Derive the relation between half life and rate constant for a first order reaction.

Answer: \(t_{1/2} = \dfrac{0.693}{k}\)

The integrated rate law for a first order reaction is \(k = \dfrac{2.303}{t}\log_{10}\dfrac{[\mathrm{A}]_0}{[\mathrm{A}]_t}\).

At \(t = t_{1/2}\), \([\mathrm{A}]_t = \dfrac{[\mathrm{A}]_0}{2}\). Substituting:

\(k = \dfrac{2.303}{t_{1/2}}\log_{10}\dfrac{[\mathrm{A}]_0}{[\mathrm{A}]_0/2} = \dfrac{2.303}{t_{1/2}}\log_{10}2 = \dfrac{2.303 \times 0.3010}{t_{1/2}}\)

\(\therefore t_{1/2} = \dfrac{0.693}{k}\)

Thus the half-life of a first order reaction is inversely proportional to the rate constant and independent of the initial concentration.