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:
C6H6(l) + 15O2(g) → 6CO2(g) + 3H2O(l) 2
在25℃100kPa下,0.25mol苯在氧气中完全燃烧放出817kJ的热量,求C6H6的标准摩尔燃烧焓∆cH m和该燃烧反应的∆rU m。
解: ξ = νB-1∆nB = (-0.25 mol) / ( -1) = 0.25 mol
∆cH m = ∆rH m = ∆rH= -817 kJ / 0.25 mol
= -3268 kJ⋅mol-1
∆rU m = ∆rH m - ∆ngRT
= -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∆rH m。
(1) Fe3O4(s) + 4H2(g) → 3Fe(s) + 4H2O(g)
(2) 2NaOH(s) + CO2(g) → Na2CO3(s) + H2O(l)
(3) 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)
(4) CH3COOH(l) + 2O2(g) → 2CO2(g) + 2H2O(l)
解: (1) ∆rH m = [4 ⨯ (-241.818) - (-1118.4)] kJ⋅mol-1
= 151.1 kJ⋅mol-1
(2) ∆rH m = [(-285.830) + (-1130.68) - (-393.509) - 2 ⨯ (-425.609)] kJ⋅mol-1
= -171.78 kJ⋅mol-1
(3) ∆rH m = [6 ⨯ (-241.818) + 4 ⨯ 90.25 - 4 ⨯ (-46.11)] kJ⋅mol-1
= -905.5 kJ⋅mol-1
(4) ∆rH 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(C2H2,g)的标准摩尔生成焓∆ fH m。
(1) C2H2(g) + 5/2O2(g) → 2CO2(g) + H2O(g) ∆ r H m= -1246.2 kJ⋅mol-1
(2) C(s) + 2H2O(g) → CO2(g) + 2H2(g) ∆ r H m = +90.9 kJ⋅mol-1
(3) 2H2O(g) → 2H2(g) + O2(g) ∆ r H m = +483.6 kJ⋅mol-1
解:反应2 ⨯ (2) - (1) - 2.5 ⨯ (3)为:
2C(s) + H2(g) → C2H2(g)
∆ f H m (C2H2) = 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)+Cu2+(aq)→Fe2+(aq)+Cu(s)
(2) AgCl(s)+Br-(aq)→AgBr(s)+Cl-(aq)
(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∆rH m,∆rS m和∆rG 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
(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
(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
(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
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: ∆rG m(1) = [2 ⨯ (-228.575) - (-466.14)] kJ⋅mol-1
= 8.99 kJ⋅mol-1
∆rG 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:
∆rH m(1) = [2 ⨯ (-241.818) - (-520.03)] kJ⋅mol-1
= 36.39 kJ⋅mol-1
∆rS m(1) = [2 ⨯ 188.825 + 32.01 - 2 ⨯ 130.684 - 53.05] J⋅mol-1⋅K-1
= 95.24 J⋅mol-1⋅K-1
∆rH m(1) - T1∆rS m(1) = 0
T1 = 36.39 kJ⋅mol-1 / (95.24 ⨯ 10-3 kJ⋅mol-1⋅K-1)
= 382.1K
∆rH m(2) = [2 ⨯ (-110.525) - (-520.03)] kJ⋅mol-1
= 298.98 kJ⋅mol-1
∆rS m(2) = [2 ⨯ 197.674 + 32.01 - 2 ⨯ 5.740 - 53.05] J⋅mol-1⋅K-1
= 362.28 J⋅mol-1⋅K-1
∆rH m(2) - T1∆rS m(2) = 0
T2 = 298.98 kJ⋅mol-1 / (362.28 ⨯ 10-3 kJ⋅mol-1⋅K-1)
= 825.27 K
T1
(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
2-8 计算25℃100kPa下反应CaCO3(s) →CaO(s)+CO2(g)的∆rH 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: ∆rH m = [-393.509 - 635.09 + 1206.92] kJ⋅mol-1
= 178.32 kJ⋅mol-1
∆rS m = [213.74 + 39.75 - 92.9] J⋅mol-1⋅K-1
= 160.6 J⋅mol-1⋅K-1
