基础化学第六章后习题解答

1、第六章后习题解答 习题1.略2. 计算以下系统内能的变化：(1)系统放出2.5kJ的热量，并且对环境作功500J。(2)系统放出650J的热量，环境对系统作功350J。解 (1) U = Q+W = -2.5kJ+(-50010-3kJ)=-3.0kJ(2) U=-650J+350J=-300J3.略4. 已知反应：A+B=C+D =-40.0kJmol-1C+D=E =60.0kJmol-1求以下各反应的。(1)C+D=A+B(2)2C+2D=2A+2B(3)A+B=E解 (1)C+D=A+B =- =40.0kJmol-1(2)2C+2D=2A+2B =2 (-)=80.0kJmol-1(

2、3)A+B=E =+=-40.0 kJmol-1+60.0 kJmol-1=20.0 kJmol-15.在一定温度下，4.0molH2(g)与2.0molO2(g)混合，经一定时间反应后，生成了0.6molH2O(g)，请按以下两个不同反应式计算反应进度。(1)2H2(g)+O2(g)=2H2O(g)(2)H2(g)+O2(g)=H2O(g)解 t=0 =0 n(H2)=4.0mol n(O2)=2.0mol n(H2O)=0 t=t =t n(H2)=3.4mol n(O2)=1.7mol n(H2O)=0.6mol按(1)式： 2H2(g)+O2(g)=2H2O(g)= = =0.30mo

3、l= = =0.30mol= = =0.30mol按(2)式： H2(g)+O2(g)= H2O(g)= = =0.60mol= = =0.60mol= =0.60mol6.略7. 已知以下反应的标准反应热：(1)C6H6(l)+7O2(g)=6CO2(g)+3H2O(l) =-3267.6kJmol-1(2)C(gra)+O2(g)=CO2(g) =-393.5kJmol-1(3)H2(g)+O2(g)=H2O(l) =-285.8kJmol-1求下述不直接发生反应的标准反应热：解 6C(gra)+3H2(g)=C6H6(l) =？由62+331得所求的反应 6C(g)+3H2 (g)=C6

4、H6(l)=63=6(-393.5) kJmol-13(-285.8) kJmol-1(-3267.6)kJmol-1= 49.2kJmol-18. 肼N2H4(l)是火箭的燃料，N2O4作氧化剂，其燃烧反应的产物为N2(g)和H2O(l)，假设(N2H4，l)=50.63kJmol-1，(N2O4，g)=9.16kJmol-1写出燃烧反应，并计算此反应的反应热。 解 2N2H4(l)+N2O4(g)=3N2(g)+4H2O(l)=0+4(-285.8 kJmol-1)-2(50.63 kJmol-1)+9.16 kJmol-1-1254 kJmol-19已知以下反应在298.15K,标准状态

5、下：(1) Fe2O3(s)+ 3CO(g) 2Fe(s) + 3CO2(g)； rH=-24.8kJmol-1，rG=- 29.4kJmol-1(2) 3 Fe2O3(s)+CO(g) 2Fe3O4(s) + CO2(g)；rH=-47.2kJmol-1，rG=-61.41kJmol-1(3)Fe3O4(s) + CO(g)3FeO(s)+CO2(g)；rH=19.4kJmol-1， rG=5.21kJmol-1试求(4) FeO(s) + CO(g) Fe(s)+ CO2(g)的r H、 rG和rS。解 反应式4可由反应式1、2和3组合求出：-在298.15K,标准状态下，反应： FeO(

6、s) + CO(g) Fe(s)+CO2(g)的r H、 rG和rS分别为 r H=-rH+- =-19.4kJmol-1 + =-11.0 kJmol-1 =- =-6.20 kJmol-1 =-6.20JK-1 mol-110. 甲醇的分解反应为：CH3OH(l)CH4(g)+O2(g)(1)在298.15K的标准状态下此反应能否自发进行？(2)在标准态下此反应的温度应高于多少才能自发进行？解 (1) =-74.6 kJmol-1+0- (-239.2 kJmol-1)=164.6kJmol-1=186.3JK-1mol-1+205.2 JK-1mol-1-126.8 JK-1mol-1=

