大學普化-速率決定式(r.d.s.)的問題

Chemical equation : 2IBr(g) → I2(g) + Br2(g)

a.
1. IBr(g) → I(g)+Br(g) .... (fast)
2. IBr(g) + Br(g) → I(g)+Br2(g) .... (slow)
3. I(g) + I(g) → I2(g) .... (fast)

Step 2 is the rate determining step (r.d.s.). Hence,
Rate = k'[IBr(g)][Br(g)] .... (1)

Br(g) is an intermediate of the reaction, and thus it should not appear in the rate law.
Br(g) is formed in step 1.
Refer to step 1. [Br(g)] is directly proportional to [IBr(g)]. Hence,
[Br(g)] = k"[IBr(g)] .... (2)

Sub. (2) into (1):
Rate = k'[IBr(g)]•k"[IBr(g)] = k'k"[IBr(g)]^2

Hence, the rate law is :
Rate = k[IBr(g)]^2


b.
1. IBr(g) + IBr(g) → I2Br^+(g) + Br^-(g) .... (fast)
2. I2Br+(g) → Br^+(g) + I2(g) .... (slow)
3. Br^+ (g) + Br^-(g) → Br2(g) .... (fast)

Step 2 is the r.d.s. Hence,
Rate = k'[I2Br^+(g)] .... (1)

I2Br^+(g) is an intermediate of the reaction, and thus it should not appear in the rate law.
I2Br^+(g) is formed in step 1.
Refer to step 1. [I2Br^+(g)] is directly proportional to [IBr(g)]^2. Hence,
[I2Br^+(g)] = k"[IBr(g)]^2 .... (2)

Sub. (2) into (1):
Rate = k'•k"[IBr(g)]^2 = k'k"[IBr(g)]^2

Hence, the rate law is :
Rate = k[IBr(g)]^2

反應速率的等式，寫速率決定步驟的產物即可
速率決定步驟就是反應慢的
此二反應，都是第二步驟慢
所以；
a. rate = k x [IBr(g)] x [Br(g)]
b. rate = k x [I2Br(g)]

註：此二題的速率決定步驟的產物之係數剛好都是１
　如果係數是2，則要寫上2次方
　而且如果是固體，因為固體濃度是固定的，不必寫進等式
例如：速率決定步驟的反應是：A(s) + 2B(g) + C(g) --> ....
　則 rate = k x [B]^2 x [C]

2010-03-11 07:55:37 補充：
老爺子的確很強，沒錯沒錯～。
我的解答有錯，給他忘了中間產物不能在速率等式中（因為濃度根本不知道）。
老爺子獲最佳解答實至名歸～。
最後，如果有投票時，能避免投票部隊，那就更讓人心服口服啦。