Medium
Write the differential rate expression for the following reaction and determine its order of reaction:
$H_{2}O_{2} + I^{-} \rightarrow H_{2}O + IO^{-}$
$H_{2}O_{2} + IO^{-} \rightarrow H_{2}O + I^{-} + O_{2}$
$H_{2}O_{2} + I^{-} \rightarrow H_{2}O + IO^{-}$
$H_{2}O_{2} + IO^{-} \rightarrow H_{2}O + I^{-} + O_{2}$
(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.
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 following mechanism has been proposed for the reaction of $NO$ with $Br_2$ to form $NOBr$:
$NO_{(g)} + Br_{2(g)} \rightleftharpoons NOBr_{2(g)}$
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$NO_{(g)} + Br_{2(g)} \rightleftharpoons NOBr_{2(g)}$
$NOBr_{2(g)} + NO_{(g)} \longrightarrow 2NOBr_{(g)}$
If the second step is the rate-determining step,the order of the reaction with respect to $NO_{(g)}$ is:
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