Explain the order of reaction for complex reactions by giving examples.

(N/A) Complex reactions involving more than three molecules: Reactions involving more than three molecules in the stoichiometric equation must take place in more than one step. For example:
$KClO_{3} + 6 FeSO_{4} + 3 H_{2}SO_{4} \rightarrow KCl + 3 Fe_{2}(SO_{4})_{3} + 3 H_{2}O$
This reaction,which apparently seems to be of $10^{th}$ order based on stoichiometry,is actually a second-order reaction.
$(b)$ This indicates that the reaction occurs in several steps,but the slowest step determines the rate of reaction. The overall rate of the reaction is controlled by the slowest step,known as the rate-determining step.
Example: The decomposition of hydrogen peroxide catalyzed by iodide ion in an alkaline medium:
$2 H_{2}O_{2} \xrightarrow{I^{-} / \text{alkaline medium}} 2 H_{2}O + O_{2}$
The rate equation is found to be: $\text{Rate} = k[H_{2}O_{2}][I^{-}]$
Thus,the order with respect to $H_{2}O_{2}$ is $1$,and with respect to $I^{-}$ is $1$. The overall order of reaction is $(1 + 1) = 2$.
The decomposition of $H_{2}O_{2}$ occurs in two steps:
$(i)$ $H_{2}O_{2} + I^{-} \rightarrow H_{2}O + IO^{-} \quad (\text{slow step})$
$(ii)$ $H_{2}O_{2} + IO^{-} \rightarrow H_{2}O + I^{-} + O_{2} \quad (\text{fast step})$
Overall reaction: $2 H_{2}O_{2} \rightarrow 2 H_{2}O + O_{2}$
Both steps are bimolecular elementary reactions. The species $IO^{-}$ is an intermediate. The first step,being slow,is the rate-determining step. The order of the slowest step equals the molecularity of the slowest step.