Question:medium

A first-order reaction is 25% complete in 30 minutes. What is its half-life?

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In first-order reactions, use $\displaystyle t_{1/2} = \frac{0.693}{k}$ and apply logarithmic equations carefully.
Updated On: Nov 26, 2025
  • 72.33
  • 70
  • 75.33
  • 77.66
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The Correct Option is A

Solution and Explanation

For a first-order reaction, the rate constant \( k \) is given by \[ k = \frac{2.303}{t} \log \frac{[A]_0}{[A]} \] When a reaction is 25% complete, 75% of the reactant remains. Therefore, the ratio of initial to remaining reactant concentration is \[ \frac{[A]_0}{[A]} = \frac{100}{75} = \frac{4}{3} \] and the time elapsed is \( t = 30 \text{ min} \). Substituting these values, the rate constant is \[ k = \frac{2.303}{30} \log \left( \frac{4}{3} \right) = \frac{2.303}{30} \times 0.1249 \approx 0.00958 \text{ min}^{-1} \] The half-life for a first-order reaction is calculated using the formula \[ t_{1/2} = \frac{0.693}{k} \] Using the determined rate constant, the half-life is \[ t_{1/2} = \frac{0.693}{0.00958} \approx 72.33 \ \text{min} \]
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