Chemical Kinetics: Numerical Problems Set (3)

Question

For a first-order reaction, the half-life is $30\text{ min}$. Calculate the time required for $75\%$ of the reactant to decompose.

Solution — Step by Step

Why This Works

For a first-order reaction, the half-life is constant — every $t_{1/2}$, the concentration drops to half. So after one half-life, $[A] = [A]_0/2$ ($50\%$ decomposed). After two half-lives, $[A] = [A]_0/4$ ($75\%$ decomposed). After three, $[A] = [A]_0/8$ ($87.5\%$). The time for $75\%$ decomposition is exactly $2 t_{1/2}$.

This logarithmic decay is unique to first-order kinetics. Zero-order and second-order reactions have half-lives that depend on starting concentration.

Alternative Method

Use the doubling shortcut directly. $75\%$ decomposed = $25\%$ remaining = $1/4$ of original = $(1/2)^2$. So we need 2 half-lives. $2 \times 30 = 60\text{ min}$. Done — no need for logarithms.

Common Mistake

Confusing first-order with zero-order or second-order half-life formulas. Only first-order has a half-life independent of $[A]_0$. For zero-order, $t_{1/2} = [A]_0/(2k)$. For second-order, $t_{1/2} = 1/(k[A]_0)$.