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(精选)无机化学简明教程(天津大学)课后习题参考答案

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33111-1222()(13.210-1.0110)kPa 32L9.6d 101.325kPa 400L d n p p V n p V -⨯⨯⨯===⨯⨯4.解:pV MpVT nR mR== = 318 K 44.9=℃ 5.解:根据道尔顿分压定律ii n p p n=p (N 2) = 7.6104 Pap (O 2) = 2.0104 Pa p (Ar) =1103 Pa6.解:(1)2(CO )n = 0.114mol; 2(CO )p = 42.87 10 Pa ⨯(2)222(N )(O )(CO )p p p p =--43.7910Pa =⨯ (3)4224(O )(CO ) 2.6710Pa0.2869.3310Pan p n p ⨯===⨯ 7.解:(1)p (H 2) =95.43 kPa (2)m (H 2) =pVMRT= 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 = Q p p ∆V = 0.771 kJ 10.解: (1)V 1 = 38.3⨯10-3m 3= 38.3L(2) T 2 = nRpV 2= 320 K (3)W =(p V ) = 502 J(4) U = Q + W = -758 J (5) H = Q p = -1260 J11.解:NH 3(g) +45O 2(g) 298.15K−−−−→标准态NO(g) + 23H 2O(g) m r H ∆= 226.2 kJ ·mol 1 12.解:m r H ∆= Q p = 89.5 kJ m r U ∆= mr H ∆ nRT =96.9 kJ13.解:(1)C (s) + O 2 (g) → CO 2 (g)m r H ∆ =m f H ∆(CO 2, g) = 393.509 kJ ·mol 121CO 2(g) + 21C(s) → CO(g) m r H ∆ = 86.229 kJ ·mol 1CO(g) +31Fe 2O 3(s) → 32Fe(s) + CO 2(g)m r H ∆ =8.3 kJ ·mol 1各反应m r H ∆之和m r H ∆= 315.6 kJ·mol 1。

(2)总反应方程式为23C(s) + O 2(g) + 31Fe 2O 3(s) → 23CO 2(g) + 32Fe(s) m r H ∆ = 315.5 kJ ·mol 1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: m r H ∆(3)= m r H ∆(2)×3-m r H ∆(1)×2=1266.47 kJ ·mol 115.解:(1)Q p = m r H ∆== 4 m f H ∆(Al 2O 3, s) -3m f H ∆(Fe 3O 4, s) =3347.6 kJ ·mol 1(2)Q = 4141 kJ ·mol 116.解:(1) m r H ∆ =151.1 kJ ·mol 1(2) m r H ∆ = 905.47 kJ ·mol 1(3)m r H ∆ =71.7kJ ·mol 117.解: m r H ∆=2 m f H ∆(AgCl, s)+m f H ∆(H 2O, l) m f H ∆(Ag 2O, s)2m f H ∆(HCl, g)m f H ∆(AgCl, s) =127.3 kJ ·mol 118.解:CH 4(g) + 2O 2(g) → CO 2(g) + 2H 2O(l)m r H ∆ = m f H ∆(CO 2, g) + 2m f H ∆(H 2O, l)m f H ∆(CH 4, g)= 890.36 kJ ·mo 1Q p =3.69104kJ第2章 化学反应的方向、速率和限度1.解:m r H ∆ =3347.6 kJ ·mol 1;m r S ∆ =216.64 J ·mol 1·K 1;m r G ∆ =3283.0 kJ ·mol1< 0该反应在298.15K 及标准态下可自发向右进行。

2.解:m r G ∆ = 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) m r H ∆ = 146.0 kJ ·mol 1;m r S ∆ = 110.45 J ·mol 1·K 1;m r G ∆ = 68.7kJ ·mol 1> 0该反应在700 K 、标准态下不能自发进行。

3.解: m r H ∆ = 70.81 kJ ·mol 1;m r S ∆ = 43.2 J ·mol 1·K 1;m r G ∆ = 43.9 kJ·mol 1(2)由以上计算可知:m r H ∆(298.15 K) = 70.81 kJ ·mol 1;m r S ∆(298.15 K) = 43.2 J ·mol 1·K 1m r G ∆ = m r H ∆ T ·m r S ∆ ≤ 0 T ≥K)(298.15K) (298.15m r m rS H ∆∆ = 1639 K4.解:(1)c K = {}O)H ( )(CH )(H (CO) 2432c c c c p K = {}O)H ( )(CH )(H (CO) 2432p p p pK = {}{}{}{}p p p p p p p p / O)H ( /)(CH / )(H / (CO) 2432(2)c K ={}{})(NH )(H )(N 3232212c c c p K ={}{})(NH )(H )(N 3232212p p pK ={}{}pp p p p p / )(NH/)(H/)(N3232212(3)c K =)(CO 2c p K =)(CO 2p K = p p /)(CO 2 (4)c K ={}{}3232 )(H O)(H c c p K ={}{}3232 )(H O)(H p pK ={}{}3232 /)(H/O)(Hpp p p5.解:设 m r H ∆、m r S ∆基本上不随温度变化。

