第二章化学热力学初步补充习题 -△G 6.7×1000 K 2.303RT2.303×8.314×1000 =0.3499 K=2.24 7.314CE03 解：C(s)+H0(g） C0(g)+H2(g) △H =-110.5-(-241.8)=131.3(kJ·mo1） △G =-137.3-(-228.6)=91.3(kJ·mo1) .△S -△H-△G ×1000 =134.1(J·mo1·K) 298 忽略△H.△S随T变化， .△G=△H-T△S 平衡时，△G=0 .T 131.3×1000 △S =979.1(K) 134.1 即T=979.1K时体系处平衡状态 8.314CE09 解：由已知得： ①△H (Br,aq)=△H =120.9(kJ·mo1） ②△G =△G(Br,aq)=-102.8(kJ·mol) ③ △S △H-△G_(-120.9)-(-102.8)×10 T 298 =-60.7J·mo1 ·K) ④△S=S (Br aq)+S (H aq)-S (Br2,1)-S (H2,g) S(Br,aq)=80.8(J·mol·K) 9.313BE08 答：常温常压下可认为K=298,P=101.325KPa即标准状况，反应是否 自发，由△G判断： △G=△G (C02,g)+2△G(HC1,g)-△G(CH2C12,1) =-517kJ·mo1 .△G<0 . 反应可自发进行. 10.313CE09解： .△G =-2.303RTK =-2.303×8.314×298 10 =308.12kJ·mo1 △G=△H -T△S △S -(△H-△G)(284.24-308.12)×10 T 298 =-80.12J·mo1·K 9第二章 化学热力学初步补充习题 9 K = -△G 2.303RT = 6.7×1000 2.303×8.314×1000 =0.3499 K =2.24 7. 314CE03 解: C(s)+H2O(g) CO(g)+H2(g) △H =-110.5-(-241.8)=131.3(kJ·mol ) △G =-137.3-(-228.6)=91.3(kJ·mol ) ∴ △S = △H -△G ×1000 298 =134.1(J·mol ·K ) 忽略△H.△S 随 T 变化, ∵ △G=△H-T△S 平衡时,△G=0 ∴ T= △H △S = 131.3×1000 134.1 =979.1(K) 即 T=979.1K 时体系处平衡状态. 8. 314CE09 解: 由已知得: ① △H (Br ,aq)=△H =120.9(kJ·mol ) ② △G =△G (Br ,aq)=-102.8(kJ·mol ) ③ △S = △H -△G T = ((-120.9)-(-102.8))×10 298 =-60.7J·mol ·K ) ④ △S =S (Br ,aq)+S (H ,aq)- S (Br2,l)- S (H2,g) S (Br ,aq)=80.8(J·mol ·K ) 9. 313BE08 答: 常温常压下可认为 K=298,P=101.325KPa 即标准状况,反应是否 自发,由△G 判断: △G =△G (CO2,g)+2△G (HCl,g)- △G (CH2Cl2,l) =-517 kJ·mol ∵ △G <0 ∴ 反应可自发进行. 10. 313CE09 解: ∵ △G =-2.303RT K =-2.303×8.314×298 10 =308.12kJ·mol △G =△H -T△S ∴△S = (△H -△G ) T = (284.24-308.12)×10 298 =-80.12J·mol ·K