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(D) For a first order reaction,the time $t$ required for completion is given by $t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_t}$.For $90 \%$ completion,

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$3t$

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(D) For a first order reaction,the time $t$ required for completion is given by $t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_t}$.For $90 \%$ completion,$[A]_t = [A]_0 - 0.9[A]_0 = 0.1[A]_0$. Thus,$t = \frac{2.303}{k} \log \frac{[A]_0}{0.1[A]_0} = \frac{2.303}{k} \log 10 = \frac{2.303}{k}$.For $99.9 \%$ completion,$[A]_t = [A]_0 - 0.999[A]_0 = 0.001[A]_0$. Thus,$t_{99.9 \%} = \frac{2.303}{k} \log \frac{[A]_0}{0.001[A]_0} = \frac{2.303}{k} \log 10^3 = 3 	imes \frac{2.303}{k}$.Substituting $t$ into the equation,we get $t_{99.9 \%} = 3t$.

Time required for $90 \%$ completion of a first order reaction is $t$. What is the time required for $99.9 \%$ completion of reaction?

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