Question:medium

For a first-order reaction, the time required to reduce the concentration of the reactant to one-fourth of its initial value is 20 minutes. What is the half-life of the reaction?

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In first-order kinetics, if a concentration reduces to \( \frac{1}{2^n} \) in time \( t \), then \( t = n \cdot t_{1/2} \).
Updated On: May 17, 2026
  • 5 min
  • 10 min
  • 20 min
  • 40 min
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The Correct Option is B

Solution and Explanation

Step 1: Apply the first-order decay equation.
The relationship for a first-order reaction is given by: \[ \frac{[A]}{[A]_0} = \left(\frac{1}{2}\right)^{t/t_{1/2}} \]
Step 2: Substitute known values into the equation.
The concentration is observed to be one-fourth of the initial concentration: \[ \left(\frac{1}{2}\right)^{t/t_{1/2}} = \frac{1}{4} = \left(\frac{1}{2}\right)^2 \Rightarrow \frac{t}{t_{1/2}} = 2 \Rightarrow t_{1/2} = \frac{t}{2} = \frac{20}{2} = 10 \text{ min} \]
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