无机及分析化学答案（第二版）第二章

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 ? = ?nB / ?B

= 3.8 g / 342 g?mol?1 = 1.11 ? 10?2 mol

W有用功 = 30%? rG? = 30%? rG?m??

= 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。

1Ea1??1?1解： ln4.0????? K ?1?118.314 J?mol?K?301278?48Ea = 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?EaRT

?84.8kJ?mol?1?3?1?1= 4.2 ? 1012 h?1 ? e8.314?10kJ?mol?K?(273.15?37)K = 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：

lnk～1/T图0-1-2-3-4-5-60.00130.001350.0014lnk = -16774(1/T) + 20.9442R = 0.9993lnk0.001450.00150.001550.00161/T 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 =??B?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 =??B?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t1= 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%