For a reaction $2{\mathrm{H}}_2{\mathrm{O}}_2 \underset{\mathrm{Alkaline} \mathrm{medium}}{\overset{{\mathrm{I}}^{-}}{\to }} 2{\mathrm{H}}_2\mathrm{O}+{\mathrm{O}}_2$, the proposed mechanism is as given below:
$\begin{array}{l}\left(1\right) {\mathrm{H}}_2{\mathrm{O}}_2+{\mathrm{I}}^{-}\to {\mathrm{H}}_2\mathrm{O}+{\mathrm{IO}}^{-}\text{ (slow) }\\ \left(2\right) {\mathrm{H}}_2{\mathrm{O}}_2+{\mathrm{IO}}^{-}\to {\mathrm{H}}_2\mathrm{O}+{\mathrm{I}}^{-}+{\mathrm{O}}_2\text{ (fast) }\end{array}$
(i) Write rate law for the reaction.
(ii) Write the overall order of reaction.
(iii) Out of steps (1) and (2), which one is rate determining step?

(i) To find the rate expression, we consider rate determining step, which is step (I) for this reaction. Then according to the rate law, Rate=k[H₂O₂][I^-] 

(ii) Since, two reagents are involved in the rate-limiting step, thus overall rate of reaction is proportional to the concentration of those reagents, that is, 2. 

(iii) The slowest step in this reaction is called the rate-determining step, that is step (1).