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(A) For a first order reaction,the integrated rate law is given by $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.At half-life,$t = t_{1/2}$ and $[A]_

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Independent of concentration

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(A) For a first order reaction,the integrated rate law is given by $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.At half-life,$t = t_{1/2}$ and $[A]_t = \frac{[A]_0}{2}$.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,$t_{1/2} = \frac{0.693}{k}$.This expression shows that the half-life of a first order reaction is independent of the initial concentration of the reactant.

Among the following statements,the correct statement about the half-life period for a first order reaction is

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Assertion : For a first order reaction,$t_{1/2}$ is independent of initial concentration.
Reason : For a first order reaction,rate constant $k \propto [R]$.

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