## Pre-equilibrium

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

Cowasjee_Sanaea_3E
Posts: 27
Joined: Fri Jul 22, 2016 3:00 am

### Pre-equilibrium

for the pre-equilibrium approach are you always trying to rewrite the rate law of the fast reaction in terms of the slow reaction to find the overall reaction?

Hannah_El-Sabrout_2K
Posts: 20
Joined: Wed Sep 21, 2016 2:58 pm

### Re: Pre-equilibrium

I don't think I quite understand what you're saying, but from what I do understand I think so. You are essentially trying to write a rate law (of the slow step) by using the fast step to get rid of the intermediates. In the end, you shouldn't have any intermediates or any parts of the fast reactions since the slow step is the rate-determining step. I hope this is helpful!

Noelle Min-1N
Posts: 10
Joined: Wed Sep 21, 2016 2:58 pm

### Re: Pre-equilibrium

What is the purpose of the pre-equilibrium approach? I still don't understand what I would use this for exactly...

Minu Reddy
Posts: 31
Joined: Sat Jul 09, 2016 3:00 am

### Re: Pre-equilibrium

The pre equilibrium approach is used when you are trying to write the rate law for a mechanism and there is a fast step followed by a slow step. You know that the overall rate law is dependent on the rate law of the slowest step, but since the slowest step is preceded by a fast step then there may be an intermediate included in the rate law of the slow step. You cannot have intermediates in your rate laws therefore you consider the preceding fast step to have reached equilibrium and develop an expression for the intermediate concentration using the equilibrium constant. You can then rewrite your rate law for the slow step without having it written in terms of the intermediate concentration but in terms of the equilibrium constant.

Hope this helps!