Question:medium

A first-order reaction has a half-life of 20 minutes. The time required for \(87.5%\) completion of the reaction is:

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For first-order reactions, memorize: \(50%\) completion = 1 half-life, \(75%\) completion = 2 half-lives, \(87.5%\) completion = 3 half-lives, \(93.75%\) completion = 4 half-lives.
Updated On: Jun 17, 2026
  • \(40\) min
  • \(60\) min
  • \(80\) min
  • \(100\) min
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The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
For a first-order reaction, the amount of reactant remaining decreases by half during each successive half-life period (\(t_{1/2}\)).
Step 2: Key Formula or Approach:
If the reaction is 87.5% complete, the amount remaining is \(100% - 87.5% = 12.5%\).
We can express the remaining amount as a fraction of the initial amount:
\(12.5% = \frac{12.5}{100} = \frac{1}{8} = (\frac{1}{2})^3 \).
This indicates that the reaction requires 3 half-lives to reach 87.5% completion.
Step 3: Detailed Explanation:

- After 1 half-life (20 min): 50% remains.
- After 2 half-lives (40 min): 25% remains.
- After 3 half-lives (60 min): 12.5% remains.
Therefore, the total time required is:
\[ \text{Total time} = 3 \times t_{1/2} = 3 \times 20 \text{ min} = 60 \text{ min} \]
Step 4: Final Answer:
The time required for 87.5% completion is 60 minutes.
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