Rate constant for first order reaction is 40 s − 1 . Calculate the time required to reduce the concentration to 1/20 th of its initial value.

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For a first order reaction, the time required is given by the integrated rate equation:

$t = \frac{2.303}{k} lo g (\frac{[ A ] _{0}}{[ A ]})$

Given $k = 40 s^{- 1}$ and $[A] = \frac{[ A ] _{0}}{20}$

$t = \frac{2.303}{40} lo g (20)$

Using $lo g 20 = lo g 2 + lo g 10 \approx 0.3 + 1 = 1.3$

$t = \frac{2.303 \times 1.3}{40}$

$t = \frac{2.9939}{40} = 0.07484 s$

$t = 7.48 \times 1 0^{- 2} s$

Answer: $7.48 \times 1 0^{- 2}$ s

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