If the half-life of a reaction is halved when | vedclass

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(B) The relationship between half-life $(t_{1/2})$ and initial concentration $(a)$ for a reaction of order $n$ is given by: $t_{1/2} \propto \frac{1}{

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$2$

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(B) The relationship between half-life $(t_{1/2})$ and initial concentration $(a)$ for a reaction of order $n$ is given by: $t_{1/2} \propto \frac{1}{a^{n-1}}$.Given that when initial concentration $a_2 = 2a_1$,the half-life becomes $(t_{1/2})_2 = \frac{(t_{1/2})_1}{2}$.Substituting these values into the ratio: $\frac{(t_{1/2})_1}{(t_{1/2})_2} = \left( \frac{a_2}{a_1} \right)^{n-1}$.$\frac{(t_{1/2})_1}{(t_{1/2})_1 / 2} = \left( \frac{2a_1}{a_1} \right)^{n-1}$.$2 = (2)^{n-1}$.Comparing the exponents: $n - 1 = 1$,which gives $n = 2$.Therefore,the order of the reaction is $2$.

If the half-life of a reaction is halved when the initial concentration of the reactant is doubled,what is the order of the reaction?

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