## Pre-equiliibrium?

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

Sophia Bozone 2G
Posts: 51
Joined: Fri Sep 29, 2017 7:07 am

### Pre-equiliibrium?

the textbook says: "A rate law is often derived from a proposed mechanism by imposing the steady-state approximation or assuming that there is a pre-equilibrium." What is meant by "there is a pre-equilibrium?"

Minie 1G
Posts: 63
Joined: Fri Sep 29, 2017 7:04 am

### Re: Pre-equiliibrium?

pre-equilibrium: basically when the process is not at equilibrium. Since the rate of reaction measures how fast the reaction progresses UNTIL equilibrium (before it reaches equilibrium)

lindsay lathrop 2C
Posts: 29
Joined: Sat Jul 22, 2017 3:00 am

### Re: Pre-equiliibrium?

Pre-equilibrium is when the reactants and intermediates are in equilibrium. We are assuming that the consumption of the intermediate in the slow step is insignificant relative to its rate of formation/decomposition in the first step.

lindsay lathrop 2C
Posts: 29
Joined: Sat Jul 22, 2017 3:00 am

### Re: Pre-equiliibrium?

Looking more into it a simple way to think about it is that the first step is producing so much intermediate that the rate determining step (slow step) cannot keep up, resulting in a sort of equilibrium taking place within the first step, where the extra intermediate is being formed and then turned back into the reactants. We cannot say it is an equilibrium because the slow step is still occurring, using up some of that intermediate being formed. That is why it is considered a "pre-equilibrium."