Steady State vs. Pre-Equilibrium

[K Stefanescu 2I]

I am still a bit confused regarding these two ways of determining the reaction mechanism. What would a problem/question involving a steady state calculation look like? How is the pre-equilibrium approach "less flexible" than the steady-state approach?

[Kailey Brodeur 1J]

In steady state calculations, there is a constant intermediate concentration in the rate limiting step which differs from the pre equilibrium approach which has a buildup of the intermediate in the fast step (not the determining step) resulting in an equilibrium of the fast step. For the steady state approach to a reaction mechanism, d[intermediate]/dt = 0. The pre equilibrium approach has a more lengthy procedure which was outlined in lecture. Lavelle said that we will be focusing on this approach.

[Sirajbir Sodhi 2K]

I'm pretty sure we won't be asked to do a steady state calculation.

But to answer your question, a problem involving steady state could be structured exactly the same way as a problem that involves the pre-equilibrium approach. The steady-state approach is based on the assumption that the concentration of intermediates remains low and constant throughout the reaction. It is more flexible because you don't necessarily need to know the rate-determining step (if there is not one) to determine the rate constants.