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Q. In the first order reaction, half of the reaction is completed in $100$ second. The time for $99\%$ reaction to occur will be

Rajasthan PMTRajasthan PMT 2003Chemical Kinetics

Solution:

Half of the reaction is completed in $100$ second
$\therefore t_{1 / 2}=100 \,sec . $
$\therefore K =\frac{0.693}{100} sec ^{-1}$
For a first order reaction
$K =\frac{2.303}{t} \log \frac{a}{a-x} $
$a=100, \,\,\,x=99, \,\,\,t_{99 \%}=? $
$t_{99 \%}=\frac{2.303}{ K } \log \frac{100}{100-99}$
$=\frac{2.303 \times 100}{0.693} \times \log 100 sec$
$=\frac{2.303 \times 100 \times 2}{0.693}$
$=664.64\, sec$