01-习题-2008.ppt

第1章习题,工科化学校级精品课程,解：查表有：2NH3(g)=N2(g)+3H2(g)fGm(298.15K)/(kJmol-1)-16.4500fHm(298.15K)/(kJmol-1)-46.1100Sm(298.15K)/(Jmol-1K-1)192.45191.50130.684(1)rGm(298.15K)=10+30-2(-16.45)=32.9kJmol-1rHm(298.15K)=10+30-2(-46.11)=92.22kJmol-1rSm(298.15K)=1191.50+3130.684-2192.45=198.652Jmol-1K-1rGm(398.15K)=rHm(298.15K)-TrSm(298.15K)=92.22398.15198.65210-3=13.1kJmol-1,P33习题1-6：,P33习题1-6：,2NH3(g)=N2(g)+3H2(g)rGm(TK)=rHm(298.15K)-TrSm(298.15K)=92.22T198.65210-3464K(2)rGm(573.15K)=rGm(573.15K)+RTlnQ=92.22573.15198.65210-3+8.314573.1510-3ln100=0.30kJmol-1,解：2Ag2O(s)=4Ag(s)+O2(g)fHm(298.15K)/(kJmol-1)-30.0500Sm(298.15K)/(Jmol-1K-1)121.342.55205.138rHm(298.15K)=40+10-2(-30.05)=60.1kJmol-1rSm(298.15K)=442.55+1205.138-2121.3=132.738Jmol-1K-1rGm(298.15K)=rHm(298.15K)-TrSm(298.15K)=60.1298.15132.73810-3=20.52kJmol-1,P33习题1-7：,rGm(498.15K)=rHm(298.15K)-TrSm(298.15K)=60.1498.15132.73810-3=-6.02kJmol-1,P33习题1-7：,rGm(TK)=rHm(298.15K)-TrSm(298.15K)=60.1T132.73810-3=0T=453K,P33习题1-7：,rGm(TK)=rGm(TK)+RTlnQ=60.1T132.73810-3+8.314T10-3ln101.32520.95%/100413K,P34习题1-8：,解：N2(g)+O2(g)=2NO(g)fHm(298.15K)/(kJmol-1)0090.25Sm(298.15K)/(Jmol-1K-1)191.50205.138210.761rHm(298.15K)=290.25-10-10=180.5kJmol-1rSm(298.15K)=2210.761-1191.50-1205.138=24.884Jmol-1K-1rGm(1573.15K)=rHm(298.15K)-TrSm(298.15K)=180.51573.1524.88410-3=141.35kJmol-1,P34习题1-9：,解：反应2NO(g)+Cl2(g)=2NOCl(g)是元反应（1）反应的速率方程为：v