工程化学基础习题解答_浙大版

298.15)=[(-634.9)- (-393.5)-(-1207.6)] kJ×mol-1Р =179.2 kJ×mol-1Р DrSm(298.15)=(38.1+213.8-91.7) J×mol-1×K-1Р =160.2 J×mol-1×K-1РΔrGmθ(1222K) DrHm(298.15) -TDrSm(298.15)Р= 179.2 kJ×mol-1-1222K×160.2 J×mol-1×K-1Р = -16.56 kJ×mol-1РΔrGmθ(1222K) <0,能自发进行。Р5. SiO2(s) + 2C(s) = Si(s) + 2CO(g)Р(298.15K)/kJ·mol-1 -910.7 0 0 -110.5Р(298.15K)/J·mol-1·K-1 41.5 5.7 18.8 -197.7Р(298.15K)/kJ mol-1 -856.3 0 0 -137.2Р(1) Р(2) Р或Р (3) Р不能自发。Р(4) ,自发,所以:Р6.(1)大于零;(2)大于零;(3)小于零;(4)小于零。Р7. C(s) + H2O (g) = CO(s) + H2(g)Р(298.15K)/kJ·mol-1 0 -241.8 -110.5 0 Р(298.15K)/J·mol-1·K-1 5.7 188.8 197.7 130.7Р(298.15K)/kJ mol-1 0 -228.6 -137.2 0Р(1) Р 不能向正方向进行。Р(2) Р因此,升高温度能向正方向进行。Р(3) Р8.已知DrHm(298.15)=-402.0kJ×mol-1,DrGm=-345.7kJ.mol-1,则 298.15K时的DrSm(298.15)值可以从下式求出:Р DrHm(298.15)-298.15K×DrSm(298.15)=DrGm(298.15)