Iodide and hypochlorite ion react in aqueous solution according to the reaction below: I^- + OCl^- → OI^- + Cl^-
If the concentration of OH^- in the solution is doubled,the rate of the reaction is halved.What is the order of the reaction with respect to OH^-?

Question

Quizplus · Accepted Answer

The rate law equation for the given reaction can be written as:
rate = k[IO^-][OCl^-]
where k is the rate constant and the order of the reaction with respect to each reactant is 1. 
Let's assume that the initial concentration of OCl^- is constant and equal to [OCl^-]0. 
If the concentration of OH^- is doubled, then the rate of the reaction becomes: 
rate' = k[IO^-][2OH^-][OCl^-]0
Since the rate' is halved, we can write: 
1/2 rate = k[IO^-][OH^-][OCl^-]0
Dividing this equation by the previous one, we get:
(1/2 rate)/rate' = ([IO^-][OH^-])/([IO^-][2OH^-]) = 1/2
Therefore, the order of the reaction with respect to OH^- is 1 (since doubling its concentration leads to a doubling of the rate), which corresponds to choice B.

Iodide and Hypochlorite Ion React in Aqueous Solution According to the Reaction