For a first-order reaction, what percentage of the reactant has been consumed after one half-life has elapsed?

• A. 75%
• B. 50%
• C. 25%
• D. 100%

Answer: (B)

Verified step by step guidance

1. Recall that for a first-order reaction, the half-life (\(t_{1/2}\)) is the time required for the concentration of the reactant to decrease to half of its initial value.
2. Express the relationship between the concentration at time \(t\) and the initial concentration \([A]_0\) using the first-order integrated rate law: \([A] = [A]_0 \times e^{-kt}\), where \(k\) is the rate constant.
3. At one half-life (\(t = t_{1/2}\)), substitute into the equation: \([A] = [A]_0 \times e^{-k t_{1/2}}\).
4. Use the definition of half-life for a first-order reaction: \(t_{1/2} = \frac{\ln 2}{k}\). Substitute this into the equation to get \([A] = [A]_0 \times e^{-k \times \frac{\ln 2}{k}} = [A]_0 \times e^{-\ln 2} = [A]_0 \times \frac{1}{2}\).
5. Since the concentration is half of the initial concentration, the percentage of reactant consumed is \$100\% - 50\% = 50\%$.