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In a first – order reaction A $\to$ B. if k is rate constant and initial concentration of the reactant A is 0.5 M, then the half – life is

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$\frac{log 2}{k}$

$\frac{log 2}{k \sqrt{0.5}}$

$\frac{ln 2}{k}$

$\frac{0.693}{0.5k}$

answer is C.

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Detailed Solution

For a first order reaction

$k = \frac{2.303}{t} log \frac{a}{a - x}$

${when t = t}_{\frac{1}{2}}$

$k = \frac{2.303}{t_{\frac{1}{2}}} log \frac{a}{a - a/2}$

${or t}_{1 /_{2}} = \frac{2.303}{k} log 2 = \frac{ln 2}{k}$

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