Achieve Question 13

[005917072]

This was how I solved this Achieve Question, and it is correct but I was unsure about including OH-? From what I can tell it's an intermediate and I thought you're not supposed to include intermediates in rate laws (which is why HClO had to be replaced with ClO-/OH- in the first place), so I don't understand why it's okay to have OH- in this rate law?

[305920501]

In this case, we are asked to write the rate law for the formation of IO- and we must look at the slowest step because this is the rate-determining step. I agree it is weird we have to replace HClO as an intermediate and not OH-. I thought when doing this problem that maybe it has to do with the fact that OH- is not immediately used by the next step to carry on the reaction, because my understanding is an intermediate is something formed by a previous step that is necessary for the succession of the next step, but in this case OH- is not needed. I think this may mean it is not treated the same way as HClO.

[Brian Cho 3E]

To write the rate law for the formation of IO-, you need to first look at the slow step and write the rate law based on that first. You get rate=k[I-][HClO]. As [HClO] is an intermediate, you need to find a way to rewrite this value. [HClO] can also be found in the first step. Write the rate law of the forward and reverse reaction and set them equal to each other as it is fast in both directions. [HClO]=[ClO][H2O]/[OH-]. [H2O] can be cancelled out as it is a catalyst. The resulting rate law is k=[ClO-][I-]/[OH-].