## HW 15.101b

$V_{0} = \frac{V_{max}[S]}{K_{M} + [S]}; K_{M} = \frac{[E][S]}{[ES]}$

Grace Ramey 2K
Posts: 57
Joined: Thu Jul 27, 2017 3:01 am

### HW 15.101b

The question asks for the rate law for the following mechanism:
ClO- + H2O ----> HClO + OH- (fast equilibrium)
HClO + I- ----> HIO + Cl- (very slow)
HIO + OH- -----> IO- + H2O (fast equilibrium)

However, the rate in the solutions manual is
rate = (k2k1/k1') [OCl-][I-]/[OH-]

Isn't OH- an intermediate in the mechanism, and should therefore not be included in the rate? Or is it somehow working as an enzyme?

Austin Ho 1E
Posts: 50
Joined: Fri Sep 29, 2017 7:04 am
Been upvoted: 2 times

### Re: HW 15.101b

You are correct that intermediates should not be in the rate law. However, what we are doing here is substituting the concentration of [HClO] for something that is known. In the first equation, we know that k1[ClO-]=k1'[OH-][HClO]. Thus, [HClO]=k1[ClO-]/k1'[OH-]. This is just then substituted in for [HClO] in the rate of the rate determining step, or the second equation. This gives us the final rate = (k2k2/k1') [OCl-][I-]/[OH-]

The reason why we do this is because HClO itself is an intermediate, thus cannot be in the final rate law. Thus we do this substitution method. The byproduct is that [OH-] appears in the rate law, but only to the -1 power. Additionally, because OH- is a product (and likewise is taken to the -1 power in the rate law), I think you can generally disregard it for the actual rate law because we only use initial concentrations.

Hope this helps!