93.75% of a first order reaction is completed in 20 minutes. The specific reaction rate is (Given that log 2 = 0.3)

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$0.14 \min^{- 1}$

$0.32 \min^{- 1}$

$0.18 \min^{- 1}$

$0.28 \min^{- 1}$

answer is B.

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Detailed Solution

$K = \frac{2.303}{t} \log \frac{{[R]}_{0}}{[R]} = \frac{2.303}{20} \log \frac{100}{100 - 93.75}$
$= \frac{2.303}{20} \log \frac{100}{6.25} = \frac{2.303}{20} \log 16 = 0.14 \min^{- 1}$

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