天津大学无机化学第五版习题答案解析
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2.解: = 113.4 kJ·mol1> 0
该反应在常温(298.15 K)、标准态下不能自发进行。
(2) = 146.0 kJ·mol1; = 110.45 J·mol1·K1; = 68.7 kJ·mol1> 0
该反应在700 K、标准态下不能自发进行。
3.解: =70.81 kJ·mol1; =43.2 J·mol1·K1; =43.9 kJ·mol1
=315.5 kJ·mol1
由上看出:(1)与(2)计算结果基本相等。所以可得出如下结论:反应的热效应只与反应的始、终态有关,而与反应的途径无关。
14.解: (3)= (2)×3- (1)×2=1266.47 kJ·mol1
15.解:(1)Qp= == 4 (Al2O3, s) -3 (Fe3O4, s) =3347.6 kJ·mol1
(3) (3) = 2 (NH3, g) =32.90 kJ·mol1
= 5.76, 故 = 5.8105
由以上计算看出:选择合成氨固氮反应最好。
8.解: = (CO2, g) (CO, g) (NO, g)
=343.94 kJ·mol1< 0,所以该反应从理论上讲是可行的。
9.解: (298.15 K) = (NO, g) = 90.25 kJ·mol1
= (CO2, g) =393.509 kJ·mol1
CO2(g) + C(s) → CO(g)
= 86.229 kJ·mol1
CO(g) + Fe2O3(s) → Fe(s) + CO2(g)
=8.3 kJ·mol1
各反应 之和 =315.6 kJ·mol1。
(2)总反应方程式为
C(s) + O2(g) + Fe2O3(s)→ CO2(g) + Fe(s)
(2) (298.15 K) = 5.76, (298.15 K) = 5.8105
7.解:(1) (l) = 2 (NO, g) = 173.1 kJ·mol1
= =30.32,故 = 4.81031
(2) (2) = 2 (N2O, g) =208.4 kJ·mol1
= =36.50,故 = 3.21037
(298.15 K) =243.03 kJ·mol1
(298.15 K) = 40.92, 故 (298.15 K) = 8.31040
(373.15 K) = 34.02,故 (373.15 K) = 1.01034
6.解:(1) =2 (NH3, g) =32.90 kJ·mol1<0
该反应在298.15 K、标准态下能自发进行。
= (CO2, g) + 2 (H2O, l) (CH4, g)
=890.36 kJ·mo1
Qp=3.69104kJ
第2章 化学反应的方向、速率和限度 习题参考答案
1.解: =3347.6 kJ·mol1; =216.64 J·mol1·K1; =3283.0 kJ·mol1< 0
该反应在298.15K及标准态下可自发向右进行。
起始分压/105Pa 1.01 2.02 1.01 0.34
J= 0.168, = 1>0.168 =J,故反应正向进行。
12.解:(1)NH4HS(s)NH3(g) + H2S(g)
平衡分压/kPa
第1章 化学反应中的质量关系和能量关系 习题参考答案
1.解:1.00吨氨气可制取2.47吨硝酸。
2.解:氯气质量为2.9×103g。
3.解:一瓶氧气可用天数
4.解:
= 318 K ℃
5.解:根据道尔顿分压定律
p(N2) = 7.6104Pa
p(O2) = 2.0104Pa
p(Ar) =1103Pa
6.解:(1) 0.114mol;
= 55.3
χ= 2290.12
p(HI) = 2χkPa = 4580.24 kPa
n= = 3.15 mol
11.解:p(CO) = 1.01105Pa,p(H2O) = 2.02105Pa
p(CO2) = 1.01105Pa,p(H2) = 0.34105Pa
CO(g) + H2O(g)CO2(g) + H2(g)
(2)由以上计算可知:
(298.15 K) =70.81 kJ·mol1; (298.15 K) =43.2 J·mol1·K1
= T· ≤0
T≥ = 1639 K
4.解:(1) = =
=
(2) = =
=
(3) = =
=
(4) = =
=
5.解:设 、 基本上不随温度变化。
= T·
(298.15 K) =233.60 kJ·mol1
(298.15 K) = 12.39 J·mol1·K1
(1573.15K)≈ (298.15 K)1573.15 (298.15 K)
= 70759 J ·mol1
(1573.15 K) =2.349, (1573.15 K) = 4.48103
10.解:H2(g) + I2(g) 2HI(g)
平衡分压/kPa 2905.74χ 2905.74χ 2χ
(2wk.baidu.comQ =4141 kJ·mol1
16.解:(1) =151.1 kJ·mol1(2) =905.47 kJ·mol1(3) =71.7 kJ·mol1
17.解: =2 (AgCl, s)+ (H2O, l) (Ag2O, s)2 (HCl, g)
(AgCl, s) =127.3 kJ·mol1
18.解:CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
(2)T2= = 320 K
(3)W=(pV)=502 J
(4)U=Q+W= -758 J
(5)H=Qp= -1260 J
11.解:NH3(g) + O2(g) NO(g) + H2O(g) =226.2 kJ·mol1
12.解: = Qp=89.5 kJ
= nRT
=96.9 kJ
13.解:(1)C (s) + O2(g) → CO2(g)
(2)
(3)
7.解:(1)p(H2) =95.43 kPa
(2)m(H2) = = 0.194 g
8.解:(1)=5.0 mol
(2)=2.5 mol
结论: 反应进度()的值与选用反应式中的哪个物质的量的变化来进行计算无关,但与反应式的写法有关。
