## Pre-equilibrium Condition

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

Posts: 59
Joined: Fri Sep 29, 2017 7:07 am

### Pre-equilibrium Condition

Can someone elaborate on this concept? I don't really understand the way that they described in in the textbook.

"The assumption that the rate of consumption of the intermediate in the slow step is insignificant relative to its rates of formation and decomposition in the first step is called a pre-equilibrium condition. A pre-equilibrium arises when an inter- mediate is formed and sustained in a rapid formation reaction and its reverse"

Julian Krzysiak 2K
Posts: 49
Joined: Fri Sep 29, 2017 7:07 am

### Re: Pre-equilibrium Condition

It's saying that when the first step is fast, and the second step is slow, then when the first step creates the intermediate, it does this at a much faster rate than the second step can consume it. So a bottleneck is created, where you have a bunch of intermediate waiting to be consumed in the slow second step. So while it's waiting to be consumed, it can turn back into the reactants of the fast first step, and this would create an equilibrium of forward and reverse reactions for the first step. But the whole reaction isn't at equilibrium yet, hence the term Pre-equilibrium.

Priyanka Bhakta 1L
Posts: 50
Joined: Fri Sep 29, 2017 7:04 am

### Re: Pre-equilibrium Condition

In order to solve a problem with this situation, make sure to follow the procedure Dr. Lavelle lined out in lecture.

Salman Azfar 1K
Posts: 50
Joined: Thu Jul 13, 2017 3:00 am

### Re: Pre-equilibrium Condition

Probably the simplest way to explain this is that the first step is so fast that you end up with overflow.

Imagine I'm baking a huge number of cakes for a bunch of birthday parties. It takes me 5 minutes to prepare batter but an hour to make the cake. I'm going to end up with a ton of batter (which is sort of an intermediate assuming my "reactants" are raw ingredients) and basically from there the slow step of making cake kicks in.