无机化学课后习题答案(天津大学第四版)
第2章 化学反应的方向、速率和限度 习题参考答案
1.解:
??rHm = ?3347.6 kJ·mol?1;
r?m??rSm =
?216.64 J·mol?1·K?1;?G = ?3283.0 kJ·mol?1 < 0
该反应在298.15K及标准态下可自发向右
进行。
2.解:
??rGm = 113.4 kJ·mol?1 > 0
该反应在常温(298.15 K)、标准态下不能自发进行。
(2)?H = 146.0 kJ·mol?1;?S = 110.45
r?mr?mJ·mol?1·K?1;?G = 68.7 kJ·mol?1 > 0
r?m 该反应在700 K、标准态下不能自发进行。 3.解:?H = ?70.81 kJ·mol?1 ;?S = ?43.2
r?mr?mJ·mol?1·K?1;?G = ?43.9 kJ·mol?1
r?m (2)由以上计算可知:
?H(298.15 K) = ?70.81 kJ·mol?1;?S(298.15 K) = ?43.2 J·mol?1·K?1
?S ≤ 0 ?G = ?H ? T ·
r?mr?mr?mr?mr?m
T ≥?Sr??rHm(298.15 K)?m(298.15 K) = 1639 K
34.解:(1)K =
cc (CO) ? c (H2) ?c (CH4) c (H2O) K =
pp (CO) ? p (H2) ?p (CH4) p (H2O)3
?
K
?p (CO) / p?? p (H) / p ? = ? p (CH)/p?? p (HO) / p???32??42 (2)
?p (N2)?? p (H2) ?p (NH3) ?12 32Kc =
?c (N2)?? c (H2) ?c (NH3) 12 32
Kp =
2)/1?2p K = ?p (Nc
2?? p (HKpp (NH3) /p2)/?p ??
232 (3)K =c (CO)
K?
c =p (CO)
? p (H2O) ? ? p (H2) ?3 =p (CO2)/p?3
p (4)K = K? ? c (H2O) ? ? c (H2) ?3 K =
? ?333
= ? ? p (H?mr p (H2O)/p? ?2)/p5.解:设?H、?S基本上不随温度变化。
r?m??rGm??rGm??rGm =
??rHm?S ? T ·
r?m(298.15 K) = ?233.60 kJ·mol?1 (298.15 K) = ?243.03 kJ·mol?1 (298.15 K) = 40.92, 故
K?
lgK?(298.15 K)
= 8.3?1040
lgK?(373.15 K) = 34.02,故
K?
(373.15 K) =
1.0?1034 6.解:(1) <0
该反应在298.15 K、标准态下能自发进行。
(2) K) = 5.8?105 7. 解:(1) kJ·mol?1
lg?K1??rGm=2?G(NH3, g) = ?32.90 kJ·mol?1
f?mlgK?(298.15 K) = 5.76,
K?
(298.15
??rGm(l) = 2?G(NO, g) = 173.1
f?m=
???fGm(1)2.303 RT = ?30.32, 故
?K1=
4.8?10?31
(2)?G(2) = 2?G(N2O, g) =208.4
r?mf?mkJ·mol?1
lg?K2=
???fGm(2)2.303 RT = ?36.50, 故
?K2=
3.2?10?37
(3)?G(3) = 2?G(NH3, g) = ?32.90
r?mf?mkJ·mol?1
?lgK3= 5.76, 故
?K3= 5.8?105
由以上计算看出:选择合成氨固氮反应最好。 8.解:?G =
r?m??fGm??fGm(CO2, g) ?
??fGm(CO, g)?
(NO, g)
= ?343.94 kJ·mol?1< 0,所以该反应从理论上讲是可行的。 9.解:
??rHm(298.15 K) =
??fHm(NO, g) = 90.25
kJ·mol?1
?S(298.15 K) = 12.39 J·mol?1·K?1 ?G(1573.15K)≈?H(298.15
r?mr?mr?mK)
?1573.15?S(298.15 K)
r?m = 70759 J ·mol?1
lgK(1573.15 K) = ?2.349, K(1573.15 K) =
??4.48?10?3
10. 解: H2(g) + I2(g) 2HI(g) 平衡分
(2x)2(2905.74?x)2压/kPa
2905.74 ?χ 2905.74 ?χ 2χ
= 55.3
χ= 2290.12
无机化学课后习题答案(天津大学第四版)
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