无机及分析化学答案(第二版)第二章

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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。

2-6 由软锰矿二氧化锰制备金属锰可采取下列两种方法:
(1) MnO2(s) + 2H2(g) → Mn(s) + 2H2O(g)
(2) MnO2(s) + 2C(s) → Mn(s) + 2CO(g)
上述两个反应在25℃,100 kPa下是否能自发进行?如果考虑工作温度愈低愈好的话,则制备锰采用哪一种方法比较好?
解:∆r G m(1) = [2 ⨯ (-228.575) - (-466.14)] kJ⋅mol-1
= 8.99 kJ⋅mol-1
∆r G m(2) = [2 ⨯ (-137.168) - (-466.14)] kJ⋅mol-1
= 191.80 kJ⋅mol-1
两反应在标准状态、298.15K均不能自发进行;
计算欲使其自发进行的温度:
∆r H m(1) = [2 ⨯ (-241.818) - (-520.03)] kJ⋅mol-1
= 36.39 kJ⋅mol-1
∆r S m(1) = [2 ⨯ 188.825 + 32.01 - 2 ⨯ 130.684 - 53.05] J⋅mol-1⋅K-1
= 95.24 J⋅mol-1⋅K-1
∆r H m(1) -T1∆r S m(1) = 0
T1 = 36.39 kJ⋅mol-1 / (95.24 ⨯ 10-3 kJ⋅mol-1⋅K-1)
= 382.1K
∆r H m(2) = [2 ⨯ (-110.525) - (-520.03)] kJ⋅mol-1
= 298.98 kJ⋅mol-1
∆r S m(2) = [2 ⨯ 197.674 + 32.01 - 2 ⨯ 5.740 - 53.05] J⋅mol-1⋅K-1
= 362.28 J⋅mol-1⋅K-1
∆r H m(2) -T1∆r S m(2) = 0
T2 = 298.98 kJ⋅mol-1 / (362.28 ⨯ 10-3 kJ⋅mol-1⋅K-1)
= 825.27 K
T1 <T2,反应(1)可在较低温度下使其自发进行,能耗较低,所以反应(1)更合适。

2-7 不用热力学数据定性判断下列反应的∆r S m是大于零还是小于零。

(1) Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)
(2) CaCO3(s) → CaO(s) + CO2(g)
(3) NH3(g) + HCl(g) → NH4Cl(s)
(4) CuO(s) + H2(g) → Cu(s) + H2O(l)
解:反应(1)、(2) 均有气体产生,为气体分子数增加的反应,∆r S m > 0;
反应(3)、(4)的气体反应后分别生成固体与液体,∆r S m < 0。

2-8 计算25℃100kPa下反应CaCO3(s) →CaO(s)+CO2(g)的∆r H m和∆r S m,并判断:
(1) 上述反应能否自发进行?
(2) 对上述反应,是升高温度有利?还是降低温度有利?
(3) 计算使上述反应自发进行的温度条件。

解:∆r H m = [-393.509 - 635.09 + 1206.92] kJ⋅mol-1
= 178.32 kJ⋅mol-1
∆r S m = [213.74 + 39.75 - 92.9] J⋅mol-1⋅K-1
= 160.6 J⋅mol-1⋅K-1
(1) ∆r G m= [178.32 - 298.15 ⨯ 160.6 ⨯ 10-3] kJ⋅mol-1
= 130.44 kJ⋅mol-1 > 0
反应不能自发进行;
(2) ∆r H m > 0,∆r S m > 0,升高温度对反应有利,有利于∆r G m < 0。

(3) 自发反应的条件为T >∆r H m /∆r S m
= [178.32 / 160.6⨯10-3] K
= 1110 K
2-9 写出下列各化学反应的平衡常数K 表达式。

