The half life period of a gaseous reactant&nb | vedclass

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Question

$\mathrm{P}(\mathrm{mm}) \quad 250300 \quad 400450$

$\mathrm{t}_{1 / 2}(\min ) 136112.585 \quad 75.5$

Now,

$\mathrm{t}_{1 / 2} \propto \frac{

Vedclass · Accepted Answer

$2$

Vedclass · Answer

$\mathrm{P}(\mathrm{mm}) \quad 250300 \quad 400450$

$\mathrm{t}_{1 / 2}(\min ) 136112.585 \quad 75.5$

Now,

$\mathrm{t}_{1 / 2} \propto \frac{1}{\left[\mathrm{A}_{0}ight]^{\mathrm{n}-1}}$

$\frac{t_{1 / 2}(1)}{t_{1 / 2}(2)}=\frac{\left[A_{0}ight]_{2}^{n-1}}{\left[A_{0}ight]_{1}^{n-1}}$

$\frac{112.5}{85}=\left(\frac{400}{300}ight)^{\mathrm{n}-1}=\left(\frac{4}{3}ight)^{\mathrm{n}-1}$

$1.323=(1.333)^{\mathrm{n}-1} \Rightarrow \mathrm{n}-1=1 \Rightarrow \mathrm{n}=2$

$	herefore$ Order of reaction, $\mathrm{n}=2$

The half life period of a gaseous reactant undergoing thermal decomposition was measured for various initial pressures $'p_0'$ as follows :

$\begin{array}{|l|l|l|} \hline P_0\,\,(mmHg) & 250 & 300 \\ \hline t_{1/2}\,\,(minutes) & 135 & 112.5 \\ \hline \end{array}$

The order of reaction is -

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