无机及分析化学宣贵达课后答案

21．25℃时，反应2H2O2(g)2H2O(g)+O2(g)的?rH?m为?210.9kJ?mol?1，?rS?m为

131.8J?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?298.15K?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?373.15K?131.8?10?3 kJ?mol?1?K?1

= ?260.1 kJ?mol?1

lgK? = ??rG?m/(2.303RT)= 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。 22．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的最低分解温度和在该温度下的p(O2)分压是多少？

解： ?rH?m= ??fH?m(Ag2O)= 31.05kJ?mol?1

?rS?m=[2?42.5+205.138/2?121.3]J?mol?1?K?1=66.27J?mol?1?K?1

T= ?rH?m/?rS?m=31.05kJ?mol?1/66.27?10?3kJ?mol?1?K?1= 468.5 K

此时，?rG?m=0 kJ?mol?1 ， K? =1 ， K? =(p(O2)/p?)1/2 ， p(O2)=100kPa 。

23．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，求该温度范围内反应的?rH?m。

? ΔrH?K11?m?1解： ln??????

R?T1T2?K2

?5ΔrH1.0?1011m?? ln?????2?3?1427?273.15527?273.15??1.1?108.314?10kJ?mol

?rH?m= ?3.2?102 kJ?mol?1

24．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 = k2c(H2)?(2c)2(NO) = 8v1

25．某基元反应 A + B? C，在1.20L溶液中，当A为4.0 mol，B为3.0mol时，v为

0.0042mol?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=0.0042mol?dm?3s?1/[(4.0 mol/1.20L)?(3.0mol)/1.20L]=5.0?10?4 mol?1?L?s?1

26．在301K时鲜牛奶大约4.0h变酸，但在278K的冰箱中可保持48h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

18.314J?mol?1?K?1?301278?48Ea=7.5?104J?mol?1=75kJ?mol?1

27．已知青霉素G的分解反应为一级反应，37℃时其活化能为84.8kJ?mol?1，指前因子A为

4.2?1012h?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?18.314?10?3kJ?mol?1?K?1?(273.15?37)K= 4.2?10h?e =4.2?1012h?1?5.2?10?15 =2.2?10?2 h?1

28．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： ln12?1

Ea1???1?1? ???1.258.314?10?3kJ?mol?1?K?1?310K313K?Ea=60.0 kJ?mol?1

30．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?=2.28?10?5.

31．From reactions (1)-(5) below, select, without any thermodynamic

calculations those reactions which have: (1) large negative standar entropy changes,(2) large positive standar entropy changes,(3) small entropy changes which might be either positive or negative. (1) Mg(s) + Cl2(g) = MgCl2(s) (4) Al2O3(s) + 3C(s) + 3Cl2(g) = 2AlCl3(g) + 3CO(g) (2) Mg(s) + I2(s) = MgI2(s) (5) 2NO(g) + Cl2(g) = 2NOCl(g) (3) C(s) + O2(g) = CO2(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)

32．Calculate the value of the thermodynamic decomposition temperature (Td)

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

=284.758J?mol?1?K?1 T = ?rH?m/?rS?m

=176.01 kJ?mol?1/284.758?10?3kJ?mol?1?K?1 =618.10K

33．Calculate ?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< 0

The reaction is spontaneous.

34．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?3920kJ?mol?1=[6?(? 393.509)+ 6?(?285.830) ??fH?m(C6H12(l))] kJ?mol?1

?fH?m(C6H12(l))=156kJ?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)= ?166kJ?mol?1

35．The following gas phase reaction follows first-order kinetics:

FClO2 ? FClO + O

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?20s?1

(2) ln6.76?10?4s?16.00?10?2186?103J?mol?1?11?????? ?1?1322?273.15T8.314J?mol?K??T=676K

第三章

1．已知分析天平能称准至±0.1 mg，滴定管能读准至±0.01 mL，若要求分析结果达到

0.1%的准确度，问至少应用分析天平称取多少克试样？滴定时所用溶液体积至少要

多少毫升？

解：称取试样时通常需称取两次，因此分析天平的称量误差为±0.2mg，为使分析结果

的相对误差达到0.1%，则至少应称取的试样质量m为：

?3?0.2?10g?0.1% =

mm=0.2 g

同样地，滴定管的读数误差为±0.02 mL，为使分析结果的相对误差达到0.1%，则滴定时所需的体积V至少为：

?0.02mL VV=20 mL

2．在NaOH的标定时，要求消耗0.1 mol?L?1NaOH溶液体积为20~30 mL，问：

(1)应称取邻苯二甲酸氢钾基准物质(KHC8H4O4)多少克？ (2)如果改用草酸(H2C2O4·2H2O)作基准物质，又该称多少克？

(3)若分析天平的称量误差为?0.0002g，试计算以上两种试剂称量的相对误差。 (4)计算结果说明了什么问题？

解：(1) NaOH + KHC8H4O4 = KNaC8H4O4 + H2O

滴定时消耗0.1 mol?L?1NaOH溶液体积为20 mL所需称取的KHC8H4O4量为：

m1=0.1 mol?L?1?20mL?10?3?204 g?mol?1=0.4g

滴定时消耗0.1 mol?L?1NaOH溶液体积为30 mL所需称取的KHC8H4O4量为：m2=0.1

mol?L?1?30mL?10?3?204 g?mol?1=0.6g

因此，应称取KHC8H4O4基准物质0.4~0.6g。 (2) 2NaOH + H2C2O4 = Na2C2O4 + 2H2O

滴定时消耗0.1 mol?L?1NaOH溶液体积为20和30 mL，则所需称的草酸基准物质的质量分别为：

?0.1% =

m1=m2=

1?0.1 mol?L?1?20mL?10-3?126 g?mol?1=0.1g 21?0.1 mol?L?1?30mL?10-3?126g?mol?1=0.2g 2(3) 若分析天平的称量误差为?0.0002g，则用邻苯二甲酸氢钾作基准物质时，其称量的相对误差为：

RE1=

?0.0002g0.4g= ?0.05%

RE2=

?0.0002g= ?0.03%

0.6g用草酸作基准物质时，其称量的相对误差为：

RE1=

?0.0002g0.1g= ?0.2%

RE2=

?0.0002g= ?0.1%

0.2g