Answer: Option (d) is the correct answer.
Explanation:
The given equations as as follows.
[tex]H_{2}O_{2} + I^{-} \rightarrow H_{2}O + IO^{-}[/tex] (slow)
[tex]H_{2}O_{2} + IO^{-} \rightarrow H_{2}O + O_{2} + I^{-}[/tex]
Therefore, overall reaction equation will be as follows.
[tex]2H_{2}O_{2} + I^{-} + IO^{-} \rightarrow 2H_{2}O + O_{2} + IO^{-} + I^{-}[/tex]
So, cancelling the spectator ions then the equation will be as follows.
[tex]H_{2}O_{2} \rightarrow 2H_{2}O + O_{2}[/tex]
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.
[tex]H_{2}O_{2} + I^{-} \rightarrow H_{2}O + IO^{-}[/tex]
Rate law = [tex]k[H_{2}O_{2}][I^{-}][/tex]
Hence, the reaction is first order with respect to [tex][I^{-}][/tex] and it is also first order reaction with respect to [tex][H_{2}O_{2}][/tex].
Also, [tex][I^{-}][/tex] acts as a catalyst in the reaction.
Thus, we can conclude that the incorrect statement is [tex]IO^{-}[/tex] is a catalyst.