(1) ∆rG 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) ∆rH m > 0,∆ r S m > 0,升高温度对反应有利,有利于∆rG m
(3) 自发反应的条件为 T > ∆rH 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)
B2Mn2+(aq) + 5SO42-(aq) + 3H2O(l) 解:(1) K =∏(pB/ p )νB= 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) + H2O(g) CO2(g) + H2(g) 在750K时其K = 2.6,求:
(1) 当原料气中H2O(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H2O(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) + H2O(g) CO2(g) + H2(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 1-xp总 1-xp总 xp总 xp总 2222
K = (p(H2) / p )⋅(p(CO2) / p )⋅(p(H2O) / p )-1⋅(p(CO) / p )-1 2.6 = (x)2⋅(1-x)-2 22
x = 0.62
α(CO) = 62%
(2) V、T不变 CO(g) + H2O(g) CO2(g) + H2(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 1-x4-xxxp总 p总 p总 p总 2.6 = (x / 5)2⋅ [(1 - x) / 5]-1⋅[(4 - x) / 5]-1
x = 0.90
α(CO) = 90%
H2O(g)浓度增大,CO(g)转化率增大,利用廉价的H2O(g),使CO(g)反应完全。
2-14 在317K,反应 N2O4(g) 2NO2(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N2O4(g)的平衡转化率,并解释计算结果。
解:总压为400 kPa时 N2O4(g) 2NO2(g)
起始n / mol 1 0
平衡n / mol 1-x 2x
平衡相对分压 1-x4002x400 ⨯⨯1+x1001+x100
⎛8.00x⎫⎛4.00(1-x)⎫ ⎪⋅ ⎪⎝1+x⎭⎝1+x⎭2-1=1.00
x = 0.243
α(N2O4) = 24.3%
⎛16.0x⎫⎛8.00(1-x)⎫总压为800kPa时 ⎪⋅ ⎪1+x1+x⎝⎭⎝⎭2-1=1.00
x = 0.174
α(N2O4) = 17.4%
增大压力,平衡向气体分子数减少的方向移动,α(N2O4)下降。
2-15 已知尿素CO(NH2)2的∆fG 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 。
2NH3(g) + CO2(g) H2O(g) + CO(NH2)2(s)
解: ∆rG m = [-197.15 - 228.575 + 394.359 + 2 ⨯ 16.45] kJ⋅mol-1
= 1.53 kJ⋅mol-1
lgK = -∆rG m / (2.303RT)
= -1.53 ⨯ 103 / (2.303 ⨯ 8.314 ⨯ 298.15)
= -0.268
K = 0.540
2-16 25℃时,反应2H2O2(g)2H2O(g) + O2(g)的∆rH m为 -210.9 kJ⋅mol-1,∆rS 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: ∆rG m = ∆rH m - T∆rS m
∆rG 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
lgK = -∆rG m / (2.303RT)
= 250.2 ⨯ 103 / (2.303 ⨯ 8.314 ⨯ 298.15)
= 43.83
K 298.15K = 6.7 ⨯ 1043
∆rG 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
lgK = 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Ag2O的分解反应为 Ag2O(s) 2Ag(s) + 1/2O2(g)。假定反应的∆rH m,∆rS m不随温度的变化而改变,估算Ag2O在标准状态的最低分解温度?
解: ∆rH m = -∆fH m(Ag2O) = 31.05 kJ⋅mol-1
∆rS m = [2 ⨯ 42.5 + 205.138 / 2 - 121.3] J⋅mol-1⋅K-1
= 66.3 J⋅mol-1⋅K-1
T = ∆rH m /∆rS m
= 31.05 kJ⋅mol-1 /( 66.3 ⨯ 10-3 kJ⋅mol-1⋅K-1)
= 468 K
2-18 已知反应 2SO2(g) + O2(g) → 2SO3(g) 在427℃和527℃时的K 值分别为1.0 ⨯ 105和1.1 ⨯ 102,求该反应的∆rH m。
K1 ΔrHm ⎛11⎫ ln =--⎪ ⎪ RT⎝1T2⎭ 解: K2
ΔrH 1.0⨯10511⎛⎫mln=-- ⎪1.1⨯1028.314⨯10-3kJ⋅mol-1⎝427+273527+273⎭
∆rH m = -3.2 ⨯ 102 kJ⋅mol-1
2-19 已知反应 2H2(g) + 2NO(g) → 2H2O(g) + N2(g) 的速率方程 v = k c(H2)⋅c2(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?
