The Rate Law 速率定律式問題

2014-04-06 7:48 am

Assume that the formation of nitrogen dioxide:

2NO(g) + O2(g) ------------> 2NO(g)

is an elementary reaction.
(a) Write the rate law for this reaction.

(b) A sample of air at a certain temperature is contaminated with 2.0 ppm of NO by volume. Under these conditions, can the rate law be simplified? If so, write the simplified rate law.

(c) Under the conditions described in (b), the half-life of there action has been estimated to be 6.4 x 103 min. What would the half-life be if the initial concentration of NO were 10 ppm?

更新1:

不好意思　第二題和第三題能具體的用中文解釋一下嗎　感恩！！

更新2:

第三題的公式我記得是　1/[A0] + kt =1/[At] 大大沒有把1/[A0]寫出來是因為相除的時候會消掉所以省略嗎!!?

回答 (4)

胡雪8°

2014-04-06 5:21 pm

✔ 最佳答案

The equation should be :
2NO(g) + O2(g) → 2NO2(g)

(a)
The mechanism of the reaction contains only one elementary step :
2NO(g) + O2(g) → 2NO2(g)

The rate law of the reaction :
Rate = k [NO]² [O2]

(b)
In the sample of air, [NO] =2 ppm and [O2] ≈ 20%
[O2] >> [NO]
Throughout the reaction, the change in [O2] is negligible.
Hence, [O2] is nearly constant, and thus k [O2] ≈ k'
where k' is a constant.

The simplified rate law of the reaction :
Rate = k' [NO]2

(c)
The above simplified rate law shows a (pseudo-) second order reaction.

For a second-order reaction :
Half-life, t1/2 = 1 / (k' [A]o)

When [NO] = 2.0 ppm, t1/2 = 6.4 x 10³ min :
6.4 x 10³ = 1 / (k' x 2.0) ......[1]

When [NO] = 10 ppm, t1/2 = ? :
t1/2 = 1 / (k' x 10) ...... [2]

[2] / [1] :
t1/2 / (6.4 x 10³) = 2.0 / 10
New half-life, t1/2 = 1.28 x 10³ min

參考: 胡雪

👍 0 👎 0

[匿名]

2014-04-28 8:15 am

下面的網址應該對你有幫助

http://phi008780426.pixnet.net/blog

👍 0 👎 0

[匿名]

2014-04-27 11:56 pm

下面的網址應該對你有幫助

http://phi008780426.pixnet.net/blog

👍 0 👎 0

[匿名]

2014-04-24 2:43 pm