Winter 2016 final 4B

$K = \frac{k_{forward}}{k_{reverse}}$

Jineava_To_3N
Posts: 35
Joined: Wed Sep 21, 2016 2:57 pm

Winter 2016 final 4B

Given:
Step 1: ClO+O2->ClO2+O
Step 2: ClO+O->ClO2
How do I go about finding the rate law for step 2? I think we are suppose to get rid of the oxygen in the final rate law because it is an intermediate but how do I do so? Thanks in advance.

Theresa Dinh 3F
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Joined: Fri Jul 22, 2016 3:00 am

Re: Winter 2016 final 4B

O is a catalyst so it's not actually involved in the reaction

Helen_Onuffer_1A
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Re: Winter 2016 final 4B

Isn't O an intermediate?

ntruong2H
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Re: Winter 2016 final 4B

Helen_Onuffer_1A wrote:Isn't O an intermediate?

You are correct. O is an intermediate. This is essentially a pre equilibrium problem. You are trying to see if setting step 2 as the slow step will give you the rate law you wanted and calculated in part A of this question. First, find the rate law for the slow step. This will give you rate=k2[O][ClO]. Because O is an intermediate, which cannot be included in the rate law for a reaction, you must use the pre equilibrium approach to "get rid of" the O. To do so, find what [O] is from step 1. Because you assume that step 1 is a fast step at equilibrium, you now have:

K(equilibrium constant, not rate constant)= [O][ClO2]/[ClO][O2]

Solving for [O], you get that [O]=(K[ClO][O2])/[ClO2]

Plug this back into your slow step. Now, you should have that rate=k2*(K[ClO][O2])/[ClO2])[ClO]

This rate law is not consistent with the one you found from part A. Therefore, Student A's proposed mechanism is not valid.

ntruong2H
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Re: Winter 2016 final 4B

Theresa Dinh 3F wrote:O is a catalyst so it's not actually involved in the reaction

O is not a catalyst. Catalysts must first appear as a reactant in the first step and finish as a product in the last step. Intermediates are produced during the first step and used up as a reactant during the subsequent step.