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Unformatted text preview: CMpTéf A0 W 20.4 If we refer to the equation in the answer to Review Question 20.3, we can see that if Eﬁnax is larger ﬁxing/'2’.“ en AB mu“ be POSIUVE, by deﬁnition. Thus, AE for an endothermic process is 20.12 A spontaneous change, in thermodynamic terms, is one for which the sign of AG is negative. It is
a process that occurs by itself, without continued outside asmstance. 20.19 Entropy is a measure of the randomn '
ess of a system. An e uivalent statement ' '
measure of the statistical probability of a system. q Is that 'emmpy IS a 20.20 (a) negative (b) negative (c) positive
_ ((1) negative (e) negative (f) positive 20.22 The entropy of the universe increases when a spontaneous event occurs. . 20.23 Two examples might be (1) a disorganized pile of bricks and boards jumping up to produce a
house, or (2) a pile of bricks falling off a cliff, and landing in the form of a perfect cube. Both of
these examples are accompanied by enormous decreases in entropy (randomness), and hence are not realistic spontaneous events. 20.24 A spontaneous event occurs when AG is negative. Since AG = AH — TAS, even if the entropy is
negative for a process, the enthalpy factor, AH may be’negative enough at a given temperature to
allow the overall change in free energy to be negative. ‘ ’ 20.25 This is the statement that the entropy of a perfect crystalline solid at 0 K is equal to zero: S = 0 at 0 K. 20.27 Entropy increases with increasing temperaturebecause vibrations and movements a solid 1 d gr 3 I‘ I‘ CI e we . IHUdIICCS ("C U
ea “if .e U at GI t p a S. M S 20.30 (a) ‘ A change is spontaneous at all temperatures only if AH is negative and AS is positive.
(b) A change is spontaneous at low temperatures but not at high temperatures only if AH is
negative and AS is negative. ' _
(c) A change is spontaneous at high temperatures but not at low temperatures only if AH is
~ positive and AS is positive. 20.31 A change is nonspontaneous regardless of the temperature, if AH is positive and AS is negative. be obtained from any process. 20.36 As with other processes that are not carried out in a reversible fashion, the energy is lost to the
environment and becomes unavailable for use. In addition, much of the energy is lost as heat which is used to maintain our body temperature. 20.38 At equilibrium, the value of AG is zero. 20.42 See Figure 20.12. word, the rate of reaction may be too slow at normal temperatures to
be observed. 20.46 As theternperature is raised, °G' will become less negative, if AH" is negative and AS" is negative.
Accordingly, less product will be present at equilibrium. 20.47 AG = AG° + RT an 20.54 £=q+w=3001+700l=+10001 The overall process is endothermic, meaning that the internal energy of the system increases.
Notice that both terms, q and W, contribute to the increase in internal energy of the system; the
system gains heat (+q) and has work done on it (+w). 20.60 In general, we have the equation: AH° = (sum AH; [productsD — (sum A
(a) AH° = {AHi [CaC03(S)]} — {AHi [C02(g)] + AH; [CaO(S)l}
AH° = {1 mol x (——1207 kJ/mol)} — {1 mol X (—394 kJ/mol) + 1 mol X (—635.5 kJ/mol)}
