化学反应动力学第二章习题参考答案
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化学反应动力学 第二章习题
1、The first-order gas reaction SO 2Cl 2 → SO 2 + Cl 2 has k = 2.20 ⨯ 10-5 s -1 at 593K,
(1) What percent of a sample of SO 2Cl 2 would be decomposed by heating at 593K for 1 hour?
(2) How long will it take for half the SO 2Cl 2 to decompose? 解:一级反应动力学方程为:
t k e Cl SO Cl SO ⋅-⋅=ο][][2222 ⇒
t k e Cl SO Cl SO ⋅-=ο
][]
[2222
(1) 反应达1小时时:
60601020.222225][]
[⨯⨯⨯--=e Cl SO Cl SO ο
=0.924=92.4%
已分解的百分数为:100%-92.4%=7.6% (2) 当
21][][2222=οCl SO Cl SO 时,7.315062
1
ln 1=-=k t s
5
21102.2693
.0-⨯=
t = 31500 s = 8.75 hour
2、T-butyl bromide is converted into t-butyl alcohol in a solvent containing 90 percent acetone and 10 percent water. The reaction is given by
(CH 3)3CBr + H 2O → (CH 3)3COH + HBr
The following table gives the data for the concentration of t-utyl bromide versus time:
T(min) 0 9 18 24 40 54 72 105
(CH 3)CBr (mol/L) 0.1056 0.0961 0.0856 0.0767 0.0645 0.0536 0.0432 0.0270
(1) What is the order of the reaction?
(2) What is the rate constant of the reaction? (3) What is the half-life of the reaction? 解: (1) 设反应级数为 n ,则 n A k dt
A d ][]
[=-
⇒
kt A A n n =---11][1][1ο 若 n=1,则 ]
[][ln 1A A t k ο
=
t = 9 01047.00961.01056.0ln
91==k , t = 18 01167.00856.01056
.0ln 181==k t = 24 01332.00767.01056.0ln 241==
k , t = 40 01232.00645
.01056.0ln 401==k t = 54 01256.0=k , t = 72 01241.0=k , t = 105 01299.0=k
若 n=2,则 )][1
][1(
1ο
A A t k -= t : 9 18 24 40 54 k : 0.1040 0.1229 0.1487 0.1509 0.1701 若 n=1.5
t : 9 18 24 k : 0.0165 0.0189 0.0222 若 n=3
t : 9 18 24 k : 2.067 2.60 3.46
反应为一级。
(2) k = 0.0123 min -1= 2.05×10-4 s -1
(3)0123.0693
.021=t = 56.3 min = 3378 s
3、已知复杂反应:
的速率方程为]][[][]
[321111A A k A k dt A d --=-,推导其动力学方程。要求写出详细的推导过程。
解:设 0=t 时, ο][][11A A = ,ο][][22A A = ,ο][][33A A =
t t = 时, x A A -=ο][][11 ,x A A +=ο][][22 ,x A A +=ο][][33 代入 ]][[][]
[321111A A k A k dt
A d --=- 得:
)])([]([)]([32111x A x A k x A k dt
dx
++--=-οοο 212131321111][][][][][x k x A k x A k A A k x k A k ---------=οοοοο 212131132111)][][(][][][x k x A k A k k A A k A k -----++--=οοοοο 令 α = οοο][][][32111A A k A k -- , β = οο][][21311A k A k k --++ , γ = 1--k
则
2x x dt
dx
γβα++= , 移项积分: ⎰⎰=++x t dt x x dx
002
γβα
⎰
=-----+--
x
t x x dx
2
2
)
24)(24(γ
αγ
ββγαγββ
A 1A 2 + A 3
k k -1