(B) Given,The reaction is $PCl_5 \longrightarrow PCl_3 + Cl_2$. Rate $= 1.02 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$. Rate constant $(k) = 3.4 \times

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(B) Given,The reaction is $PCl_5 \longrightarrow PCl_3 + Cl_2$. Rate $= 1.02 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$. Rate constant $(k) = 3.4 \times 10^{-5} \ s^{-1}$. The rate law for this first-order reaction is: $\text{Rate} = k [PCl_5]$ Substituting the given values: $1.02 \times 10^{-4} = 3.4 \times 10^{-5} \times [PCl_5]$ $[PCl_5] = \frac{1.02 \times 10^{-4}}{3.4 \times 10^{-5}}$ $[PCl_5] = 3 \ mol \ L^{-1}$.

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

Medium

For the reaction,$PCl_5 \longrightarrow PCl_3 + Cl_2$,the rate and rate constant are $1.02 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$ and $3.4 \times 10^{-5} \ s^{-1}$ respectively at a given instant. The molar concentration of $PCl_5$ at that instant is

KCET 2024 All KCET

A
$8.0 \ mol \ L^{-1}$
B
$3.0 \ mol \ L^{-1}$
C
$0.2 \ mol \ L^{-1}$
D
$2.0 \ mol \ L^{-1}$

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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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If the order of the reaction $x + y \xrightarrow{hv} xy$ is zero,it means that the rate of

Medium

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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 $0.1 \ mol \ L^{-1}$.

MediumAIIMS 2011

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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:

DifficultAIPMT 2011

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

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

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

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

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

Medium

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The results given in the below table were obtained during kinetic studies of the following reaction:
$2 A + B \longrightarrow C + D$

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

$I$ $0.1$ $0.1$ $6.00 \times 10^{-3}$

$II$ $0.1$ $0.2$ $2.40 \times 10^{-2}$

$III$ $0.2$ $0.1$ $1.20 \times 10^{-2}$

$IV$ $X$ $0.2$ $7.20 \times 10^{-2}$

$V$ $0.3$ $Y$ $2.88 \times 10^{-1}$

$X$ and $Y$ in the given table are respectively :

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

Experiment	$[A] / mol \ L^{-1}$	$[B] / mol \ L^{-1}$	Initial rate / $mol \ L^{-1} \ min^{-1}$
$I$	$0.1$	$0.1$	$6.00 \times 10^{-3}$
$II$	$0.1$	$0.2$	$2.40 \times 10^{-2}$
$III$	$0.2$	$0.1$	$1.20 \times 10^{-2}$
$IV$	$X$	$0.2$	$7.20 \times 10^{-2}$
$V$	$0.3$	$Y$	$2.88 \times 10^{-1}$

For the reaction,$PCl_5 \longrightarrow PCl_3 + Cl_2$,the rate and rate constant are $1.02 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$ and $3.4 \times 10^{-5} \ s^{-1}$ respectively at a given instant. The molar concentration of $PCl_5$ at that instant is

Similar Questions

If the order of the reaction $x + y \xrightarrow{hv} xy$ is zero,it means that the rate of

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 $0.1 \ mol \ L^{-1}$.

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:

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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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:

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

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

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

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

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