## Slow reaction

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

Emily Glaser 1F
Posts: 156
Joined: Thu Jul 27, 2017 3:01 am

### Slow reaction

How do I know what to make the approximation to?

In my slow reaction, I have:

Fast 1st step:
forward O3 rate decomp= k1[O3]
reaverse form rate= kprime1[O2][O]

Slow 2nd Step:
forward O3 consume rate = k2[O][O3]

Why is it k2[O][O3] << kprime1[O2][O]

and not

k2[O][O3] << k1[O3]

Justin Lau 1D
Posts: 51
Joined: Sat Jul 22, 2017 3:00 am

### Re: Slow reaction

I think it's because there's an equilibrium for the first step due to the bottleneck effect, the two rates are equal. Also, O is an intermediate in this reaction so I think they're comparing that to eliminate the intermediate.

aTirumalai-1I
Posts: 55
Joined: Thu Jul 27, 2017 3:00 am
Been upvoted: 1 time

### Re: Slow reaction

Also remember that Dr. Lavelle outlined 2 methods to write the rate law of a mechanism: steady-state approximations and pre-equilibrium conditions. He told us that we would be using the pre-equilibrium method in 14B, so I think approaching this problem with that method will make it a bit easier.