6th edition 15.49

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Maddy Mackenzie
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6th edition 15.49

Postby Maddy Mackenzie » Tue Mar 12, 2019 9:37 pm

For this question, why can we say that the rate for step 2 is rate=k[HBr][HOBr] when [HOBr] is an intermediate? I thought that you could not include intermediates in the rate law.

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Re: 6th edition 15.49

Postby CaminaB_1D » Tue Mar 12, 2019 9:43 pm

someone correct me if I'm wrong but if it asks for the rate law of that specific step in the rxn I think you do include intermediates. However it is not included in the rate law of the overall reaction. The slow step indicates the overall rate law so if there was an intermediate in the reactants of the slow step then you would not include it

Sophia Ding 1B
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Re: 6th edition 15.49

Postby Sophia Ding 1B » Wed Mar 13, 2019 3:29 pm

I think that holds true; for the steps of the rate law itself, you can include the intermediate. This allows you to substitute a factor of the later equations to find the overall rate law, which should not include the intermediate.

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