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Rate law in multiple step reactions

Posted: Sat Mar 17, 2018 12:37 am
by Paula Sing 1J
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?

Re: Rate law in multiple step reactions  [ENDORSED]

Posted: Sat Mar 17, 2018 1:48 am
by GabrielGarciaDiscussion1i
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.

Re: Rate law in multiple step reactions

Posted: Sat Mar 16, 2019 4:15 pm
by Destiny Diaz 4D
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.

Re: Rate law in multiple step reactions

Posted: Sat Mar 16, 2019 10:58 pm
by Madison Hurst
You must replace the intermediates with their equal rate law expressions using the molar ratios when needed.