For a reaction $\mathrm{A} ightarrow \mathrm{B}$, it is found that $[\mathrm{A}]_{0} \sqrt{t_{1 / 2}}$ is constant. The order of the reaction with respect to $\mathrm{A}$ is

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Aicharya · Accepted Answer

$t_{1 / 2} \propto 1 /[\mathrm{A}]_{0}^{n-1}$ where $n$ is the order of the reaction. If $n=3$, then $t_{1 / 2} \propto 1 /[\mathrm{A}]_{0}^{2}$ or $\left(t_{1 / 2}ight)\left([\mathrm{A}]_{0}ight)^{2}=$ constant ,

For a reaction $\mathrm{A} ightarrow \mathrm{B}$, it is found that $[\mathrm{A}]_{0} \sqrt{t_{1 / 2}}$ is…

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