t½ = 0.693/k. So k = 0.693/693 = 0.001 s⁻¹. The 0.693 = ln2. Units of k for first order.

Half-Life of First Order Reaction is 693 s — Find Rate Constant

Question

The half-life of a first-order reaction is 693 s. Calculate the rate constant of the reaction.

Solution — Step by Step

For first-order reactions, the half-life is related to the rate constant by:

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

The 0.693 comes from $\ln 2 = 0.6931...$ , which appears when we set the integrated rate law equal to half the initial concentration. This is a fixed relationship — unlike zero or second order, the half-life of a first-order reaction doesn’t depend on the initial concentration.

We need $k$ , so rearrange:

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

This is the only rearrangement needed. Plug in $t_{1/2} = 693$ s:

k = \frac{0.693}{693 \text{ s}}

k = 0.001 \text{ s}^{-1} = 1 \times 10^{-3} \text{ s}^{-1}

The units of $k$ for a first-order reaction are always $\text{s}^{-1}$ (or $\text{min}^{-1}$ , $\text{hr}^{-1}$ depending on the time unit given). This is a key fact examiners test directly.

Answer: $k = 1 \times 10^{-3} \text{ s}^{-1}$

Why This Works

The integrated rate law for a first-order reaction is $\ln[A] = \ln[A]_0 - kt$ . At $t = t_{1/2}$ , the concentration drops to $[A]_0 / 2$ . Substituting gives $\ln(1/2) = -kt_{1/2}$ , which simplifies to $kt_{1/2} = \ln 2 = 0.693$ .

This is why 0.693 appears — it’s not a magic number someone chose. It’s the natural log of 2, locked in by the mathematics of exponential decay.

The fact that $t_{1/2}$ is independent of $[A]_0$ is what makes first-order reactions special. You don’t need to know the starting concentration to find $k$ — just the half-life. This is heavily used in radioactive decay calculations in both chemistry and physics chapters.

In JEE and NEET, you’ll often see half-life given in minutes or hours but needing $k$ in $\text{s}^{-1}$ , or vice versa. Always check the time units before dividing. Convert first, then calculate.

Alternative Method

You can work backwards from the integrated rate law directly without memorising the $t_{1/2}$ formula.

Start with: $[A] = [A]_0 \cdot e^{-kt}$

At half-life, $[A] = \frac{[A]_0}{2}$ :

\frac{[A]_0}{2} = [A]_0 \cdot e^{-k \cdot t_{1/2}}

\frac{1}{2} = e^{-k \cdot 693}

Taking natural log on both sides:

\ln\left(\frac{1}{2}\right) = -k \times 693

-0.693 = -693k

k = \frac{0.693}{693} = 1 \times 10^{-3} \text{ s}^{-1}

Same answer. Deriving it this way is slower but shows you understand the origin of the formula — which NCERT-based questions sometimes ask for as a short derivation (2 marks in board exams).

Common Mistake

Using $t_{1/2} = \frac{0.693}{k}$ for a zero-order or second-order reaction. This formula is exclusive to first order. For zero-order, $t_{1/2} = \frac{[A]_0}{2k}$ , which depends on the initial concentration. If the question doesn’t explicitly say “first order”, identify the order first. Applying the wrong formula here is a full-marks loss in boards.

A second common slip: writing $k = 693 / 0.693$ instead of $k = 0.693 / 693$ . When numbers are close, it’s easy to flip the fraction. Always keep $t_{1/2}$ in the denominator when solving for $k$ .

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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