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 :

34.1 s

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44.1 s

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54.1 s

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24.1 s

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Solution

The correct option is D $24.1 s$
Given,
$T_{1} = 10 s and T_{2} = 20 s$
therefore,
$T_{2} - T_{1} = 10 s$

For first order reaction,
rate $\propto$ concentration

$∴ \frac{r_{1}}{r_{2}} = \frac{C_{1}}{C_{2}} = \frac{0.04}{0.03} = \frac{4}{3}$

$k = \frac{2.303}{T_{2} - T_{1}} l o g \frac{C_{1}}{C_{2}}$
$k = \frac{2.303 \times 0.125}{10}$
$k = 0.0287 s^{- 1}$

Halflife period, $t_{1 / 2} = \frac{0.693}{k}$
$t_{1 / 2} = \frac{0.693}{0.0287} t_{1 / 2} = 24.14 s$

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The rate of a first order reaction is 0.04 mol L−1s−1 at 10 seconds and 0.03 mol L−1s−1 at 20 seconds after initiation of the reaction. The half-life period of the reaction is :

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