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(C) For a first order reaction,the rate constant $K$ is given by $K = \frac{2.303}{t} \log \left(\frac{[A]_0}{[A]_t}\right)$.For $99.9 \%$ completion,

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$3$

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(C) For a first order reaction,the rate constant $K$ is given by $K = \frac{2.303}{t} \log \left(\frac{[A]_0}{[A]_t}ight)$.For $99.9 \%$ completion,$[A]_t = 0.1 \% 	ext{ of } [A]_0$,so $[A]_t = 0.001 [A]_0$. Thus,$t_{99.9 \%} = \frac{2.303}{K} \log \left(\frac{100}{0.1}ight) = \frac{2.303}{K} \log(10^3) = \frac{2.303}{K} 	imes 3$.For $90 \%$ completion,$[A]_t = 10 \% 	ext{ of } [A]_0$,so $[A]_t = 0.1 [A]_0$. Thus,$t_{90 \%} = \frac{2.303}{K} \log \left(\frac{100}{10}ight) = \frac{2.303}{K} \log(10) = \frac{2.303}{K} 	imes 1$.Taking the ratio,$\frac{t_{99.9 \%}}{t_{90 \%}} = \frac{3}{1} = 3$.Therefore,the time required for $99.9 \%$ completion is $3$ times the time required for $90 \%$ completion.

$S$.No.	Time	Total pressure in $Pa$
$1$.	At $t = 10 \ min$	$300$
$2$.	At $t = \infty$	$200$

Time required for $99.9 \%$ completion of a first order reaction is . . . . . . . times the time required for completion of $90 \%$ reaction. (nearest integer).

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