大学化学(第二版)部分习题参考答案.pptx

第一章部分习题参考答案

=0-2×(-16.45)=32.90(kJ·mol-1)[P33:6题]解题思路解:查附录4中有关数据如下：2NH3(g)=N2(g)+3H2(g)ΔfHm?(298.15K)/(kJ·mol-1)-46.1100Sm?(298.15K)/(J·mol-1·K-1)192.45191.50130.684ΔfGm?(298.15K)/(kJ·mol-1)-16.4500△rGm?(298.15K)=∑vi△fGm?(生成物)-∑vi△fGm?(反应物)解:(1)T=298.15K

T=398.15KΔrGm?(T)≈ΔrHm?(298.15K)-T·ΔrSm?(298.15K)=92.22-398.15×198.65×10-3=13.13（kJ·mol-1）根据:ΔrGm?(T)≈ΔrHm?(298.15K)–T?ΔrSm?(298.15K)计算.△rSm?(298.15K)＝∑viSm?(生成物)-∑viSm?(反应物)＝[191.50+(3×130.684)]-(2×192.45）=198.65(J·mol-1·K-1)△rHm?(298.15K)＝∑vi△fHm?(生成物)-∑vi△fHm?(反应物)＝0–2×(-46.11）=92.22(kJ·mol-1)

2NH3(g)=N2(g)+3H2(g)(3)T=300℃=300+273.15K=573.15K根据ΔrGm(T)=ΔrGm?(T)+2.303RTlgQ计算.ΔrGm?(T)=ΔrHm?(298.15K)-T·ΔrSm?(298.15K)=92.22-573.15×198.65×10-3=-21.63(kJ·mol-1)Q===100(PNH3/P?)2(PN2/P?)1·(PH2/P?)3(1000/100)1·(1000/100)3(1000/100)2ΔrGm(T)=ΔrGm?(T)+2.303RTlgQ=-21.63+2.303×8.314×10-3×573.15×lg(100)=0.318(kJ·mol-1)

p(O2)=101.325Kpa×21％=21.28KpaP(O2)P?1P(O2)/P?21.28Kpa100Kpa解：查表可知：Sn(s)+O2(g)=SnO2(s)ΔfGm?(298.15K)/(kJ·mol-1)00-519.7[P33:3题]解题思路Q====4.699当T=298.15K时，ΔrGm?(T)=△rGm?(298.15K)△rGm?(298.15K)=∑vi△fGm?(生成物)-∑vi△fGm?(反应物)=(-519.7)-0=-519