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. 计算下列系统内能的变化:(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+(-500×10-3kJ=-3.0kJ(2 U =-650J+350J=-300J 3.(略 4. 已知反应:A+B=C+D m,1r H =-40.0kJ mol -1 C+D=E r m,2H =60.0kJ mol-1求下列各反应的r m H 。(1C+D=A+B (22C+2D=2A+2B (3A+B=E解 (1C+D=A+B m r H =-m,1r H

2、=40.0kJmol -1 (22C+2D=2A+2B m r H =2 ×(-m,1r H =80.0kJmol -1(3A+B=E m r H =m,1r H +m,2r H =-40.0 kJmol -1+60.0 kJmol -1=20.0 kJmol -15.在一定温度下,4.0molH 2(g与2.0molO 2(g混合,经一定时间反应后,生成了0.6molH 2O(g,请按下列两个不同反应式计算反应进度。(12H 2(g+O 2(g=2H 2O(g (2H 2(g+21O 2(g=H 2O(g 解 t =0 =0 n (H 2=4.0mol n (O 2=2.0mol

3、n (H 2O=0 t =t =t n (H 2=3.4mol n (O 2=1.7mol n (H 2O=0.6mol 按(1式: 2H 2(g+O 2(g=2H 2O(g=OH (O H (22n = 20mol 60.0-=0.30mol=H (H (22n =2mol0.4.4mol 3- =0.30mol=O (O (22n = 1mol0.2.7mol 1-=0.30mol按(2式: H 2(g+21O 2(g= H 2O(g =OH (O H (22n = 10mol 60.0-=0.60mol=H (H (22n =1mol0.4.4mol 3- =0.60mol=O (O (

4、22n =21mol0.2.7mol 1-=0.60mol6.(略7. 已知下列反应的标准反应热: (1C 6H 6(l+721O 2(g=6CO 2(g+3H 2O(l 1,m r H =-3267.6kJ mol -1 (2C(gra+O 2(g=CO 2(g 2,m r H =-393.5kJ mol -1 (3H 2(g+21O 2(g=H 2O(l 1,r m H =-285.8kJ mol -1 求下述不直接发生反应的标准反应热:解 6C(gra+3H 2(g=C 6H 6(l m r H =?由6×(2+3×(3-(1得所求的反应 6C(g+3H 2 (g=C

5、 6H 6(lmr H =62,m r H +33,m r H -1,m r H =6×(-393.5 kJmol -1+3×(-285.8 kJmol -1-(-3267.6kJmol -1 = 49.2kJmol -18. 肼N 2H 4(l是火箭的燃料,N 2O 4作氧化剂,其燃烧反应的产物为N 2(g和H 2O(l,若m f H (N 2H 4,l=50.63kJ mol -1,m f H (N 2O 4,g=9.16kJ mol -1写出燃烧反应,并计算此反应的反应热m r H 。解 2N 2H 4(l+N 2O 4(g=3N 2(g+4H 2O(lmr H =B

6、m f B (B H =0+4×(-285.8 kJmol -1-2×(50.63 kJmol -1+9.16 kJmol -1 -1254 kJmol -19.已知下列反应在298.15K,标准状态下:(1 Fe 2O 3(s+ 3CO(g 2Fe(s + 3CO 2(g; r H 1m =-24.8kJ·mol -1,r G 1m =- 29.4kJ·mol -1 (2 3 Fe 2O 3(s+CO(g 2Fe 3O 4(s + CO 2(g;r H 2m =-47.2kJ·mol -1,r G 2m =-61.41kJ·mol

7、-1 (3Fe 3O 4(s + CO(g3FeO(s+CO 2(g;r H 3m =19.4kJ·mol -1, r G 3m =5.21kJ·mol -1 试求(4 FeO(s + CO(g Fe(s+ CO 2(g的r H 4m 、 r G 4m 和r S 4m 。解 反应式(4可由反应式(1、(2和(3组合求出:-2(611(213(31-+ 在298.15K,标准状态下,反应: FeO(s + CO(g Fe(s+CO 2(g的r H 4m 、 r G 4m 和r S 4m 分别为r H 4m =-31×r H 3m +m1r 21H -m2r 61H

