## Slow Step

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

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Alexandra Carpenter 1G
Posts: 32
Joined: Sat Jul 22, 2017 3:00 am

### Slow Step

Just to be clear, if a mechanism has two steps, can the first one be the slow step and the second one be treated at equilibrium?

404995677
Posts: 82
Joined: Fri Sep 29, 2017 7:07 am

### Re: Slow Step

No I dont think the second step would be treated under equilibrium. The rate would just be determined by the slow first step. The second step does not go to equilibrium because the first slow step occurs then the second faster step happens.

Equilibrium happens when a later step or second step is slower, so the first step is hindered or in a paused state, so over time the first (faster) step goes to equilibrium.

Hope this helps!

Sarkis Sislyan 1D
Posts: 31
Joined: Thu Jul 27, 2017 3:00 am

### Re: Slow Step

In mechanisms, everything that comes after the slow-step is not used when determining the overall rate law of the proposed mechanism. It's only when there's a step before the slow, rate-determining step that we may use the pre-equilibrium approach.

Justin Lai 1C
Posts: 50
Joined: Fri Sep 29, 2017 7:04 am

### Re: Slow Step

I don't think the second step would be considered to be in equilibrium. If the first step is the slow step, the second step would just proceed normally. The rate is determined by the slow step so just the first.

The reason why when the second step is slower, that the first step is considered to be in equilibrium is because of the bottleneck. The first step is so much faster than the second that first step is considered to have reached equilibrium already as the second step is proceeding.

Angel R Morales Dis1G
Posts: 50
Joined: Sat Jul 22, 2017 3:01 am

### Re: Slow Step

The step treated at equilibrium is the slow step, and any step after that doesn't really matter. All we care about is the slow step and the steps that come before it.

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