m r G ∆ = m r H ∆ T · m r S ∆m r G ∆(298.15 K) = 233.60 kJ ·mol 1 m r G ∆(298.15 K) = 243.03 kJ ·mol 1K lg (298.15 K) = 40.92, 故 K (298.15 K) = 8.31040 K lg (373.15 K) = 34.02,故 K (373.15 K) = 1.010346.解:(1) m r G ∆=2m f G ∆(NH 3, g) = 32.90 kJ ·mol 1<0该反应在298.15 K 、标准态下能自发进行。

(2) K lg (298.15 K) = 5.76, K (298.15 K) = 5.81057. 解:(1) m r G ∆(l) = 2 m f G ∆(NO, g) = 173.1 kJ ·mol 11lgK =RTG 303.2)1(m f∆- = 30.32, 故1K = 4.81031(2) m r G ∆(2) = 2m f G ∆(N 2O, g) =208.4 kJ ·mol 12lgK =RTG 303.2)2(m f∆- = 36.50, 故2K = 3.21037(3) m r G ∆(3) = 2m f G ∆(NH 3, g) = 32.90 kJ ·mol13lg K = 5.76, 故3K = 5.8105由以上计算看出:选择合成氨固氮反应最好。

8.解: m r G ∆ = m f G ∆(CO 2, g) m f G ∆(CO, g)m f 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: m r H ∆(298.15 K) = m f H ∆(NO, g) = 90.25 kJ ·mol 1m r S ∆(298.15 K) = 12.39 J ·mol 1·K 1m r G ∆(1573.15K)≈ m r H ∆(298.15 K) 1573.15 m r S ∆(298.15 K)= 70759 J ·mol 1K lg (1573.15 K) = 2.349, K (1573.15 K) = 4.4810310. 解: H 2(g) + I 2(g) 2HI(g)平衡分压/kPa 2905.74 χ 2905.74 χ 2χ22)74.2905()2(x x -= 55.3χ= 2290.12p (HI) = 2χkPa = 4580.24 kPan =pVRT= 3.15 mol 11.解:p (CO) = 1.01105 Pa, p (H 2O) = 2.02105Pap (CO 2) = 1.01105 Pa, p (H 2) = 0.34105PaCO(g) + H 2O(g) CO 2(g) + H 2(g)起始分压/105Pa 1.01 2.02 1.01 0.34 J = 0.168, p K = 1>0.168 = J,故反应正向进行。

12.解:(1) NH 4HS(s)NH 3(g) + H 2S(g)平衡分压/kPa x xK ={}{}/ S)(H / )(NH 23 p p p p = 0.070 则 x = 0.26100 kPa = 26 kPa 平衡时该气体混合物的总压为52 kPa(2)T 不变, K 不变。

NH 4HS(s) NH 3(g) + H 2S(g)平衡分压/kPa 25.3+y yK ={}{}/ / ) 25.3 p y p y +( = 0.070y = 17 kPa13.解:(1) PCl 5(g) PCl 3(g) + Cl 2(g)平衡浓度/(mol ·L 1)0.250.070.0- 0.250.0 0.250.0c K = )PCl ()Cl ()PCl (523c c c = 0.62mol · L 1, α(PCl 5) = 71%PCl5(g) PCl3(g) + Cl2(g)平衡分压 0.20V RT 0.5V RT 0.5V RTK={}{}{}p p p p p p /)(PCl/) (Cl /)(PCl 523= 27.2(2) PCl 5(g)PCl 3(g) + Cl 2(g)新平衡浓度/(mol ·L 1) 0.10 + y 0.25 y 0.25 +y -210.0 c K =)10.0()30.0)(25.0(y y y +--mol ·L 1 = 0.62mol · L 1(T 不变,c K 不变)y =0.01 mol ·L 1,α(PCl 5) = 68%(3) PCl 5(g) PCl 3(g) + Cl 2(g)平衡浓度/(mol ·L 1) z -35.0 z 0.050 +zc K =zz z -+35.0)050.0(= 0.62 mol ·L 1z = 0.24 mol ·L 1,α(PCl 5) = 68%比较(2)、(3)结果,说明最终浓度及转化率只与始、终态有关,与加入过程无关。

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