Chemical Kinetics: Tricky Questions Solved (7)

Question

A first-order reaction has a rate constant $k = 0.0693$ min$^{-1}$. Find: (a) the half-life, (b) the time required for $75\%$ of the reactant to react.

Solution — Step by Step

Why This Works

For a first-order reaction, the half-life is independent of initial concentration — a defining feature. Each successive half-life takes the same time, so:

• $50\%$ remaining after $1 \cdot t_{1/2}$
• $25\%$ remaining after $2 \cdot t_{1/2}$
• $12.5\%$ remaining after $3 \cdot t_{1/2}$
• …

The integrated rate law $\ln([A]_0/[A]) = kt$ comes from integrating $-d[A]/dt = k[A]$.

Alternative Method

Use the $n$-half-life trick: $75\%$ reacted = $25\%$ remaining = $(1/2)^2$, so $n = 2$. Time = $2 \cdot t_{1/2} = 20$ min.

This shortcut works for any percentage that’s a clean power of $1/2$ ($50\%$, $75\%$, $87.5\%$, $93.75\%$). Otherwise use the integrated rate law.

Common Mistake

Students sometimes use the second-order half-life formula ($t_{1/2} = 1/(k[A]_0)$) for a first-order problem. Always confirm the order of the reaction first — read the question carefully or check the units of $k$:

• First-order: $k$ has units of time$^{-1}$.
• Second-order: $k$ has units of M$^{-1}$ time$^{-1}$.

Another trap: confusing $t_{1/2}$ (half-life) with $\tau$ (mean lifetime). For first-order, $\tau = 1/k$ and $t_{1/2} = 0.693/k$. They differ by a factor of $\ln 2$.