(1) CaCO3(s)CaO(s) + CO2(g)
(2) 2SO2(g) + O2(g)2SO3(g)
(3) C(s) + H2O(g)CO(g) + H2(g)
(4) AgCl(s)Ag+(aq) + Cl-(aq)
(5) HAc(aq)H+(aq) + Ac-(aq)
(6) SiO2(s) + 6HF(aq)H2[SiF6](aq) + 2H2O(l)
(7) Hb(aq)(血红蛋白) + O2(g)HbO2(aq)(氧合血红蛋白)
(8) 2MnO4-(aq) + 5SO32-(aq) + 6H+(aq)2Mn2+(aq) + 5SO42-(aq) + 3H2O(l)
解:(1) K =∏
B
B
(p/ p )Bν= p(CO2)/p
(2) K = (p(SO3)/p )2⋅(p(O2)/p )-1⋅(p(SO2) /p )-2
(3) K = (p(CO)/p )⋅(p(H2)/p )⋅(p(H2O)/p )-1
(4) K = (c(Ag+)/c )⋅(c(Cl-)/c )
(5) K = (c(H+)/c )⋅(c(Ac-)/c )⋅(c(HAc)/c )-1
(6) K = (c(H2[SiF6])/c )⋅(c(HF)/c )-6
(7) K = (c(HbO2)/c )⋅(c(Hb)/c )-1⋅(p(O2)/p )-1
(8) K =(c(Mn2+)/c )2⋅(c(SO42-)/c )5⋅(c(MnO4-)/c )-2⋅(c(SO32-)/c )-5⋅(c(H+)/c )-6
2-10 已知下列化学反应在298.15K时的平衡常数:
(1) CuO(s) + H2(g)Cu(s) + H2O (g) K 1 = 2⨯1015
(2) 1/2O2(g) + H2(g)H2O(g) K 2 = 5⨯1022
计算反应CuO(s)Cu(s) + 1/2O2(g) 的平衡常数K 。

解:反应(1) - (2)为所求反应,根据多重平衡规则:
K = K 1 / K 2
= 2 ⨯ 1015 / 5 ⨯ 1022 = 4 ⨯ 10-8
2-11 已知下列反应在298.15K的平衡常数:
(1) SnO2(s) + 2H2(g)2H2O(g) + Sn(s) K 1 = 21
(2) H2O(g) + CO (g)H2(g) + CO2(g);K 2 = 0.034
计算反应2CO(g) + SnO2(s)Sn(s) + 2CO2 (g)在298.15K时的平衡常数K 。

解:反应(1) + 2 ⨯ (2)为所求反应,所以
K = K 1⨯ (K 2)2
= 21 ⨯ 0.0342 = 2.4 ⨯ 10-2
2-12 密闭容器中反应2NO(g) + O2(g)2NO2(g) 在1500K条件下达到平衡。

若始态p(NO) = 150 kPa,p(O2) = 450 kPa,p(NO2) = 0;平衡时p(NO2) = 25 kPa。

试计算平衡时p(NO),p(O2)的分压及平衡常数K 。

解:V、T不变,p ∝n,各平衡分压为:
p(NO) =150 kPa - 25 kPa
= 125 kPa
p(O2) = 450 kPa - (25 / 2) kPa
= 437.5 kPa
K = (p(NO2) / p )2(p(NO) / p )-2(p(O2) / p )-1
= (25 / 100) 2(125 / 100) -2(437.5 / 100) -1
= 9.1 ⨯ 10-3
2-13 密闭容器中的反应 CO(g) + H 2O(g) CO 2(g) + H 2(g) 在750K 时其K = 2.6,求:
(1) 当原料气中H 2O(g)和CO(g)的物质的量之比为1:1时,CO(g)的平衡转化率为多少?
(2) 当原料气中H 2O(g):CO(g)为4:1时,CO(g)的平衡转化率为多少?说明什么问题? 解:(1) V 、T 不变 CO(g) + H 2O(g) CO 2(g) + H 2(g)
起始n / mol 1 1 0 0
平衡n / mol 1-x 1-x x x
∑n = 2(1 - x ) + 2x = 2
平衡分压 21x -p 总 21x -p 总 2x p 总 2
x p 总 K = (p (H 2) / p )⋅(p (CO 2) / p )⋅(p (H 2O) / p )-1⋅(p (CO) / p )-1 2.6 = (2x )2⋅(2
1x -)-2 x = 0.62
α(CO) = 62%
(2) V 、T 不变 CO(g) + H 2O(g) CO 2(g) + H 2(g)
起始n / mol 1 4 0 0
平衡n / mol 1-x 4-x x x ∑n = 5
平衡分压 51x -p 总 5
4x -p 总 5x p 总 5x p 总 2.6 = (x / 5)2⋅ [(1 - x ) / 5]-1⋅[(4 - x ) / 5]-1
x = 0.90
α(CO) = 90%
H 2O(g)浓度增大,CO(g)转化率增大,利用廉价的H 2O(g),使CO(g)反应完全。

