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

A reaction proceeds by first order, $75 \%$ of this reaction was completed in 32 min. The time required for $50 \%$ completion is
  1. $8 \mathrm{~min}$
  2. $16 \mathrm{~min}$
  3. $20 \mathrm{~min}$
  4. $24 \mathrm{~min}$

Solution

Given: $75 \%$ reaction gets completed in 32 $\min$
Thus, $k=\frac{2.303}{t} \log \frac{a}{(a-x)}$
$=\frac{2.303}{32} \log \frac{100}{(100-75)}$
$=\frac{2.303}{32} \log 4=0.0433 \mathrm{~min}^{-1}$
Now we can use this value of $k$ to get the value of time required for $50 \%$ completion of reaction
$t=\frac{2.303}{k} \log \frac{a}{(a-x)}=\frac{2.303}{0.0433} \log \frac{100}{50}$
$=\frac{2.303}{0.0433} \log 2=16 \mathrm{~min}$

Asked in: JEE-TOPICTESTS-CHEMISTRY

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