The time of completion of 90% of a first orde | vedclass

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(C) For a first order reaction,the rate constant is given by $K = \frac{2.303}{t} \log \frac{a}{a-x}$.For $90\%$ completion,$x = 0.9a$,so $a-x = 0.1a$

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$3.3$ times that of half life

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(C) For a first order reaction,the rate constant is given by $K = \frac{2.303}{t} \log \frac{a}{a-x}$.For $90\%$ completion,$x = 0.9a$,so $a-x = 0.1a$. Thus,$t_{90\%} = \frac{2.303}{K} \log \frac{a}{0.1a} = \frac{2.303}{K} \log 10 = \frac{2.303}{K}$.We know that the half-life is $t_{1/2} = \frac{0.693}{K}$,which implies $K = \frac{0.693}{t_{1/2}}$.Substituting $K$ in the expression for $t_{90\%}$:$t_{90\%} = \frac{2.303}{0.693 / t_{1/2}} = \frac{2.303}{0.693} 	imes t_{1/2} \approx 3.32 	imes t_{1/2}$.Therefore,the time is approximately $3.3$ times the half-life.

Experiment	$[A]$ $(mol \ L^{-1})$	$[B]$ $(mol \ L^{-1})$	Initial rate $(mol \ L^{-1} \ min^{-1})$
$I$	$0.10$	$0.20$	$6.93 \times 10^{-3}$
$II$	$0.10$	$0.25$	$6.93 \times 10^{-3}$
$III$	$0.20$	$0.30$	$1.386 \times 10^{-2}$

The time of completion of $90\%$ of a first order reaction is approximately

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