Solution (A)
Step 1 (Formula & Given Data): [0.5 Mark]
For a first-order reaction, the rate constant $k$ is given by the integrated rate equation:
$$k = \frac{2.303}{t} \log \frac{[R]_0}{[R]_t}$$Given data:
- Initial concentration $[R]_0 = 0.6 \text{ mol L}^{-1}$
- Final concentration $[R]_t = 0.2 \text{ mol L}^{-1}$
- Time $t = 5 \text{ minutes}$
Step 2 (Step-by-Step Mathematical Working): [1.0 Mark]
Substituting the values into the formula:
$$k = \frac{2.303}{5} \log \left( \frac{0.6}{0.2} \right)$$ $$k = \frac{2.303}{5} \log(3)$$Using the given value $\log 3 = 0.48$:
$$k = \frac{2.303 \times 0.48}{5}$$ $$k = \frac{1.10544}{5}$$Step 3 (Final Concluded Answer): [0.5 Mark]
$$k = 0.2211 \text{ min}^{-1}$$Solution (B)
Step 1 (Formula & Given Data): [0.5 Mark]
For a first-order reaction, the time $t$ required is:
$$t = \frac{2.303}{k} \log \frac{[R]_0}{[R]_t}$$Given data:
- $k = 2.54 \times 10^{-3} \text{ s}^{-1}$
- If $3/4$ of the reactant decomposes, the remaining amount $[R]_t = [R]_0 - \frac{3}{4}[R]_0 = \frac{1}{4}[R]_0$
Step 2 (Step-by-Step Mathematical Working): [1.0 Mark]
Substituting the values:
$$t = \frac{2.303}{2.54 \times 10^{-3}} \log \left( \frac{[R]_0}{[R]_0 / 4} \right)$$ $$t = \frac{2.303}{2.54 \times 10^{-3}} \log(4)$$Using the given value $\log 4 = 0.60$:
$$t = \frac{2.303 \times 0.60}{2.54 \times 10^{-3}}$$ $$t = \frac{1.3818}{2.54 \times 10^{-3}}$$ $$t = 0.544 \times 10^3 \text{ s}$$Step 3 (Final Concluded Answer): [0.5 Mark]
$$t = 544 \text{ seconds}$$💡 EL10 Concept Insight: Think of a first-order reaction like a leaky bucket where the amount of water leaking out depends only on how much is currently inside; because the "leak" slows down as the bucket empties, we use logarithms to calculate exactly how long it takes to reach a