(A) Applying the steady state approximation to the intermediate $B$:The rate of change of concentration of $B$ is given by: $\frac{d[B]}{dt} = k_1 [A]

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$\left( \frac{k_1}{k_2} \right) [A]$

(A) Applying the steady state approximation to the intermediate $B$:The rate of change of concentration of $B$ is given by: $\frac{d[B]}{dt} = k_1 [A] - k_2 [B]$According to the steady state approximation,the rate of formation of $B$ is set to zero: $\frac{d[B]}{dt} = 0$Therefore,$0 = k_1 [A] - k_2 [B]$Rearranging the equation to solve for $[B]$:$k_2 [B] = k_1 [A]$$[B] = \left( \frac{k_1}{k_2} \right) [A]$

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

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For a reaction scheme $A$ $\xrightarrow{k_1} B$ $\xrightarrow{k_2} C$,if the rate of formation of $B$ is set to be zero,then the concentration of $B$ is given by:

JEE MAIN 2019 All JEE MAIN

A
$\left( \frac{k_1}{k_2} \right) [A]$
B
$\left( k_1 - k_2 \right) [A]$
C
$k_1 k_2 [A]$
D
$\left( k_1 + k_2 \right) [A]$

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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

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For the reaction system $2NO_{(g)} + O_{2(g)} \to 2NO_{2(g)}$,the volume is suddenly reduced to half its initial value by increasing the pressure. If the reaction is of first order with respect to $O_2$ and second order with respect to $NO$,the rate of reaction will:

EasyAIEEE 2003

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The reaction between $X$ and $Y$ is first order with respect to $X$ and zero order with respect to $Y$.

$Experiment$ $[X] / (mol \ L^{-1})$ $[Y] / (mol \ L^{-1})$ $\text{Initial rate} / (mol \ L^{-1} \ min^{-1})$

$I$ $0.1$ $0.1$ $2 \times 10^{-3}$

$II$ $0.2$ $0.2$ $4 \times 10^{-3}$

$III$ $0.4$ $0.4$ $M \times 10^{-3}$

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

Examine the data of the table and calculate the ratio of the numerical value of $M$ to $0.2$.

MediumJEE MAIN 2022

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The rate law for the reaction below is given by the expression $Rate = k[A][B]$.
$A + B \to \text{Product}$
If the concentration of $B$ is increased from $0.1 \ M$ to $0.3 \ M$,keeping the concentration of $A$ constant at $0.1 \ M$,the rate constant $(k)$ will be:

DifficultJEE MAIN 2016

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The rate equation for the reaction $2 A + B \longrightarrow$ products is $\text{rate} = k[A][B]^2$. If $k$ at $T \ K$ is $5.0 \times 10^{-6} \ mol^{-2} \ L^2 \ s^{-1}$,the initial rate of the reaction,when $[A] = 0.05 \ mol \ L^{-1}$ and $[B] = 0.1 \ mol \ L^{-1}$ is:

DifficultAP EAMCET 2022

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The 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:

MediumAIEEE 2012

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$Experiment$	$[X] / (mol \ L^{-1})$	$[Y] / (mol \ L^{-1})$	$\text{Initial rate} / (mol \ L^{-1} \ min^{-1})$
$I$	$0.1$	$0.1$	$2 \times 10^{-3}$
$II$	$0.2$	$0.2$	$4 \times 10^{-3}$
$III$	$0.4$	$0.4$	$M \times 10^{-3}$
$IV$	$0.1$	$0.2$	$2 \times 10^{-3}$

For a reaction scheme $A$ $\xrightarrow{k_1} B$ $\xrightarrow{k_2} C$,if the rate of formation of $B$ is set to be zero,then the concentration of $B$ is given by:

Similar Questions

For the reaction system $2NO_{(g)} + O_{2(g)} \to 2NO_{2(g)}$,the volume is suddenly reduced to half its initial value by increasing the pressure. If the reaction is of first order with respect to $O_2$ and second order with respect to $NO$,the rate of reaction will:

The rate law for the reaction below is given by the expression $Rate = k[A][B]$.$A + B \to \text{Product}$If the concentration of $B$ is increased from $0.1 \ M$ to $0.3 \ M$,keeping the concentration of $A$ constant at $0.1 \ M$,the rate constant $(k)$ will be:

The rate equation for the reaction $2 A + B \longrightarrow$ products is $\text{rate} = k[A][B]^2$. If $k$ at $T \ K$ is $5.0 \times 10^{-6} \ mol^{-2} \ L^2 \ s^{-1}$,the initial rate of the reaction,when $[A] = 0.05 \ mol \ L^{-1}$ and $[B] = 0.1 \ mol \ L^{-1}$ is:

The 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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The rate law for the reaction below is given by the expression $Rate = k[A][B]$.
$A + B \to \text{Product}$
If the concentration of $B$ is increased from $0.1 \ M$ to $0.3 \ M$,keeping the concentration of $A$ constant at $0.1 \ M$,the rate constant $(k)$ will be: