## fast step before slow step

$aR \to bP, Rate = -\frac{1}{a} \frac{d[R]}{dt} = \frac{1}{b}\frac{d[P]}{dt}$

marcus lin 1E
Posts: 54
Joined: Fri Sep 28, 2018 12:28 am

### fast step before slow step

Why do we have to assume that the slow step comes after the fast step?

Matthew Tran 1H
Posts: 165
Joined: Fri Sep 28, 2018 12:16 am

### Re: fast step before slow step

This is not a general assumption. It's most likely that we will be given whether a step is slow or fast. Fast step before the slow step is a condition that must be met in order to do the pre-equilbrium approach. If the slow step is the first step, then it is the rate determining step and its rate law is the overall rate law.

Briana Lopez 4K
Posts: 44
Joined: Wed Nov 22, 2017 3:01 am

### Re: fast step before slow step

The slow step requires more energy for the reaction to achieve, and in turn, more time which is why it is the rate-determining step

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