJ1nol )1 6x[ 3.94.4kJmol ] —><[0) 124.3 124.5kJmol
.0?IE\.ﬁﬁFIuIC_ I I 2 _
= 3202 lonol 1
= 3202 kJmol 1x 111ml = 40.96 ng '
?8.18 g toogtg) >Hgom On a per gram basis. nearl}r three times as much work can be extracted from the oxidation of hydrogen than benzene. P6127) A gas mixture with 3.00 11101 ofAr..1' moles of Ne. and '1‘ moles oer is prepared at a pressure of
1 bar and a temperature of 298 K. The total number (if moles in the mixture is four times that {if AI. in terms of .i'. At what value of 1' does the magnitude of AG have its "15.1115 1Write an expression for AGMME minimum value‘? Calculate AG for this value of .1'. m 111'! lg Iftlie number of moles ofNe is .r. the number of moles (Jer is_1' = 8 —.1'. 56mm; = HRTZ .11.. 111 .11..
r' _ _ . .\
=11RT 1h14 J In} I9 £1119 A
. 4 12 12 12 12 (JAG.. " _ I g _ I 3 . \
$=HRT illlilil— i111g 1II9 1 1 [111:0
(111' \12 12 12 .1' 12 12 12 9 .1' r' _ _ \ _
11RT 1 ln 3 I1 1 In9 A 1  MRI—1.11 3‘ O
.12 12 12 12 ,. 12 9 .1;
.' 9
1‘ =1: ‘1:—
9 1 2
f 3 ES 3‘
AGWW, = HR?" 11114 I —ln— I 3111—
""° . 4 S 8 S 8,. =12mol >< 8.3.14 Jmol'l K'lx 298.15 Kx 6.05.4 32.21403 J P628) In Example Problem 6.9. KP for the reaction CC)ng — H2013} —> C‘Oﬁg)  H;(g} was calculated to be 3.32 x 103 at 298.15 K. At what temperature does KP = 2.50 x 103? What is the highest value that KP can have by changing the temperature? Assume that AH is independent of temperature. ” \
' AH” _
h1K_U(I__r}:]HKP{298.15K) l 1 a ,‘If safmm=AH;.[col.g)IAHHHEg} minim} aa_;.[co.g)
= 3935x103 .Tmol'l — 0 +285.8x10‘ .tmol'1 +110.5x10‘ .Tntol'1 = 2.80x103 Jmol':
i =—1 —R In —KP(T";
I}. 298.155: KP[298.15K)
1 8.314JK'111101'1 7.50x10‘ _ _ a _1 x111 1 _9.342x1t1'‘K 1
298.13 K 2.80x10' .Tmol 3.32x10‘ .T, =1.D'.'><1U3 K Because the reaction is endothermic. KP increases with temperature. As I —> "13.
1 R In K‘Dlrf _} _
298.15}: AH” KP[J298.1SK) rea t'n'oai 0 K; (1} +131} 2.80x1o31nmt1 h1,—=—=_—
KP[298.15K} RxZQSlSK 3.3.14J K"ntol'1><298.15K =1.1295?
KP (If _> "20) = 3.094x Kpf298.15 K) = 3.094x 3.33103 =1.t13»<10+ P640} Under anaerobic conditions. glucose is broken down in muscle tissue to fonn lactic acid according to the reaction: C‘ﬁHDOE —> 2CH3C‘HOHC‘OOH[.5'}. Thermodynamic data at I: 298 K for glucose and lactic acid are given below. [:kJmOlll QUE: 111011) ‘S'JlJK'l 11101;]I Glucose(s) 12?3.1 219.2 209.2 Lactic Acid(s}  3.6 12?.6 193.1 Calculate AG’ at I = 298 K and I = 310. K. Assume all heat capacities are constant un this temperature
interval.
AHR [298.15 — BAH} [lactic acid] — AH; {glucose} — 2x [—6.736 qu111ol'1:]+12?5.1kJmol'1
— 44.1 id 11101—1 353(29815  2.5"" [lactic acid] — 3'" [glucose]
— 2x192.1JK'1 11101—1 —209.2 J K'1 mol'l —175 JK'I' lnol'l AGﬁlilFJSJS K‘J— AHR(295.15 K‘J—msﬁ{298.15K]——126 kJ mol‘1
AHR[310.K]—&HR{298.15 K]+AC'PAT — —74.1kJnml"+[2x12?.6 J K'1 mol'i —219.2 J K'1 11101'1:]>< {510. K—298.15 K}
——73.? Id 11101"1 AGR[310.K]— aHR{31O.K]—310.IixASRBIO. K] ‘ y 310. K
—&H?f310.K]—310.Kx asﬁtzgstKHagm— 293.15 K
. . 310.1’
——73.? k111101'1—310.Kx ms ,T K'1 11101" +3010 J K" Illol'lxln—x
298.15 K — —73.T' kl 111c1'1 —54.? kJ 111014 — —128 LU 11101'1 ...
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 Fall '08
 OHRN
 Physical chemistry, pH, total pressure

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