同济大学普通化学第一章二章习题答案详细

普通化学（新教材）习题参考答案 第一章 化学反应的基本规律 (习题P50-52）16解（1） H2O( l ) = H2O(g)DfH/ kJ×mol-1 -285.83 -241.82S/ J×mol-1×k-1 69.91 188.83 DrH(298k) = -241.82-(-285.83) kJ×mol-1 = 44.01 kJ×mol-1Dr S(298k) = (188.83-69.91) J×mol-1×k-1 = 118.92 J×mol-1×k-1 ( 2 ) 是等温等压变化 Qp = DrH(298k) ´ N = 44.01 kJ×mol-1 ´ 2mol = 88.02 kJ W = -P×DV = -nRT = -2 ´ 8.315 J×k-1×mol-1 ´ 298k = -4955.7 J = -4.956 kJ (或 -4.96kJ ) DU = Qp + W = 88.02 kJ - 4.96kJ = 83.06 kJ17解（1） N2 （g）+ 2O2 （g） = 2 NO2 (g) DfH/ kJ×mol-1 0 0 33.2 S/ J×mol-1×k-1 191.6 205.14 240.1 DrH(298k) = 33.2 kJ×mol-1 ´ 2 = 66.4 kJ×mol-1 Dr S(298k) = ( 240.1 J×mol-1×k-1 ) ´ 2 -(205.14 J×mol-1×k-1 ) ´ 2 - 191.6 J×mol-1×k-1 = - 121.68 J×mol-1×k-1 (2) 3 Fe(s) + 4H2O (l) = Fe3O4 (s ) + 4 H2 (g)DfH/ kJ×mol-1 0 -285.83 -1118.4 0S/ J×mol-1×k-1 27.3 69.91 146.4 130.68DrH(298k) = -1118.4 - (-285.83 ´ 4 ) kJ×mol-1 = 24.92 kJ×mol-1Dr S(298k) = (130.68 ´ 4 + 146.4 ) - (27.3 ´ 3 + 69.91 ´ 4 ) J×mol-1×k-1= ( 669.12 - 361.54 ) J×mol-1×k-1 = 307.58 J×mol-1×k-118. 解： 2Fe2O3 (s) + 3C (s ,石墨) = 4 Fe (s) + 3 CO2 (g)DfH(298k)/ kJ×mol-1 - 824.2S(298k)/ J×mol-1×k-1 87.4 5.74 27.3DfG(298k)/ kJ×mol-1 -742.2 DrG = DrH - T Dr S 301.32 kJ×mol-1 = 467.87 kJ×mol-1 - 298 kDr S Dr S= 558.89 J×mol-1×k-1 Dr S= 3 S( CO2(g) 298k) + 27.3 J×mol-1×k-1´ 4 -87.4 J×mol-1×k-1´ 2 - 5.74 J×mol-1×k-1´ 3S( CO2(g) 298k) = 1/3 (558.89 +192.02 - 109.2 ) J×mol-1×k-1 = 213.90 J×mol-1×k-1DfH(298k, C (s ,石墨)=0 DfG(298k, C (s ,石墨)=0DfH（298k, Fe (s)）=0 DfG（298k, Fe (s)）=0DrH=3DfH(298k, CO2(g) ) -2DfH(298k, Fe2O3 (s) )Þ 467.87 kJ×mol-1 =3DfH(298k, CO2(g) ) -2 ´ (- 824.2 kJ×mol-1) DfH(298k, CO2(g) ) = 1/3 (467.87-1648.4) kJ×mol-1 = -393.51 kJ×mol-1同理 DrG=3DfG(298k, CO2(g) ) -2DfG(298k, Fe2O3 (s) )Þ 301.32 kJ×mol-1 = 3DfG(298k, CO2(g) ) -2 ´ (-742.2 kJ×mol-1 ) DfG(298k, CO2(g) ) = 1/3 (301.32 - 1484.4 ) kJ×mol-1 = -394.36 kJ×mol-119.