2-14 在317K ,反应 N 2O 4(g) 2NO 2(g) 的平衡常数K = 1.00。

分别计算当系统总压为400 kPa 和800 kPa 时N 2O 4(g)的平衡转化率,并解释计算结果。

解:总压为400 kPa 时 N 2O 4(g) 2NO 2(g)
起始n / mol 1 0
平衡n / mol 1-x 2x
平衡相对分压 10040011⨯+-x x 100
40012⨯+x x 00.11)1(00.4100.812=⎪⎭⎫ ⎝⎛+-⋅⎪⎭⎫ ⎝⎛+-x x x x
x = 0.243
α(N 2O 4) = 24.3%
总压为800kPa 时 00.11)1(00.810.1612=⎪⎭⎫ ⎝⎛+-⋅⎪⎭⎫ ⎝⎛+-x x x x
x = 0.174
α(N 2O 4) = 17.4%
增大压力,平衡向气体分子数减少的方向移动,α(N 2O 4)下降。

2-15 已知尿素CO(NH 2)2的∆f G m = -197.15 kJ ⋅mol -1,求尿素的合成反应在298.15 K 时的∆ r G m 和K 。

2NH 3(g) + CO 2(g) H 2O(g) + CO(NH 2)2(s)
解: ∆r G m = [-197.15 - 228.575 + 394.359 + 2 ⨯ 16.45] kJ ⋅mol -1
= 1.53 kJ ⋅mol -1
lg K = -∆r G m / (2.303RT )
= -1.53 ⨯ 103 / (2.303 ⨯ 8.314 ⨯ 298.15)
= -0.268
K = 0.540
2-16 25℃时,反应2H 2O 2(g)2H 2O(g) + O 2(g)的∆r H m 为 -210.9 kJ ⋅mol -1,∆r S m 为131.8 J ⋅mol -1⋅K -1。

试计算该反应在25℃和100℃时的K ,计算结果说明什么问题?。

解: ∆r G m = ∆r H m - T ∆r S m
∆r G m,298.15K = -210.9 kJ ⋅mol -1 - 298.15 K ⨯ 131.8 ⨯ 10-3 kJ ⋅mol -1⋅K -1
= -250.2 kJ ⋅mol -1
lg K = -∆r G m / (2.303RT )
= 250.2 ⨯ 103 / (2.303 ⨯ 8.314 ⨯ 298.15)
= 43.83
K 298.15K = 6.7 ⨯ 1043
∆r G m,373.15K = -210.9 kJ ⋅mol -1 -373.15 K ⨯ 131.8 ⨯ 10-3 kJ ⋅mol -1⋅K -1
= -260.1 kJ ⋅mol -1
lg K = 260.1 ⨯ 103 / (2.303 ⨯ 8.314 ⨯ 373.15)
= 36.40
K 373.15K = 2.5 ⨯ 1036
该反应为放热反应,对放热反应,温度升高,K 下降。

2-17 在一定温度下Ag 2O 的分解反应为 Ag 2O(s) 2Ag(s) + 1/2O 2(g)。

假定反应的∆r H m ,∆r S m 不随温度的变化而改变,估算Ag 2O 在标准状态的最低分解温度?
解: ∆r H m = -∆f H m (Ag 2O) = 31.05 kJ ⋅mol -1
∆r S m = [2 ⨯ 42.5 + 205.138 / 2 - 121.3] J ⋅mol -1⋅K -1
= 66.3 J ⋅mol -1⋅K -1
T = ∆r H m /∆r S m
= 31.05 kJ ⋅mol -1 /( 66.3 ⨯ 10-3 kJ ⋅mol -1⋅K -1)
= 468 K
2-18 已知反应 2SO 2(g) + O 2(g) → 2SO 3(g) 在427℃和527℃时的K 值分别为1.0 ⨯ 105和1.1 ⨯ 102,求该反应的∆r H m 。

