Correct Option: (B) $5.0 \times 10^{-4} \text{ s}^{-1}$

Step 1 (Formula / Governing Law & Given Data): [1/3 Mark]

For a first-order reaction, the relationship between the rate constant ($k$) and the half-life ($t_{1/2}$) is given by the formula:

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

Given data:

• $t_{1/2} = 1386 \text{ s}$

Step 2 (Step-by-Step Mathematical Working): [1/3 Mark]

Substituting the given value into the formula:

$$k = \frac{0.693}{1386}$$

To simplify the calculation, express $0.693$ as $693 \times 10^{-3}$:

$$k = \frac{693 \times 10^{-3}}{1386}$$

Since $1386 = 2 \times 693$, we can simplify the fraction:

$$k = \frac{1}{2} \times 10^{-3}$$ $$k = 0.5 \times 10^{-3} \text{ s}^{-1}$$

Adjusting to standard scientific notation to match the options:

$$k = 5.0 \times 10^{-4} \text{ s}^{-1}$$

Step 3 (Final Concluded Answer): [1/3 Mark]

The value of the rate constant is $5.0 \times 10^{-4} \text{ s}^{-1}$. Therefore, the correct option is (B).

💡 EL10 Concept Insight: Think of the half-life as a "speedometer" for a reaction; because the rate constant is inversely proportional to the time it takes for half the substance to disappear, a longer half-life simply means the reaction is moving slower, resulting in a smaller rate constant value.