L AH° = —178 kJ favored.  H; [reactantsD‘ (b) °={APEICZH6(g)1}~{Amic2H2<g)]+2AmiIiz(g)1}
AH" = {1 mol X (—84.5 kJ/mol)}  {1 mol X (227 kJ/mol) + 2 mo] X (0.0 kJ/mol)}
° = ~311 kJ _ favored. (C) ° = {AHi [1362030)] + 34H? [Ca(S)]} _
 {2 AH? [136(3)] + 3 AH? {(330601}
AH° = {1 mol X (—822.2 kJ/mol) + 3 mol X (0.0 kJ/mol)} — {2 mol X (0.0 kJ/mol) + 3 mol x (—635.5 kJ/mol)}
AH° = +1084.3 kl not favorable from the standpoint of enthalpy alone. — ’ >1}
AH° = AH° [H 0(1)] + AH? [Ca0(S)]} {AH‘E [C§(0H)2(S
(d) ° = :1 mbl x2(—285.9 kJ/mol) + 1 mol X (—635.5 kJ/rnol)} — 1)}
— {1 mol X( 986.59 kJ/mo ' V
AH° = +652 kJ not favored from the standpomt of enthalpy alone. (6) 43° r {2 AH? [HC1(g)l + AH} [N62304(S)l}
— {2 AH? [NaCKSH + AH? {11230401} — .5 kJ/mol)} ° = 1X —92.30 kJ/mol) + 1 mol X ( 1384 AH {2 m0 ( {2 mol X (~411.0 kJ/mol) + 1 mol x (—811.32 kit/11101)}
AH" % +642 kJ not favored from the standpoint of enthalpy alone. 20.6 ‘ ' I
“I 4 (a) negative — Since the number of moles of gaseous material decreases. Eb; negative ~ since the number of moles of gaseous material decreases
(: negatlve — smce the number of moles of gas decreases. ‘
) posmve — Since a gas appears where there forlnerly was none. 20.66 AS" = (sum S°[products]) — (sum S°[reactants]) (21) AS" = {28°[NHs(g)]}  {3S°[Hz(g)] + S°[N2(g)]}
AS° = {2 moi x (192.5 J mol”1'K'1)}  {3 moi x (130.6 J moi"1 K“)
+ 1 moi x (191.5 J moi'1 K'1)}
AS° = —198.3» J/K not spontaneous from thestandpoint of entropy. (b) A ° = {S°[CH30H(I)]} ~ {28°[Hz(g)] + S°[CO(g)]}
AS° = {1 mol x (126.8 J mol" K“)}
— {2 moi x (130.6 J moi‘l K‘1)+ 1 mol x (197.9 J moi'1 K")} A ° = ~332.3 J/K not favored from the standpoint of entropy alone. (0) AS° = {6S°[HzO(g)] + 4S°[Coz(g)]}  {7 S°[Oz(g)] + 2S°lC2H6(g)]} 
AS° = {6 moi x (188.7 J moi‘1 K‘1)+ 4 moi x (213.6] moi1 K“)}
— {7 mol x (205.0 J mor‘ K‘1)+ 2 mol x (229.5 J moi" 1(1)}
AS° = +926 J/K favorable ﬁom the standpoint of entropy alone. (d) AS° =‘{ZS°[HzO(l)] + S°[CaSO4(s)]}
— {S°[H2804(l)] + S°[Ca(OH)2(s)]}
AS° = {2 mol x (69.96 J mol‘1 1(4) + 1 mol x (107 J moi1 1(4)}
— {1 mol x (157 J mol'lK'1)+ 1 moi x (76.1 J moi"1K“)} ‘ AS° = +14 ‘J/K favorable from the standpoint of entropy alone. (6) AS" = {23°[N2(g)] + S°[Soz(g)]} — {28°[N20(g)] + S°[S(s)l}
AS° = {2 mol x (191.5, J mol"‘ K") + 1 mol x (248 J moi"l 1(4)} ,
— {2 mol x (220.0 J mor‘ K"‘) + 1 mol x (31.9 J mor‘ K"1)} AS° = +159 J/K favorable from the standpoint of entropy alone. i I S . l 1 .
.  ' their 5 .
111016 Of 3‘ subsmce ls formed from elements m f a single pure substance, the umts understood to correspond to the reaction forming one mole 0 may be written either I K“1 or J mol"1 K'l. 20 +2H(g)—9CH4(g)
(a) ASE”: {3°lézH4(g)l‘:  {23°[C(S)l + 28°[H2(g)l} AS° = {1 moi x (219.8 J moi‘1 K“)}
' — {2 moi x (5.69 J moi‘1 K’ AS" = ~52.8 J/K or —52.8 J moi‘1 K~1 1) + 2 mol x (130.6 J mor1 K“)} N )+ 1/20 (g) —9 N20(g)
(b) A§° = {3°lN:0(g)]}  {3°lNz(g)l + 1/25°l02(g)l} 133° = {1 mol >'< (220.0 J mol“ K71»
— {1 moi x (191.5 J moi1 K“) + 1/2 mol x (205.0 J moi~1 K“)} A ° = —74.0 J/K or —74.0 J moi”1 K‘1 (c) Na(s) + 1/2C12(g) ——> NaC1(s)
AS° = {3°maCi(s)]} —{1/ZS°[Clz(g)] + S°[Na(s)]} .
