Sapling Week 9 and 10 Question 13

[Amanda Bueno-Kling]

I am confused on how to do sapling 13, which is about rate laws. However, what I have tried is not working and I am a little stumped. This is the advice the problem gives me to help solve it.

"The rate of the overall reaction is governed by the rate of the slowest step in the mechanism. In this proposed mechanism, the slowest step is step 2. For an elementary step, its rate law can be written from its chemical equation. Therefore, the rate law for the slowest step, where k*slow
is the rate constant, is

rate=k*slow[I−][HClO]"

I understand this part, but not the next part.

"However, HClO is an intermediate that does not appear in the overall reaction, as it is the product of one step in the mechanism and a reactant in a subsequent step. Therefore, it should not be included in the rate law, and a substitute expression is needed for HClO."


HClO also appears in step 1 of the proposed mechanism. Write rate laws for the forward and reverse reactions in step 1. At equilibrium, the rates of these two reactions are equal, so you can set the rate laws equal to each other. Then, rearrange the expression to solve for HClO and substitute it into the rate law.

I do not understand how to do these steps and I was wondering if anyone could help explain?

[Valerie Doan 3I]

The rate determining step gives us the rate law rate=k[I-][HClO], however that can't be our overall rate law because [HClO] is an intermediate. This means we have to utilize the pre-equilibrium approach and assume that the preceding step is at equilibrium. Therefore, we can write K=[OH-][HClO]/[ClO-] and rearrange so that [HClO]=K[ClO-]/[OH-] but also remember that K=k forward/ k reverse which means [HClO]=(k forward/k reverse) * [ClO-]/[OH-] . So, we plug that into our rate law rate=k[I-][HClO] and we should then get a rate law that is expressed in [I-], [ClO-], and [OH-].

[Amanda Bueno-Kling]

Thank you for explaining Valerie! I tried to follow the steps you had and did the following.
k=[OH-][HClO]/[ClO-]

[HClO]=k[ClO-]/[OH-]

k[I-][HClO]
=k[I-]*([ClO-]/[OH-])
=k[I-][ClO-]/[OH-]

Thank you for explaining to me how to arrive to this answer!