Derive the half-life $t_{1/2}$ of a first-order reaction.

(N/A) For a first-order reaction,the rate constant $(k)$ is given by the equation: $k = \frac{2.303}{t} \log \frac{[R]_{0}}{[R]}$ ...$(i)$
At half-life,$t = t_{1/2}$ and the concentration of the reactant $[R] = \frac{[R]_{0}}{2}$.
Substituting these values into equation $(i)$:
$k = \frac{2.303}{t_{1/2}} \log \frac{[R]_{0}}{[R]_{0}/2}$
$k = \frac{2.303}{t_{1/2}} \log 2$
Since $\log 2 \approx 0.3010$:
$k = \frac{2.303 \times 0.3010}{t_{1/2}}$
$k = \frac{0.693}{t_{1/2}}$
Therefore,$t_{1/2} = \frac{0.693}{k}$.
Conclusion: For a first-order reaction,the half-life is independent of the initial concentration of the reactant.