AS° = {1 mol x (72.38 J moi‘l K4» — {1/2 moi x 223.0 J moi"1 K‘1)+ 1 moi x (51.0 J
' mol"’ K‘l)}
A ° = —90.1 J/K or «90.1 J mor1 K“ (d) Ca(s) + 3(9) + 302(g) + 2H2(g) —> (321304211200) 
AS° = {S°[CaSO42H20(s)1}~ {28°[H2(g)] + 3S°IOz(g)1+ S°[S(s)1+ S°[Ca(s)]}
AS" = {1 11101 x (194.0 J mol’1 1(4)} — {2 11101 x (130.6 J mar1 K‘l)
+ 3 mol x (205.0 J mar1 K‘1)+ 1 mol x (31.8 J mor1 K’1)+ 1 mol x (154.8 J
mol’1 14—1)} .
AS° = —868.9 J/K or —868.9 J mol’1 K“1 (e) 2H2(g) + 2C(S) + 02(g) ‘> HC2H302(Z)
A ° = {S°[HC2H302(I)1}~{ZS°[Hz(g)] + 28°[C(s)1 + S°[Oz(g)]}
AS° = {1 11101 x (160 J 11101—1 K"‘)} ~ {2 mo1x (130.6 J mor‘ K“)
+ 2 m0] x (5.69 J marl K“’) + 1 mol x (205.0 J mol‘1 1(4)}
A ° =—318 J/K 0r—318 Jmorl K'l quation in which one mole of pure phosgene is produced from 20.72 The quantity AG; applies to the e
the naturally occurriﬂg forms of the elements: > o  ‘7
C(5) + 1/202(g) + C12(g) —> C0C12(g), AGf   We can determine AG} if we can values for AH; and AS; , because: AG° = AH° —— TAS° The value of AS; is determined using S° for phosgene in the followmg way: AS? = {3°[C0C12(g)]} — {S°[C(S)1+ 1/23°[02(g)] + S°[C12(g)]} —1 n1
AS;={1m01x(284]m01_1K")}—{1molX(5.69Jmol K 31 1
+1/2m61x(205.0Jmo1—11<—1)+1m61x(223.01m61 K )} ' As; = —47 J mor‘ K“ or —47 UK = AH; — TAS; = 223 kJ/mol — (298 K)(—0.047 kJ/mol K) = .209 kJ/mol
20.74 AG° = (sum AG} [productsD — (sum AG; [reactantsD AG} (a) AG° = {M}? 111230401}  MGM—120(1)]+ AG; 13030301}
AG° = {1 11101 x (~689.9 kJ/mol)} —
{1 mol x (.2372 kJ/mol) + 1 mol x (—370 kJ/mol)} AG" = 82.3 kJ '03) AG" = {2M}? [NH3(g)] + AG; [1120(1)] + AG? [CaC12(S)]}
V  {AG} [(3800)] + 2M3} [NEC1(S)]}
AG° =_ {2 11101 X (—16.7 kJ/mol) + 1 11101 x (~237.2 kJ/mol)
+ 1 11101 x («750.2 kJ/mol)} — {1 mo1>< (.6042 kJ/mol)
+ 2 mol x (.2039 kJ/mol)}
AG° = ~8.8 kJ (Q ' AG“ {AGE 9123040)] + AG; 1021012011} — {Ac}; [CaSO4(s)] + AG; [HCKgm
AG° = {1 m01X(—689.9 kJ/mol) + 1 m01X(——750.2 kJ/m01)}
4 {1 11101 x (—1320.3 kJ/mol) + 2 11101 x (—95.27 kJ/mol)}
AG° = +707 1d (0) AG9 = {AGE [C2H50H(l)]} — {AG} [IIZO(g)] + AG; [szgm
AG° = {1 11101 x (—174.8 kJ/mol)} ~ {1 mol X («228.6 kJ/mol) + 1 mol x (68.12 kJ/mol)} _ AG° = —14.3 k] v (e) AG° = {2M}; [1120(1)] + AG? [502(3)] + AG; [Ca304(s)1}
 {2M}; [1123040)] + AG? [Ca(S)]}
AG° = {2 mol x (——237.2 kJ/mol) + 1 12110] X (—300 kJ/mol)
+ 1 11101 X (~1320.3 kJ/mol)} —— {2 11101 X (489.9 kJ/mol)
+ 1 mol X (0.0 kJ/mol)}
AG° = ~715 k1 COC12(g) + H20(g) —> COz(g) + 2HC1(g)
