Reaction rate depending on steps

[Natasha Sutkin 2B]

Is there a difference in calculating the reaction rate for a reaction with the slow step first versus one with the slow step second? If so, how do you go about solving these problems?

[Audrey Hanna]

I'd recommend referring to today's lecture (3/13/23) as Dr. Lavelle discussed it in class today!

[Sophia C]

I don't think that there is a difference in the way that you solve them. However it is important to note that when coming up with the rate law of a reaction with multiple elementary steps, the slow step is the one that determines this rate law. Also, remember that intermediates are not included in these laws and will need to be substituted by another step using the pre-equilibrium approach(discussed in the most recent lecture).

[Alyson Chou 3H]

There is no difference, but you must note that intermediates should not be added to the rate law even if it is in the slowest step. In order to remove the intermediate, you can use the pre-equilibrium approach.

[Natalie Bender]

There's no difference since the slow step is always the one that determine the rxn rate, but if there is an intermediate in your rate law then you have to look at the fast rxn (if there are 2 fast rxns then look at the rxn that is fast and equilibrium) and determine what the K would be for that fast rxn. From there you use the K expression to solve for the concentration of the intermediate. Once you have that expression you would substitute it in for the intermediate in the rate law you found for the slow step.