Question

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?

• A. 5 min
• B. 10 min
• C. 20 min
• D. 40 min

Hint:

In first-order kinetics, if a concentration reduces to \( \frac{1}{2^n} \) in time \( t \), then \( t = n \cdot t_{1/2} \).

Answer 1

The Correct Option is B

Step 1: Use the first-order decay formula.
For a first-order reaction: \[ \frac{[A]}{[A]_0} = \left(\frac{1}{2}\right)^{t/t_{1/2}} \]
Step 2: Plug in the given values.
We are told the concentration is reduced to one-fourth of the initial: \[ \left(\frac{1}{2}\right)^{t/t_{1/2}} = \frac{1}{4} = \left(\frac{1}{2}\right)^2 \Rightarrow \frac{t}{t_{1/2}} = 2 \Rightarrow t_{1/2} = \frac{t}{2} = \frac{20}{2} = 10 \text{ min} \]