The following mechanism was proposed for the reaction of H2 (g)
with I2 (g) to give HI (g).
Step 1: I2 (g) ＜＝＝＞2I (g) (fast)
Step 2: H2 (g) + 2I(g)＜＝＝＞2HI (g) (slow)
a) Write an expression for the equilibrium constant K,for Step 1.
b) Based on this mechanism, show that the reaction of H2 (g) with
I2 (g) is first order with respect to H2 (g) and I2 (g).
Phy chem~
2009-03-01 11:29 pm
回答 (1)
2009-03-02 1:21 am
✔ 最佳答案
a)K = [I(g)]2/[I2(g)]
b)
Step 2 is the rate determining step. Hence,
Rate = k1[H2(g)][I(g)]2 …… (i)
From a)
[I(g)]2 = K[I2(g)] …… (ii)
Substitute (ii) into (i):
Rate = k1[H2(g)](K[I2(g)])
Rate = k1K[H2(g)][I2(g)]
Rate = k[H2(g)][I2(g)] -- where k = k1K = constant
Refer to the above rate equation,
The reaction is first order with respect to H2(g) and I2(g).
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收錄日期: 2021-04-29 18:21:35
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