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Q. The time required for $60\%$ completion of a first order reaction is $50$ min. The time required for $93.6\%$ completion of the same reaction will be

KCETKCET 2020Chemical Kinetics

Solution:

For a first order reaction the, the rate constant is given by

$k=\frac{2.303}{t} \log \frac{\left[R_{0}\right]}{[R]}$

Given, at $50 \,min , 60 \%$ of the reaction is completed

$\therefore k=\frac{2.303}{t} \log \frac{\left[R_{0}\right]}{[R]}$

$=\frac{2.303}{50} \log \frac{100}{40}$

$=\frac{2.303}{50} \times 0.397$

So, when $93.6 \%$ of the reaction is completed,

$\Rightarrow \frac{2.303}{50} \times 0.397$

$=\frac{2.303}{t} \log \frac{100}{6.4}$

$\Rightarrow \frac{2.303}{50} \times 0.397$

$=\frac{2.303}{t} \times 1.19$

$\Rightarrow t \approx 150\, min$