## Pre- equilibrium approach [ENDORSED]

$\frac{d[R]}{dt}=-k[R]^{2}; \frac{1}{[R]}=kt + \frac{1}{[R]_{0}}; t_{\frac{1}{2}}=\frac{1}{k[R]_{0}}$

Kira_Maszewski_1B
Posts: 17
Joined: Wed Sep 21, 2016 2:57 pm

### Pre- equilibrium approach

When are we supposed to use the pre equilibrium approach? If there is an intermediate in the first step and the first step is also the slow step, would we need to use the pre-equilibrium approach to find the rate law and get rid of the intermediate?

Theresa Dinh 3F
Posts: 26
Joined: Fri Jul 22, 2016 3:00 am

### Re: Pre- equilibrium approach

Usually when there's an intermediate involved, you have to use the pre equilibrium approach.

divya_devineni_2E
Posts: 34
Joined: Wed Sep 21, 2016 2:59 pm
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### Re: Pre- equilibrium approach

You do not include the intermediate if it is not in the slowest step

Vincent Tse 2B
Posts: 30
Joined: Fri Jul 22, 2016 3:00 am
Been upvoted: 3 times

### Re: Pre- equilibrium approach  [ENDORSED]

You apply the pre-equilibrium approach when the mechanism has its first step as fast; this will create a "bottleneck" effect which results in a build-up of intermediates. You can then assume the first step reaction to be at an equilibrium, and use it to help you get rid of the intermediate concentrations in your slow-step rate law.

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