8、=-31×19.4kJ·mol -1 +11mol 47.2kJ (61mol kJ 8.24(21-=-11.0 kJ·mol -1m2r m1r m3r m4r 612131G G G G -+-= =-111mol 61.4kJ (61mol 29.4kJ (21mol 5.2kJ 31-+ =-6.20 kJ·mol -1 K15.298mol J 1020.6(mol J 1099.12(K 15.2981313m4r m4r m4r -=-=G H S=-6.20J·K -1· mol -1 10. 甲醇的分解反应为:CH

9、 3OH(lCH 4(g+21O 2(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 m r H =-74.6 kJmol -1+21×0- (-239.2 kJmol -1 =164.6kJmol -1mr S =186.3JK -1mol -1+21×205.2 JK -1mol -1-126.8 JK -1mol -1 =162.1JK -1mol -1mr G =m r H -T m r S =164.6kJmol -1- 298.15K×162.1JK -1mol -1 =116.3

10、 kJmol -1>0(或m r G =Bm f B(B G =-50.5kJmol -1+0- (-166.6 kJmol -1=116.3kJmol -1>025和标准状态下反应不能自发进行。(2T m,298.15K r m,298.15K r S H =1113mol K 162.1J mol J 10164.6-=1015.42K(742.37 11. 试计算298.15K ,标准态下的反应:H 2O(g+CO(g=H 2(g+CO 2(g的mr m r ,G H 和m r S ,并计算298.15K 时H 2O(g的m S 。解 m r H = 0+(-393.5kJm

11、ol -1+(-241.8kJmol -1-(-110.5 kJmol -1=-41.2kJmol -1mr G = 0+(-394.4kJmol -1+(-228.6kJmol -1-(-137.2 kJmol -1 =-28.6kJmol -1mr S =K 15.298m r m r G H -=298.15K mol J 1028.6(mol J 1041.21313-=-42.26 JK -1mol -1g O,H (2m S =g ,H (2m S +g ,CO (2m S -g CO,(m S -mr S =130.7 JK -1mol -1+213.8 JK -1mol -1-

12、197.77 JK -1mol -1 -(-42.26 JK -1mol -1 =189.1 JK -1mol -112. 计算下列反应在298.15K 标准态下的m r G ,判断自发进行的方向,求出标准平衡常数K 。(1 H 2(g + O 2H 2O(g1 2(2 N 2 (g + O 2(3 3C 2H 26H 6(l 2(g + N 2(g(5 C 6H 12O 62H 5OH(l + 2CO 2(g 解 (1m r G = -228.6 kJmol -1-0-21×0=-228.6 kJmol -1<0 正向反应可以自发进行。K =RTG em r -=298.15

13、K mol K 8.314J mol J 10228.6(1113-e =1.1×1040(2m r G = 2×87.6kJmol -1-0-0=175.2kJmol -1K =RTG emr -=298.15Kmol K 8.314J mol J 102.7511113-e=2.04×10-30(3 m r G = 124.5 kJmol -1-3×209.9 kJmol -1=-505.2kJmol -1K =RTG emr -=298.15Kmol K 8.314J mol J 102.055(13-e=3.13×1088(4 m r

14、G = (-394.4 kJmol -1+21×0-(-137.2 kJmol -1+87.6 kJmol -1 =-344.8 kJmol -1<0K =RTG em r -=298.15Kmol K 8.314J mol J 108.443(1113-e =2.5×1060(5 m r G = 2×(-394.4kJmol -1+2×(-174.8 kJmol -1-(-910.6 kJmol -1=-227.8 kJmol -1K =RTG emr -=298.15Kmol K 8.314J mol J 10227.8(1113-e=8.01

