Time required for completion of 99.9 % of a F | vedclass

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Question

(C) For a first order reaction,the time $t$ is given by $t = \frac{2.303}{k} \log \left( \frac{a}{a-x} \right)$.For $99.9 \%$ completion,$x = 0.999a$,

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(C) For a first order reaction,the time $t$ is given by $t = \frac{2.303}{k} \log \left( \frac{a}{a-x} ight)$.For $99.9 \%$ completion,$x = 0.999a$,so $a-x = 0.001a = \frac{a}{1000}$.Thus,$t_{99.9 \%} = \frac{2.303}{k} \log \left( \frac{a}{a/1000} ight) = \frac{2.303}{k} \log(10^3) = \frac{2.303 	imes 3}{k}$.The half-life is $t_{1/2} = \frac{0.693}{k} = \frac{2.303 	imes 0.301}{k}$.Taking the ratio: $\frac{t_{99.9 \%}}{t_{1/2}} = \frac{2.303 	imes 3 / k}{2.303 	imes 0.301 / k} \approx \frac{3}{0.3} = 10$.

$t/min$	Pressure of system at time $t/mm \ Hg$
$10$	$160$
$\infty$	$240$

Time required for completion of $99.9 \%$ of a First order reaction is . . . . . . times of half life $(t_{1/2})$ of the reaction.

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$10$ $160$
$\infty$ $240$
Which of the following option is incorrect?

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