Rate law in multiple step reactions  [ENDORSED]

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Paula Sing 1J
Posts: 36
Joined: Fri Sep 29, 2017 7:06 am

Rate law in multiple step reactions

Postby Paula Sing 1J » Sat Mar 17, 2018 12:37 am

How are we supposed to differentiate the cases in which the overall rate law is equal to the rate law of the slow step vs when it is not?

GabrielGarciaDiscussion1i
Posts: 71
Joined: Fri Sep 29, 2017 7:04 am
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Re: Rate law in multiple step reactions  [ENDORSED]

Postby GabrielGarciaDiscussion1i » Sat Mar 17, 2018 1:48 am

This question was worded a little tricky but I will try my best to answer it.

The overall rate law should ALWAYS be equal to the rate law of the slow step. However, I believe what you may mean to ask is "How do we differentiate between the instances where the rate law is simply:
A+B-->C rate=k[A][B]

versus the complicated k1/k'1 stuff we do. If the rate law of the slow step contains an intermediate, we CANNOT have that and so we rearrange to get it replaced with ratios of reaction constants. HOWEVER, it is still the same as the overall rate law, which is still the rate law of the slowest step.

Destiny Diaz 4D
Posts: 51
Joined: Fri Sep 28, 2018 12:28 am

Re: Rate law in multiple step reactions

Postby Destiny Diaz 4D » Sat Mar 16, 2019 4:15 pm

so the overall rate law is dependent on its slowest step, so it is always going to be the determinate in the overall rate. To help remember this, I relate this to the quote "you are only as strong as your weakest link" meaning the reaction is only as fast as its slowest step.

Madison Hurst
Posts: 62
Joined: Fri Sep 28, 2018 12:26 am

Re: Rate law in multiple step reactions

Postby Madison Hurst » Sat Mar 16, 2019 10:58 pm

You must replace the intermediates with their equal rate law expressions using the molar ratios when needed.


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