物化07-1-17(A)答案.doc

化学学院本科生2006-2007学年第一学期物理化学(2-1)课程期末考试试卷(A卷)标准答案一、选择题（每小题2分，共20分） 1. A, 2. D, 3. D, 4. C, 5. B, 6. A, 7. A, 8. C, 9. B, 10. C二、填空题(每空1分, 共10分) 1. -81.85kJmol-1; 2. 小于, 等于, 小于, 等于; 3. 1.881023; 4.p/nBeq, (c)-1; 5. 等于,大于三、计算题(共70分) 1. (15分) 解: (1)H2O(l)298.15K p(1/2)O2(g)298.15K pH2(g)298.15K p +H2O(l)298.15K PsH2O(g)298.15K PsH2O(g)298.15K p H1=BfHm(B)= fHm(H2O,g) =-241.83(kJmol-1)S1=BSm(B) =188.72-130.59-(1/2) 205.03=-44.385(Jmol-1K-1) G1=H1-TS1=-241.83-298.15(-44.38510-3)=-228.60(kJmol-1)G2=Vgdp=RTln(ps/p） =8.314298.15ln(3167.74/101325) =-8.59(kJmol-1) G3=0G4=Vldp= Vm(H2O,l)（p-ps） =1810-6(101325-3167.74) =1.77 (kJmol-1)rGm=G1+G2+G3+G4 =-237.188(kJmol-1)因为 rGm0, 所以是正偏差。或：AR= 1.51971, BR= 1.63131 所以是正偏差。5. (15分) 解： (1) rHm=BfHm(B) =-1130.9-241.8-393.5-2 (-947.7)=129.2(kJmol-1)rSm=BSm(B) =136.6+188.7+213.6-2 102.1=334.7(Jmol-1K-1) rGm=rHm-TrSm=129.2-298 334.710-3=29.46(kJmol-1) rGm=-RTlnKp Kp=exp(-rGm/RT)=6.8510-6设平衡总压力为p, 平衡时：peq(H2O)=peq(CO2)=(1/2)p Kp=( peq(H2O)/ p) (peq(CO2)/ p) =(1/4) ( p / p)2 =6.8510-6p=(4Kp)1/2 p=530.4(Pa)(2) p=101325Pa时，Kp(T2)=(1/4) ( p / p)2 =0.25因为Cp=0, 所以, rHm, rSm与温度无关rGm(T2)=rHm-T2rSm=-RT2lnKp(T2) T2=rHm/rSm-RlnKp(T2) =129.2103/334.7-8.314ln0.25 =373.17K或：lnKp(T2)/ Kp(T1)= (rHm/R)(1/T1-1/T2) 1/T2=1/T1- (R /rHm) lnKp(T2)/ Kp(T1) =0.003356-0.000676 =0.00268(K-1)T2=373.13K(3) p (H2O)=3167Pa, p (CO2)=30.4Pa时 Qp=( p(H2O)/