i. The integrated rate law for the first-order reaction is given by:

\[k = \frac{2.303}{t} \log_{10} \left(\frac{[A]_0}{[A]_t}\right)\]

Where [A]₀ is the initial concentration of the reactant at \(t = 0\). The concentration falls to [A]_t at time \(t\) after the start of the reaction. The concentration of the reactant falls to \(\frac{[A]_0}{2}\) at time \(t_{1/2}\).

ii. The time required for [A]₀ to become \(\frac{[A]_0}{2}\) is denoted as \(t_{1/2}\), which satisfies \([A]_t = \frac{[A]_0}{2}\) at \(t = t_{1/2}\).

Putting this condition in the integrated rate law, we have:

\[k = \frac{2.303}{t_{1/2}} \log_{10} 2\]

Substituting the value of \(\log_{10} 2\),

\[k = \frac{2.303}{t_{1/2}} \cdot 0.3010\]

Therefore, \[k = 0.693/t_{1/2}\]

Therefore, \[t_{1/2} = 0.693/k\]