The rate of a first-order reaction is $0.04 m o l L^{- 1} s^{- 1}$ at $10$ seconds and $0.03 m o l L^{- 1} s^{- 1}$ at $20$ seconds after initiation of the reaction. The half-life period of the reaction is:

24.1

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34.1

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44.1

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54.1

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Solution

The correct option is B $24.1$ s
For given $1^{s t}$ order reaction, $K = \frac{1}{t_{2} - t_{1}} l n \frac{C_{1}}{C_{2}}$

where $C_{1}$ & $C_{2}$ donate concentrations at $t = 10 s$ & $t = 20 s$ respectively.

$∴ K = \frac{1}{20 - 10} l n \frac{0.04}{0.03} = 0.0288$

$∴ t_{1 / 2} = \frac{l n 2}{K} = \frac{l n 2}{0.0288} \approx 24.1 s$

Hence, the correct option is $A$

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