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Topic: Kinetics Intermediate question  (Read 1647 times)

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Offline tla

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Kinetics Intermediate question
« on: February 09, 2012, 12:14:18 PM »
So, I have been teaching AP chemistry for a couple years now and today my students stumped me.  We were going over reaction mechanisms and discussing how the rate of a reaction is determined by the slowest step.  I have always said that the rate is only determined by the species in the slowest step.  The problem below is the one we were going over when my kids stumped me:



                Step 1:   A   +    B   :rarrow: I1                (fast)
      Step 2:    A   +    I1    :rarrow:   I2       (slow)
      Step 3:    C    +    I2    :rarrow:  D                   (fast)

So I said that the rate is only dependent on the second step because it is the slowest.  And my students asked me:  Since the first step produces I1 and that is involved in the slow step, how is it that increasing the concentration of either A or B (which makes I1 ) which would increase the I1 for the slow step won't affect the rate?  I had no good answer for this, and was hoping someone could *delete me*   

Second question:  I also passed out a HW sheet that had a rate determining step with an intermediate in it.  The students asked if it was ok to write the rate law with an intermediate in it?  My guess is no, so how do you write a rate law if the slow step has an intermediate in it?  Thanks for any *delete me*  

Offline Schrödinger

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Re: Kinetics Intermediate question
« Reply #1 on: February 10, 2012, 07:34:25 AM »
Actually the rate depends on all reactants that appear in and above the slowest step. The intermediate I1 can be eliminated. That's all. The rate of a reaction depends on the slowest step doesnt imply that if there is a fast step before it, you should ignore it because it doesnt make a difference. It does. The fast one after the slow one doesnt matter
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