Steady-state vs. Pre-equilibrium [ENDORSED]

[Kate Zeile 2D]

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]

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]

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.