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Q.
The following graph shows how $t_{1/2}$ (half-life) of a reactant $R$ changes with the initial reactant concentration $a_0$. The order of the reaction will be
Half-life period $t_{1 / 2} \propto \frac{1}{a_{0}^{n-1}} \ldots$(i)
where, $a_{0}=$ initial concentration and $n=$ order of reaction since graph betweent $_{1 / 2}$ and $\frac{1}{a_{0}}$ is a straight line,
$t_{1 / 2} \propto \frac{1}{a_{0}} \ldots$(ii)
From Eq (i) and (ii),
$\Rightarrow \frac{1}{a_{0}^{n-1}}=\frac{1}{a_{0}}$
or $n - 1 = 1 $
$\therefore n = 2$
Hence, the reaction is of second order.