解: ∆r H m = -3.2 ⨯ 102 kJ ⋅mol -1
2-19 已知反应 2H 2(g) + 2NO(g) → 2H 2O(g) + N 2(g) 的速率方程 v = k c (H 2)⋅c 2(NO),在一定温度下,若使容器体积缩小到原来的1/2时,问反应速率如何变化?
⎪⎪⎭⎫ ⎝⎛--=21r 2
111Δln m T T R H K K ⎪⎭⎫ ⎝⎛+-+⋅⨯-=⨯⨯--27352712734271mol kJ 10314.8Δ101.1100.1ln 13m r 25H
解: 体积缩小为1/ 2,浓度增大2倍:
v 2 = k 2c 1(H 2)⋅(2c 1)2(NO)
= 8 k ⋅ c 1(H 2)⋅(c 1)2(NO)
= 8v 1
2-20 某基元反应 A + B → C ,在1.20 L 溶液中,当A 为4.0 mol ,B 为3.0 mol 时,v 为0.0042 mol ⋅L -1s -1,计算该反应的速率常数,并写出该反应的速率方程式。

解:v = kc A c B
k = 0.0042 mol ⋅L -1s -1 / [(4.0 mol / 1.20 L) ⨯ (3.0 mol) / 1.20 L]
= 5.0 ⨯ 10-4 mol -1⋅L ⋅s -1
2-21 某一级反应,若反应物浓度从1.0 mol ⋅L -1降到0.20 mol ⋅L -1需30min ,问:
(1) 该反应的速率常数k 是多少?
(2) 反应物浓度从0.20 mol ⋅L -1降到0.040 mol ⋅L -1需用多少分钟?
解: (1) kt c c -=0
B ln ln(0.20 /1.0) = -k ⋅30 min
k = 0.054 min -1 ;
(2) ln(0.040 / 0.20) = - 0.054 min -1t
t = 30 min
2-22 From reactions(1)~(5)below, select, without any thermodynamic calculations those reactions which have: (a) large negative standar entropy changes, (b) large positive standar entropy changes, (c) small entropy changes which might be either positive or negative.
(1) Mg(s) + Cl 2(g) = MgCl 2(s)
(2) Mg(s) + I 2(s) = MgI 2(s)
(3) C(s) + O 2(g) = CO 2(g)
(4 Al 2O 3(s) + 3C(s) + 3Cl 2(g) = 2AlCl 3(g) + 3CO(g)
(5) 2NO(g) + Cl 2(g) = 2NOCl(g)
Solution: (1) large negative standar entropy changes:(1) ,(5).
(2) large positive standar entropy changes:(4).
(3) small entropy changes which might be either positive or
negative (2),(3).
2-23 Calculate the value of the thermodynamic decomposition temperature (T d ) for the reaction NH 4Cl(s).= NH 3(g) + HCl(g) at the standard state.
Solution: ∆r H m = [- 46.11- 92.307 + 314.43] kJ ⋅mol -1
= 176.01 kJ ⋅mol -1
∆r S m = [192.45 + 186.908 - 94.6] J ⋅mol -1⋅K -1
= 284.8 J ⋅mol -1⋅K -1
T = ∆r H m /∆r S m
=176.01 kJ ⋅mol -1 / 284.758 ⨯ 10-3 kJ ⋅mol -1⋅K -1
= 618.0 K
2-24Calculate ∆r G m at 298.15K for the reaction 2NO 2(g)→N 2O 4(g). Is this reaction spontaneous?
Solution: ∆r G m = [97.89 -2 ⨯ 51.31] kJ ⋅mol -1
= - 4.73 kJ ⋅mol -1 < 0
The reaction is spontaneous.
2-25 The following gas phase reaction follows first -order kinetics:
FClO 2(g) → FClO(g) + O(g)
The activation energy of th is reaction is measured to be 186 kJ ⋅mol -1. The value of k at 322℃ is determined to be 6.76⨯10-4s -1.
(1) What would be the value of k for this reaction at 25℃?
(2) At what temperature would this reaction have a k value of 6.00⨯10-2s -1?
Solution: (1) ⎪⎭
⎫ ⎝⎛-+⋅⋅⋅⨯-=⨯-----15.298115.2733221K mol J 314.8mol J 10186s 1076.6ln 1113214k k 2 = 3.70⨯10-20 s -1 (2) ⎪⎭
⎫ ⎝⎛-+⋅⋅⋅⨯-=⨯⨯-----T 115.2733221K mol J 314.8mol J 101861000.61076.6ln 111324 T = 676 K
2-26 某理想气体在恒定外压(101.3 kPa)下吸热膨胀,其体积从80 L 变到160 L ,同时吸收25 kJ 的热量,试计算系统热力学能的变化。

