研招試題-考試組卷答案

一、1. 2、 3、不 4.、如 5、不*A6、A +A7、 8、 10、 14、 15、 16、Pt|Br2(l)|Br||Cl|Cl2(p)|1、解：fH（C2H4,g）=[(－393.5)×2+(－285.8)×2-(－1411.0)]kJ·mol-1=52.4kJ·mol-2、解：cH[(CH3)2Og]=fH(C2H5OHl)+cH(C2H5OHl)－fH[(CH3)2O=[－27769+(－1366.8)－(－1841)]kJ·mol-=－14604kJ·mol-解：SSASBSA=nACpmAln(T2/T1)+nARln(pA1/pA2)=nACpmAln(T2/T1)+nARln(p1/p2SB=nBCpmBln(T2/T1)+nBRln(p1/p2（nACpmAnBCpmB）ln(T2/T1nA+nBRln(p1/p2T2T1(p2/p1)6R/17RT1(p2/300K(100/600)6/17159.4解：(1)x(CH3OH)=0.33、x(CH3OH)=0.15、y(CH3OH)=0.52y(CH3OH)=0.52、x(CH3OH)=0.15 反復蒸餾或精LL(2)生成的化合物的分子式：Mg2Sn(C)。（3）六解：（1）rH(298K)=[-241.84+52.28-（-235.3）]kJmol1=45.74kJmolrS(298K)=（188.74+219.56-282.0）JK1mol1=126.3JK1molrCp,m,B=（40.04+60.11-94.67）JK1mol1=5.48JK1molrH(633K)=rH(298K)+633K dT=47.58kJmol298K

r (633K)=r (298K)+rCp,m,Bln(633/298)=130.43JK1molrG(633K)=－34.98kJmol34980Jmol-K=exp(8.314JK-1mol-1633K)=(2)增1

dCYkC

，k1CACB=k-1CF,CF=k1C

,

k1k2cc k 2Fk

A

ABE1+E2－E-2、解：p=2pA0=2×48.4kPa=96.8

cA0=pA0/RT=8.314JK1mol1(518273)K7.36

k1

1ptA

cA0

pt/k/AAkA =406/ k

0.335dm3mol1s1(7.36103)moldm3A m4r33r=0.621×10-4pplnrp

2MRT 272.75103Nm118.021038.3145JK1mol1293.15K0.9982103kgm30.621106=1.735×10-所以r=（1）E(298K)=RTln0.01=（0.05916lg0.1）V=0.05916 （2）電池反應：H2(p1=0.1pH2(p2=0.01pE(305K)=2

ln0.1

=(8.314350ln0.1)V=0.034722I=2

bz

2 BB (3×0.001×1+0.001×32B=