AG° = (sum AG} [productsD — (sum AG} [reactantsD
AG° = {2133; [HCKgH + AG} [C02(g)]} “ MG; [H2003] + AG; [COC12(g)]} AG° == {2 mol X (—95.27 kJ/mol) + 1 mol X (—3944 kJ/mol)} ’
— {1 mol X (—2286 kJ/mol) + 1 11101 X (—210 Isl/11101)} AG° = —l46 k1 20.77 20.79 Add the reverse of theﬁrst equation to the second equation plus twice the third equation: CO(NH2)2(S) + 2NH4C1(S) —> COC12(g) + 4NH3(g), AG" = +3320 kJ COC12(g) + H200) ——> C02(g) + 2HCl(g), AG" = —l41.8 kJ 2NH3(g) + 2HCl(g) —> 2NH4C1(S), AG° = —183.9 k]
AG° = +6.3 kJ CO(NH2)2(S) + H200) —> 2NH3(g) + (302%), 20.80 The work obtainable ﬁom a reaction is equal in magnitude to the value of AG for the
reaction. Thus, we need only determine AG" for the process: AG" = (sum AG; [products]) —' (sum AG; [reactants])
ACT" = {3 AG; [H2009] + 213G; {(302091} " {3 AG; [02%)] + AG; [QHSOHWB
AG° = {3 mol x (—2286 kJ/mol) + 2 mol x (—3944 kJ/mol)}
— {3 11101 x (0.0 kJ/moi) 1 11101 x (—174.23 kJ/mol)}
AG° =—1299.8 kJ v ~ ' Teq = Ali/AS, and assuming that ' '
’1 ‘ — 333 K Teq = (31.4 x 103 J moi")+ (942 J mol K“ ) — 20.85 We proceed as in the answer to Review Problem 20.84: AS = All/Tm: (31.9 X 103 I/mol) + (56.2 + 273.15 K) .
AS = 96.9 i moi“1 K‘1
2086 The reaction is spontaneous if its V ' alue for AG° is negative. ‘
AG° = (suni AG; [productsD — (sum AG? [reactantsD 4’ ,
AG° = {AGE [Horizon] + AG? [moan + AG} [Now] + (AG; [N02(g)]}
 {AGE [02144491+ 2M3} [HNOnDB ‘ ‘
—237.2 kJ/niol) °= 1 1x .3925 kJ/mol)+lmolx( 1
AG { m0 ( 1molX(Sl.84 kJ/mol)} + 1 mol X (86.69 kJ/mol) +
— {1 11101 X (68.12 kJ/mol) + 2 11101 X (—79.9 kJ/mol)} AG" = —399.5 kl Yes, the reaction is spontaneous. AH° = {3 AH} [HzO(g)] + 5 AH? [502(g)1 + 24H? [N0(g)i}  {5 AH? [50369] + 24H? M3691}
° = {3 mol X (—242 kJ/mol) + 5 mol X (—297 kJ/mol)
+ 2 11101 X (90.4 kJ/mol)} — {5 mol X (—396 kJ/mol)
+ 2 11101 X (—46.0 kJ/mol)} 20.89 AH° = +42 k1 43° = {390420091 + 53°[302(g)] + 23°[N0(g)l} — {58°[SOs(g)] + ZS°[NH3(g)]} l
AS° = {3 11101 x (188.7 J mot1 K") + 5 11101 x (248.5 J mol" K” ) + 2 mol x (210.6 J mol“ K")} — {5 mol x (256.2 J mgr1 K") _
+ 2 mol x (192.5 J mot‘ K")} AS° = +564 J/K = 0.564 kJ/K AG" =AH° — TAS°
AG°373 = 42 kl —— (373 K)(0.562 kJ/K) == —l68 kJ AG" = {2 X AG; [POC13(g)]} —— {2 X AG} [PC13(g)] 2 X AG; [02(g)]} 20.90 (a)
AG° = {2 mol X (~1019 kJ/mol)}
— {2 mol x (—2678 kJ/mol) + 1 11101 x (0 kJ/rnol)}
A68 = 1502 kl = —1.502 ><106 J
—1.502 x 106 J = Rr1nKp = 48.3 14 J/K mol)(298 K) x 111K?
