大学化学部分习题参考的答案

单击此处编辑母版标题样式,单击此处编辑母版文本样式,第二级,第三级,第四级,第五级,*,*,第一章部分习题参考答案,IP33:6题1解题思路,解:查附录4中有关数据如下:,2NH3(g)=N2(g)+3H2(g,AHnQ(298.15K/(K J.mol-1)-46.11,Sm(29815K)/(J.mol-lK1)192.45191.50130.684,AGm(29815K/(k j mol1)-16.45,解:(1)T=298.15K,Gmn(298.15K)=v2Gn(生成物)-v2Gn(反应物),=0-2(-1645)=3290(k mol1),(2)T=398.15K,根据:Gm()AHm(29815K)-TSmn(29815K)计算,Hm(29815K)=v2Hn(生成物)-vHm(反应物),0-2(-4611),92.22(kJ mol-),Sn(29815K)=vSn(生成物)-vSn(反应物),191.50+(3130.684)-(2192.45),198.65(mol-.K-),AGn(D)AHm(298.15K)-TASn(298.15kK),=92.22-398.1519865103,13.13(kJ mol1),2NI3(g)=N2(g)+3H2(g),(3)T=300=300+273.15K=573.15K,根据Gm(T)=Gm(T+2303RTgQ计算,Gm(I)=,Hm(298.15K)-TSn(298.15K),=9222-573.1519865103,21.63(kJ mol-),(,)(Pm2P)3(1000/100)1(1000/100)3,(1000/100),A Gm(T)=AGm(T)+2.303RTIgQ,2163+23038314103573.15lg(100),0.318(kJ mol-),P33:3题解题思路,解:查表可知,Sn(s)+O,(g)=Sno,(s),A Gm(29815K)(kJmol-)0,5197,(1)根据Gm(T)=Gm(T+2.303RTgQ计算,当T=29815K时,GnT)=Gm(29815K),Gmn(29815K)=vGn(生成物)-v;人Gm(反应物),=(-5197)-0=-519.7(k Jmol1),P(O2)=101325Kpa21%=21.28Kpa,100Kp,4.699,P(O2)/P,P(O2),21.28Kpa,