第二章 化学反应一般原理2-1 苯和氧按下式反应： C6H6(l) + O2(g) 6CO2(g) + 3H2O(l)在25100kPa下，0.25mol苯在氧气中完全燃烧放出817kJ的热量，求C6H6的标准摩尔燃烧焓DcHym和该燃烧反应的DrUym。 解： x = nB-1DnB = (-0.25 mol) / ( -1) = 0.25 molyDcHym = DrHym = = -817 kJ / 0.25 mol= -3268 kJmol-1 DrUym = DrHym - DngRT= -3268 kJmol-1 - (6 -15 / 2) 8.314 10-3 298.15 kJmol-1= -3264 kJmol-1 2-2 利用附录III的数据，计算下列反应的DrHym。(1) Fe3O4(s) + 4H2(g) 3Fe(s) + 4H2O(g)(2) 2NaOH(s) + CO2(g) Na2CO3(s) + H2O(l)(3) 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)(4) CH3COOH(l) + 2O2(g) 2CO2(g) + 2H2O(l)解： (1) DrHym = 4 (-241.818) - (-1118.4) kJmol-1 = 151.1 kJmol-1(2) DrHym = (-285.830) + (-1130.68) - (-393.509) - 2 (-425.609) kJmol-1= -171.78 kJmol-1(3) DrHym = 6 (-241.818) + 4 90.25 - 4 (-46.11) kJmol-1= -905.5 kJmol-1(4) DrHym = 2(-285.830) + 2(-393.509) - (-484.5) kJmol-1 = -874.1 kJmol-12-3 已知下列化学反应的标准摩尔反应焓变，求乙炔(C2H2，g)的标准摩尔生成焓D fHym。(1) C2H2(g) + 5/2O2(g) 2CO2(g) + H2O(g) D r Hym= -1246.2 kJmol-1(2) C(s) + 2H2O(g) CO2(g) + 2H2(g) D r Hym = +90.9 kJmol-1(3) 2H2O(g) 2H2(g) + O2(g) D r Hym = +483.6 kJmol-1解：反应2 (2) - (1) - 2.5 (3)为：2C(s) + H2(g) C2H2(g)D f Hym (C2H2) = 2 D r Hym(2) - D r Hym(1) - 2.5D r Hym(3)= 2 90.9 - ( -1246.2) - 2.5 483.6 kJmol-1 = 219.0 kJmol-1 2-4 求下列反应在298.15 K的标准摩尔反应焓变D r Hym。(1) Fe(s)+Cu2+(aq)Fe2+(aq)+Cu(s)(2) AgCl(s)+Br-(aq)AgBr(s)+Cl-(aq)(3) Fe2O3(s)+6H+(aq)2Fe3+(aq)+3H2O(l)(4) Cu2+(aq)+Zn(s) Cu(s)+Zn2+(aq)解： D r Hym(1) = -89.1-64.77 kJmol-1= -153.9 kJmol-1 D r Hym(2) = -167.159 -100.37 - (-121.55) - (-127.068) kJmol-1= -18.91 kJmol-1D r Hym(3) = 2 (-48.5) + 3 (-285.830) + 824.2 kJmol-1= -130.3 kJmol-1D r Hym(4) = (-153.89) - 64.77 kJmol-1= -218.66 kJmol-12-5 计算下列反应在298.15K的DrHym，DrSym和DrGym，并判断哪些反应能自发向右进行。(1) 2CO(g) + O2(g) 2CO2(g)(2) 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)(3) Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g) (4) 2SO2(g) + O2(g) 2SO3(g)解：(1) D r Hym = 2 (-393.509) - 2 (-110.525) kJmol-1= -565.968 kJmol-1D r Sym = 2 213.74 - 2 197.674 - 205.138 Jmol-1K-1= -173.01 Jmol-1K-1D r Gym = D r Hym - TD r Sym= -565.968 - 298.15 (-173.01 10-3) kJmol-1K-1= -514.385kJmol-1 0，反应自发。(2) D r Hym = 6 (-241.818) + 4 90.25 - 4 (-46.11) kJmol-1= -905.5 kJmol-1D r Sym = 6 188.825 + 4 210.761 - 5 205.138 - 4 192.45 Jmol-1K-1= 180.50 Jmol-1K-1D r Gym = D r Hym - TD r Sym = -905.5 - 298.15 180.50 10-3 kJmol-1= -959.3kJmol-1 0，反应自发。(3) D r Hym = 3 (-393.509) - 3 (-110.525) - (-824.2) kJmol-1= -24.8 kJmol-1D r Sym = 3 213.74 + 2 27.28 - 3 197.674 - 87.4 Jmol-1K-1= 15.4 Jmol-1K-1D r Gym = D r Hym - TD r Sym = -24.8 - 298.15 15.4 10-3 kJmol-1= -29.4kJmol-1 0，反应自发。(4) D r Hym = 2 (-395.72) - 2 (-296.830) kJmol-1= -197.78kJmol-1D r Sym = 2 256.76 - 205.138 - 2 248.22 Jmol-1K-1= -188.06Jmol-1K-1D r Gym = D r Hym - TD r Sym = -197.78 - 298.15 (-188.06 10-3) kJmol-1= -141.71kJmol-1 0，反应自发。2-6 由软锰矿二氧化锰制备金属锰可采取下列两种方法： (1) MnO2(s) + 2H2(g) Mn(s) + 2H2O(g)(2) MnO2(s) + 2C(s) Mn(s) + 2CO(g) 上述两个反应在25，100 kPa下是否能自发进行？如果考虑工作温度愈低愈好的话，则制备锰采用哪一种方法比较好？解： DrGym(1) = 2 (-228.575) - (-466.14) kJmol-1 = 8.99 kJmol-1 DrGym(2) = 2 (-137.168) - (-466.14) kJmol-1 = 191.80 kJmol-1 两反应在标准状态、298.15K均不能自发进行；计算欲使其自发进行的温度：DrHym(1) = 2 (-241.818) - (-520.03) kJmol-1 = 36.39 kJmol-1DrSym(1) = 2 188.825 + 32.01 - 2 130.684 - 53.05 Jmol-1K-1 = 95.24 Jmol-1K-1 DrHym(1) - T1DrSym(1) = 0 T1 = 36.39 kJmol-1 / (95.24 10-3 kJmol-1K-1)= 382.1KDrHym(2) = 2 (-110.525) - (-520.03) kJmol-1 = 298.98 kJmol-1DrSym(2) = 2 197.674 + 32.01 - 2 5.740 - 53.05 Jmol-1K-1 = 362.28 Jmol-1K-1 DrHym(2) - T1DrSym(2) = 0T2 = 298.98 kJmol-1 / (362.28 10-3 kJmol-1K-1) = 825.27 KT1 0；反应(3)、(4)的气体反应后分别生成固体与液体，D r Sym 0 反应不能自发进行；(2) DrHym 0，D r Sym 0，升高温度对反应有利，有利于DrGym DrHym /D r Sym= 178