Chemical Kinetics - Result Question 35
####35. If $60 %$ of a first order reaction was completed in 60 minutes, $50 %$ of the same reaction would be completed in aproximately
(a) 45 minutes
(b) 60 minutes
(c) 40 minutes
(d) 50 minutes
[2007] $(\log 4=0.60, \log 5=0.69)$
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Solution:
- (a) For a first order reaction
$k=\frac{2.303}{t} \log \frac{a}{a-x}$
when $t=60$ and $x=60 %$
$k=\frac{2.303}{60} \log \frac{100}{100-60}=\frac{2.303}{60} \log \frac{100}{40}=0.0153$ Now,
$ \begin{aligned} t _{1 / 2} & =\frac{2.303}{0.0153} \log \frac{100}{100-50}=\frac{2.303}{0.0153} \times \log 2 \\ & =\frac{2.303}{0.0153} \times 0.3010=45.31 min . \end{aligned} $
