Final Review

[Erik Buetow 1F]

For reaction mechanisms that have an intermediate in the rate law, I don't completely understand the concept behind multiplying rates 1 and 2 to determine whether or not that mechanism adheres to the determined rate law. Can someone explain how we multiply these rates?

[sbeall_1C]

I'm not sure if this is the what you are getting at, but when there are intermediates in the rate-determining step (i.e the slow step), you need to replace the intermediates with the correct reactants to match the determined rate law. In order to do this, you need to solve for the intermediates using the other equations, like in the pre-equilibrium approach. Doing this will result in multiple rate constants in the final rate law. You can start by writing a capital K which will equal to the classic equilibrium constant of reactants over products, and then each elementary step will have a rate constant of little k and going in the reverse reaction will lend a k prime. Multiplying these all together will result in your final K value for your rate law.

[Daniel Yu 1E]

You aren't multiplying rate 1 and 2. You are assuming that Step 2 is the slow step. Using the pre-equilibrium approach, we treat reaction 1 as if it is in equilibrium. Since reaction 1 is in equilibrium, the forward rate is equal to the reverse reaction rate. Using this, we can get rid of the intermediate in the reaction mechanism for the 2nd step (because you don't want intermediates since they are instantly consumed). To get rid of the intermediate, you just set up step 1 forward rate and reverse rate, set up equal to each other, divide k1 by k1' to get K1, then rearrange until the intermediate is by itself. Then, substitute the equation that you just got for the intermediate in the Step 2. This is the differential rate for step 2 using the pre-equilibrium approach.