mKp = 606 It)ng = 263, and K1) .= 10263.
(b) AG" = {2 >< AG} [802(8)] + 1>< AG? [0209]} ~ {2 X AG¥[303(g)]}
(370 kJ/rnol)} AG° = {2 m1 x (—300 kJ/mol) + 1 11101 x (0 kJ/mol)} — {2 mol >< AG" =’ 140 kJ = 1.40 x105 I
1.40 x 105 = —RT1nK., = {8.314 J/K mol)(298 K) >4pr 1nKP = —56.5 and Kp = 2.90 x 10‘25
20.93 79.8 x 103 J = —(8.3 14 J/K mol)(673 K) x In Kp
1n Kp = —14.3 Kp = 6.40 x 107 we start hy calculating the reaction quotient, Q. Be sure to determine the pressure of the gases
usmg the ideal gas law. ’ (3.8 x 103 11101100821" iﬁﬁn K) 2.50 L
Q = _
(0.040 molﬁ0821 2:3(X673 K) (0.022 mol)@.0821 ﬁﬁw K)
2.50 L 2.50 L = 4.32 Since the value on is larger than the value of K9, the system must shift to the left in order to
reach equilibrium. ' start with pure products, the value of Q will be inﬁnite (there are zero ‘ft towards the reactants, i.e., the pure products ' 20.96 lfAG" = 0, Kc =1. Ifwe
the equilibrium will sh1 reactants) and, since Q > K,
will decompose to their elements. The sum of all the bond energies in the molecule should be equal to the atomization energy:
AH°mm = 1(CC bond) + 5(CH bonds) + 1(OH bond) + 1(CO bond) We use values ﬁorn Table 20.4:
3225.2 U = 1(348 kl) + 5(412 kl) + 1(463 k1) + 1(C~O bond) C—O bond energy = 354 kJ/mol 20.101 The energy released during the formation of 1 mol of acetone is equal to the sum of all of the bond
energies in the molecule:
for the six C—H bonds: 412 kJ/mol X 6 mol C—H bonds
for the two CeC bonds: 348 kJ/mol x 2 mol C—C bonds
for the one C=O bond: 743 kJ/mol X 1 mol C=O bonds Adding the above contributions we get 3.91 X 103k] released per mole of acetone formed. 20.102 The heat of formation for ethanol vapor describes the following change:
2C(s) + 3H2(g) +1/202(g) > C2H50H(g) This can be arrived at by adding the following thermochemical equations, using data from Table 20.3:
3H2(g) 9 6H(g) AH1° = (6)217.89 kl : 1307.34 kl
2C(s) —> 2C(g) AH2° = (2)716.67 k1 = 1,433.34 k]
1/202(g) 9 O(g) AH3° = (1)249.17 k] = 249.17 1d
6H(g) + 2C(g_) + 0(g) —> C_2H_50H(g) ‘ ___° = x 3H2(g) + 2C(s) + 1/202(g) ) C2H50H(g) AHfo = (2989.85 + x) 121
Since Ali? is given as —235.3 k]. .. ~235.3 = 2989.85 + x x = —3225.2 kJ ARK“om is the reverse reaction, so the sign will change: AH°amm = 3225.2 k1 ...
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