Why can we not determine the order of a reaction by taking into consideration the balanced chemical equation?
(N/A) The balanced chemical equation represents the overall stoichiometry of a reaction but does not necessarily reflect the mechanism or the rate-determining step.
$1.$ The order of a reaction is an experimental quantity,whereas the stoichiometry is derived from the balanced equation.
$2.$ Many reactions are complex and occur in multiple steps. The rate of the overall reaction is determined by the slowest step in the mechanism.
$3.$ For example,consider the reaction: $CHCl_3 + Cl_2 \rightarrow CCl_4 + HCl$. Based on the stoichiometry,one might incorrectly predict a second-order reaction,but the experimental rate law is $Rate = k[CHCl_3][Cl_2]^{1/2}$,which is of order $1.5$.
$4.$ Thus,the order of a reaction cannot be determined simply by looking at the balanced chemical equation.
$1.$ The order of a reaction is an experimental quantity,whereas the stoichiometry is derived from the balanced equation.
$2.$ Many reactions are complex and occur in multiple steps. The rate of the overall reaction is determined by the slowest step in the mechanism.
$3.$ For example,consider the reaction: $CHCl_3 + Cl_2 \rightarrow CCl_4 + HCl$. Based on the stoichiometry,one might incorrectly predict a second-order reaction,but the experimental rate law is $Rate = k[CHCl_3][Cl_2]^{1/2}$,which is of order $1.5$.
$4.$ Thus,the order of a reaction cannot be determined simply by looking at the balanced chemical equation.
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Similar Questions
$A$ reaction is first order in $A$ and second order in $B$.
$(i)$ Write the differential rate equation.
$(ii)$ How is the rate affected on increasing the concentration of $B$ three times?
$(iii)$ How is the rate affected when the concentrations of both $A$ and $B$ are doubled?
$(i)$ Write the differential rate equation.
$(ii)$ How is the rate affected on increasing the concentration of $B$ three times?
$(iii)$ How is the rate affected when the concentrations of both $A$ and $B$ are doubled?
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$2B \xrightarrow{k} B_2$ $[Slow]$
$B_2 + A \to D$ $[Fast]$
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$2B \xrightarrow{k} B_2$ $[Slow]$
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The rate law expression,order with respect to $A$,order with respect to $B$ and overall order of reaction are respectively
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View SolutionThe reaction rate between two substances $A$ and $B$ is expressed as: $\text{rate} = k[A]^n[B]^m$. If the concentration of $A$ is doubled and the concentration of $B$ is halved,the ratio of the new rate to the initial rate will be:
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View SolutionFor the reaction $H_{2(g)} + Br_{2(g)} \to 2HBr_{(g)}$,the experimental data suggest,$\text{rate} = K[H_2][Br_2]^{1/2}$. The molecularity and order of the reaction are respectively:
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