4. Which step of the proposed mechanism must be slow in order to agree with this rate law?

Question

Consider the reaction, H₂ + I₂ => 2 HI , which is found to be half order in I₂ and first order in H₂. Which step of the proposed mechanism must be slow in order to agree with this rate law?

I ₂ => 2 I           

I + H₂ => H₂I         

H₂I + I => 2 HI     

Detailed explanation please

Answer 1

First, understand that the steps in a proposed reaction mechanism must add up to the overall reaction, and that the slowest elementary step is the rate limiting step. You are given a mechanism that involves 3 elementary steps:

(1) I₂ ==> 2I

(2) I + H₂ ==> H₂I

(3) H₂I + I ==> 2HI

Also understand that the predicted rate law MUST agree with the experimentally determined rate law. Here you have a predicted rate law of rate = k[H₂][I₂] but the actual rate law is rate = k[H][I₂]^1/2

If step (1) were slowest, the rate law would be rate = k[I₂] which is incorrect

If step (2) were slowest, the rate law would be rate = k[I₂]^1/2[H₂] which agrees with experimental. Thus, step (2) is slow.

NOTE: [I₂] is raised to the 1/2 power because you have only 1/2 of an I₂ molecule and the exponent of a reactant in the rate expression for a step is its coefficient in the equation for that step.

Also, another easy way to identify the slow step in this particular question is that step (2) is the ONLY elementary step that contains both I and H2 as reactants.