(B) The rate law for the reaction is given by: $r = k[A]^2 [B]^2$ Substituting the given values: $3.6 \times 10^{-2} = k(0.2)^2 (0.1)^2$ $3.6 \times 1

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$90 \ mol^{-3} \ dm^9 \ sec^{-1}$

(B) The rate law for the reaction is given by: $r = k[A]^2 [B]^2$ Substituting the given values: $3.6 \times 10^{-2} = k(0.2)^2 (0.1)^2$ $3.6 \times 10^{-2} = k(0.04)(0.01)$ $3.6 \times 10^{-2} = k(4 \times 10^{-4})$ $k = \frac{3.6 \times 10^{-2}}{4 \times 10^{-4}} = 0.9 \times 10^2 = 90 \ mol^{-3} \ dm^9 \ sec^{-1}$

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

Medium

For the reaction $A + B \rightarrow$ product,the rate of reaction is $3.6 \times 10^{-2} \ mol \ dm^{-3} \ sec^{-1}$. When $[A] = 0.2 \ mol \ dm^{-3}$ and $[B] = 0.1 \ mol \ dm^{-3}$,find the rate constant of the reaction if it is second order with respect to both reactants.

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A
$18 \ mol^{-3} \ dm^9 \ sec^{-1}$
B
$90 \ mol^{-3} \ dm^9 \ sec^{-1}$
C
$72 \ mol^{-3} \ dm^9 \ sec^{-1}$
D
$36 \ mol^{-3} \ dm^9 \ sec^{-1}$

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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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The correct difference between $1^{st}$-order and $2^{nd}$-order reaction is that:

NEET 2018

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The differential form of the rate equation is given by:
$dx/dt = k[P][Q]^{0.5}[R]^{0.5}$
Which statement about the above equation is wrong?

Medium

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In a reaction,the concentration of reactant is increased two times and three times,then the increases in the rate of reaction were four times and nine times respectively. The order of reaction is:

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Consider the reaction $2 \ A + 2 \ B \rightarrow C + 2 \ D$. If the concentration of $A$ is doubled at constant $B$,the rate increases by a factor of $4$. If the concentration of $B$ is doubled at constant $A$,the rate is doubled. What is the rate law?

MediumMHT CET 2020

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The reaction between $A$ and $B$ is first order with respect to $A$ and zero order with respect to $B$. Fill in the blanks in the following table:

Experiment $[A] / mol \, L^{-1}$ $[B] / mol \, L^{-1}$ Initial rate / $mol \, L^{-1} \, min^{-1}$

$I$ $0.1$ $0.1$ $2.0 \times 10^{-2}$

$II$ $-$ $0.2$ $4.0 \times 10^{-2}$

$III$ $0.4$ $0.4$ $-$

$IV$ $-$ $0.2$ $2.0 \times 10^{-2}$

Experiment	$[A] / mol \, L^{-1}$	$[B] / mol \, L^{-1}$	Initial rate / $mol \, L^{-1} \, min^{-1}$
$I$	$0.1$	$0.1$	$2.0 \times 10^{-2}$
$II$	$-$	$0.2$	$4.0 \times 10^{-2}$
$III$	$0.4$	$0.4$	$-$
$IV$	$-$	$0.2$	$2.0 \times 10^{-2}$

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For the reaction $A + B \rightarrow$ product,the rate of reaction is $3.6 \times 10^{-2} \ mol \ dm^{-3} \ sec^{-1}$. When $[A] = 0.2 \ mol \ dm^{-3}$ and $[B] = 0.1 \ mol \ dm^{-3}$,find the rate constant of the reaction if it is second order with respect to both reactants.

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The correct difference between $1^{st}$-order and $2^{nd}$-order reaction is that:

The differential form of the rate equation is given by:$dx/dt = k[P][Q]^{0.5}[R]^{0.5}$Which statement about the above equation is wrong?

In a reaction,the concentration of reactant is increased two times and three times,then the increases in the rate of reaction were four times and nine times respectively. The order of reaction is:

Consider the reaction $2 \ A + 2 \ B \rightarrow C + 2 \ D$. If the concentration of $A$ is doubled at constant $B$,the rate increases by a factor of $4$. If the concentration of $B$ is doubled at constant $A$,the rate is doubled. What is the rate law?

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The differential form of the rate equation is given by:
$dx/dt = k[P][Q]^{0.5}[R]^{0.5}$
Which statement about the above equation is wrong?