$99 %$ at a first-order reaction was completed in $32 m i n .$ when will $99.9 %$ of the reaction complete?

48 m i n

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46 m i n

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50 m i n

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45 m i n

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Solution

The correct option is A $48 m i n$
$k \times 32 = l n (\begin{matrix} \frac{100}{1} \end{matrix}) . . . (i)$
$k \times t = l n (\begin{matrix} \frac{100}{0.1} \end{matrix}) . . . (i i)$
$e q . (i i) / (i)$
$\frac{t}{32} = \frac{3 l n 10}{2 l n 10}$
$t = 48 m i n$

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