The rate constant for a first order reaction is 60 s − 1 . How much time will it take to reduce the concentration of the reactant to 1/2 0 th of its initial value?

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

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

Given $k = 60 s^{- 1}$ and the final concentration $[R] = \frac{[ R ] _{0}}{20}$ .

Substituting the values into the equation:

$t = \frac{2.303}{60} lo g \frac{[ R ] _{0}}{\frac{[ R ] _{0}}{20}}$

$t = \frac{2.303}{60} lo g (20)$

Using $lo g (20) = lo g (2 \times 10) = lo g 2 + lo g 10 = 0.3010 + 1 = 1.3010$ :

$t = \frac{2.303 \times 1.3010}{60}$

$t = \frac{2.9962}{60}$

$t \approx 0.0499 s$

Answer: $0.0499$ s

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