物理化学第五版上册习题答案 下载本文

=3 1 xa t2t1

xa,2(1?xa,1)xa,1(1?xa,2) 对于二级反应有: t87.5%t50% 1

ca,01?xa ? xa

?kt t?

kca,01?xa ? ? ?

0.875?(1?0.5)0.5?(1?0.875) =7

11-6 偶氮甲烷分解反应:ch3nnch3 (g) c2h6(g)+n2(g) 为一级反应。287℃时,一密闭恒容容器中ch3nnch3初始压力为21.332 kpa,1000s后总压力为22.732 kpa,求k及t1。 2

解:

?c2h6(g)+n2(g) ch3nnch3 (g)??

t=t pyp0-py p0-py p总=2p0-py py=19.932 kpa 或t=tp0-pxpxpx p总=p0+pxpx=p总-p0=1.4 kpa k=ln t1

p02p0?p总 =ln t 1

p0p0?px =

11000 ln

21.33221.332?1.4 =

0.067881000 -5 -1 s

或 k=ln t 1

p0py ? =

11000 ln

21.33219.932 11?xa 11000

10.93437 或 ?= pxp0

1.421.332

=0.06563k=ln ln2 = ln

-5 -1 s t= 2

ln2k =

6.788?10 ?5

=10211.4 s=170.2 min

11-7 硝基乙酸在酸性溶液中的分解反应 (no2)ch2cooh(l)

ch3no2(l)+co2(g) 为一级反应。25℃、101.3kpa下,测定不同反应时间产生的体积如下:

反应不是从t=0开始的。求速率常数k。 解: (no2)ch2cooh(l) ch3no2(l)+co2(g)

t=0 ca,0v0 t=t ca,tvt t=∞ 0 v∞ ca,0∝(v∞-v0)ca,t∝(v∞-vt) ln

ca,0ca =ln

v??v0v??vt

=kt 作ln(v??vt)?t图如下: k= - - - 2 - - -

ca,0ca =4 k=ln t

ln2k

ln20.02310 1

ca,0ca =

ln460

=0.02310 min-1 r0k t1= 2 =

=30 min ca,0= = 1?10 ?3

0.02310 235238

等于多少?(a是时间单位年的符号) 解: c238,0c238 c238,0c235,0 =k238tln c235,0c235 =k235t ln(

ln c238,0

c235,0c238 ?

c235

)=(k238-k235)t -10

c238,0c235,0 =26.96

或x238?1?e?1.52?0.2=0.2621x235?1?e?9.72?0.2=0.8569 c238,0c235,0 1?x2351?x238 0.14310.7379 =26.96 r0ca,0 2 -3

时,反应的初始速率为-( dcadt )t?0 - - - =

5?100.2 2 ?2 - - - -

1ca0,1?x ?xa

?kt k= a 1

tca,01?xa ? xa

= 0.25

10?1?(1?0.25)

11-12 在oh离子的作用下,硝基苯甲酸乙酯的水解反应: no2c6h4cooc2h5+h2o no2c6h4cooh+c2h5 oh

解: - 作 1? xa

xa1?xa ?ca,0kt ca,01?xa ?t图如下: 1ca,0

k=11-13 二级气相反应 2a(ga2(g)在恒温恒容下的总压p数据如下。求ka。

解: 2a(g) a2(g)

t=0pa,0 0 p0=pa,0 t=t pa,t 12

(pa,0-pa,t) pt= 12 12

(pa,0+pa,t) 12

t=∞0

pa,0 p∞=pa,0