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(B) For a first order reaction,the time $t$ required for completion is given by $t = \frac{1}{k} \ln \left(\frac{[A]_0}{[A]_t}\right)$.For $67 \, \%$

Vedclass · Accepted Answer

$16$

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(B) For a first order reaction,the time $t$ required for completion is given by $t = \frac{1}{k} \ln \left(\frac{[A]_0}{[A]_t}ight)$.For $67 \, \%$ completion,$[A]_t = [A]_0 - 0.67[A]_0 = 0.33[A]_0 \approx \frac{1}{3}[A]_0$.Thus,$t_{67 \, \%} = \frac{1}{k} \ln \left(\frac{1}{1/3}ight) = \frac{\ln 3}{k}$.The half-life is $t_{1/2} = \frac{\ln 2}{k}$.Therefore,$\frac{t_{67 \, \%}}{t_{1/2}} = \frac{\ln 3}{\ln 2} \approx \frac{1.0986}{0.6931} \approx 1.585$.Given $t_{67 \, \%} = (x 	imes 10^{-1}) 	imes t_{1/2}$,we have $x 	imes 10^{-1} = 1.585$,which implies $x = 15.85$.Rounding to the nearest integer,$x = 16$.

For a first order reaction $A \rightarrow B$,the rate constant,$k = 5.5 \times 10^{-14} \, s^{-1}$. The time required for $67 \, \%$ completion of reaction is $x \times 10^{-1}$ times the half life of reaction. The value of $x$ is $....$ (Nearest integer)

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