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The half-life of a 1st order reaction is 1 hr. What is the fraction of the reactant remaining after 3 hours?

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For a first-order reaction, the fraction remaining after time \( t \) is given by the equation \( \frac{[A]_t}{[A]_0} = e^{-kt} \), where \( k \) is the rate constant and \( t \) is the time. The half-life for a first-order reaction is constant and is independent of the initial concentration.
Updated On: Nov 26, 2025
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Solution and Explanation

For a first-order reaction, the rate is directly proportional to the reactant concentration. The half-life (\( t_{1/2} \)) of a first-order reaction is constant, irrespective of the initial concentration. The fraction of reactant remaining at time \( t \) is described by:\[\frac{[A]_t}{[A]_0} = e^{-kt}\]where:
- \( [A]_t \) denotes the reactant concentration at time \( t \),
- \( [A]_0 \) is the initial concentration,
- \( k \) represents the rate constant,
- \( t \) is the elapsed time.
The relationship between half-life and rate constant for a first-order reaction is:
\[t_{1/2} = \frac{0.693}{k}.\]Given a half-life \( t_{1/2} = 1 \, \text{hr} \), the rate constant \( k \) can be determined as:\[k = \frac{0.693}{t_{1/2}} = \frac{0.693}{1} = 0.693 \, \text{hr}^{-1}.\]To ascertain the fraction remaining after 3 hours, we utilize the formula for the fraction of reactant remaining:\[\frac{[A]_t}{[A]_0} = e^{-0.693 \times 3} = e^{-2.079} \approx \frac{1}{8}.\]Consequently, the fraction of reactant remaining after 3 hours is \( \frac{1}{8} \).
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