Step 1: Understanding the Concept:
A first-order chemical reaction is a reaction whose rate depends linearly on the concentration of only one reactant. A key characteristic of first-order reactions is that their half-life ($t_{1/2}$) is entirely constant and independent of the initial reactant concentration. This means it always takes the exact same amount of time for any given concentration to break down by half.
Step 2: Key Formula or Approach:
The integrated rate equation for a first-order reaction is:
$$ k = \frac{2.303}{t} \log\left(\frac{[A]_0}{[A]_t}\right) $$
Where:
- $[A]_0$ is the initial reactant concentration (100%).
- $[A]_t$ is the remaining reactant concentration at time $t$.
- For 75% completion, the remaining concentration is $[A]_t = 100% - 75% = 25%$.
- The constant half-life formula is derived as:
$$ t_{1/2} = \frac{0.693}{k} $$
Step 3: Detailed Explanation:
Let's find the relationship between the time taken for 75% completion ($t_{75%}$) and the half-life ($t_{50%}$):
1. Method 1: Stepwise Half-Life Progression
- Let the initial reactant concentration be $100%$.
- After the first half-life ($t_{1/2}$), the concentration drops by half: $100% \to 50%$.
- After a second half-life ($t_{1/2}$), the remaining 50% concentration drops by half again: $50% \to 25%$.
- When the concentration reaches 25%, exactly $100% - 25% = 75%$ of the reaction has completed.
- Therefore, the total time required to reach 75% completion is exactly equal to two successive half-lives:
$$ t_{75%} = 2 \times t_{1/2} $$
2. Calculate the numerical value:
- We are given that $t_{75%} = 40 \text{ minutes}$.
- Substitute this value into our relationship:
$$ 40 \text{ min} = 2 \times t_{1/2} $$
$$ t_{1/2} = \frac{40}{2} = 20 \text{ minutes} $$
This result matches option (B).
Step 4: Final Answer:
The half-life of the reaction is 20 minutes.