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Q. What is the two third life of a first order reaction having $ k=5.48\times 10^{-14}s^{-1} $

J & K CETJ & K CET 2006Chemical Kinetics

Solution:

For two-third of a reaction:
${[A]_{0}=a,[A]=a-\frac{2}{3} a=\frac{a}{3}} $
$t_{2 / 3}=\frac{2.303}{k} \log \frac{[A]_{0}}{[A]}$
$=\frac{2.303}{k} \log \frac{a}{\frac{a}{3}}=\frac{2.303}{k} \log 3 $
$t_{2 / 3}=\frac{2.303 \times 0.4771}{5.48 \times 10^{-14}}$
$=2.01 \times 10^{13} s$