For first-order decay with k = 0.231 h^-1 and initial concentration [A]0 = 0.100 M, what is [A] after 3 h?

• A. 0.10 M
• B. 0.050 M
• C. 0.23 M
• D. 0.027 M

Answer: (B)

First-order decay means the concentration drops exponentially with time according to [A] = [A]0 e^{-kt}. With k = 0.231 h^-1 and t = 3 h, the exponent is kt = 0.231 × 3 = 0.693, which is ln 2. That tells us the amount halves over this 3-hour interval. So [A] = 0.100 × e^{-0.693} ≈ 0.100 × 0.500 ≈ 0.050 M. The half-life here is ln 2 / k ≈ 3 h, so after 3 hours you’re at half the initial concentration. The other numbers come from either applying the decay formula incorrectly (giving a larger value as if the concentration grew) or from using more than one half-life.