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

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

What is the difference between "imposing the steady-state approximation" and "assuming that there is a pre-equiibrium" for determining rate laws?

Aya Shokair- Dis 2H
Posts: 58
Joined: Fri Sep 29, 2017 7:07 am

### Re: Steady-State v Pre equilibrium?

It explains it well in the book.
What I understood from it, is that steady state is assuming that the intermediate formed in the reaction mechanism stays at a constant, low concentration. Most times when you write the rate law for each elementary reaction, and then combine them to form the overall rate law. You find a intermediate in the rate law. This is not allowed. So, to accommodate for that you find the net rate of formation and make that equal to zero. The justification is that the intermediate is so reactive that it reacts as soon as it is formed because the concentration of intermediate is constant, the net change in formation is equal to zero. Then you solve for the concentration of that intermediate. Then you subsitute it in for your over all rate law.

allyz1F
Posts: 58
Joined: Sat Jul 22, 2017 3:00 am

### Re: Steady-State v Pre equilibrium?

^^However, if both overall reaction rates obtained from the above two methods match the experimentally determined rate law, then there should be no difference in which method you use, just that the pre-equilibrium approach is faster and sometimes less accurate in complex reactions.