发布时间 : 星期日 文章物理化学期末复习试题更新完毕开始阅读4f3125e1856a561252d36f1b
研究生入学考试物理化学模拟试卷(二)参考答案
一、单项选择题
1.(D) 2. (C) 3. (A) 4. (C) 5. (B) 6. (C) 7. (B) 8. (C) 9. (B) 10. (C) 11. (C) 12. (C) 13. (C)
14. (B) 15. (D) 16. (B) 17. (B) 18. (B) 19. (D) 20. (B) 二、简答题答案略
三、解 (1)该过程为不可逆相变过程,设计如下过程
θG、?A、?S、?H、?U C6H6?l,p1?pθ??????????C6H6(g,p2?0.9p)
?G1?0?A1,?H1 ?G2,?H2?0
?A2
C6H6(g,p)
?G??G1??G2?0?RTlnp20.9??-1-1??8.314?353.3ln?J?mol=-309.3J?mol p1?1??A??A1??A2
?A1??G1?p?V??pVg??RT???8.314?353.3?J?mol=-2937J?mol ?A2??G2???pV???309.3J?mol
-1θ-1-1?A??A1??A2???2937+(-309.3)?J?mol=-3246.3J?mol
-1-1?H=?H1??H2?30.75kJ?mol
-1?S??H??GT?30.75?103?(?309.3)?-1-1-1-1???J?K?mol?87.91J?K?mol
353.3???U??H???pV???H?p2Vg??H?RT
??30.75?8.314?353.3?10?3?kJ?mol-1=27.81kJ?mol
-1四、解 20℃时,HCl在苯中的亨利系数
M(HCl)=36.5,M(苯)=78
p(HCl)x(HCl)101.3250.0425k=?kPa?2384kPa
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当HCl和笨蒸气的总压p为101.325kPa时,
p?p??HCl?pC6H6?kxHCl?pxC6H6?kxHCl?p(1?xHCl)
则 xp?p?101.32?5HCl?k?p??2384?10?100.038 5WHCl由 x36.5HCl?W?0.038 5HCl10036.5?78求得 W0.0385?100/78HCl????36.5??1?0.0385?g=1.87g
?五、 解 反应 BaCO3(s)→BaO(s)+CO2(g)
?θθθθrHm (298.15K)= ?fHm (BaO,s)+ ?fHm (CO2,g)﹣?fHm (BaCO3,s)
={(﹣393.51﹣556.47)﹣(﹣1217.5)}kJ·mol-1
=267.52 kJ·mol-1
?Gθθθθrm (298.15K)= ?fGm (BaO,s)+ ?fGm (CO2,g)﹣?fGm (BaCO3,s)
={(﹣394.38﹣527.18) ﹣(﹣1138.0)} kJ·mol-1 =216.44 kJ·mol-1
△rCp,m=Cp,m(BaO,s) +Cp,m(CO2,g)﹣Cp,m(BaCO3,s)
={37.66+43.51﹣90.79} J·K-1·mol-1
=﹣9.62 J·K-1·mol-1
由基尔霍夫定律(??HdT)??Cp ,不定积分得
?H??H0??rCpT (1)
T=298.15K, ?θrHm (298.15K)代入上式,可得
?H0?270.39kJ?mol-1
将?H0、?rCp代入式(1),可得
?θ?3rHm(T)/kJ·mol-1?270.39?9.62?10T/K
六、证明 (1)dU?TdS?pdV
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??U???T??S????p?p??V????p?T?T??VT???T?? V? 由pV?RT??,则p
p?RTRV??,??p????T?? ?VV??所以,??U???T(R)?p??V??TV???p?p?0 p2??S?p2(2)?S????V?p2??p?dp??????Tp1??T?dp ?pp2 ???RRlnp2p1pdp??p
1?Rlnp1p
2七、解 由公式 RTlncrc?2?M0?r
代入相关的数据,得 lncrc?2?0.?20.080900?0.1?10?6?0.5?8.314?298.2
lncrc?0.285,则
0crc?1.33
0八、解 对于二级反应 k1t?1c?1Ac,则
A01?? k1?11?t??11??cAcA?????1??1????mol?1?L?min?10?60??0.070.1???7.14?10?2mol?1?L?min?1
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Ea由 k?Ae?RT,
Ea,2得 k?Ea,1RT1?Ae,kRT2?Ae?
则
Ea,1?Ea,RT?2ln?k2??k?
?1?3代入已知数据得
10.4?6108.314?373?lnk27.14?10?2 求得 k2?2.08mol?1?L?min?1 由 k12t?1c?
BcB0代入相关数据,求得 t?22.4m in九、解(1)电池的表示式为
Pb(s)PbSO4(s)H2SO4(aq)PbSO4(s)PbO2(s)
正极反应:PbO(2s)+SO2-4(aq)+4H+(aq)+2e- ? PbSO4(s)+2H2O(l)
负极反应:Pb(s)+SO2-4(aq)?PbSO-4(s)+2e
(2)电池总反应:PbO+2-2(s)+Pb(s)+4H(aq)+2SO4(aq)?2PbSO4(s)+2H2O(l) E=Eθ?RT2Fln1(a22?Eθ?2RTlna? ?)F 1.918?82.?024?18.31?4298.29.64?8410a? ln 求得, a??0.0927 (3) ?rGm??2FE,可得
?rGm??2FE?{?2?9.64?8410?1.91??83810?}- 1kJmol??370.3kJ?mol-1
(4) ?θθrGm??2FE ?θrGm??RTlnKθ
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