解: ∆U = Q + W = Q - p ∆V
= 25 kJ - 101.3 kPa ⨯ (160 - 80) ⨯ 10-3 m 3
= 25 kJ - 8.104 kJ
= 17 kJ
2-27 蔗糖(C 12H 22O 11)在人体内的代谢反应为:
C 12H 22O 11(s) + 12O 2(g) → 12CO 2(g) + 11H 2O(l)
假设在标准状态时其反应热有30%可转化为有用功,试计算体重为70kg 的人登上3000m 高的山(按有效功计算),若其能量完全由蔗糖转换,需消耗多少蔗糖?(∆f H m (C 12H 22O 11) = -2222kJ ⋅mol -1)
解: W = -70 kg ⨯ 3000 m
= -2.1 ⨯ 105 kg ⋅m
= -2.1 ⨯ 105 ⨯ 9.8 J = -2.1 ⨯ 103 kJ
∆r H = -2.1 ⨯ 103 kJ / 30%
= -7.0 ⨯ 103 kJ
∆ r H m = 11 ⨯ (-285.830 kJ ⋅mol -1) + 12 ⨯ (-393.509 kJ ⋅mol -1) - (-2222 kJ ⋅mol -1)
= -5644 kJ ⋅mol -1
ξ = ∆r H / ∆r H m
= (-7.0 ⨯ 103) kJ / (-5644) kJ ⋅mol -1
= 1.2 mol
m (C 12H 22O 11) = n (C 12H 22O 11) / M (C 12H 22O 11)
= 1.2 mol ⨯ 342.3 g ⋅mol -1
= 4.2 ⨯ 102 g
2-28 人体靠下列一系列反应去除体内酒精影响:
CH 3CH 2OH(l)−−→−2O CH 3CHO(l)−−→−2O CH 3COOH(l)−−→−2O CO 2(g)
已知∆f H m (CH 3CHO, g) = -166.4 kJ ⋅mol -1,计算人体去除1 mol C 2H 5OH(l)时各步反应的∆r H m 及总反应的∆r H m (假设T = 298.15 K)。

解: CH 3CH 2OH(l) + 1/2O 2(g) → CH 3CHO(l) + H 2O(l)
∆r H m (1) = [-285.830 -166.4 + 277.69] kJ ⋅mol -1
= -174.5 kJ ⋅mol -1
CH3CHO(l) + 1/2O2(g) → CH3COOH(l)
∆r H m(2) = [-484.5 + 166.4] kJ⋅mol-1
= -318.1 kJ⋅mol-1
CH3COOH(l) + O2(g) → 2CO2 + 2H2O(l)
∆r H m(3) = [2 ⨯ (-285.830) + 2 ⨯ (-393.509) + 484.5] kJ⋅mol-1
= -874.2 kJ⋅mol-1
∆r H m(总) = ∆r H m(1) + ∆r H m(2) + ∆r H m(3)
= [-174.5 -318.1-874.2] kJ⋅mol-1
= -1366.8 kJ⋅mol-1
2-29 Calculate the values of ∆r H m, ∆r S m, ∆r G m and K at 298.15K for the reaction NH4HCO3(s) = NH3 (g) + H2O(g) + CO2(g) 。

