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(B) The half-life $(t_{1/2})$ of an $n^{th}$ order reaction is related to the initial concentration $(a)$ of the reactant as $t_{1/2} \propto a^{1-n}$

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$2$

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(B) The half-life $(t_{1/2})$ of an $n^{th}$ order reaction is related to the initial concentration $(a)$ of the reactant as $t_{1/2} \propto a^{1-n}$.For two different initial concentrations,we have the relation: $\frac{(t_{1/2})_1}{(t_{1/2})_2} = \left(\frac{a_2}{a_1}ight)^{n-1}$.Given: $(t_{1/2})_1 = 5 \ 	ext{min}$,$a_1 = 0.1 \ M$ and $(t_{1/2})_2 = 50 \ 	ext{min}$,$a_2 = 0.01 \ M$.Substituting the values: $\frac{5}{50} = \left(\frac{0.01}{0.1}ight)^{n-1}$.$\frac{1}{10} = (0.1)^{n-1}$.$0.1^1 = (0.1)^{n-1}$.Comparing the exponents: $1 = n - 1$,which gives $n = 2$.

Half-life periods for a reaction at initial concentrations of $0.1 \ M$ and $0.01 \ M$ are $5$ and $50$ minutes,respectively. The order of reaction is

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