百度文库- 让每个人平等地提升自我11 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) fH m / kJ mol 1S m / J mol 1 k 1 rH m (298k) = [( ] kJ mol 1 = kJ mol 1 r S m (298k) = J mol1 k1 = J mol1 k1 ( 2 ) 是等温等压变化∴ Qp = rH m(298k) N = kJ mol 1 2mol = kJ W = PV = nRT = 2 J k1 mol1 298k = J = kJ (或 ) ∴ U = Qp + W = kJ = kJ 17 解( 1)N2 (g)+ 2O2 (g) == 2 NO2 (g) fH m / kJ mol 1 0 0 S m / J mol 1 k 1 ∴ rH m(298k) = kJ mol1 2 = kJ mol1r S m(298k) = ( J mol 1 k 1 ) 2 J mol 1 k 1 ) 2 J mol 1 k 1= J mol1 k1 (2) 3 Fe(s) + 4H2O (l) == Fe3O4 (s ) + 4 H2(g) fH m / kJ mol 1 0 0 S m / J mol 1 k 1∴rH m(298k) = [ ( 4 ) ] kJ mol1 = kJ mol1 r S m (298k) = [ 4 + ) 3 + 4 )] J mol 1 k 1= ( ) J mol1 k1 = J mol1 k1 百度文库- 让每个人平等地提升自我22 18. 解:2Fe2O3 (s) + 3C (s ,石墨 ) == 4 Fe (s) + 3 CO2 (g) fH m (298k)/ kJ mol 1S m (298k)/ J mol 1 k 1fG m (298k)/ kJ mol 1 rG m = rH m T ?r S m∴kJ mol 1 = kJ mol1 298 k?r Sm∴r S m= J mol 1 k 1 ∴r Sm = 3 S m( CO2(g) 298k) + J mol 1 k 1 4 J mol 1 k 1 2 J mol 1 k 1 3 ∴S m( CO2(g) 298k) = 1/3 + ) J mol 1 k 1 = J mol 1 k 1fH m (298k, C (s ,石墨 ))=0 fG m (298k, C (s ,石墨 ))=0 fH m (298k, Fe (s))=0 fG m(298k, Fe (s))=0 rH m =3fH m (298k, CO2(g) ) 2fH m (298k, Fe2O3 (s) ) kJ mol 1 =3fH m(298k, CO2(g) ) 2 (kJ mol 1) ∴fH m(298k, CO2(g) ) = 1/3 kJ mol1 = kJ mol1 同理rG m =3fG m (298k, CO2(g) ) 2fG m (298k, Fe2O3 (s) ) kJ mol 1 = 3fG m(298k, CO2(g) ) 2 ( kJ mol 1 ) ∴fG m(298k, CO2(g) ) = 1/3 ) kJ mol1 = kJ mol1 19.解6CO2(g) + 6H2O(l) == C6H12O6 (s) +...