解: 体积缩小为1/ 2,浓度增大2倍:
v2 = k 2c1(H2)⋅(2c1)2(NO)
= 8 k⋅ c1(H2)⋅(c1)2(NO)
= 8v1
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 = kcAcB
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?
c解: (1) lnB=-kt c0
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) + Cl2(g) = MgCl2(s)
(2) Mg(s) + I2(s) = MgI2(s)
(3) C(s) + O2(g) = CO2(g)
(4 Al2O3(s) + 3C(s) + 3Cl2(g) = 2AlCl3(g) + 3CO(g)
(5) 2NO(g) + Cl2(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 (Td) for the reaction NH4Cl(s).= NH3(g) + HCl(g) at the standard state.
Solution: ∆rH m = [- 46.11- 92.307 + 314.43] kJ⋅mol-1
= 176.01 kJ⋅mol-1
∆rS m = [192.45 + 186.908 - 94.6] J⋅mol-1⋅K-1
= 284.8 J⋅mol-1⋅K-1
T = ∆rH m /∆rS m
=176.01 kJ⋅mol-1 / 284.758 ⨯ 10-3 kJ⋅mol-1⋅K-1
= 618.0 K
2-24Calculate ∆rG m at 298.15K for the reaction 2NO2(g)→N2O4(g). Is this reaction spontaneous?
Solution: ∆rG m = [97.89 -2 ⨯ 51.31] kJ⋅mol-1
= - 4.73 kJ⋅mol-1
The reaction is spontaneous.
2-25 The following gas phase reaction follows first-order kinetics:
FClO2(g) → FClO(g) + O(g)
The activation energy of this 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?
6.76⨯10-4s-1186⨯103J⋅mol-1⎛11⎫=--Solution: (1) ln ⎪ k28.314J⋅mol-1⋅K-1⎝322+273.15298.15⎭
k2 = 3.70⨯10-20 s-1 6.76⨯10-4186⨯103J⋅mol-1⎛11⎫=--(2) ln ⎪ 6.00⨯10-28.314J⋅mol-1⋅K-1⎝322+273.15T⎭
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 m3
= 25 kJ - 8.104 kJ
= 17 kJ
2-27 蔗糖(C12H22O11)在人体内的代谢反应为:
C12H22O11(s) + 12O2(g) → 12CO2(g) + 11H2O(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?(∆fH m(C12H22O11) = -2222kJ⋅mol-1)
解: W = -70 kg ⨯ 3000 m
= -2.1 ⨯ 105 kg⋅m
= -2.1 ⨯ 105 ⨯ 9.8 J = -2.1 ⨯ 103 kJ
∆rH = -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 = ∆rH / ∆rH m
= (-7.0 ⨯ 103) kJ / (-5644) kJ⋅mol-1
= 1.2 mol
m(C12H22O11) = n(C12H22O11) / M(C12H22O11)
= 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:
222−→−−→−−→−CH3CH2OH(l)−OCH3CHO(l)−OCH3COOH(l)−OCO2(g)
已知∆fH m(CH3CHO, g) = -166.4 kJ⋅mol-1,计算人体去除1 mol C2H5OH(l)时各步反应的∆rH m及总反应的∆rH m(假设T = 298.15 K)。
解: CH3CH2OH(l) + 1/2O2(g) → CH3CHO(l) + H2O(l)
∆rH m(1) = [-285.830 -166.4 + 277.69] kJ⋅mol-1
= -174.5 kJ⋅mol-1
CH3CHO(l) + 1/2O2(g) → CH3COOH(l)
∆rH m(2) = [-484.5 + 166.4] kJ⋅mol-1
= -318.1 kJ⋅mol-1
CH3COOH(l) + O2(g) → 2CO2 + 2H2O(l)
∆rH m(3) = [2 ⨯ (-285.830) + 2 ⨯ (-393.509) + 484.5] kJ⋅mol-1
= -874.2 kJ⋅mol-1
∆rH m(总) = ∆rH m(1) + ∆rH m(2) + ∆rH m(3)
= [-174.5 -318.1-874.2] kJ⋅mol-1
= -1366.8 kJ⋅mol-1
2-29 Calculate the values of ∆rH m, ∆rS m, ∆rG m and K at 298.15K for the reaction NH4HCO3(s) = NH3 (g) + H2O(g) + CO2(g) 。
Solution: ∆rH m=[- 46.11 - 241.818 -393.509 + 849.4 ] kJ⋅mol-1
= 168.0 kJ⋅mol-1
∆rS m =[192.45 + 188.825 + 213.74 - 120.9 ] J⋅mol-1⋅K-1
