(A) Let the rate law be $Rate = k[A]^x[B]^y$.From the data:$1$) $1.2 \times 10^{-3} = k(0.05)^x(0.05)^y$$2$) $2.4 \times 10^{-3} = k(0.10)^x(0.05)^y$$

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(A) Let the rate law be $Rate = k[A]^x[B]^y$.From the data:$1$) $1.2 \times 10^{-3} = k(0.05)^x(0.05)^y$$2$) $2.4 \times 10^{-3} = k(0.10)^x(0.05)^y$$3$) $1.2 \times 10^{-3} = k(0.05)^x(0.10)^y$Dividing equation $(2)$ by $(1)$:$\frac{2.4 \times 10^{-3}}{1.2 \times 10^{-3}} = (\frac{0.10}{0.05})^x$ $\Rightarrow 2 = 2^x$ $\Rightarrow x = 1$.Dividing equation $(3)$ by $(1)$:$\frac{1.2 \times 10^{-3}}{1.2 \times 10^{-3}} = (\frac{0.10}{0.05})^y$ $\Rightarrow 1 = 2^y$ $\Rightarrow y = 0$.Thus,the order with respect to $A$ is $1$ and with respect to $B$ is $0$.

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

Easy

Calculate the order of the reaction with respect to $A$ and $B$ based on the following data:

$[A] \ (mol/L)$ $[B] \ (mol/L)$ Rate $(mol/L \cdot s)$

$0.05$ $0.05$ $1.2 \times 10^{-3}$

$0.10$ $0.05$ $2.4 \times 10^{-3}$

$0.05$ $0.10$ $1.2 \times 10^{-3}$

A
$1$ and $0$
B
$1$ and $1$
C
$0$ and $1$
D
None of these

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Rate law , Rate constant , Order of Reaction and Molecularity

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MediumAIIMS 2011

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MediumMHT CET 2020

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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.

MediumMHT CET 2021

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Mechanism of a hypothetical reaction $X_2 + Y_2 \rightarrow 2XY$ is given below:
$(i)$ $X_2 \rightarrow X + X$ (fast)
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$[A] \ (mol/L)$	$[B] \ (mol/L)$	Rate $(mol/L \cdot s)$
$0.05$	$0.05$	$1.2 \times 10^{-3}$
$0.10$	$0.05$	$2.4 \times 10^{-3}$
$0.05$	$0.10$	$1.2 \times 10^{-3}$

Calculate the order of the reaction with respect to $A$ and $B$ based on the following data: $[A] \ (mol/L)$ $[B] \ (mol/L)$ Rate $(mol/L \cdot s)$ $0.05$ $0.05$ $1.2 \times 10^{-3}$ $0.10$ $0.05$ $2.4 \times 10^{-3}$ $0.05$ $0.10$ $1.2 \times 10^{-3}$

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The rate constant for a reaction is $10.8 \times 10^{-5} \ mol \ L^{-1} \ s^{-1}$. The order of the reaction is .......

The rate constant for the reaction,$2N_2O_5 \to 4NO_2 + O_2$ is $3.0 \times 10^{-4} \ s^{-1}$. If the reaction starts with $1.0 \ mol \ L^{-1}$ of $N_2O_5$,calculate the rate of formation of $NO_2$ at the moment when the concentration of $O_2$ is $0.1 \ mol \ L^{-1}$.

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Calculate the order of the reaction with respect to $A$ and $B$ based on the following data:

$[A] \ (mol/L)$ $[B] \ (mol/L)$ Rate $(mol/L \cdot s)$

$0.05$ $0.05$ $1.2 \times 10^{-3}$

$0.10$ $0.05$ $2.4 \times 10^{-3}$

$0.05$ $0.10$ $1.2 \times 10^{-3}$

What is the molecularity and order of the following reaction if the rate law is $\text{rate} = k[O_3][O]$ respectively?
$O_{3(g)} + O_{(g)} \longrightarrow 2O_{2(g)}$

Mechanism of a hypothetical reaction $X_2 + Y_2 \rightarrow 2XY$ is given below:
$(i)$ $X_2 \rightarrow X + X$ (fast)
$(ii)$ $X + Y_2 \rightleftharpoons XY + Y$ (slow)
$(iii)$ $X + Y \rightarrow XY$ (fast)
The overall order of the reaction will be