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

Belle Calforda3f
Posts: 67
Joined: Fri Sep 29, 2017 7:07 am

do we need to know these to concepts and if so can someone explain them?

Curtis Tam 1J
Posts: 105
Joined: Thu Jul 13, 2017 3:00 am

Re: steady state and pre equilibrium

The steady state is a much longer approach to finding the correct reaction mechanism. This involves setting the rate of reaction of the intermediate equal to zero and then plugging it into another rate law for another elementary reaction. For preequilibrium the approach is much simpler and simply involves finding the equilibrium, K, for the fast equilibrium step and plugging it into the rate law for the slower elementary step to get it to match the overall rate law. Since we didn't cover steady state, I doubt it will play a huge part on the test.

Lauren Seidl 1D
Posts: 51
Joined: Fri Sep 29, 2017 7:06 am

Re: steady state and pre equilibrium

My TA told us that steady state will not be on the final because Dr. Lavelle didn't really cover it in class, and that we will only have to use the pre equilibrium approximation approach.