Write the differential rate expression for the following reaction and determine its order of reaction:
$H_{2}O_{2} + I^{-} \rightarrow H_{2}O + IO^{-}$
$H_{2}O_{2} + IO^{-} \rightarrow H_{2}O + I^{-} + O_{2}$

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(N/A) The overall reaction is the sum of the two elementary steps: $2H_{2}O_{2} \rightarrow 2H_{2}O + O_{2}$.
The rate of the reaction is determined by the slow step,which is the first step: $H_{2}O_{2} + I^{-} \rightarrow H_{2}O + IO^{-}$.
The differential rate expression is given by: $Rate = -\frac{d[H_{2}O_{2}]}{dt} = k[H_{2}O_{2}][I^{-}]$.
The order of the reaction is the sum of the powers of the concentration terms in the rate law: $1 + 1 = 2$. Thus,the reaction is of second order.

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The order of a reaction can have:

Consider the following data for the given reaction $2 HI_{(g)} \rightarrow H_{2_{(g)}} + I_{2_{(g)}}$. The order of the reaction is:
Experiment $1$ $2$ $3$
$[HI] \ (mol \ L^{-1})$ $0.005$ $0.01$ $0.02$
Rate $(mol \ L^{-1} \ s^{-1})$ $7.5 \times 10^{-4}$ $3.0 \times 10^{-3}$ $1.2 \times 10^{-2}$

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