Slow Step in Reaction Mechanisms

[Andrew Nguyen 2I]

Just wanted to clarify - in a multi-step reaction, if the first step is a slow step does that mean the overall reaction rate should be the rate of the slow step? And if the first step is a slow step and a latter step is the slow step we have to take into account both the first and slow step? If I'm saying anything wrong here, please correct me and it would be appreciated if you could explain why.

[Johann Park 2B]

If I am understanding your question correctly:

I believe that if the early step is the slow step, yes, the overall reaction rate is the rate of that first slow step.

If the early step is fast, however, and the second step is the slow step, there is a "bottleneck" effect and a build-up of intermediate results in the 1st step being at near equilibrium. So from then, we take the pre-equilibrium approach and create a "new" step 1 and step 2 that accounts for this equilibrium.

[lindsay lathrop 2C]

If i read it correctly Johann was also trying to ask what happens when the first reaction is a slow, but it is not the slow step: Do we treat as a pre-equilibrium because it is relatively "fast"
or can we not assume this?

[Gurkriti Ahluwalia 1K]

let's say that the first step of a 2 step mechanism is the forward and reverse reaction, where both are fast and at equilibrium. and the second step is the slow step. does the first step also not affect the rate law? if not, then why? if anyone is curious im referring to problem 15.53 (c).

[Cassandra Mullen 1E]

Your overall rate law for the reaction will be based on the slow step, since the slow step is the rate-determining step. If the slow step is not the first step in the mechanism, then you'll have intermediates that you'll need to get rid of through substitution.

[Michael Downs 1L]

All steps before the slow step should be taken into account, but any steps listed after should not be accounted for

[Clarissa Molina 1D]

How do we know when to use the pre-equilibrium condition?

[Cynthia Bui 2H]

If the slow step is first, the reaction is determined by the slow step but if it is second, you use the pre-equilibrium steps we learned.