7、162.1JK-1mol-1=-T=164.6kJmol-1- 298.15K162.1JK-1mol-1=116.3 kJmol-10或=-50.5kJmol-1+0- (-166.6 kJmol-1)=116.3kJmol-1025和标准状态下反应不能自发进行。2T =1015.42K(742.37)11. 试计算298.15K，标准态下的反应：H2O(g)+CO(g)=H2(g)+CO2(g)的和，并计算298.15K时H2O(g)的。解 = 0+(-393.5kJmol-1)+(-241.8kJmol-1)-(-110.5 kJmol-1)=-41.2kJmol-1= 0+(-394.

8、4kJmol-1)+(-228.6kJmol-1)-(-137.2 kJmol-1)=-28.6kJmol-1=-42.26 JK-1mol-1=+-=130.7 JK-1mol-1213.8 JK-1mol-1197.77 JK-1mol-1-42.26 JK-1mol-1=189.1 JK-1mol-112. 计算以下反应在298.15K标准态下的，判断自发进行的方向，求出标准平衡常数。(可用于汽车尾气的无害化)解 1= -228.6 kJmol-100=-228.6 kJmol-10正向反应可以自发进行。=1.110402= 287.6kJmol-100=175.2kJmol-1=2.0

9、410-30(3) = 124.5 kJmol-13209.9 kJmol-1=-505.2kJmol-1=3.131088(4) = (-394.4 kJmol-1+0)(-137.2 kJmol-1+87.6 kJmol-1)=-344.8 kJmol-10=2.51060(5) = 2(-394.4kJmol-1)+2(-174.8 kJmol-1)(-910.6 kJmol-1)=-227.8 kJmol-1=8.01103913. 某病人平均每天需要6300kJ能量以维持生命。假设每天只能吃250g牛奶(燃烧值为3.0kJ.g-1)和50g面包(燃烧值为12kJ.g-1)，问每天还需

10、给他输入多少升质量浓度为50.0gL-1的葡萄糖(燃烧值为15.6kJg-1)溶液？解 设每天需输入50.0gL-1葡萄糖的体积为V3.0 kJg-1250g+12 kJg-150g+50gL-1V15.6 kJg-1=6300kJV=6.31L14. 糖代谢的总反应为：C12H22O11(s) + 12O2(g) = 12CO2(g)+ 11H2O(l)(1)从附表的热力学数据求298.15 K，标准态下的和。(2)如果在体内只有30%的自由能变转化为非体积功，求在37下，1.00molC12H22O11(s)进行代谢时可以得到多少非体积功。解 (1) =-393.5 kJmol-112+

11、(-285.8 kJmol-1)11-012-(-2226.1 kJmol-1)-5640 kJmol-1= 213.8JK-1mol-112+70.0 JK-1mol-111-205.2 JK-1mol-112-360.2 JK-1mol-1= 513JK-1mol-1=-T =-5640kJmol-1-298.15K51310-3kJK-1mol-1=-5793 kJ mol-1或 = =-394.4 kJ mol-112+(-237.1 kJ mol-1)11-012-(-1544.6 kJmol-1)=-5796 kJ mol-1(2) ，310.5 =-T =-5640kJmol-1

12、-310.5K51310-3kJK-1mol-1=-5799.11kJ mol-1Wf =(310.5K)30%=-5799.11kJ mol-130%=-1740 kJ mol-115在823K，标准态下以下反应的：(1)CO2(g)+H2(g)=CO(g)+H2O(g) (2)CoO(s)+H2(g)=Co(s)+H2O(g) 试求823K，标准态下反应(3)的：(3)CoO(s)+CO(g)=Co(s)+CO2(g)并求反应(2)和(3)的，比较CO(g)和H2(g)对CoO(s)的复原能力谁更强些。解 (2)- (1)为(3) 反应(2) =-28.78 kJ mol-1反应(3) =