9.解: U=Qpp V= 0.771 kJ
10.解:(1)V1= 38.3 10-3m3= 38.3L
该反应在常温(298.15 K)、标准态下不能自发进行。
(2) = 146.0 kJ·mol1; = 110.45 J·mol1·K1; = 68.7 kJ·mol1> 0
该反应在700 K、标准态下不能自发进行。
3.解: =70.81 kJ·mol1; =43.2 J·mol1·K1; =43.9 kJ·mol1
=315.5 kJ·mol1
由上看出:(1)与(2)计算结果基本相等。所以可得出如下结论:反应的热效应只与反应的始、终态有关,而与反应的途径无关。
14.解: (3)= (2)×3- (1)×2=1266.47 kJ·mol1
15.解:(1)Qp= == 4 (Al2O3, s) -3 (Fe3O4, s) =3347.6 kJ·mol1
(3) (3) = 2 (NH3, g) =32.90 kJ·mol1
= 5.76, 故 = 5.8105
由以上计算看出:选择合成氨固氮反应最好。
8.解: = (CO2, g) (CO, g) (NO, g)
=343.94 kJ·mol1< 0,所以该反应从理论上讲是可行的。
9.解: (298.15 K) = (NO, g) = 90.25 kJ·mol1
= (CO2, g) =393.509 kJ·mol1
CO2(g) + C(s) → CO(g)
= 86.229 kJ·mol1
CO(g) + Fe2O3(s) → Fe(s) + CO2(g)
=8.3 kJ·mol1
各反应 之和 =315.6 kJ·mol1。
(2)总反应方程式为
C(s) + O2(g) + Fe2O3(s)→ CO2(g) + Fe(s)
(2) (298.15 K) = 5.76, (298.15 K) = 5.8105
7.解:(1) (l) = 2 (NO, g) = 173.1 kJ·mol1
= =30.32,故 = 4.81031
(2) (2) = 2 (N2O, g) =208.4 kJ·mol1
= =36.50,故 = 3.21037
(298.15 K) =243.03 kJ·mol1
(298.15 K) = 40.92, 故 (298.15 K) = 8.31040
(373.15 K) = 34.02,故 (373.15 K) = 1.01034
6.解:(1) =2 (NH3, g) =32.90 kJ·mol1<0
该反应在298.15 K、标准态下能自发进行。
= (CO2, g) + 2 (H2O, l) (CH4, g)
=890.36 kJ·mo1
Qp=3.69104kJ
第2章 化学反应的方向、速率和限度 习题参考答案
1.解: =3347.6 kJ·mol1; =216.64 J·mol1·K1; =3283.0 kJ·mol1< 0
该反应在298.15K及标准态下可自发向右进行。
起始分压/105Pa 1.01 2.02 1.01 0.34
J= 0.168, = 1>0.168 =J,故反应正向进行。
12.解:(1)NH4HS(s)NH3(g) + H2S(g)
平衡分压/kPa
第1章 化学反应中的质量关系和能量关系 习题参考答案
1.解:1.00吨氨气可制取2.47吨硝酸。
2.解:氯气质量为2.9×103g。
3.解:一瓶氧气可用天数
4.解:
= 318 K ℃
5.解:根据道尔顿分压定律
p(N2) = 7.6104Pa
p(O2) = 2.0104Pa
p(Ar) =1103Pa
6.解:(1) 0.114mol;
= 55.3
χ= 2290.12
p(HI) = 2χkPa = 4580.24 kPa
n= = 3.15 mol
11.解:p(CO) = 1.01105Pa,p(H2O) = 2.02105Pa
p(CO2) = 1.01105Pa,p(H2) = 0.34105Pa
CO(g) + H2O(g)CO2(g) + H2(g)
(2)由以上计算可知:
(298.15 K) =70.81 kJ·mol1; (298.15 K) =43.2 J·mol1·K1
= T· ≤0
T≥ = 1639 K
4.解:(1) = =
=
(2) = =
=
(3) = =
=
(4) = =
=
5.解:设 、 基本上不随温度变化。
= T·
(298.15 K) =233.60 kJ·mol1
(298.15 K) = 12.39 J·mol1·K1
(1573.15K)≈ (298.15 K)1573.15 (298.15 K)
= 70759 J ·mol1
(1573.15 K) =2.349, (1573.15 K) = 4.48103
10.解:H2(g) + I2(g) 2HI(g)
平衡分压/kPa 2905.74χ 2905.74χ 2χ
(2wk.baidu.comQ =4141 kJ·mol1
16.解:(1) =151.1 kJ·mol1(2) =905.47 kJ·mol1(3) =71.7 kJ·mol1
17.解: =2 (AgCl, s)+ (H2O, l) (Ag2O, s)2 (HCl, g)
(AgCl, s) =127.3 kJ·mol1
18.解:CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
(2)T2= = 320 K
(3)W=(pV)=502 J
(4)U=Q+W= -758 J
(5)H=Qp= -1260 J
11.解:NH3(g) + O2(g) NO(g) + H2O(g) =226.2 kJ·mol1
12.解: = Qp=89.5 kJ
= nRT
=96.9 kJ
13.解:(1)C (s) + O2(g) → CO2(g)
(2)
(3)
7.解:(1)p(H2) =95.43 kPa
(2)m(H2) = = 0.194 g
8.解:(1)=5.0 mol
(2)=2.5 mol
结论: 反应进度()的值与选用反应式中的哪个物质的量的变化来进行计算无关,但与反应式的写法有关。
9.解: U=Qpp V= 0.771 kJ
10.解:(1)V1= 38.3 10-3m3= 38.3L