解 6CO2(g) + 6H2O（l） = C6H12O6 (s) + 6O2（g） DfG(298k)/ kJ×mol-1 -394.36 -237.18 902.9 0 DrG(298k) = 902.9 - (-237.18 ´ 6 ) - (-394.36 ´ 6 ) kJ×mol-1 = 4692.14 kJ×mol-1 >0 所以这个反应不能自发进行。20解(1) 4NH3(g) + 5O2(g) = 4NO(g) + 6H2O(l)Df G(298k) /kJ×mol-1 -16.4 0 86.57 -237.18DrG(298k) = (-237.18) ´6 + 86.57´ 4 - (-16.4) ´4 kJ×mol-1 = -1011.2 kJ×mol-1<0 此反应能自发进行。(2) 2SO3(g) = 2SO2(g) + O2(g) Df G(298k) / kJ×mol-1 -371.1 -300.19 0 DrG(298k) = (-300.19) ´2 - (-371.1) ´ 2 kJ×mol-1= 141.82 kJ×mol-1 > 0 此反应不能自发进行。21解 （1） MgCO3(s) = MgO(s) + CO2(g)DfH(298k)/ kJ×mol-1 -1111.88 -601.6 -393.51 S(298k)/ J×mol-1×k-1 65.6 27.0 213.8 Df G(298k) / kJ×mol-1 -1028.28 -569.3 -394.36 DrH(298k) = -601.6 + (-393.51) - (-1111.88) = 116.77 kJ×mol-1 DrS(298k) = 213.8+ 27.0 - 65.6 = 175.2 J×mol-1×k-1 DrG(298K) = (-394.36) +(-569.3)-(-1028.28) = 64.62 kJ×mol-1(2) DrG(1123K) = DrH(298k)-T× DrS(298k) = 116.77 kJ×mol-1 - 1123k ´175.2 J×mol-1×k-1 = 116.77 kJ×mol-1 -196.75 kJ×mol-1 = -79.98 kJ×mol-1 又 RTlnKq(1123k)= - DrG(1123k) 8.315 J×mol-1×k-1´1123 k×ln Kq(1123k) = -(-79.98) kJ×mol-1 Kq(1123k) = 5.25 ´ 103 (3) 刚刚分解时 DrG(T) =DrH(298k)-T× DrS(298k) =0 分解温度T可求: 分解最低温度为666.5 k22.解法一: Kq (298k)=5.0 ´ 1016 DrG(298k) = -RTlnKq(298k)= -8.315 J×mol-1×k-1´298k×ln(5.0 ´ 1016) = -95.26 kJ×mol-1 DrG(298k) = DrH(298k)-298k× DrS(298k) -95.26 kJ×mol-1 = -92.31 kJ×mol-1-298k×DrS(298k) DrS(298k) =9.90 J×mol-1×k-1 DrG(500k) = DrH(298k)-500k× DrS(298k) = -92.31 kJ×mol-1-500k´9.90 J×mol-1×k-1= -97.26 kJ×mol-1 而 DrG(500k) = -RTlnKq(500k)= -8.315 J×mol-1×k-1´ 500k×ln Kq(500k) ln Kq(500k)= - = = 23.40 Kq(500k) = 1.45 ´ 1010解法二： ln = = -15.05 = 2.9 ´ 10-7 Kq(500k) =2.9 ´ 10-7 ´ Kq(298k) = 2.9 ´ 10-7 ´ ( 5.0 ´ 1016 ) = 1.45 ´ 101023.解： N2(g) + 3H2(g) = 2NH3(g) DfH(298k)/ kJ×mol-1 0 0 -45.9 S(298k)/ J×mol-1×k-1 191.6 130.68 192.8 DrH(298k) = 2´(-45.9) kJ×mol-1 = -91.8 kJ×mol-1 S(298k) = (2´192.8 -191.6 -3´130.68 ) J×mol-1×k-1= -198.04 J×mol-1×k-1 DrG(T) = DrH(298k) -T× DrS(298k) =0 = -91.8 kJ×mol-1 -T× (-198.04 J×mol-1×k-1 ) =0 T =