For a zero-order reaction,the ratio of t_{1/2 | vedclass

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(C) For a zero-order reaction,the integrated rate equation is $[A]_t = [A]_0 - kt$.$t_{1/2}$ is the time when $[A]_t = \frac{[A]_0}{2}$,so $t_{1/2} =

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(C) For a zero-order reaction,the integrated rate equation is $[A]_t = [A]_0 - kt$.$t_{1/2}$ is the time when $[A]_t = \frac{[A]_0}{2}$,so $t_{1/2} = \frac{[A]_0}{2k}$.$t_{3/4}$ is the time when $75\%$ of the reactant is consumed,so $[A]_t = [A]_0 - \frac{3}{4}[A]_0 = \frac{[A]_0}{4}$.Substituting this into the rate equation: $\frac{[A]_0}{4} = [A]_0 - kt_{3/4}$,which gives $kt_{3/4} = \frac{3[A]_0}{4}$,so $t_{3/4} = \frac{3[A]_0}{4k}$.The ratio $\frac{t_{1/2}}{t_{3/4}} = \frac{[A]_0 / 2k}{3[A]_0 / 4k} = \frac{1}{2} 	imes \frac{4}{3} = \frac{2}{3}$.

Time $(s)$	Concentration of the reactant $(mol \ L^{-1})$
$0$	$0.1$
$15$	$0.05$
$20$	$0.1 - x$

For a zero-order reaction,the ratio of $t_{1/2}$ to $t_{3/4}$ is ....

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