WORD格式
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）
1 fH m / kJ mol
1 1 k S m / J mol
H2O( l )
285.83
== H2O(g) 241.82
69.91 188.83
1
1 = 44.01 kJ mol rH m (298k) = [ 241.82 ( 285.83) ] kJ mol
1 1 1 1
k = 118.92 J mol k r Sm (298k) = (188.83 69.91) J mol
( 2 )
∵是等温等压变化 ∴Qp =
H m (298k)
1 N = 44.01 kJ mol 2
2mol = 88.02 kJ
r
W = P V = nRT =
1 1
mol 298k = 4955.7 J 8.315 Jk
= 4.956 kJ (或 4.96kJ )
∴ U
17 解（1）
f
= Qp + W = 88.02 kJ 4.96kJ = 83.06 kJ
2O2 （g） == 2 NO2 (g)
0
N2 （g）+
1 H m / kJ mol
33.2
0
1 1
k S m / J mol
191.6 205.14
1
240.1
1 ∴ rH m (298k) = 33.2 kJ mol
2 = 66.4 kJ mol
(2)
1 1 1 1 1 1
k ) 2 (205.14 J mol k ) 2 191.6 Jmol k mol r S m (298k) = ( 240.1 J
=
3 Fe(s) + 4H 1 fH m / kJ mol
1 1
k S m / J mol
1 1
k 121.68 Jmol
2O (l) == Fe3O4 (s ) + 4 H2 (g)
0 27.3
285.83 69.91
1118.4 146.4
0 130.68
∴ rH ( 285.83 4 ) ] kJ m (298k) = [ 1118.4
mol
1 1 = 24.92 kJ mol
1 1
k r Sm (298k) = [(130.68 4 + 146.4 ) (27.3 3 + 69.91 4 )] J
mol
1 1 1 1 k = 307.58 J mol k = ( 669.12 361.54 ) J mol
1
专业资料
WORD格式
18. 解：
1 fH m (298k)/ kJ mol
1 1
k S m (298k)/ J mol
2Fe2O3 (s) + 3C (s ,石墨) == 4 Fe (s) + 3 CO2 (g) 824.2
87.4 742.2
5.74 27.3
1 fG m (298k)/ kJ mol
∵ rGm = rH m
T ? r S m
1 1
= 467.87 kJ mol 298 k? r S ∴301.32 kJ mol m
1 1
k ∴ r S m = 558.89 J mol
1 1 1 1 1 1 k 4 87.4 Jmol k 2 5.74 J mol k 3 ∴ r Sm = 3 S m ( CO2(g) 298k) + 27.3 J mol
1 1 1 1
k = 213.90 Jmol k ∴S m ( CO2(g) 298k) = 1/3 (558.89 +192.02 109.2 ) J mol
f
H m (298k, C (s ,石墨))=0 H m （298k, Fe (s)）=0
f
Gm (298k, C (s ,石墨))=0 Gm （298k, Fe (s)）=0
ff
r
H m =3 fH m (298k, CO2(g) ) 2 fH m (298k, Fe2O3 (s) )
1
1
467.87 kJ mol =3 fH
∴
)
1
( 824.2 kJ mol=3 m (298k, CO2(g) ) 2
1
H ) f
1 1
= 393.51 kJ mol mol fH m (298k, CO2(g) ) = 1/3 (467.87 1648.4) kJ
同理
r
G m =3 fGm (298k, CO2(g) ) 2 fGm (298k, Fe2O3 (s) )
1
1
301.32 kJ mol = 3 fG
∴
)
1
(298k, CO2(g) ) 2 ( 742.2 kJ mol= m
1
3 fG )
1 1
= 394.36 kJ mol fG m (298k, CO2(g) ) = 1/3 (301.32 1484.4 ) kJ mol
19.解
1 fG m (298k) kJ mol
6CO2(g) + 6H2O（l） == C6H12O6 (s) + 6O2（g） 394.36
237.18
902.9
0
∴ rG m (298k) = [ 902.9 mol
1 1 = 4692.14 kJ mol >0 ( 237.18 6 ) ( 394.36 6 ) ] kJ
所以这个反应不能自发进行。
2
专业资料
WORD格式
20．解(1)
1 f G m (298k) /kJ mol
4NH3(g) + 5O2(g) == 4NO(g) + 6H2O(l )
16.4
0
86.57
237.18
1 1
= 1011.2 kJ mol <0
∴ rG ( 16.4) 4 ] kJ m (298k) =[ ( 237.18) 6 + 86.57 4
mol
∴ 此反应能自发进行。 (2)
1 f Gm (298k) / kJ mol
2SO3(g) == 2SO2(g) + O2(g) 371.1
300.19
0
∴ rGm (298k) = [( 300.19) 2
11
= 141.82 kJ mol > 0 ( 371.1) 2] kJ mol
∴ 此反应不能自发进行。
21．解 （1）
1 fH m (298k)/ kJ mol
1 1
k Sm (298k)/ J mol
MgCO3(s) == MgO (s) + CO2(g)
1111.88
601.6
393.51
65.6 27.0 213.8
f
1 G m (298k) / kJ mol
1028.28 569.3 394.36
1
∴ rH m (298k) = [ 601.6 + ( 393.51) ( 1111.88)] = 116.77 kJmol
1 1
k 65.6] = 175.2 J mol rS m (298k) = [ 213.8+ 27.0
) = [ ( 394.36) +( 569.3) ( 1028.28)] = 64.62 kJ rG m (298K mol ) = rG m (1123K
H m (298k) T
Sm (298k) = 116.77 kJ mol
1
1
(2)
r r
1123k 175.2 Jmol k
1 1
1 1
1 196.75 kJ mol = 79.98 kJ mol = 116.77 kJ mol
又 ∵ RT lnK (1123k)=
r
G m (1123k)
1 1 1
k 1123 k ln K (1123k) = ( 79.98) kJ mol ∴8.315 J mol
专业资料
WORD格式
3
专业资料
WORD格式
∴ K (1123k) = 5.25
3 10
(3) ∵ 刚刚分解时
G m (T) = rH m (298k) T
1 H ( 298k) r m
∴ 分解温度 T 可求: T
r
r
Sm (298k) =0 116.77kJ mol
666.5
1 k
1
k
S ( 298k ) r m
175. 2 J mol
∴ 分解最低温度为 666.5 k
16
20.解法一:
K (298k)=5.0 10
10) = 95.26 kJ mol
16
1
1 1
∴ rG k 298k ln(5.0 8.315 J mol m (298k) = RTlnK (298k)=
∵ rG m (298k) =
r
H m (298k) 298k
r
Sm (298k)
1 1 ∴ 95.26 kJ mol = 92.31 kJ mol 298k rS m (298k)
1 1 ∴ rS k m (298k) =9.90 J mol
∴ rG m (500k) = 而
r
H m (298k) 500k
r
Sm (298k)
1 1 11
500k 9.90 Jmol k = 97.26 kJ mol = 92.31 kJ mol
1 1
k 500k ln K (500k) 8.315 J mol rG m (500k) = RTlnK (500k)=
∴ ln K (500k)=
k ) r Gm (500
=
RT
3 1
467.88 J mol = 23.40
k 1
301.33J
10
(500k ) 1
mol
10
∴ K (500k) = 1.45 10 解法二：
1 H m ( 298k ) 1
∵ ln ) r
500 298 K (298k) = (
R
3 1 202
( 92.31 10 J mol ) )k = 15.05
500 298 = (
1 1
8.315 J mol k
专业资料
K (500k )
同济大学普通化学第一章、二章习题答案(详细)