15、×103913. 某病人平均每天需要6300kJ 能量以维持生命。若每天只能吃250g 牛奶(燃烧值为3.0kJ.g -1和50g 面包(燃烧值为12kJ.g -1,问每天还需给他输入多少升质量浓度为50.0g·L -1的葡萄糖(燃烧值为15.6kJ·g -1溶液?解 设每天需输入50.0g·L -1葡萄糖的体积为V3.0 kJg -1×250g+12 kJg -1×50g+50gL -1×V ×15.6 kJ·g -1=6300kJV =6.31L14. 糖代谢的总反应为:(21C 12H 22O 11

16、(s + 12O 2(g = 12CO 2(g+ 11H 2O(l(1从附表的热力学数据求298.15 K ,标准态下的mr m r ,H G 和m r S 。 (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,1.00molC 12H 22O 11(s进行代谢时可以得到多少非体积功。解 (1 m r H =-393.5 kJmol -1×12+ (-285.8 kJmol -1×11-0×12-(-2226.1 kJmol -1 -5640 kJmol -1mr S = 213.8JK -1mol -1×12+70.0 JK -1mol -

17、1×11-205.2 JK -1mol -1×12 -360.2 JK -1mol -1 = 513JK -1mol -1mr G =m r H -T m r S =-5640kJmol -1-298.15K×513×10-3kJK -1mol -1=-5793 kJ mol -1或 m r G =Bm f B(B G =-394.4 kJ mol -1×12+(-237.1 kJ mol -1×11-0×12-(-1544.6 kJmol -1=-5796 kJ mol -1(2 m r G ,310.5 =m r H -

18、T m r S =-5640kJmol -1-310.5K×513×10-3kJK -1mol -1=-5799.11kJ mol -1W f =m r G (310.5K×30%=-5799.11kJ mol -1×30%=-1740 kJ mol -1 15.在823K ,标准态下下列反应的K : (1CO 2(g+H 2(g=CO(g+H 2O(g 14.01=K (2CoO(s+H 2(g=Co(s+H 2O(g 672=K试求823K ,标准态下反应(3的3K :(3CoO(s+CO(g=Co(s+CO 2(g并求反应(2和(3的r m,823

19、G ,比较CO(g和H 2(g对CoO(s的还原能力谁更强些。 解 (2- (1为(3 212CO CO 3108.447814.067/2=K K p p p p K 反应(2 ln67K 823mol K J 314.8ln 112m,823K r -=-=-K RT G=-28.78 kJ mol -1反应(3 10ln(4.8K 823mol K J 314.8ln 2113m,823K r -=-=-K RT G=-42.25 kJ mol -1反应(3的自由能比(2更负,所以CO(g对CoO 的还原能力大于H 2(g对CoO 的还原能力。 16. 在某细胞内ADP 和H 3PO 4

20、浓度分别为3.0mmol·L -1和1.0mmol·L -1。ATP 的水解反应为:ADP+H 3PO 4 在310.15时,r m G =-31.05kJ·mol-1,试求ATP 在细胞内的平衡浓度;如实际上A TP 的浓度是10mmol·L -1,求反应的r m G 。解 ADP 与H 3PO 4浓度即可以看成是平衡浓度,也可以看成任意时刻浓度。6K15.103mol K8.314J mol J 1005.13(107.11113mr =-e eK RTGAT P PO ADPH 43=K1-9643L m m o l 108.1107.110001

21、/10001(1/10003(PO ADPH ATP =-K /(/(/(lnATP ADP PO H m r m r 43c c c c c c RT G G +=1/100010(1/10003(1/10001(ln 15.310314.8mol kJ 05.311+-=-=-51.97 kJ mol -117. 欲用MnO 2与HCl 溶液反应制备Cl 2(g,已知该反应的方程式为:MnO 2(s+4H +(aq+2Cl -Mn 2+(aq+Cl 2(g+2H 2O(l (1写出此反应的标准平衡常数K 的表达式。(2根据附表中的热力学数据,求出298.15K 标准状态下此反应的m G r

