普通化学（新教材）习题参考答案
第一章 化学反应的基本规律 (习题P50-52） 16解（1） H2O( l ) == H2O(g)
?1?fH?m/ kJ?mol ?285.83 ?241.82
?1?1 69.91 188.83 S?m/ J?mol?k
?1 = 44.01 kJ?mol?1 ?rH?m(298k) = [?241.82?(?285.83) ] kJ?mol
?1?1 = 118.92 J?mol?1?k?1 ?r S?m(298k) = (188.83?69.91) J?mol?k
( 2 ) ∵是等温等压变化
?1 ∴ Qp = ?rH?m(298k) ? N = 44.01 kJ?mol ? 2mol = 88.02 kJ
W = ?P??V = ?nRT = ?2 ? 8.315 J?k?1?mol?1 ? 298k = ?4955.7 J
= ?4.956 kJ (或 ?4.96kJ )
∴ ?U = Qp + W = 88.02 kJ ? 4.96kJ = 83.06 kJ
17解（1） N2 （g）+ 2O2 （g） == 2 NO2 (g)
?1 0 0 33.2 ?fH?m/ kJ?mol
?1?1 191.6 205.14 240.1 S?m/ J?mol?k
?1 ? 2 = 66.4 kJ?mol?1 ∴ ?rH?m(298k) = 33.2 kJ?mol
?1?1?1?1?1?1 ?r S?m(298k) = ( 240.1 J?mol?k ) ? 2 ?(205.14 J?mol?k ) ? 2 ? 191.6 J?mol?k
= ? 121.68 J?mol?1?k?1
(2) 3 Fe(s) + 4H2O (l) == Fe3O4 (s ) + 4 H2 (g)
?1 0 ?285.83 ?1118.4 0 ?fH?m/ kJ?mol
?1?1S?m/ J?mol?k 27.3 69.91 146.4 130.68 ?1?1 ∴?rH?m(298k) = [?1118.4 ? (?285.83 ? 4 ) ] kJ?mol = 24.92 kJ?mol?1?1?r S?m(298k) = [(130.68 ? 4 + 146.4 ) ? (27.3 ? 3 + 69.91 ? 4 )] J?mol?k
= ( 669.12 ? 361.54 ) J?mol?1?k?1 = 307.58 J?mol?1?k?1
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18. 解： 2Fe2O3 (s) + 3C (s ,石墨) == 4 Fe (s) + 3 CO2 (g)
?1?fH?m(298k)/ kJ?mol ? 824.2
?1?1S?m(298k)/ J?mol?k 87.4 5.74 27.3
?1 ?742.2 ?fG?m(298k)/ kJ?mol
?? ∵ ?rG?m = ?rHm ? T ? ?r Sm
∴ 301.32 kJ?mol?1 = 467.87 kJ?mol?1 ? 298 k??r S?m
?1?1 ∴?r S?m= 558.89 J?mol?k
?1?1?1?1?1?1∴ ?r S?3 S?m= m( CO2(g) 298k) + 27.3 J?mol?k? 4 ?87.4 J?mol?k? 2 ? 5.74 J?mol?k? 3 ?1?1?1?1∴S?m( CO2(g) 298k) = 1/3 (558.89 +192.02 ? 109.2 ) J?mol?k = 213.90 J?mol?k
??fH?m(298k, C (s ,石墨))=0 ?fGm(298k, C (s ,石墨))=0
??fH?m（298k, Fe (s)）=0 ?fGm（298k, Fe (s)）=0
???rH?m=3?fHm(298k, CO2(g) ) ?2?fHm(298k, Fe2O3 (s) )
?1? 467.87 kJ?mol?1 =3?fH?m(298k, CO2(g) ) ?2 ? (? 824.2 kJ?mol) ?1?1 ∴ ?fH?m(298k, CO2(g) ) = 1/3 (467.87?1648.4) kJ?mol = ?393.51 kJ?mol
??同理 ?rG?m=3?fGm(298k, CO2(g) ) ?2?fGm(298k, Fe2O3 (s) )
?1? 301.32 kJ?mol?1 = 3?fG?m(298k, CO2(g) ) ?2 ? (?742.2 kJ?mol ) ?1 = ?394.36 kJ?mol?1 ∴ ?fG?m(298k, CO2(g) ) = 1/3 (301.32 ? 1484.4 ) kJ?mol
19.解 6CO2(g) + 6H2O（l） == C6H12O6 (s) + 6O2（g）
?1 ?fG?m(298k)? kJ?mol ?394.36 ?237.18 902.9 0
?1?1 >0 ∴ ?rG?m(298k) = [ 902.9 ? (?237.18 ? 6 ) ? (?394.36 ? 6 ) ] kJ?mol = 4692.14 kJ?mol
所以这个反应不能自发进行。
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20．解(1) 4NH3(g) + 5O2(g) == 4NO(g) + 6H2O(l)
?1?f G?m(298k) /kJ?mol ?16.4 0 86.57 ?237.18
?1?1∴?rG?m(298k) =[ (?237.18) ?6 + 86.57? 4 ? (?16.4) ?4 ] kJ?mol = ?1011.2 kJ?mol<0
∴ 此反应能自发进行。
(2) 2SO3(g) == 2SO2(g) + O2(g)
?1 ?f G?m(298k) / kJ?mol ?371.1 ?300.19 0
