7.23b

$K = \frac{k_{forward}}{k_{reverse}}$

JamieVu_2C
Posts: 108
Joined: Thu Jul 25, 2019 12:16 am

7.23b

ClO- + H2O $\rightleftharpoons$ HClO + OH- (fast equilibrium)
HClO + I- --> HIO + Cl- (very slow)
HIO + OH- $\rightleftharpoons$ +H2O (fast equilibrium)

(b) Write the rate law based on this mechanism.

Since step 2 is slow, the rate law is rate = k2[HClO][I-]. Then, the solutions manual shows that you should substitute [HClO] because it is an intermediate. Then, they get: $K = \frac{k_{1}}{k_{1}^{'}} = \frac{[HOCl][OH-]}{[ClO-]}$. How does k1 = [HOCl][OH-] and k'1 = [ClO-] when k1 should equal the rate of the forward reaction to get k1 = [ClO-][H2O] and k'1 should equal the rate of the reverse reaction to get k'1 = [HClO][OH-}?
Also, why did the book not include [H2O] in the equation?

Ryan 1K
Posts: 101
Joined: Fri Aug 09, 2019 12:15 am

Re: 7.23b

H20 is not included in the reaction since it is a liquid, and liquids are not included in equilibrium constant equations.

The rate of the forward reaction would be $rate = k1[ClO-]$, and the rate of the reverse reaction would be $rate = k1'[HClO][OH-]$. Since the rate of the forward reaction is equal to the rate of the reverse reaction at equilibrium, you can set these two equations equal to each other. As a result $k1[ClO-]=k1'[HClO][OH-]$. Divide both sides by k1' and $[ClO-]$ to get $K = \frac{k1}{k1'}=\frac{[HClO][OH-]}{[ClO-]}$. Keep in mind that the rate constants are not necessarily equal to the concentration values, which is why you were confused I believe.