(B) Let the rate law be $r = k[A]^x[B]^y$. According to the problem,when the concentration of $B$ is doubled,the rate doubles: $2r = k[A]^x(2[B])^y$.

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(B) Let the rate law be $r = k[A]^x[B]^y$. According to the problem,when the concentration of $B$ is doubled,the rate doubles: $2r = k[A]^x(2[B])^y$. This implies $2^y = 2$,so $y = 1$. When the concentration of both $A$ and $B$ is doubled,the rate increases by a factor of $8$: $8r = k(2[A])^x(2[B])^y$. Substituting $y = 1$,we get $8r = k(2^x[A]^x)(2[B]) = 2^{x+1}k[A]^x[B]$. Since $r = k[A]^x[B]$,we have $8 = 2^{x+1}$,which means $2^3 = 2^{x+1}$. Therefore,$x + 1 = 3$,so $x = 2$. The rate law is $r = k[A]^2[B]$.

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

Medium

For a reaction $2A + B \to \text{Products}$,doubling the initial concentration of both the reactants increases the rate by a factor of $8$,and doubling the concentration of $B$ alone doubles the rate. The rate law for the reaction is

A
$r = k[A][B]^2$
B
$r = k[A]^2[B]$
C
$r = k[A][B]$
D
$r = k[A]^2[B]^2$

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Chemical Kinetics

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Rate law , Rate constant , Order of Reaction and Molecularity

For 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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For a certain reaction $A \to P$,the half-life for different initial concentrations of $A$ is mentioned below:
$[A_0]$ $0.1 \ M$ $0.025 \ M$
$t_{1/2} \ (s)$ $100 \ s$ $50 \ s$

Which of the following option$(s)$ is/are correct?

Medium

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Consider the following reaction: $A \longrightarrow \text{Products}$. This reaction is completed in $100 \ min$. The rate constant of this reaction at $t_1 = 10 \ min$ is $10^{-2} \ min^{-1}$. What is the rate constant (in $min^{-1}$) at $t_2 = 20 \ min$?

MediumAP EAMCET 2022

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Reaction : $2Br^{-} + H_2O_2 + 2H^{+} \to Br_2 + 2H_2O$
takes place in two steps :
$(a)$ $Br^{-} + H^{+} + H_2O_2 \xrightarrow{slow} HOBr + H_2O$
$(b)$ $HOBr + Br^{-} + H^{+} \xrightarrow{fast} H_2O + Br_2$
The order of the reaction is

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The rate constant for the reaction, $2 \,N_2O_{5(g)} \rightarrow 2 \,N_2O_{4(g)} + O_{2(g)}$ is $4.98 \times 10^{-4} \,s^{-1}$. What is the order of the reaction?

EasyMHT CET 2025

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For a reaction $2A + B \to \text{Products}$,doubling the initial concentration of both the reactants increases the rate by a factor of $8$,and doubling the concentration of $B$ alone doubles the rate. The rate law for the reaction is

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For 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:

For a certain reaction $A \to P$,the half-life for different initial concentrations of $A$ is mentioned below:$[A_0]$$0.1 \ M$$0.025 \ M$$t_{1/2} \ (s)$$100 \ s$$50 \ s$Which of the following option$(s)$ is/are correct?

Consider the following reaction: $A \longrightarrow \text{Products}$. This reaction is completed in $100 \ min$. The rate constant of this reaction at $t_1 = 10 \ min$ is $10^{-2} \ min^{-1}$. What is the rate constant (in $min^{-1}$) at $t_2 = 20 \ min$?

Reaction : $2Br^{-} + H_2O_2 + 2H^{+} \to Br_2 + 2H_2O$takes place in two steps :$(a)$ $Br^{-} + H^{+} + H_2O_2 \xrightarrow{slow} HOBr + H_2O$$(b)$ $HOBr + Br^{-} + H^{+} \xrightarrow{fast} H_2O + Br_2$The order of the reaction is

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For a certain reaction $A \to P$,the half-life for different initial concentrations of $A$ is mentioned below:
$[A_0]$ $0.1 \ M$ $0.025 \ M$
$t_{1/2} \ (s)$ $100 \ s$ $50 \ s$

Which of the following option$(s)$ is/are correct?

Reaction : $2Br^{-} + H_2O_2 + 2H^{+} \to Br_2 + 2H_2O$
takes place in two steps :
$(a)$ $Br^{-} + H^{+} + H_2O_2 \xrightarrow{slow} HOBr + H_2O$
$(b)$ $HOBr + Br^{-} + H^{+} \xrightarrow{fast} H_2O + Br_2$
The order of the reaction is