The half-life of a first order chemical reaction is 6.93 minutes. The time required for the completion of 99% of the reaction will be (Given: log 2 = 0.301)

A
46.06 minutes
B
460.6 minutes
C
230.6 minutes
D
23.03 minutes

Solution:

$\begin{array}{l} k = \frac{0.693}{t_{1 / 2}} = \frac{0.693}{6.93 \min} = 0.1 \min^{- 1} \\ t = - \frac{2.303}{k} \log \frac{[A]_{t}}{[A]_{0}} = - \frac{2.303}{0.1 \min^{- 1}} \log (\frac{1}{100}) = \frac{2 \times 2.303}{0.1 \min^{- 1}} = 46.06 \min \end{array}$

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