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In the first order reaction, the concentration of the reactants is reduced to 25% in one hour, the half-life of the reaction is

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$\frac{1}{2}$ hrs

$\frac{1}{4}$ hrs

2 hrs

answer is C.

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Detailed Solution

$K = \frac{2.303}{t} \log (\frac{a}{a - x}) k = \frac{2.303}{1} \log (\frac{100}{25}) = 2 \times 0.693 t_{1 / 2} = \frac{0.693}{k} = \frac{0.693}{2 \times 0.693} = \frac{1}{2} h r s$

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