## HW 15.49

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

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Joined: Fri Sep 25, 2015 3:00 am

### HW 15.49

For part b of this question it asks for the rate law and the answer in the solutions manual says that the rate for step 2 equals k2[HBR][HOBR] however HOBR is an intermediate. I thought you weren't suppose to include the intermediate in the rate laws???

Marvin Lu 1E
Posts: 28
Joined: Fri Sep 25, 2015 3:00 am

### Re: HW 15.49

While the intermediate is part of rate 2, it is not the overall rate of the reaction. The intermediate is apart of the fast step 2, and we know that the slowest step determines the rate of the overall reaction. It is when the intermediate is apart of the overall rate that we must substitute it out.

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Joined: Thu Jul 27, 2017 3:01 am

### Re: HW 15.49

So, if it was the slow step, would we then disregard HOBr? And it's because the slow step determines the overall rate of the reaction right?

KayleeMcCord1F
Posts: 31
Joined: Fri Sep 29, 2017 7:05 am

### Re: HW 15.49

What's an example of a situation when an intermediate would be considered in the overall rate law?

Matthew Lin 2C
Posts: 30
Joined: Fri Sep 29, 2017 7:06 am

### Re: HW 15.49

KayleeMcCord1F wrote:What's an example of a situation when an intermediate would be considered in the overall rate law?

An intermediate would never be in the overall rate law. You would have to substitute it out using the pre-equilibirium approach.

Gurkriti Ahluwalia 1K
Posts: 37
Joined: Fri Sep 29, 2017 7:07 am

### Re: HW 15.49

since the question is still asking you for the rate law of each step, it is appropriate to include the intermediate molecule in the rate law since it is part of the reaction in step 2.