## Quiz 2 Question

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

Elle_Bertuccelli_1B
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Joined: Sat Jul 09, 2016 3:00 am

### Quiz 2 Question

I'm confused about #3, part a on Quiz 2 (where we are asked to determine the rate limiting step):

It makes sense that k1 << k2 implies that k1 is the slow step. However, I don't understand why the reaction at equilibrium is considered the slow step. Usually the slow step isn't at equilibrium and the fast step is at equilibrium. Help!

Jenny2G
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Joined: Wed Sep 21, 2016 2:56 pm

### Re: Quiz 2 Question

The show step has two double sided arrows signifying that they can be reversible, one sided arrows means that they are at equilibrium therefore the slow step can have two arrows. It is not possible for the slow step to be at equilibrium.

Chem_Mod
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### Re: Quiz 2 Question

Jenny2G wrote:The show step has two double sided arrows signifying that they can be reversible, one sided arrows means that they are at equilibrium therefore the slow step can have two arrows. It is not possible for the slow step to be at equilibrium.

This is incorrect. Stacked forward and reverse arrows indicate equilibrium in chemistry. The double headed vs. single headed reactions arrows are just an artifact of word processing of the document. Just because a reaction is at equilibrium, doesn't indicate whether it is fast or slow. It could take the reaction a long time to reach equilibrium (k1 is small) or it could be nearly instantaneous (k1 is large). Either way, if it is the slow step, then your rate law is determined by that step.