22、 及K 值,并指出此反应能否自发进行?(3若HCl 溶液浓度为12.0mol·L -1,其它物质仍 为标准态,反应在298K 时能否自发进行? 解 (1 2Cl 4H Cl Mn /22c c c c p p c c K -+=(2 =Bm f Bm,298.15K r (B G G = (-237.1 kJmol -1×2+0+(-228.1×1-(-465.1 kJmol -1 +4×0+(-131.2 kJmol -1×2 = 25.2 kJmol -1>0在298.15K 时标态下不能自发进行。(3 Q RT G G ln m r

23、 m r += 25.2 kJmol -1+ 8.314JK -1mol -1×298K×241/12(1/12(100/100(1/1(ln=-11.77 kJmol -1<0此时上面的反应能自发进行。 18. 已知下列反应2SO 2(g+O 2(g=2SO 3(g在800K 时的K =910,试求900K 时此反应的K 。假设温度对此反应的m r H 的影响可以忽略。解 m,298.15K r H = (-395.7kJmol -1×2-0+(-296.8kJmol -1×2= -197.78kJmol -1900800800900(8.314

24、2.303910K 900(lg K 800(K 900(lg m,298.15K r -=H K K K900800100314.8303.2100078.197(910lg K900(lg -+=KK =33.4519. 试由298.15K 时下述反应的热力学数据求AgCl 的sp K 。(sp K K =Ag +(aq+Cl -(aq 解 m,298.15K r G = (77.1 kJmol -1 + (-131.2kJmol -1- (-109.8kJmol -1=55.7kJmol -1mr G =-RT ln (AgClln sp K RT K -=10sp 1075.1(AgC

25、l-=K20.用有关热力学函数计算Ag 2CO 3在298.15K 和373.15K 时的溶度积常数(假设r H m 、r S m 不随温度变化。解 Ag 2CO 3(s2Ag +(aq + CO +23(aq r G m / kJ·mol -1 -437.2 77.1 -527.8 r H m / kJ·mol -1 -505.8 105.6 -667.0 S m / J·K -1·mol -1 167.4 72.7 -56.8 298.15K 时,m r G = mfG(产物 -mfG(反应物=2×77.1 kJ·mol -1+(

26、-527.8 kJ·mol -1-(-437.2kJ·mol -1=63.6 kJ·mol -1 r G m =-2.303RT lg K sp (Ag 2CO 3,298.15Klg K sp (Ag 2CO 3,298.15K=RTG 303.2,298.15K CO Ag (32m r -=298.15Kmol K 8.314J 2.303mol J 1063.61113- =-11.1K sp (Ag 2CO 3,298.15K=7.94×10-12373.15K 时,m r H (298.15K=m f H (产物 - mfH(反应物=2

27、15;105.6 kJ·mol -1 +(-667.0 kJ·mol -1 (-505.8 kJ·mol -1 =50.0 kJ·mol -1m r S (298.15K=m S (产物 -m S (反应物= 2×72.7 J·mol -1+(-56.8 J·mol -1-167.4 J·mol -1 =-78.8 J·mol -1根据 T m,r G =T m,r H T m,r S m,298.15r H -T m,298.15r S m r G (373.15K = 50.0 kJ·mol

28、 -1-373.15×(-78.8×10-3 kJ·mol -1=79.4 kJ·mol -1lg K sp (Ag 2CO 3,373.15K=RTG 303.2K 15.373,CO Ag (32m r -=373.15Kmol K kJ 108.3142.303mol 79.4kJ 1131-=-11.11 K sp (Ag 2CO 3,373.15K=7.76×10-1221.甲醇的分解反应为:CH 3OH(l CH 4(g +g (O 212,利用有关热力学数据求算在298.15K 、标准状态下的mr G ,并判断反应能否自发进行。如

