(B) The rate of reaction is defined as:$-\frac{1}{2} \frac{d[N_2O_5]}{dt} = \frac{1}{4} \frac{d[NO_2]}{dt} = \frac{d[O_2]}{dt}$Substituting the given

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(B) The rate of reaction is defined as:$-\frac{1}{2} \frac{d[N_2O_5]}{dt} = \frac{1}{4} \frac{d[NO_2]}{dt} = \frac{d[O_2]}{dt}$Substituting the given rate expressions:$\frac{1}{2} k[N_2O_5] = \frac{1}{4} k'[N_2O_5] = k''[N_2O_5]$Dividing by $[N_2O_5]$:$\frac{k}{2} = \frac{k'}{4} = k''$From $\frac{k}{2} = \frac{k'}{4}$,we get $k' = 2k$.From $\frac{k}{2} = k''$,we get $k'' = \frac{k}{2}$.

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

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The rate of the reaction: $2N_2O_5 \rightarrow 4NO_2 + O_2$ can be written in three ways.
$-\frac{d[N_2O_5]}{dt} = k[N_2O_5]$
$\frac{d[NO_2]}{dt} = k'[N_2O_5]$ ; $\frac{d[O_2]}{dt} = k''[N_2O_5]$
The relationship between $k$ and $k'$ and between $k$ and $k''$ are:

AIPMT 2011 All AIPMT

A
$k' = 2k$ ; $k'' = k$
B
$k' = 2k$ ; $k'' = k/2$
C
$k' = 2k$ ; $k'' = 2k$
D
$k' = k$ ; $k'' = k$

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Chemical Kinetics

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Rate law , Rate constant , Order of Reaction and Molecularity

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AIPMT 2011 Paper All AIPMT years →

For conversion of compound $A \rightarrow B$,the rate constant of the reaction was found to be $4.6 \times 10^{-5} \ L \ mol^{-1} \ s^{-1}$. The order of the reaction is $..........$

MediumJEE MAIN 2023

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The value of $\frac{t_{0.875}}{t_{0.50}}$ for $n^{th}$ order reaction is

Difficult

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Calculate the overall order of a reaction which has the rate expression:
$(a)$ $\text{Rate} = k[A]^{1/2}[B]^{3/2}$
$(b)$ $\text{Rate} = k[A]^{3/2}[B]^{-1}$

Medium

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Why can we not determine the order of a reaction by taking into consideration the balanced chemical equation?

Easy

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The experimental data for the reaction $2A + B_2 \longrightarrow 2AB$ is given below:

Exp. $[A] \ (mol \ L^{-1})$ $[B_2] \ (mol \ L^{-1})$ Rate $(mol \ L^{-1} \ S^{-1})$

$1$ $0.50$ $0.50$ $1.6 \times 10^{-4}$

$2$ $0.50$ $1.00$ $3.2 \times 10^{-4}$

$3$ $1.00$ $1.00$ $3.2 \times 10^{-4}$

Determine the rate law for the reaction.

Exp.	$[A] \ (mol \ L^{-1})$	$[B_2] \ (mol \ L^{-1})$	Rate $(mol \ L^{-1} \ S^{-1})$
$1$	$0.50$	$0.50$	$1.6 \times 10^{-4}$
$2$	$0.50$	$1.00$	$3.2 \times 10^{-4}$
$3$	$1.00$	$1.00$	$3.2 \times 10^{-4}$

Difficult

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The rate of the reaction: $2N_2O_5 \rightarrow 4NO_2 + O_2$ can be written in three ways.$-\frac{d[N_2O_5]}{dt} = k[N_2O_5]$$\frac{d[NO_2]}{dt} = k'[N_2O_5]$ ; $\frac{d[O_2]}{dt} = k''[N_2O_5]$The relationship between $k$ and $k'$ and between $k$ and $k''$ are:

Similar Questions

For conversion of compound $A \rightarrow B$,the rate constant of the reaction was found to be $4.6 \times 10^{-5} \ L \ mol^{-1} \ s^{-1}$. The order of the reaction is $..........$

The value of $\frac{t_{0.875}}{t_{0.50}}$ for $n^{th}$ order reaction is

Calculate the overall order of a reaction which has the rate expression:$(a)$ $\text{Rate} = k[A]^{1/2}[B]^{3/2}$$(b)$ $\text{Rate} = k[A]^{3/2}[B]^{-1}$

Why can we not determine the order of a reaction by taking into consideration the balanced chemical equation?

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The rate of the reaction: $2N_2O_5 \rightarrow 4NO_2 + O_2$ can be written in three ways.
$-\frac{d[N_2O_5]}{dt} = k[N_2O_5]$
$\frac{d[NO_2]}{dt} = k'[N_2O_5]$ ; $\frac{d[O_2]}{dt} = k''[N_2O_5]$
The relationship between $k$ and $k'$ and between $k$ and $k''$ are:

Calculate the overall order of a reaction which has the rate expression:
$(a)$ $\text{Rate} = k[A]^{1/2}[B]^{3/2}$
$(b)$ $\text{Rate} = k[A]^{3/2}[B]^{-1}$