Step 1: Understanding the Question:
The question describes a first-order reaction and gives its half-life (the time for 50% completion). It then asks for the total percentage of reactant that has reacted after a period of two half-lives.
Step 2: Key Formula or Approach:
For a first-order reaction, the half-life (t\(_{1/2}\)) is constant. This means that in every half-life period, the amount of reactant decreases by 50% of its value at the beginning of that period.
Given: t\(_{1/2}\) = 16 minutes (time for 50% completion).
We need to find the percentage reacted after 32 minutes.
Notice that 32 minutes = 2 \( \times \) 16 minutes = 2 \( \times \) t\(_{1/2}\).
Step 3: Detailed Explanation:
Let the initial amount of reactant be 100%.
After the first half-life (16 minutes):
- Amount of reactant reacted = 50% of the initial amount.
- Amount of reactant remaining = 100% - 50% = 50% of the initial amount.
After the second half-life (another 16 minutes, for a total of 32 minutes):
- The reaction will proceed by another 50%, but this is 50% of the amount {remaining} after the first half-life.
- Amount of reactant remaining after 32 mins = 50% of (50% of initial amount) = 0.5 \( \times \) 50% = 25% of the initial amount.
The question asks for the percentage of reactant reacting in 32 minutes.
- Total percentage reacted = Initial percentage - Final percentage remaining
- Total percentage reacted = 100% - 25% = 75%.
Step 4: Final Answer:
After 32 minutes (two half-lives), 75% of the reactant has reacted. This corresponds to option (D).