Question

Below is a proposed mechanism for the decomposition of H2O2. H2O2 + I– → H2O + IO– slow H2O2 + IO– → H2O + O2 + I– fast Which of the following statements is incorrect? a. IO– is a catalyst. b. The reaction is first-order with respect to [I–]. c. The reaction is first-order with respect to [H2O2]. d. The net reaction is 2H2O2 → 2H2O + O2. e. I– is a catalyst.

Answer 1

Answer: Option (d) is the correct answer.

Explanation:

The given equations as as follows.

       $H_{2}O_{2} + I^{-} \rightarrow H_{2}O + IO^{-}$           (slow)

       $H_{2}O_{2} + IO^{-} \rightarrow H_{2}O + O_{2} + I^{-}$

Therefore, overall reaction equation will be as follows.

     $2H_{2}O_{2} + I^{-} + IO^{-} \rightarrow 2H_{2}O + O_{2} + IO^{-} + I^{-}$

So, cancelling the spectator ions then the equation will be as follows.

         $H_{2}O_{2} \rightarrow 2H_{2}O + O_{2}$

As, it is known that slow step of a reaction is the rate determining step. Therefore, rate law for the slow step will be as follows.

        $H_{2}O_{2} + I^{-} \rightarrow H_{2}O + IO^{-}$

                 Rate law = $k[H_{2}O_{2}][I^{-}]$

Hence, the reaction is first order with respect to $[I^{-}]$ and it is also first order reaction with respect to $[H_{2}O_{2}]$.

Also, $[I^{-}]$ acts as a catalyst in the reaction.

Thus, we can conclude that the incorrect statement is $IO^{-}$ is a catalyst.