7.23b


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

7.23b

Postby JamieVu_2C » Sun Mar 08, 2020 6:56 pm

ClO- + H2O HClO + OH- (fast equilibrium)
HClO + I- --> HIO + Cl- (very slow)
HIO + OH- +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: . 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

Postby Ryan 1K » Sun Mar 08, 2020 9:32 pm

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 , and the rate of the reverse reaction would be . 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 . Divide both sides by k1' and to get . Keep in mind that the rate constants are not necessarily equal to the concentration values, which is why you were confused I believe.


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