(C) The initial rate is $R_1 = k[A]^n[B]^m$. When the concentration of $A$ is doubled $([A]' = 2[A])$ and the concentration of $B$ is halved $([B]' =

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(C) The initial rate is $R_1 = k[A]^n[B]^m$. When the concentration of $A$ is doubled $([A]' = 2[A])$ and the concentration of $B$ is halved $([B]' = \frac{1}{2}[B])$,the new rate $R_2$ is: $R_2 = k(2[A])^n(\frac{1}{2}[B])^m$ $R_2 = k \times 2^n \times [A]^n \times (\frac{1}{2})^m \times [B]^m$ $R_2 = 2^n \times 2^{-m} \times k[A]^n[B]^m$ $R_2 = 2^{(n-m)} \times R_1$ Therefore,the ratio of the new rate to the initial rate is $\frac{R_2}{R_1} = 2^{(n-m)}$.

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

Easy

The rate law for the reaction between substances $A$ and $B$ is given by $\text{Rate} = k[A]^n[B]^m$. If the concentration of $A$ is doubled and the concentration of $B$ is halved,what is the ratio of the new rate to the initial rate?

A
$2^{(n+m)}$
B
$\frac{1}{2^{(m-n)}}$
C
$2^{(n-m)}$
D
$2^{(m-n)}$

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Class 12

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Chemical Kinetics

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Rate law , Rate constant , Order of Reaction and Molecularity

Rate law for the reaction,$C_2H_5I_{(g)} \rightarrow C_2H_{4_{(g)}} + HI_{(g)}$ is $r = k[C_2H_5I]$. What is the order and molecularity of this reaction?

EasyMHT CET 2025

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The following results are obtained for the reaction $S + Nu \rightarrow \text{product}$. By which reaction mechanism does this reaction occur?

Experiment $[S]$ $[Nu]$ Rate

$1$ $0.1$ $0.1$ $2.2 \times 10^{-3}$

$2$ $0.2$ $0.1$ $4.4 \times 10^{-3}$

$3$ $0.1$ $0.2$ $4.4 \times 10^{-3}$

Experiment	$[S]$	$[Nu]$	Rate
$1$	$0.1$	$0.1$	$2.2 \times 10^{-3}$
$2$	$0.2$	$0.1$	$4.4 \times 10^{-3}$
$3$	$0.1$	$0.2$	$4.4 \times 10^{-3}$

EasyGUJCET 2023

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For a reaction between substances $A$ and $B$,the rate law is given by: $\text{Rate} = K[A]^n[B]^m$. If the concentration of $A$ is doubled and the concentration of $B$ is halved,what will be the ratio of the new rate to the original rate?

Medium

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If the rate law of a reaction $n A \rightarrow B$ is expressed as Rate $= -\frac{1}{n} \frac{d[A]}{dt} = \frac{d[B]}{dt} = k[A]^{x}$,the unit of rate constant $k$ will be:

Medium

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Identify the reaction order from each of the following rate constants.
$(i)$ $k = 2.3 \times 10^{-5} \, L \, mol^{-1} \, s^{-1}$
$(ii)$ $k = 3 \times 10^{-4} \, s^{-1}$

Medium

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The rate law for the reaction between substances $A$ and $B$ is given by $\text{Rate} = k[A]^n[B]^m$. If the concentration of $A$ is doubled and the concentration of $B$ is halved,what is the ratio of the new rate to the initial rate?

Similar Questions

Rate law for the reaction,$C_2H_5I_{(g)} \rightarrow C_2H_{4_{(g)}} + HI_{(g)}$ is $r = k[C_2H_5I]$. What is the order and molecularity of this reaction?

The following results are obtained for the reaction $S + Nu \rightarrow \text{product}$. By which reaction mechanism does this reaction occur? Experiment $[S]$ $[Nu]$ Rate $1$ $0.1$ $0.1$ $2.2 \times 10^{-3}$ $2$ $0.2$ $0.1$ $4.4 \times 10^{-3}$ $3$ $0.1$ $0.2$ $4.4 \times 10^{-3}$

For a reaction between substances $A$ and $B$,the rate law is given by: $\text{Rate} = K[A]^n[B]^m$. If the concentration of $A$ is doubled and the concentration of $B$ is halved,what will be the ratio of the new rate to the original rate?

If the rate law of a reaction $n A \rightarrow B$ is expressed as Rate $= -\frac{1}{n} \frac{d[A]}{dt} = \frac{d[B]}{dt} = k[A]^{x}$,the unit of rate constant $k$ will be:

Identify the reaction order from each of the following rate constants.$(i)$ $k = 2.3 \times 10^{-5} \, L \, mol^{-1} \, s^{-1}$$(ii)$ $k = 3 \times 10^{-4} \, s^{-1}$

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The following results are obtained for the reaction $S + Nu \rightarrow \text{product}$. By which reaction mechanism does this reaction occur?

Experiment $[S]$ $[Nu]$ Rate

$1$ $0.1$ $0.1$ $2.2 \times 10^{-3}$

$2$ $0.2$ $0.1$ $4.4 \times 10^{-3}$

$3$ $0.1$ $0.2$ $4.4 \times 10^{-3}$

Identify the reaction order from each of the following rate constants.
$(i)$ $k = 2.3 \times 10^{-5} \, L \, mol^{-1} \, s^{-1}$
$(ii)$ $k = 3 \times 10^{-4} \, s^{-1}$