= 474.1J⋅mol-1⋅K-1
∆rG m=[-16.45 - 228.572 - 394.359 + 665.9] kJ⋅mol-1
= 26.5 kJ⋅mol-1
lgK = -∆rG 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∆rG 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∆fH m = -2222 kJ⋅mol-1, S m = 360.2 J⋅mol-1⋅K-1)
解: C12H22O11(s) + 12O2(g) → 12CO2(g) + 11H2O(l)
∆fH m / kJ⋅mol-1 -2222 0 -393.509 -285.830
S m / J⋅mol-1⋅K-1 360.2 205.138 213.74 69.91
∆rH m = [11 ⨯ (-285.830) +12 ⨯ (-393.509) - ( -2222)] kJ⋅mol-1
= -5644 kJ⋅mol-1
∆rS m = [11 ⨯ 69.91 + 12 ⨯ 213.74 - 12 ⨯ 205.138 - 360.2] J⋅mol-1⋅K-1
= 512.0 J⋅mol-1⋅K-1
∆ rG m = ∆rH m - T∆rS m
= -5644 kJ⋅mol-1 - 310.15K ⨯ 512.0 ⨯ 10-3 kJ⋅mol-1⋅K-1
= -5803 kJ⋅mol-1
ξ = ∆nB / νB
= 3.8 g / 342 g⋅mol-1
= 1.11 ⨯ 10-2 mol
W有用功 = 30%∆ rG = 30%∆ rG 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 Vi / V = ni / n = xi pi = xi p
K = (p(NO) / p )2(p(O2) / p )-1(p(N2) / 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(cB / c0) = -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。
1
Ea1⎫-1⎛1解: ln=-- ⎪ K -1-118.314 J⋅mol⋅K⎝301278⎭
48
Ea = 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: k=A⋅e-Ea
RT
-84.8kJ⋅mol-1
8.314⨯10kJ⋅mol⋅K⨯(273.15+37)K= 4.2 ⨯ 10h⨯ 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?
Ea11⎫⎛1=-⨯-解: ln ⎪ -3-1-11.25310K313K8.314⨯10kJ⋅mol⋅K⎝⎭
Ea = 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lnk~1/T图计算反应活化能Ea;
(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
lnk -5.09 -3.82 -2.56 -1.37
用Excel 作图: 12 -1
从图中直线方程得: -Ea/R = -16.774 K
Ea = 8.314J⋅mol-1⋅K-1 ⨯16774K
= 1.39⨯105 J⋅mol-1 = 139 kJ⋅mol-1
(2) lnk = -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 ∆fH m values for these compounds cannot be measured directly. For many organic compounds, it is easier to measure the standard enthalpy of combustion ∆cH m by reaction of the compounds with excess O2(g) to form CO2(g) and H2O(l). From the following standard enthalpies of combustion at 298.15K, determine ∆fH m for the compound.
(1) cyclohexane, C6H12(l), a useful organic solvent: ∆cH m= -3920kJ⋅mol-1
(2) phenol, C6H5OH(s), used as a disinfectant and in the production of thermo-setting plastics : ∆cH m= -3053kJ⋅mol-1
Solution: (1) C6H12(l) + 9O2(g) = 6CO2(g) + 6H2O(l)
∆ rH m = ∆cH m =∑νB∆fH m(B)
B
-3920 kJ⋅mol-1=[6⨯(-393.509) + 6 ⨯ (-285.830) -∆fH m(C6H12(l))] kJ⋅mol-1
∆fH m(C6H12(l)) =156 kJ⋅mol-1
(2) C6H5OH(s) +O2(g) = 6CO2(g)+3H2O(l)
∆ rH m = ∆cH m =∑νB∆fH m(B)
B
-3053kJ⋅mol-1=[6⨯(- 393.509)+ 3⨯(-285.830) -∆fH m(C6H5OH(s))]kJ⋅mol-1
∆fH m(C6H5OH(s) = -166 kJ⋅mol-1
2-38 Tb(铽)的同位素16165Tb的半衰期t= 6.9 d,求10 d后该同位素样品所剩百分数。 2
解:同位素的衰变为一级反应
t1/2 = 0.693 / k
k = 0.693 / 6.9 d = 0.10 d-1
lncB / c0 = -k t
= -0.10 d-1 ⨯ 10 d = 1.0
cB / c0 = 0.37 即还剩37%