For the first order reaction half-life is 14 sec, the time required for the initial concentration to reduce to 1/8 of its value is:

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${(14)}^{2}$ sec

${(14)}^{3}$ sec

28 sec

42 sec

answer is C.

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Detailed Solution

$t_{1 / 2} = 14 \sec, N = \frac{N_{0}}{8}$

$\begin{array}{l} \frac{N}{N_{o}} = {(\frac{1}{2})}^{n} \\ \frac{N}{8 N_{o}} = {(\frac{1}{2})}^{n} \Rightarrow {(\frac{1}{2})}^{3} = {(\frac{1}{2})}^{n} \Rightarrow n = 3 \\ t = {nt}_{1 / 2} = 3 \times 14 = 42 \sec \end{array}$

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