## Using Steady State Approx/Pre-Equilibrium Approx

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

Sjeffrey_1C
Posts: 108
Joined: Wed Feb 20, 2019 12:17 am

### Using Steady State Approx/Pre-Equilibrium Approx

I'm confused as to when we would use a steady state approximation or equilibrium approximation, versus when we would just use the rate of the slow step?

For example, question 7C.7 asks that we find the rate law for the formation of the product, and the answer provided is just the rate law of the slow elementary step. Why wouldn't you use one the approximations?

Sahil Jog 1F
Posts: 126
Joined: Thu Jul 11, 2019 12:16 am

### Re: Using Steady State Approx/Pre-Equilibrium Approx

The pre equilibrium approach basically accounts for the rate-limiting reaction to be the rate law for the overall reaction. Furthermore, Dr. Lavelle says not to worry about the steady state approach, since it would not be covered on the syllabus.