First-order reaction — derive half-life expression and solve numerical

Question

For a first-order reaction, derive the expression for half-life ( $t_{1/2}$ ). Then solve: A first-order reaction has a rate constant $k = 1.386 \times 10^{-2}\ \text{min}^{-1}$ . Calculate the half-life and find how much of the reactant remains after 200 minutes if the initial concentration is 0.8 mol/L.

Solution — Step by Step

The integrated rate law for a first-order reaction is:

\ln [A]_t = \ln [A]_0 - kt

Or equivalently: $\ln \dfrac{[A]_0}{[A]_t} = kt$

We need this form because half-life is defined as the time when $[A]_t = \dfrac{[A]_0}{2}$ .

At $t = t_{1/2}$ , the concentration drops to half its initial value. Substituting:

\ln \frac{[A]_0}{[A]_0/2} = k \cdot t_{1/2}

\ln 2 = k \cdot t_{1/2}

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

The key insight here: $t_{1/2}$ for a first-order reaction is independent of initial concentration. This is what separates first-order from zero- and second-order reactions.

Given $k = 1.386 \times 10^{-2}\ \text{min}^{-1}$ :

t_{1/2} = \frac{0.693}{1.386 \times 10^{-2}} = 50\ \text{min}

Notice that $1.386 \approx 2 \times 0.693$ , so the division is clean. NCERT loves to set up $k$ this way.

200 minutes = 4 half-lives (since $t_{1/2} = 50$ min).

After each half-life, concentration halves:

[A]_{200} = [A]_0 \times \left(\frac{1}{2}\right)^4 = 0.8 \times \frac{1}{16} = 0.05\ \text{mol/L}

Remaining concentration = 0.05 mol/L

Why This Works

The independence of $t_{1/2}$ from initial concentration is a direct consequence of the mathematics. When we substitute $[A]_t = [A]_0/2$ , the $[A]_0$ cancels out completely — it doesn’t matter whether you started with 1 mol/L or 10 mol/L.

This is not true for zero-order reactions (where $t_{1/2} = [A]_0/2k$ , so it decreases as the reaction proceeds) or second-order reactions. In board exams and JEE Main, a common MCQ type gives you two different initial concentrations and asks which reaction order has the same $t_{1/2}$ — the answer is always first-order.

The factor 0.693 is simply $\ln 2$ . Keep this memorised — the examiner will never give you time to derive it from scratch in the exam hall.

Alternative Method

Instead of the half-life shortcut, we can use the integrated rate law directly for the 200-minute part.

\ln \frac{[A]_0}{[A]_t} = kt = (1.386 \times 10^{-2}) \times 200 = 2.772

\frac{[A]_0}{[A]_t} = e^{2.772} = e^{4 \ln 2} = 2^4 = 16

[A]_t = \frac{0.8}{16} = 0.05\ \text{mol/L}

Same answer. The “n half-lives” method is faster when the time is an exact multiple of $t_{1/2}$ , but in JEE Advanced, the time is often not a clean multiple — use the integrated rate law then.

When $kt$ comes out as a multiple of $\ln 2$ (like $2\ln 2$ , $4\ln 2$ ), it means the time is an exact number of half-lives. Always check this first — it saves you from messy exponential calculations.

Common Mistake

Students often write $t_{1/2} = \frac{0.693}{k}$ and then assume this applies to all reaction orders. It does not. This formula is only valid for first-order reactions. For zero-order: $t_{1/2} = \frac{[A]_0}{2k}$ . For second-order: $t_{1/2} = \frac{1}{k[A]_0}$ . In CBSE 2023 and JEE Main 2024, questions specifically tested whether students could identify the order from the $t_{1/2}$ expression — don’t mix them up.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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