Solution: ∆r H m=[- 46.11 - 241.818 -393.509 + 849.4 ] kJ⋅mol-1
= 168.0 kJ⋅mol-1
∆r S m =[192.45 + 188.825 + 213.74 - 120.9 ] J⋅mol-1⋅K-1
= 474.1J⋅mol-1⋅K-1
∆r G m=[-16.45 - 228.572 - 394.359 + 665.9] kJ⋅mol-1
= 26.5 kJ⋅mol-1
lg K = -∆r G m / (2.303RT)
= -26.5 kJ⋅mol-1 / (2.303 ⨯ 8.314 ⨯ 10-3 kJ⋅mol-1⋅K-1 ⨯ 298.15K)
= - 4.64
K = 2.3 ⨯ 10-5.
2-30 糖在人体中的新陈代谢过程如下:
C12H22O11(s) + 12O2(g) → 12CO2(g) + 11H2O(l)
若反应的吉布斯函数变∆r G m只有30%能转化为有用功,则一匙糖(~3.8g)在体温37℃时进行新陈代谢,可得多少有用功?(已知C12H22O11的∆f H m = -2222 kJ⋅mol-1,S m = 360.2 J⋅mol-1⋅K-1)
解:C12H22O11(s) + 12O2(g) → 12CO2(g) + 11H2O(l)
∆f H m / kJ⋅mol-1 -2222 0 -393.509 -285.830
S m / J⋅mol-1⋅K-1360.2 205.138 213.74 69.91
∆r H m = [11 ⨯ (-285.830) +12 ⨯ (-393.509) - (-2222)] kJ⋅mol-1
=-5644 kJ⋅mol-1
∆r S m = [11 ⨯ 69.91 + 12 ⨯ 213.74 - 12 ⨯ 205.138 - 360.2] J⋅mol-1⋅K-1
= 512.0 J⋅mol-1⋅K-1
∆r G m = ∆r H m -T∆r S m
= -5644 kJ⋅mol-1 - 310.15K ⨯ 512.0 ⨯ 10-3 kJ⋅mol-1⋅K-1
= -5803 kJ⋅mol-1
ξ= ∆n B / νB
=3.8 g / 342 g⋅mol-1
= 1.11 ⨯ 10-2 mol
W有用功= 30%∆r G = 30%∆r G m⋅ξ
= 30% ⨯ (-5803 kJ⋅mol-1) ⨯ 1.11⨯10-2 mol
= -19 kJ (负号表示系统对环境做功) 2-31 在2033 K和3000 K的温度条件下混和等摩尔的N2和O2,发生如下反应:
N2(g) + O2(g)2NO(g)
平衡混合物中NO 的体积百分数分别是0.80%和4.5%。

计算两种温度下反应的K ,并判断该反应是吸热反应还是放热反应。

解: 体积分数等于摩尔分数 V i / V = n i / n = x i p i = x i p
K = (p (NO) / p )2(p (O 2) / p )-1(p (N 2) / p )-1
K 2033K = 0.00802 ⨯ [(1 - 0.0080) / 2]-2
= 2.6 ⨯ 10-4
K 3000K = 0.0452 ⨯ [(1 - 0.045) / 2]-2
= 8.9⨯10-3
T 升高,K 增大,该反应为吸热反应。

2-32 14C 的半衰期为5730y(y :年的时间单位)。

考古测定某古墓木质样品的14C 含量为原来的63.8%。

问此古墓距今已有多少年?
解:放射性同位素的衰变为一级反应
ln(c B / c 0) = -k ⋅t
ln50% = - k ⨯ 5730 y
k = 1.210 ⨯ 10-4 y -1
ln63.8% = -⋅1.210 ⨯ 10-4 y -1⋅t
t = 3714 y
2-33 在301 K 时鲜牛奶大约4.0 h 变酸,但在278 K 的冰箱中可保持48 h 时。

