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

Kate Zeile 2D
Posts: 39
Joined: Sat Jul 22, 2017 3:01 am

In lecture, Dr. Lavelle said that the pre-equilibrium approach is simpler than the steady-state approach, but it is also less flexible. What did he mean by "less flexible"?

Rakhi Ratanjee 1D
Posts: 59
Joined: Fri Sep 29, 2017 7:04 am

During lecture, he mentioned that the steady-state approach assumes a constant intermediate concentration in the rate-limiting step and that the pre-equilibrium approach relies on the reaction before the rate-limiting step is at equilibrium, requiring the equilbrium constant. Therefore, it may be less flexible if the concentration of the intermediate changes due to loss to the surroundings.

Alexia Joseph 2B
Posts: 56
Joined: Thu Jul 27, 2017 3:01 am

### Re: Steady-state vs. Pre-equilibrium  [ENDORSED]

In a steady state approach, a constant intermediate concentration (the d[intermediate]/dt = 0). The pre-equilibrium approach means the reaction before the rate-limiting step is at equilibrium, so you can use the equilibrium constant K. The pre-equilibrium approach is less flexible because it applies to a a reaction in which there is a fast step followed by the slow step.