第一章习题解 1-13

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 ? 10/ (2.303 ? 8.314 ? 298.15)

= ?0.268

K= 0.540

?

3

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? 298.15 K ? 131.8 ? 10 kJ?mol?K

= ?250.2 kJ?mol?1

lgK?= ??rG?m / (2.303RT)

= 250.2 ? 10/ (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 3

lgK?= 260.1 ? 10/ (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/( 66.3 ? 10kJ?mol?K) = 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 ? 10和1.1 ? 10，求该反应的?rH?m。

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1-14 第一章习题解

?

ΔrHK1

ln???RK2

?

?1?

m ?1解：

ln1.0?101.1?10521?????T??1T2?

??ΔrHm8.314?10?3kJ?mol11?????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?Ls，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？ 解： (1) lncBc0??kt

?1?1

ln(0.20 /1.0) = ?k?30 min

?1

k = 0.054 min；

(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)

第一章习题解 1-15

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.

?1

Solution: ?rH?m = [? 46.11? 92.307 + 314.43] kJ?mol

= 176.01 kJ?mol?1

?1?1

?rS?m = [192.45 + 186.908 ? 94.6] J?mol?K

= 284.8 J?mol?K T = ?rH?m /?rS?m

=176.01 kJ?mol/ 284.758 ? 10kJ?mol?K= 618.0 K

2-24Calculate ?rG?m at 298.15K for the reaction 2NO2(g)→N2O4(g). Is this reaction spontaneous?

?1

Solution: ?rG?m = [97.89 ?2 ? 51.31] kJ?mol

= ? 4.73 kJ?mol< 0

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 ?1?4?1

kJ?mol. The value of k at 322℃ is determined to be 6.76?10s.

(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)

ln6.76?10k2?4?1?1

?1 ?3 ?1?1

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s?1??186?10J?mol8.314J?mol?13?1?1?K11?????

322?273.15298.15???20 ?1

k2 = 3.70?10

(2) ln6.76?106.00?10?4?23s

?1?1??186?10J?mol8.314J?mol?1?K11???? ??322?273.15T?T = 676 K

提高题

1-16 第一章习题解

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 ? 10? 9.8 J = ?2.1 ? 10kJ

?rH? = ?2.1 ? 103 kJ / 30%

= ?7.0 ? 103 kJ ?1?1?1

? r H?m = 11 ? (?285.830 kJ?mol) + 12 ? (?393.509 kJ?mol) ? (?2222 kJ?mol)

= ?5644 kJ?mol

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? = ?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：

O2O22CH3CH2OH(l)?O???CH3CHO(l)????CH3COOH(l)????CO2(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

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)

?1

?rH?m(3) = [2 ? (?285.830) + 2 ? (?393.509) + 484.5] kJ?mol

= ?874.2 kJ?mol?1

?rH?m(总) = ?rH?m(1) + ?rH?m(2) + ?rH?m(3)

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