Question

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)
(Given : $\log 3=0.4771$ )

Answer 1

$t _{67 \%}=\frac{1}{ k } \ln \left(\frac{1}{1-0.67}\right)=\frac{ t _{1 / 2}}{\ln 2} \times \ln \left(\frac{1}{1-\frac{2}{3}}\right)$
$t _{67 \%}=\frac{ t _{1 / 2}}{\log 2} \times \log 3=\frac{ t _{1 / 2} \times 0.4771}{0.301}$
$\Rightarrow t _{67 \%}=1.585 \times t _{1 / 2}$
$X \times 10^{-1}=1.585$
$\Rightarrow X =15.85$
Ans.16