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(A) The half-life period $(t_{1/2})$ of a reaction of order $n$ is related to the initial concentration $([A]_0)$ as: $t_{1/2} \propto [A]_0^{1-n}$.Gi

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(A) The half-life period $(t_{1/2})$ of a reaction of order $n$ is related to the initial concentration $([A]_0)$ as: $t_{1/2} \propto [A]_0^{1-n}$.Given that doubling the concentration $([A]_0 	o 2[A]_0)$ doubles the half-life $(t_{1/2} 	o 2t_{1/2})$,we have:$2t_{1/2} \propto (2[A]_0)^{1-n}$.Dividing the two relations: $2 = 2^{1-n}$.This implies $1 = 1 - n$,so $n = 0$.Therefore,the reaction is of zero order.

For a reaction,doubling the concentration of the reactant doubles the half-life period. What is the order of the reaction?

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