Rate constant of first order reaction is 0.005 min⁻¹ — find half life

Question

The rate constant of a first-order reaction is $k = 0.005 \text{ min}^{-1}$ . Calculate the half-life of the reaction.

Solution — Step by Step

For a first-order reaction, the half-life $t_{1/2}$ is related to the rate constant $k$ by:

t_{1/2} = \frac{0.693}{k}

This is derived from the integrated rate law for first order: $\ln[A] = \ln[A]_0 - kt$ . At $t = t_{1/2}$ , $[A] = [A]_0/2$ , so $\ln 2 = k \cdot t_{1/2}$ , giving $t_{1/2} = 0.693/k$ .

The key property: half-life for a first-order reaction is constant — it does not depend on initial concentration.

t_{1/2} = \frac{0.693}{k} = \frac{0.693}{0.005}

t_{1/2} = 138.6 \text{ min}

\boxed{t_{1/2} \approx 138.6 \text{ min}}

$k$ has units of min $^{-1}$ , so $t_{1/2} = 0.693 / k$ has units of min. Units check out.

For reference: 138.6 min ≈ 2.31 hours. In about 2.3 hours, half the reactant will have been consumed.

Why This Works

The special property of first-order reactions is that half-life is independent of concentration. Whether you start with 1 mol/L or 0.001 mol/L, it always takes 138.6 minutes for half the reactant to disappear.

This makes intuitive sense from the rate law $r = k[A]$ : as concentration halves, so does the rate. The system slows proportionally as it proceeds, which is why each successive half is consumed in the same time.

This concentration-independence distinguishes first-order from second-order reactions, where $t_{1/2} = 1/(k[A]_0)$ depends on initial concentration.

Alternative Method

If you forget the formula, derive it quickly:

\ln \frac{[A]_0}{[A]} = kt

At $t_{1/2}$ : $[A] = [A]_0/2$ , so $\ln 2 = k \cdot t_{1/2}$

t_{1/2} = \frac{\ln 2}{k} = \frac{0.693}{0.005} = 138.6 \text{ min}

This derivation takes about 20 seconds in an exam and is worth showing even if you remember the formula — it demonstrates understanding.

Common Mistake

Do not confuse half-life formulas for different reaction orders. For zero order: $t_{1/2} = [A]_0/(2k)$ — depends on initial concentration. For first order: $t_{1/2} = 0.693/k$ — independent of concentration. For second order: $t_{1/2} = 1/(k[A]_0)$ — inversely proportional to initial concentration. Using the wrong formula is the most common error in this type of question.

A nice follow-up: after how many minutes will 75% of the reactant have decomposed? That requires two half-lives (50% gone after one, 75% gone after two). So: $2 \times 138.6 = 277.2$ min. JEE Main frequently asks this type of multi-step question.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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