1、 第五章 化學反應速率及化學平衡習題 (p111-113)參考答案：p(NO )/p pO )/ p p(HCl)/p 44 K K 22p(N O )/p p(H O)/p 22252p(H S)/p c(Zn )/c 2 K p )/p K 2c(H )/c 22c(HCl)/c c(HClO)/c 1 K K pCl )/ p c(Ag )/c cCl )/c 2 在v 0.125moldm min31 2 v 0.089mol L min1 1瞬時ABv9c c2A B C O ；2422 s ；-3-26 -1 當 ， -3 -3C O2224v 7.410 moldm s631 A

2、B； k= s-26-1vc c ) k = s2 當c =c = v= s-3-3 -1ABAB-33-1-1E = -1a 38 K 0.0307； p= -3K w(CO ) 50%(CO) 150% 50%2 =4 4K 8312a = +O 2KK21012K 4.01083510K ;5005K 255 G RTlnK CmrK e e(103) 6.05105(8.314298.15) GrmKK H T T21mCln()1r2RTT1 2 K92.210 773.15298.153 ln()6.051058.314773.15298.15 K5 7.03; c = ; K-3

3、 )2=; c=-3c )2 -3c(BrCl)/c + 2 = 0.01142c(Br )/c c(Cl )/c 0.00430.004322 7.03K 22 為 x -32+= 22-3 c(BrCl)/c (0.0114 2x)22 7.03K c(Br )/c cCl )/c (0.00430.01 )(0.0043 )xx22x= -3 -3 )=-32 )=-32 =-3 4.0103(p/ p )2p(HI)/ p 25.6102 0.16 K 54.5Jp(H )/ p p(I )/ p 2.01035.01025.6102(p/ p )(p/ p )225.6102 H -

4、 -1mrm G103由 lnkJ mol 1 44.3 K ere 2.12104 1 3 0GmRT K r1 = +O p p222Hg1 K p / p p / p p /(p ) 23342Hg2Hgp =70.6% p = AB。A n =+=總0.0 0 6 0 (x C H) 0.0 6 00.1 0 0260.047x(C H ) x(H ) 0.470.100242p(C H )/ p p(H )/ p pxC H )/p px(H )/p 0.470.47K 3.68242242pC H )/ ppxC H )/ p0.0602626 n =+總 0.0060 (x C

5、H) 0.0300.200.04726x(C H ) x(H ) 0.2350.20242pC H )/p p(H )/p pxC H )/ p px(H )/p 0.2350.235J 1.84 K242242pC H )/ ppxC H )/p0.03026260.1502.12 ;p =; p =K 11.3kPa4K9.746 G 28.59kJ1rm J K, 1184J mol K 4.0kJ m o 1K 0.65 H 180kJmolS11GmmrmGrr由 ln得 K rmG時時RTlnK 150010 ln(2.110 )kJmol3 31rrmGlnK 127310 10 )kJ3 21m由 H S 有GmrrGmrm時 H S H 1500 S 10 kJmol31rrmGrmrmrmrm時 H S H 1273 S 10 kJmol31mrmrmrmrm H kJmol S 184J mol K111r