The half-life period of a first-order reactio | vedclass

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Question

(B) For a first-order reaction,the rate constant $K$ is given by the equation: $K = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.At half-life $(t = t_{1/

Vedclass · Accepted Answer

$0.693 / K$

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(B) For a first-order reaction,the rate constant $K$ is given by the equation: $K = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.At half-life $(t = t_{1/2})$,the concentration of the reactant becomes half of its initial concentration,i.e.,$[A]_t = \frac{[A]_0}{2}$.Substituting these values into the equation: $K = \frac{2.303}{t_{1/2}} \log \frac{[A]_0}{[A]_0 / 2} = \frac{2.303}{t_{1/2}} \log 2$.Since $\log 2 \approx 0.3010$,we get $K = \frac{2.303 	imes 0.3010}{t_{1/2}} = \frac{0.693}{t_{1/2}}$.Therefore,$t_{1/2} = \frac{0.693}{K}$.

The half-life period of a first-order reaction is given by:

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