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(A) For an $n^{th}$ order reaction,the half-life $t_{1/2}$ is related to the initial concentration $[A]_0$ as $t_{1/2} \propto [A]_0^{1-n}$.Given that

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$0, 20 \ min$

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(A) For an $n^{th}$ order reaction,the half-life $t_{1/2}$ is related to the initial concentration $[A]_0$ as $t_{1/2} \propto [A]_0^{1-n}$.Given that the plot of $t_{1/2}$ versus $[A]_0$ is a straight line,this implies $t_{1/2} \propto [A]_0^1$.Comparing the powers,$1-n = 1$,which gives $n = 0$.For a zero-order reaction,$t_{1/2} = \frac{[A]_0}{2k}$.When $[A]_0 = 2 \ mol \ L^{-1}$,$t_{1/2} = 10 \ min$.So,$10 = \frac{2}{2k} \implies k = 0.1 \ mol \ L^{-1} \ min^{-1}$.When $[A]_0 = 4 \ mol \ L^{-1}$,$t = \frac{4}{2 	imes 0.1} = \frac{4}{0.2} = 20 \ min$.Thus,$n = 0$ and $t = 20 \ min$.

$[R] \text{ (molar)}$	$1.0$	$0.75$	$0.40$	$0.10$
$t \text{ (min.)}$	$0.0$	$0.05$	$0.12$	$0.18$

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