Chemical Kinetics: Diagram-Based Questions (1)

Question

For a first-order reaction, the rate constant $k = 0.0231 \text{ min}^{-1}$. Calculate (a) the half-life, (b) the time required for 75% of the reactant to decompose.

Solution — Step by Step

Final answers: $t_{1/2} = \mathbf{30 \text{ min}}$; time for 75% decomposition $= \mathbf{60 \text{ min}}$.

Why This Works

First-order kinetics has a unique signature: half-life is independent of initial concentration. This is why radioactive decay (also first-order) and many drug eliminations follow this pattern. Each half-life cuts the remaining amount in half, regardless of where you start.

Two half-lives reduce the amount to $(1/2)^2 = 1/4$ — exactly our 25% case. Three half-lives → 12.5% remains. Memorize these benchmarks for fast MCQ work.

Alternative Method

Use $[A]/[A]_0 = e^{-kt}$ form. $0.25 = e^{-0.0231 t}$. Take log: $-\ln 0.25 = 0.0231 t$, $t = 1.386/0.0231 = 60$. Same answer; pick whichever form you find faster.

Common Mistake