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Chemistry
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
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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
KCET
KCET 2020
Chemical Kinetics
A
100 min
13%
B
83.8 min
24%
C
50 min
14%
D
150 min
48%
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,
$\Rightarrow \frac{2.303}{50} \times 0.397$
$=\frac{2.303}{t} \log \frac{100}{6.4}$
$\Rightarrow \frac{2.303}{50} \times 0.397$
$=\frac{2.303}{t} \times 1.19$
$\Rightarrow t \approx 150\, min$