(B) The rate law for the reaction $A + 2B \to C + D$ can be written as $Rate = k[A]^x[B]^y$,where $x$ and $y$ are the orders with respect to $A$ and $

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(B) The rate law for the reaction $A + 2B \to C + D$ can be written as $Rate = k[A]^x[B]^y$,where $x$ and $y$ are the orders with respect to $A$ and $B$ respectively.According to the problem,when $[A]$ is increased $9$ times,the rate increases $3$ times:$3 \times Rate = k(9[A])^x[B]^y$$3 = 9^x \implies 3 = (3^2)^x \implies 3^1 = 3^{2x} \implies x = \frac{1}{2}$.When $[B]$ is increased $2$ times,the rate increases $2$ times:$2 \times Rate = k[A]^x(2[B])^y$$2 = 2^y \implies y = 1$.The total order of the reaction is $x + y = \frac{1}{2} + 1 = \frac{3}{2}$.

Class 12 Chemistry Chemical Kinetics Rate law , Rate constant , Order of Reaction and Molecularity

Medium

For a chemical reaction,$A + 2B \to C + D$,the rate of reaction increases $3$ times when the concentration of $A$ only is increased $9$ times. While when the concentration of $B$ only is increased $2$ times,the rate of reaction also increases $2$ times. The order of this reaction is:

A
$3$
B
$\frac{3}{2}$
C
$\frac{1}{2}$
D
None of these

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Class 12 Questions

Class 12

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Chemical Kinetics

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Rate law , Rate constant , Order of Reaction and Molecularity

The diazonium salt decomposes as $C_6H_5N_2^+Cl^- \to C_6H_5Cl + N_2$. At $0 \ ^\circ C$,the evolution of $N_2$ becomes two times faster when the initial concentration of the salt is doubled. Therefore,it is:

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The initial rates of decrease of $I_2$ in acetone-iodine reaction catalysed by $H^{+}$ are given in the table.
Experiment Initial $[I_2]$ $(mol \ L^{-1})$ Initial $[H^{+}]$ $(mol \ L^{-1})$ Initial $[CH_3COCH_3]$ $(mol \ L^{-1})$ Initial rate $(mol \ L^{-1} \ s^{-1})$
$1$ $0.01$ $0.1$ $0.1$ $0.096$
$2$ $0.01$ $0.2$ $0.1$ $0.192$
$3$ $0.02$ $0.2$ $0.1$ $0.192$
$4$ $0.01$ $0.2$ $0.2$ $0.384$

The order with respect to $I_2, H^{+}$,acetone and total order of the reaction respectively are:

EasyAP EAMCET 2018

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$A$ chemical reaction takes place in two steps as follows:
$(i) \ NO_2Cl_{(g)} \xrightarrow{K_1} NO_{2(g)} + Cl_{(g)}$
$(ii) \ NO_2Cl_{(g)} + Cl_{(g)} \xrightarrow{K_2} NO_{2(g)} + Cl_{2(g)}$
Identify the reaction intermediate.

MediumMHT CET 2016

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For a reaction,$A + 2B \rightarrow$ Products,when the concentration of $B$ alone is increased,the half-life remains the same. If the concentration of $A$ alone is doubled,the rate remains the same. The unit of the rate constant for the reaction is:

MediumKCET 2021

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For a reaction $A + B \rightarrow \text{Product}$,the order with respect to $A$ is $2$ and with respect to $B$ is $3$. If the concentration of both is doubled,by how much will the rate increase?

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Experiment	Initial $[I_2]$ $(mol \ L^{-1})$	Initial $[H^{+}]$ $(mol \ L^{-1})$	Initial $[CH_3COCH_3]$ $(mol \ L^{-1})$	Initial rate $(mol \ L^{-1} \ s^{-1})$
$1$	$0.01$	$0.1$	$0.1$	$0.096$
$2$	$0.01$	$0.2$	$0.1$	$0.192$
$3$	$0.02$	$0.2$	$0.1$	$0.192$
$4$	$0.01$	$0.2$	$0.2$	$0.384$

For a chemical reaction,$A + 2B \to C + D$,the rate of reaction increases $3$ times when the concentration of $A$ only is increased $9$ times. While when the concentration of $B$ only is increased $2$ times,the rate of reaction also increases $2$ times. The order of this reaction is:

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The diazonium salt decomposes as $C_6H_5N_2^+Cl^- \to C_6H_5Cl + N_2$. At $0 \ ^\circ C$,the evolution of $N_2$ becomes two times faster when the initial concentration of the salt is doubled. Therefore,it is:

$A$ chemical reaction takes place in two steps as follows:$(i) \ NO_2Cl_{(g)} \xrightarrow{K_1} NO_{2(g)} + Cl_{(g)}$$(ii) \ NO_2Cl_{(g)} + Cl_{(g)} \xrightarrow{K_2} NO_{2(g)} + Cl_{2(g)}$Identify the reaction intermediate.

For a reaction,$A + 2B \rightarrow$ Products,when the concentration of $B$ alone is increased,the half-life remains the same. If the concentration of $A$ alone is doubled,the rate remains the same. The unit of the rate constant for the reaction is:

For a reaction $A + B \rightarrow \text{Product}$,the order with respect to $A$ is $2$ and with respect to $B$ is $3$. If the concentration of both is doubled,by how much will the rate increase?

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$A$ chemical reaction takes place in two steps as follows:
$(i) \ NO_2Cl_{(g)} \xrightarrow{K_1} NO_{2(g)} + Cl_{(g)}$
$(ii) \ NO_2Cl_{(g)} + Cl_{(g)} \xrightarrow{K_2} NO_{2(g)} + Cl_{2(g)}$
Identify the reaction intermediate.