无机及分析化学答案(第二版)第二章

第二章 化学反应一般原理

2-1 苯和氧按下式反应：

C 6H 6(l) + 2

15O 2(g) → 6CO 2(g) + 3H 2O(l) 在25℃100kPa 下，0.25mol 苯在氧气中完全燃烧放出817kJ 的热量，求C 6H 6的标准摩尔燃烧焓∆c H m 和该燃烧反应的∆r U m 。

解： ξ = νB -1∆n B = (-0.25 mol) / ( -1) = 0.25 mol

∆c H m = ∆r H m = = -817 kJ / 0.25 mol

= -3268 kJ ⋅mol -1

∆r U m = ∆r H m - ∆n g RT

= -3268 kJ ⋅mol -1 - (6 -15 / 2) ⨯ 8.314 ⨯ 10-3 ⨯ 298.15 kJ ⋅mol -1

= -3264 kJ ⋅mol -1

2-2 利用附录III 的数据，计算下列反应的∆r H m 。

(1) Fe 3O 4(s) + 4H 2(g) → 3Fe(s) + 4H 2O(g)

(2) 2NaOH(s) + CO 2(g) → Na 2CO 3(s) + H 2O(l)

(3) 4NH 3(g) + 5O 2(g) → 4NO(g) + 6H 2O(g)

(4) CH 3COOH(l) + 2O 2(g) → 2CO 2(g) + 2H 2O(l)

解： (1) ∆r H m = [4 ⨯ (-241.818) - (-1118.4)] kJ ⋅mol -1

= 151.1 kJ ⋅mol -1

(2) ∆r H m = [(-285.830) + (-1130.68) - (-393.509) - 2 ⨯ (-425.609)] kJ ⋅mol -1

= -171.78 kJ ⋅mol -1

(3) ∆r H m = [6 ⨯ (-241.818) + 4 ⨯ 90.25 - 4 ⨯ (-46.11)] kJ ⋅mol -1

= -905.5 kJ ⋅mol -1

(4) ∆r H m = [2(-285.830) + 2(-393.509) - (-484.5)] kJ ⋅mol -1

= -874.1 kJ ⋅mol -1

2-3 已知下列化学反应的标准摩尔反应焓变，求乙炔(C 2H 2，g)的标准摩尔生成焓∆ f H m 。

(1) C 2H 2(g) + 5/2O 2(g) → 2CO 2(g) + H 2O(g) ∆ r H m = -1246.2 kJ ⋅mol -1

(2) C(s) + 2H 2O(g) → CO 2(g) + 2H 2(g) ∆ r H m = +90.9 kJ ⋅mol -1

(3) 2H 2O(g) → 2H 2(g) + O 2(g) ∆ r H m = +483.6 kJ ⋅mol -1

解：反应2 ⨯ (2) - (1) - 2.5 ⨯ (3)为：

2C(s) + H 2(g) → C 2H 2(g)

∆ f H m (C 2H 2) = 2 ⨯ ∆ r H m (2) - ∆ r H m (1) - 2.5∆ r H m (3)

= [2 ⨯ 90.9 - ( -1246.2) - 2.5 ⨯ 483.6] kJ ⋅mol -1

= 219.0 kJ ⋅mol -1

2-4 求下列反应在298.15 K 的标准摩尔反应焓变∆ r H m 。

(1) Fe(s)+Cu 2+(aq)→Fe 2+(aq)+Cu(s)

(2) AgCl(s)+Br -(aq)→AgBr(s)+Cl -(aq)

ξ

H r ∆

(3) Fe2O3(s)+6H+(aq)→2Fe3+(aq)+3H2O(l)

(4) Cu2+(aq)+Zn(s) →Cu(s)+Zn2+(aq)

解：∆r H m(1) = [-89.1-64.77 ] kJ⋅mol-1

= -153.9 kJ⋅mol-1

∆r H m(2) = [-167.159 -100.37 - (-121.55) - (-127.068)] kJ⋅mol-1

= -18.91 kJ⋅mol-1

∆r H m(3) = [2 ⨯ (-48.5) + 3 ⨯ (-285.830) + 824.2] kJ⋅mol-1

= -130.3 kJ⋅mol-1

∆r H m(4) = [(-153.89) - 64.77] kJ⋅mol-1

= -218.66 kJ⋅mol-1

2-5 计算下列反应在298.15K的∆r H m，∆r S m和∆r G m，并判断哪些反应能自发向右进行。

(1) 2CO(g) + O2(g) → 2CO2(g)

(2) 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)

(3) Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)

(4) 2SO2(g) + O2(g) → 2SO3(g)

解：(1) ∆r H m = [2 ⨯ (-393.509) - 2 ⨯ (-110.525)] kJ⋅mol-1

= -565.968 kJ⋅mol-1

∆r S m = [2 ⨯ 213.74 - 2 ⨯ 197.674 - 205.138] J⋅mol-1⋅K-1

= -173.01 J⋅mol-1⋅K-1

∆r G m = ∆r H m -T⋅∆r S m

= [-565.968 - 298.15 ⨯ (-173.01 ⨯ 10-3)] kJ⋅mol-1⋅K-1

= -514.385kJ⋅mol-1 < 0，反应自发。

(2) ∆r H m = [6 ⨯ (-241.818) + 4 ⨯ 90.25 - 4 ⨯ (-46.11)] kJ⋅mol-1

= -905.5 kJ⋅mol-1

∆r S m = [6 ⨯ 188.825 + 4 ⨯ 210.761 - 5 ⨯ 205.138 - 4 ⨯ 192.45] J⋅mol-1⋅K-1

= 180.50 J⋅mol-1⋅K-1

∆r G m = ∆r H m -T⋅∆r S m

= [-905.5- 298.15 ⨯ 180.50 ⨯ 10-3 ] kJ⋅mol-1

= -959.3kJ⋅mol-1 < 0，反应自发。

(3) ∆r H m = [3 ⨯ (-393.509) - 3 ⨯ (-110.525) - (-824.2)] kJ⋅mol-1

= -24.8 kJ⋅mol-1

∆r S m = [3 ⨯ 213.74 + 2 ⨯ 27.28 - 3 ⨯ 197.674 - 87.4] J⋅mol-1⋅K-1

= 15.4 J⋅mol-1⋅K-1

∆r G m = ∆r H m -T⋅∆r S m

= [-24.8 - 298.15 ⨯ 15.4 ⨯ 10-3 ] kJ⋅mol-1

= -29.4kJ⋅mol-1 < 0，反应自发。

(4) ∆r H m = [2 ⨯ (-395.72) - 2 ⨯ (-296.830)] kJ⋅mol-1

= -197.78kJ⋅mol-1

∆r S m = [2 ⨯ 256.76 - 205.138 - 2 ⨯ 248.22] J⋅mol-1⋅K-1

= -188.06J⋅mol-1⋅K-1

∆r G m = ∆r H m -T⋅∆r S m

= [-197.78 - 298.15 ⨯ (-188.06 ⨯ 10-3) ] kJ⋅mol-1

= -141.71kJ⋅mol-1< 0，反应自发。