In a first-order reaction A rightarrow B, if | vedclass

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(C) 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 $(t_{1/2})$

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$\frac{\ln 2}{k}$

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(C) 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 $(t_{1/2})$,the concentration of the reactant $[A]_t = \frac{[A]_0}{2}$. Substituting this into the equation: $t_{1/2} = \frac{2.303}{k} \log \frac{[A]_0}{[A]_0 / 2}$ $t_{1/2} = \frac{2.303}{k} \log 2$ Since $2.303 \log 2 = \ln 2$,we get: $t_{1/2} = \frac{\ln 2}{k}$ The half-life of a first-order reaction is independent of the initial concentration of the reactant.

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

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