?1?1 ∴?rG?m(298k) = [(?300.19) ?2 ? (?371.1) ? 2] kJ?mol= 141.82 kJ?mol > 0
∴ 此反应不能自发进行。
21．解 （1） MgCO3(s) == MgO(s) + CO2(g)
?1 ?1111.88 ?601.6 ?393.51 ?fH?m(298k)/ kJ?mol
?1?1 S?m(298k)/ J?mol?k 65.6 27.0 213.8
?1 ?f G?m(298k) / kJ?mol ?1028.28 ?569.3 ?394.36
?1 ∴ ?rH?m(298k) = [ ?601.6 + (?393.51) ? (?1111.88)] = 116.77 kJ?mol
?1?1 ?rS?m(298k) = [ 213.8+ 27.0 ? 65.6] = 175.2 J?mol?k
?1 ?rG?m(298K) = [ (?394.36) +(?569.3)?(?1028.28)] = 64.62 kJ?mol
???1?1?1 (2) ?rG?m(1123K) = ?rHm(298k)?T? ?rSm(298k) = 116.77 kJ?mol ? 1123k ?175.2 J?mol?k
= 116.77 kJ?mol?1 ?196.75 kJ?mol?1 = ?79.98 kJ?mol?1
又 ∵ RTlnK?(1123k)= ? ?rG?m(1123k)
∴ 8.315 J?mol?1?k?1?1123 k?ln K?(1123k) = ?(?79.98) kJ?mol?1
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∴ K?(1123k) = 5.25 ? 103
?? (3) ∵ 刚刚分解时 ?rG?m(T) =?rHm(298k)?T? ?rSm(298k) =0
116.77kJ?mol?1 ∴ 分解温度T可求: T???666.5k ??1?1?rSm(298k)175.2J?mol?k?rHm(298k)? ∴ 分解最低温度为666.5 k
22.解法一: K? (298k)=5.0 ? 1016
16??1?1?1 ∴?rG?m(298k) = ?RTlnK(298k)= ?8.315 J?mol?k?298k?ln(5.0 ? 10) = ?95.26 kJ?mol
?? ∵?rG?m(298k) = ?rHm(298k)?298k? ?rSm(298k)
∴?95.26 kJ?mol?1 = ?92.31 kJ?mol?1?298k??rS?m(298k)
?1?1 ∴?rS?m(298k) =9.90 J?mol?k
?? ∴?rG?m(500k) = ?rHm(298k)?500k? ?rSm(298k)
= ?92.31 kJ?mol?1?500k?9.90 J?mol?1?k?1= ?97.26 kJ?mol?1
??1?1? 而 ?rG?m(500k) = ?RTlnK(500k)= ?8.315 J?mol?k? 500k?ln K(500k)
??rGm(500k)97.26?103J?mol?1 = = 23.40 ?1?1RT8.315J?mol?k?(500k) ∴ ln K?(500k)= ?
∴ K?(500k) = 1.45 ? 1010 解法二：
???rHm(298k)1K?(500k)1?( ∵ ln? = ?)
R500298K(298k)?(?92.31?103J?mol?1)?202?()k= ?15.05 =
8.315J?mol?1?k?1500?298K?(500k)∴ ? = 2.9 ? 10-7
K(298k) ∴ K?(500k) =2.9 ? 10-7 ? K?(298k) = 2.9 ? 10-7 ? ( 5.0 ? 1016 ) = 1.45 ? 1010
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23.解： N2(g) + 3H2(g) == 2NH3(g)
?1 ?fH?m(298k)/ kJ?mol 0 0 ?45.9
?1?1 S?m(298k)/ J?mol?k 191.6 130.68 192.8
?1?1∴ ?rH?m(298k) = 2?(?45.9) kJ?mol = ?91.8 kJ?mol
?1?1?1?1S?m(298k) = (2?192.8 ?191.6 ?3?130.68 ) J?mol?k= ?198.04 J?mol?k
?? ?rG?m(T) = ?rHm(298k) ?T? ?rSm(298k) =0
= ?91.8 kJ?mol?1 ?T? (?198.04 J?mol?1?k?1 ) =0
?91.8?103J?mol?1 ∴ T = = 463.5 k
?198.04J?mol?1?k?1 ∴ T>463.5 k 时 反应能自发进行。
24.解：（1）?（2）得： CO2(g) +H2(g)? CO(g) + H2O(g)
根据化学平衡得多重规则，
此反应的K? (T)为： K? (T) = K1? (T) ? K2?(T) ∴ 该反应各温度下的平衡常数为: T/(k) K? (T)
973 0.618 1073 0.905 1173 1.29 1273 1.66 ???根据 ?rG?m(T) = ?RT ln K (T) = ?rHm(T) ?T? ?rSm(T)
???rHm?rSm?) lnK(T)??(RTR??m?m
???rHmK1(T)11假设 ?rH(T), ?rS(T)不变， 则ln???(?)
RT1T2K2(T)∵K1? (T)〈 K2?(T) T1〈 T2 即 T ↑ 则 K? (T) ↑ ，?rHm? > 0 ∴该正反应为吸热反应。
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同济大学普通化学 第一章、二章习题答案(详细)
