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

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(D) The half-life of a reaction of order $n$ is related to the initial concentration $[A]_0$ by the expression: $t_{1/2} \propto [A]_0^{1-n}$.Given that doubling the initial concentration doubles the half-life,we have: $2 	imes t_{1/2} \propto (2 	imes [A]_0)^{1-n}$.Comparing the proportionality,we get $2^1 = 2^{1-n}$.Equating the exponents: $1 = 1 - n$,which gives $n = 0$.Therefore,the reaction is of zero order.

If doubling the initial concentration of reactant doubles $t_{1/2}$ of reaction,the order of reaction is

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