13、-42.25 kJ mol-1反应3的自由能比2更负，所以CO(g)对CoO的复原能力大于H2(g)对CoO的复原能力。16. 在某细胞内ADP和H3PO4浓度分别为3.0mmolL-1和1.0mmolL-1。ATP的水解反应为：在310.15时，=-31.05kJmol-1,试求ATP在细胞内的平衡浓度；如实际上ATP的浓度是10mmolL-1,求反应的。 解 ADP与H3PO4浓度即可以看成是平衡浓度，也可以看成任意时刻浓度。 =-51.97 kJ mol-117. 欲用MnO2与HCl溶液反应制备Cl2(g)，已知该反应的方程式为：(1)写出此反应的标准平衡常数的表达式。(2)根据附表中

14、的热力学数据，求出298.15K标准状态下此反应的及值，并指出此反应能否自发进行？(3)假设HCl溶液浓度为12.0molL-1,其它物质仍 为标准态，反应在298K时能否自发进行？解 (1) (2) = (-237.1 kJmol-1)2+0+(-228.11)-(-465.1 kJmol-1)+40+(-131.2 kJmol-1)2= 25.2 kJmol-10在298.15K时标态下不能自发进行。(3) = 25.2 kJmol-1+ 8.314JK-1mol-1298K=-11.77 kJmol-1 0 此反应在298.15K、标准状态下不能自发进行。 (2) 由于此过程的 0，0，

15、升高温度有利于反应自发进行，T即T1015K时，此过程自发进行。Exercises1. (omit)2. (omit)3. Phosgene, COCl2, was used as a war gas during World War 1. It reacts with the moisture in the lungs to produce HCl, which causes the lungs to fill fluid, and CO2. Write an equation of the reaction and compute . For COCl2(g), (g)=-210 kJmo

16、l-1Solution COCl2(g) + H2O(l) 2HCl(g) + CO2(g)=2(-95.3kJmol-1) + (-394.4kJmol-1)- -210kJmol-1+(-237.1 kJmol-1)=-137.9kJmol-14. Given the following,4NO(g) 2 N2O(g) + O2(g) = -139.56 kJmol-12 NO(g) + O2(g) 2 NO2(g) = -69.70kJmol-1calculate for the reaction 2 N2O(g) + 3O2(g) 4 NO2(g).Solution =0.16 kJm

17、ol-15. Chloroform, formerly used as aneshetic and now belived to be a carcinogen (cancer- causing agent), has a heat of vaporization =31.4 kJmol-1 . The change, CHCl3(l) CHCl3(g), has =94.2 JK-1mol-1. At what temperature do we expect CHCl3 to boil? .Solution 6. A reaction that can convert coal to me

18、thane (the cihef component of natural gas) is C(s) +2 H2(g) = CH4(g)for which = -50.79 kJmol-1 . What is the value of K for the reaction at 25 0C ? Does this value of K suggest that studying this reaction as a mean of methane production is worth pursuing? Solution According to the result, this react

19、ion as a mean of methane production is worth pursuing .7Triglyceride is one of typical fatty acids, its metabolic reaction is :C57H104O6(s) + 80O2(g) = 57CO2(g) +52H2O(l), =-3.3510-4kJ/mol.Calculate the of Triglyceride.Solution The necessary data can be found from table in the addenda of the text bo

20、ok: (CO2 g)=-393.5 kJ/mol (H2O,l)=-285.8 kJ/mol (O2 g)=0So =57 (CO2 g)+52 (H2O,l)- (C57H104O6)-80 (O2 g) (C57H104O6)=57(-393.5)+52(-285.8)-(-3.3510-4)=-3.79103 kJ/mol8predict the sign of the entropy change for each of the following reactions:(a) O2(g)(100kPa,298K)O2(g) (10kPa,298K)(b) NH4Cl(s)NH3(g)+HCl(g)(c)CO(g)+H2O(g)CO2(g)+H2(g)Solution (a) S0; (b) S0; (c) For no change take place in the number of moles of gas, the entropy change