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(C) For a zero order reaction,the half-life is given by $t_{1/2} = \frac{[A]_0}{2k}$.Let the initial concentration be $[A]_0 = a$.Then,$t_{1/2} = \fra

Vedclass · Accepted Answer

Becomes one-fourth

Vedclass · Answer

(C) For a zero order reaction,the half-life is given by $t_{1/2} = \frac{[A]_0}{2k}$.Let the initial concentration be $[A]_0 = a$.Then,$t_{1/2} = \frac{a}{2k} \dots (1)$.When the initial concentration is reduced to one-fourth,the new concentration is $[A]_0' = \frac{1}{4}a$.The new half-life is $t_{1/2}' = \frac{\frac{1}{4}a}{2k} = \frac{1}{4} 	imes \frac{a}{2k} \dots (2)$.Comparing equation $(2)$ with $(1)$,we get $t_{1/2}' = \frac{1}{4} t_{1/2}$.Therefore,the time taken for half of the reaction to complete becomes one-fourth of the original time. Hence,option $C$ is correct.

If the initial concentration is reduced to its $1/4^{th}$ in a zero order reaction,the time taken for half of the reaction to complete

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