Half-life of a first-order reaction is:

To find the half-life ($t_{1/2}$) of a first-order reaction, we use the integrated rate law for first-order kinetics: $$k = \frac{2.303}{t} \log \frac{[A]_0}{[A]}$$ Where: - $k$ is the rate constant. - $[A]_0$ is the initial concentration. - $[A]$ is the concentration at time $t$. By definition, at the half-life ($t = t_{1/2}$), the concentration of the reactant is half of its initial value: $$[A] = \frac{[A]_0}{2}$$ Substituting this into the equation: $$k = \frac{2.303}{t_{1/2}} \log \frac{[A]_0}{[A]_0/2}$$ $$k = \frac{2.303}{t_{1/2}} \log(2)$$ Since $\log(2) \approx 0.3010$, we calculate: $$k = \frac{2.303 \times 0.3010}{t_{1/2}}$$ $$k = \frac{0.693}{t_{1/2}}$$ Rearranging for $t_{1/2}$: $$t_{1/2} = \frac{0.693}{k}$$ Thus, the half-life of a first-order reaction is independent of the initial concentration and is equal to $0.693/k$.