29、不能自发进行,温度至少要升高到多少(K 反应才能自发进行?解 (1 m r H =m f H (产物 -m f H (反应物=f H m (CH 4,g+21×f H m (O 2,g-f H m (CH 3OH,l =(-74.6kJ·mol -1 +21×0 kJ·mol -1 (-239.2 kJ·mol -1=164.6 kJ·mol -1mr S = -mm(S S产物(反应物 =m S (CH 4,g +21×m S (O 2,g -m S (CH 3OH ,l =186.3J·K·mol -

30、1 +21×205.2 J·mol -1·K -1 126.8 J·mol -1·K -1 =162.1 J·mol -1·K -1m r G =m r H -T m r S =-164.6kJ·mol -1-298.15K×162.1×10-3kJ·K -1·mol -1=116.3kJ mol -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 m r H >0,m r S >0,升高温度有利于反应自发进行,T >K

31、1015molK 162.1J mol kJ 10164.61113m,298.15r m,298.15r Tm,r Tm,r =-S H SH 即T >1015K 时,此过程自发进行。Exercises1. (omit2. (omit3. 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 equa

32、tion of the reaction and compute q D r G m . For COCl2(g, D f G m (g=-210 kJmol -1 Solution D rG m = COCl2(g + H2O(l 2HCl(g + CO2(g ån B B D f G m (B =2× (-95.3kJmol-1 + (-394.4kJmol-1- -210kJmol-1+(-237.1 kJmol-1 =-137.9kJmol-1 4. Given the following, 4NO(g 2 N2O(g + O2(g 2 NO(g + O2(g 2

33、NO2(g D r G m = -139.56 kJmol D r G m = -69.70kJmol -1 -1 calculate D r G m for the reaction 2 N2O(g + 3O2(g 4 NO2(g. Solution D r G m,3 = 2 D r G m,2 - D r G m,1 =0.16 kJmol-1 5. Chloroform, formerly used as aneshetic and now belived to be a carcinogen (cancer- causing agent, has a heat of vaporiza

34、tion D r H m =31.4 kJmol-1 . The change, CHCl3(l CHCl3(g, has D r S m =94.2 J· -1· -1. At what K mol temperature do we expect CHCl3 to boil? D r G m = D r H m - T D r S m = 0 . Solution T = DrH m D rSm = 31 . 4 ´ 1000 J × mol 94 . 2 J × mol -1 -1 -1 ×K = 333 K » 60

35、 6. A reaction that can convert coal to methane (the cihef component of natural gas is C(s +2 H2(g = CH4(g for which D r G m = -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 pursuin

36、g? - D rG m - ( - 50 . 79 ´10 J × mol 3 -1 Solution K = e K RT = e 8.314J × K -1 × mol -1 ´ 298 . 15 K = 7 . 92 ´ 10 8 According to the result, this reaction as a mean of methane production is worth pursuing . 7Triglyceride is one of typical fatty acids, its metabolic r

37、eaction is : C57H104O6(s + 80O2(g = 57CO2(g +52H2O(l, 11 D r H m =-3.35× kJ/mol. 10 -4 q Calculate the D f H m of Triglyceride. Solution The necessary data can be found from table in the addenda of the text book: D f H m (CO2 g=-393.5 kJ/mol D f H m (H2O,l=-285.8 kJ/mol D f H m (O2 g=0 q q q q So D r H m =57 D f H m (CO2 g+52 D f H m (H2O,l- D f H m (C57H104O6-80 D f H m (O2 g D f H m (C57H104O6=57× (-393.5+52× (-285.8-(-3.35× =-3.79× kJ/mol 10 10 q q q q q q -4 3 8predict the sign of the entropy change for each of