假定反应速率与变酸时间成反比,求牛奶变酸反应的活化能。

解: ln 111a K 27813011K mol J 314.848
10.41
---⎪⎭
⎫ ⎝⎛-⋅⋅-=E E a = 7.5 ⨯ 104 J ⋅mol -1 = 75 kJ ⋅mol -1
2-34 已知青霉素G 的分解反应为一级反应,37 ℃时其活化能为84.8 kJ ⋅mol -1,指前因子A 为4.2 ⨯ 1012 h -1,求37℃时青霉素G 分解反应的速率常数?
解: RT E a
e A k -⋅=
= 4.2 ⨯ 1012 h -1 ⨯ K )3715.273(K m ol kJ 10314.8m ol kJ 8.841131
+⨯⋅⋅⨯⋅-----e
= 4.2 ⨯ 1012 h -1 ⨯ 5.2 ⨯ 10-15 = 2.2 ⨯ 10-2 h -1
2-35 某病人发烧至40℃时,使体内某一酶催化反应的速率常数增大为正常体温(37℃)的1.25倍,求该酶催化反应的活化能?
解: ⎪⎭
⎫ ⎝⎛-⨯⋅⋅⨯-=---K 3131K 3101K mol kJ 108.31425.11ln 113a E E a = 60.0 kJ ⋅mol -1
2-36 某二级反应,其在不同温度下的反应速率常数如下:
T / K 645 675 715 750
k ⨯103/mol -1 L ⋅min -1 6.15 22.0 77.5 250
(1) 作ln k ~1/T 图计算反应活化能E a ;
(2) 计算700 K 时的反应速率常数k 。

解:(1) 1 / T 1.55⨯10-3 1.48⨯10-3 1.40⨯10-3 1.33⨯10-3
ln k -5.09 -3.82 -2.56 -1.37
用Excel 作图:
ln k ~1/T 图
ln k = -16774(1/T) + 20.944R 2 = 0.9993-6
-5
-4
-3
-2
-1
0.00130.001350.00140.001450.00150.001550.0016
1/T l n k
从图中直线方程得: -E a /R = -16.774 K
E a = 8.314J ⋅mol -1⋅K -1 ⨯16774K
= 1.39⨯105 J ⋅mol -1 = 139 kJ ⋅mol -1
(2) ln k = -16774K(1/T ) + 20.944
= -16774 / 700 + 20.944 = -3.02
k = 4.89 ⨯ 10-2
2-37 It is difficult to prepare many compounds directly from the elements, so ∆f H m values for these compounds cannot be measured directly. For many organic compounds, it is easier to measure the standard enthalpy of combustion ∆c H m by reaction of the compounds with excess O 2(g) to form CO 2(g) and H 2O(l). From the following standard enthalpies of combustion at 298.15K, determine ∆f H m for the compound.
(1) cyclohexane, C 6H 12(l), a useful organic solvent: ∆c H m = -3920kJ ⋅mol -1
(2) phenol, C 6H 5OH(s), used as a disinfectant and in the production of thermo-setting plastics : ∆c H m = -3053kJ ⋅mol -1
Solution: (1) C 6H 12(l) + 9O 2(g) = 6CO 2(g) + 6H 2O(l)
∆ r
H m = ∆c H m =∑B
νB ∆f H m (B) -3920 kJ ⋅mol -1=[6⨯(-393.509) + 6 ⨯ (-285.830) -∆f H m (C 6H 12(l))] kJ ⋅mol -1
∆f H m (C 6H 12(l)) =156 kJ ⋅mol -1
(2) C 6H 5OH(s) +O 2(g) = 6CO 2(g)+3H 2O(l)
∆ r
H m = ∆c H m =∑B
νB ∆f H m (B) -3053kJ ⋅mol -1=[6⨯(- 393.509)+ 3⨯(-285.830) -∆f H m (C 6H 5OH(s))]kJ ⋅mol -1
∆f H m (C 6H 5OH(s) = -166 kJ ⋅mol -1
2-38 Tb(铽)的同位素Tb 16165的半衰期2
1t = 6.9 d ,求10 d 后该同位素样品所剩百分数。

解:同位素的衰变为一级反应
t 1/2 = 0.693 / k
k = 0.693 / 6.9 d = 0.10 d -1
ln c B / c 0 = -k t
= -0.10 d -1 ⨯ 10 d = 1.0
c B / c 0 = 0.37 即还剩37%。

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