## Pre-Equilibrium Approach

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

Ajith Raja 2L
Posts: 24
Joined: Wed Nov 18, 2015 3:00 am

### Pre-Equilibrium Approach

Hello, so the way I learned pre-equilibrium approach was to first write the rate law for the slow step. If there were intermediates in the rate law then you would continue, if not you were done. So does that mean if the slow step is the first step you are done after you write the rate law since you can't have intermediates in the first step? Thank you.

Jose_Arambulo_2I
Posts: 35
Joined: Wed Sep 21, 2016 2:59 pm

### Re: Pre-Equilibrium Approach

Yes you're right. If the slow step was the first step, and there were no intermediates on the reactant side, then that rate law should be the same as the experimental rate law.

Anthony_Imperial_1G
Posts: 22
Joined: Wed Sep 21, 2016 2:59 pm

### Re: Pre-Equilibrium Approach

If the slow step is first then you can just right the rate law using concentrations in that step.

However, if there is a fast step followed by a slow step then you would have to use the pre-equilibrium approach because intermediate values would be in your overall rate law, which you cannot allow. You would need to isolate your intermediate value for the slow portion and substitute that into the overall reaction so that no intermediates exist.