How to determine a Rate Law from a reaction mechanism?

I thought the slow reaction determined the rate law. Why is this question telling me to do something other than that? So in the first part of the question I calculated the Rate Law from a table, which is "Rate = k [CH3COCH3][H+]." This is the rate mechanism (the total formula is at the top): http://tinypic.com/r/23k2gk4/5 . I'm confused on this... the middle step is the rate determining step, yet it is not in the rate law. I need to explain why this reaction mechanism is consistent with the rate law I found earlier (I'm sure it's right... I double checked). I'm not sure how to do that
Answer:

Unfortunately I can't open that pic, but when working with multiple steps there are a couple things to remember. First off, you are correct in that the slow step (or rate determining step) will determine the overall rate of the reaction. So the rate law described by this step should match the overall rate law. The most common mistake is when there is an intermediate involved where the rate law predicted from the slow step includes a species that is not in the overall equation. In this case you usually use one of the other steps to represent the intermediate in terms of what is actually present in the overall equation. More often then not the first step is used where the forward reaction and reverse reaction are set equal to one another. The intermediate is isolated on one side of the equation then plugged back into the rate determining step.

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The reason why the species in the 2nd step don't appear in the rate law is because the two organic species are intermediates that are used up in the reaction. The rate law must contain only reactants. Fortunately, you can deduce alternate expressions for the two intermediates by using the first (fast equilibrium step). If you wrote the rate law according to step 2 it would be rate = k[CH3C(=OH+)CH3][H2O] . . .but CH3C(=OH+)CH3 is an intermediate. Since step 1 is a fast, equilibrium step, then the forward rate of the reaction and the reverse rate of reaction are the same. forward rate = k1[CH3C(=O)CH3][H3O+] and the reverse rate = k-1[CH3C(=OH+)CH3][H2O] Set these two equal to each other. k1[CH3C(=O)CH3][H3O+] = k-1[CH3C(=OH+)CH3][H2O] . . . now solve for the intermediate. [CH3C(=OH+)CH3] = ((k1)[CH3C(=O)CH3][H3O+]) / ((k-1)[H2O]) Substitute the above into the very first rate equation that we wrote which is rate = k[CH3C(=OH+)CH3][H2O] rate = k((k1)[CH3C(=O)CH3][H3O+]) / ((k-1)[H2O])[H2O] . . .if we lump all the k's together as k2 ... rate = k2[CH3C(=O)CH3][H3O+] which is an acceptable rate law since it contains only reactants.

HPV

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