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  4. A reaction proceeds by first order, 75 % of this reaction was completed in 32 min. The time required for 50 % completion is

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Q. A reaction proceeds by first order, $ 75\% $ of this reaction was completed in $ 32 $ min. The time required for $ 50\% $ completion is

UPSEEUPSEE 2008

Solution:

For first order reaction
$k=\frac{2.303}{t} log_{10} \frac{a}{\left(a-x\right)}$
where, $a$ = Initial concentration
$x$ = change in concentration during time ‘t’.
If $75\%$ of the reaction was completed in $32$ min,then
$k=\frac{2.303}{32}log_{10} \frac{100}{\left(100-75\right)}$
$=\frac{2.303}{32} log_{10}\,4$
$k=0.0433\, min^{-1}$
Hence, time required for the completion of $50\% $ reaction
$t=\frac{2.303}{0.0433} log_{10} \frac{100}{\left(100-50\right)}$
$=\frac{2.303}{0.0433} log_{10}2$
$=16\,min$