## Rate Determining Step

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

ntruong2H
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Joined: Sat Jul 09, 2016 3:00 am
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### Rate Determining Step

Can someone please explain to me what happens if we have a step that is said to be "very very slow"? Does this now mean that this is the rate determining step? Also, can someone please explain what happens if there is a fast, equilibrium step that comes before the slow step? Do we now include the reactants in the fast, equilibrium step (since it came before the slow step) in the overall rate law expression? Also, if there is an intermediate in the slow step, is that reactant still reported in the overall rate law expression? Thank you.

Natalie Rotstein 3J
Posts: 20
Joined: Wed Sep 21, 2016 2:59 pm

### Re: Rate Determining Step

If a step is "very very slow," and that is the slowest specification for the rate of any step, it is the rate determining step. Because that step determines rate, a faster step prior to it would not affect the rate of the reaction, and therefore would not be included in the rate expression.

ntruong2H
Posts: 50
Joined: Sat Jul 09, 2016 3:00 am
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### Re: Rate Determining Step

Natalie Rotstein 3J wrote:If a step is "very very slow," and that is the slowest specification for the rate of any step, it is the rate determining step. Because that step determines rate, a faster step prior to it would not affect the rate of the reaction, and therefore would not be included in the rate expression.

Okay that's what I thought thanks. And what if there are intermediates in the slow step, do we include them in our rate law?

Jazmin_Morales_3J
Posts: 17
Joined: Wed Sep 21, 2016 2:59 pm

### Re: Rate Determining Step

Hi there,

Intermediates are not included in the overall rate law because intermediates are not final products. So because intermediates are not observable, they are not included in the rate law that is determined experimentally. Hope that helps!