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The rate of a first order reaction is 0.01 mol/lit-sec at 10 min and 0.03 mole/lit-sec at 20 min. After intiation, Find the half life of the reaction. [Given $\log \frac{4}{3} = 0 .1238$ ]

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detailed solution

Correct option is X

$\begin{array}{l} K = \frac{2 .303}{t} \log \frac{a}{(a - x)} \\ t = 20 - 10 = 10 min; a = 0 .04; (a - x) = 0 .03 \\ K = \frac{2 .303}{10} \log \frac{0 .04}{0 .03} K = \frac{2 .303}{10} \log (\frac{4}{3}) ∣ \\ t_{1 / 2} = \frac{0 .693}{K} = \frac{0 .693}{(\frac{2 .303}{10}) \log (4 / 3)} \\ t_{1 / 2} = \frac{0 .693}{2 .303} \times 10 \times (\frac{1}{0 .1238}) = 24 .2 \end{array}$

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detailed solution

Correct answer is 24.2

$\begin{array}{l} K = \frac{2 .303}{t} \log \frac{a}{(a - x)} \\ t = 20 - 10 = 10 min; a = 0 .04; (a - x) = 0 .03 \\ K = \frac{2 .303}{10} \log \frac{0 .04}{0 .03} K = \frac{2 .303}{10} \log (\frac{4}{3}) ∣ \\ t_{1 / 2} = \frac{0 .693}{K} = \frac{0 .693}{(\frac{2 .303}{10}) \log (4 / 3)} \\ t_{1 / 2} = \frac{0 .693}{2 .303} \times 10 \times (\frac{1}{0 .1238}) = 24 .2 \end{array}$

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