(A) The rate of a reaction is determined by the slowest step in the mechanism. The rate law for the slow step is $R = k_3[A][B]$. Since $B$ is an inte

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(A) The rate of a reaction is determined by the slowest step in the mechanism. The rate law for the slow step is $R = k_3[A][B]$. Since $B$ is an intermediate formed in the fast equilibrium step $A \underset{k_2}{\stackrel{k_1}{\rightleftharpoons}} B$,we have the equilibrium constant $K_{eq} = \frac{[B]}{[A]} = \frac{k_1}{k_2}$. This gives $[B] = \frac{k_1}{k_2}[A]$. Substituting this into the rate expression: $R = k_3[A](\frac{k_1}{k_2}[A]) = \frac{k_3 k_1}{k_2}[A]^2$. Thus,the rate is proportional to $[A]^2$.

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

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For a hypothetical reaction,$A \rightarrow C$. The mechanism is: $A \underset{k_2}{\stackrel{k_1}{\rightleftharpoons}} B$ (Fast),$A + B \xrightarrow{k_3} C$ (Slow). The rate law for this reaction is:

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A
$\propto [A]^2$
B
$\propto [A][B]$
C
$\propto [A]^2[B]^2$
D
$\propto [A]^2[B]$

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Class 12 Questions

Class 12

Chemistry Questions

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Chemical Kinetics

Subtopic

Rate law , Rate constant , Order of Reaction and Molecularity

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What is the order of reaction for $A + B \to C$?

$Observation$ $[A] \ (mol \ L^{-1})$ $[B] \ (mol \ L^{-1})$ $Rate \ (mol \ L^{-1} \ sec^{-1})$

$1$ $0.1$ $0.1$ $2 \times 10^{-3}$

$2$ $0.2$ $0.1$ $4 \times 10^{-3}$

$3$ $0.1$ $0.2$ $8 \times 10^{-3}$

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For the reaction $X + Y \rightarrow Z$,the rate law is $r = K[X][Y]$. If the concentration of $Y$ is increased significantly,what will be the order of the reaction?

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For a certain reaction,the expression for half-life is $t_{1/2} \propto \frac{1}{a^{n-1}}$. What is the order of the reaction?

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The value of the rate constant of a pseudo first order reaction:

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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 $t_2 = 20 \ min$?

MediumTS EAMCET 2001

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$Observation$	$[A] \ (mol \ L^{-1})$	$[B] \ (mol \ L^{-1})$	$Rate \ (mol \ L^{-1} \ sec^{-1})$
$1$	$0.1$	$0.1$	$2 \times 10^{-3}$
$2$	$0.2$	$0.1$	$4 \times 10^{-3}$
$3$	$0.1$	$0.2$	$8 \times 10^{-3}$

For a hypothetical reaction,$A \rightarrow C$. The mechanism is: $A \underset{k_2}{\stackrel{k_1}{\rightleftharpoons}} B$ (Fast),$A + B \xrightarrow{k_3} C$ (Slow). The rate law for this reaction is:

Similar Questions

What is the order of reaction for $A + B \to C$? $Observation$ $[A] \ (mol \ L^{-1})$ $[B] \ (mol \ L^{-1})$ $Rate \ (mol \ L^{-1} \ sec^{-1})$ $1$ $0.1$ $0.1$ $2 \times 10^{-3}$ $2$ $0.2$ $0.1$ $4 \times 10^{-3}$ $3$ $0.1$ $0.2$ $8 \times 10^{-3}$

For the reaction $X + Y \rightarrow Z$,the rate law is $r = K[X][Y]$. If the concentration of $Y$ is increased significantly,what will be the order of the reaction?

For a certain reaction,the expression for half-life is $t_{1/2} \propto \frac{1}{a^{n-1}}$. What is the order of the reaction?

The value of the rate constant of a pseudo first order reaction:

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 $t_2 = 20 \ min$?

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What is the order of reaction for $A + B \to C$?

$Observation$ $[A] \ (mol \ L^{-1})$ $[B] \ (mol \ L^{-1})$ $Rate \ (mol \ L^{-1} \ sec^{-1})$

$1$ $0.1$ $0.1$ $2 \times 10^{-3}$

$2$ $0.2$ $0.1$ $4 \times 10^{-3}$

$3$ $0.1$ $0.2$ $8 \times 10^{-3}$