(C) For an $n$-th order reaction,the half-life $t_{1/2}$ is related to the initial pressure $P_0$ as $t_{1/2} \propto \frac{1}{P_0^{n-1}}$. Therefore,

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(C) For an $n$-th order reaction,the half-life $t_{1/2}$ is related to the initial pressure $P_0$ as $t_{1/2} \propto \frac{1}{P_0^{n-1}}$. Therefore,$\frac{(t_{1/2})_1}{(t_{1/2})_2} = \left( \frac{(P_0)_2}{(P_0)_1} \right)^{n-1}$. Substituting the given values: $\frac{135}{112.5} = \left( \frac{300}{250} \right)^{n-1}$. $1.2 = (1.2)^{n-1}$. Comparing the exponents,$n - 1 = 1$,which gives $n = 2$.

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

Medium

The half-life period of a gaseous reactant undergoing thermal decomposition was measured for various initial pressures $P_0$ as follows:
$P_0 \text{ (mmHg)}$ $250$ $300$
$t_{1/2} \text{ (minutes)}$ $135$ $112.5$

The order of reaction is -

A
$0$
B
$1$
C
$2$
D
$3$

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Chemical Kinetics

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Rate law , Rate constant , Order of Reaction and Molecularity

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$P_0 \text{ (mmHg)}$	$250$	$300$
$t_{1/2} \text{ (minutes)}$	$135$	$112.5$

The half-life period of a gaseous reactant undergoing thermal decomposition was measured for various initial pressures $P_0$ as follows: $P_0 \text{ (mmHg)}$$250$$300$$t_{1/2} \text{ (minutes)}$$135$$112.5$The order of reaction is -

Similar Questions

The rate law expression for the reaction $aA + bB \to P$ is $\text{rate} = K [A]^p [B]^q$. The order of the reaction is

The rate of a gaseous reaction is given by the expression $K [A] [B]$. If the volume of the reaction vessel is suddenly reduced to $1/4$ of the initial volume,the reaction rate relating to the original rate will be

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The half-life period of a gaseous reactant undergoing thermal decomposition was measured for various initial pressures $P_0$ as follows:
$P_0 \text{ (mmHg)}$ $250$ $300$
$t_{1/2} \text{ (minutes)}$ $135$ $112.5$

The order of reaction is -