60% of the reactant decomposes in 45 minutes in a first order reaction. Calculate the half life period of the reaction. Write the relation between half life period and initial concentration for zero order reaction.

Answer: \(t_{1/2} \approx 34.0\) min; for zero order \(t_{1/2} = \dfrac{[A]_0}{2k}\)

Let \([A]_0 = 100\); after 60% decomposition \([A]_t = 40\), \(t = 45\) min.

\(k = \dfrac{2.303}{t}\log_{10}\dfrac{[A]_0}{[A]_t} = \dfrac{2.303}{45}\log_{10}\dfrac{100}{40} = \dfrac{2.303}{45}\times 0.3979\)

\(k = \dfrac{0.9164}{45} = 0.02036\) min⁻¹

\(t_{1/2} = \dfrac{0.693}{k} = \dfrac{0.693}{0.02036} = 34.03\) min

Zero order reaction: \(t_{1/2} = \dfrac{[A]_0}{2k}\); the half life is directly proportional to the initial concentration of the reactant.