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

{(14)}^{3} s e c

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28 s e c

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42 s e c

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1.75 s e c

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Solution

The correct option is C $42 s e c$
$t_{1 / 2} = 14$ sec= $\frac{I n 2}{k}$ ....[half-life for 1st order reaction]
$k = \frac{I n 2}{14}$ -(1)
For 1st order reaction, $k = \frac{1}{t} I n \frac{a_{0}}{a_{t}}$
given that $a_{t} = \frac{a_{0}}{8}$
$k = \frac{1}{t} I n \frac{a_{0}}{a_{0} / 8}$ => $k = \frac{1}{t} I n 8$ -(2)
From (1) and (2),
$\frac{I n 2}{14} = \frac{1}{t} I n (2)^{3} = \frac{3}{t} I n (2)$
$\frac{1}{14} = \frac{3}{t}$
t=42 sec.

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

The correct option is C 42 sect1/2=14sec=In2k....[half-life for 1st order reaction]k=In214-(1)For 1st order reaction, k=1tIna0atgiven that at=a08k=1tIna0a0/8=> k=1tIn8-(2)From (1) and (2),In214=1tIn(2)3=3tIn(2)114=3tt=42 sec.

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