MCAT® ExamChemical and Physical Foundations of Biological SystemsHard
A chemist is performing a reaction in a sealed container and observes that the concentration of a reactant decreases by 0.20 M over a period of 50 seconds. If the reaction follows first-order kinetics and the initial concentration was 1.0 M, what is the rate constant (k) for this reaction?
- A0.006 s⁻¹
- B0.002 s⁻¹
- C0.004 s⁻¹
- D0.005 s⁻¹
Show answer & explanationAnswer & explanation
Correct answer: C. 0.004 s⁻¹
For a first-order reaction, the integrated rate law is ln[A]t - ln[A]0 = -kt, or ln([A]t/[A]0) = -kt. Initial concentration [A]0 = 1.0 M. Final concentration [A]t = 1.0 M - 0.20 M = 0.80 M. Time (t) = 50 s. Substitute these values: ln(0.80 / 1.0) = -k * 50 s. ln(0.80) = -k * 50 s. -0.223 = -k * 50 s. k = 0.223 / 50 s = 0.00446 s⁻¹. Rounding to one significant figure for the given options, the closest is 0.004 s⁻¹.
Why the other options are wrong
- A. Incorrect. This might result from an incorrect formula or calculation.
- B. Incorrect. This might result from using incorrect concentrations or an arithmetic error.
- D. Incorrect. This might result from misinterpreting the concentration change or an arithmetic error.
First-Order Kinetics
In first-order kinetics, the reaction rate is directly proportional to the concentration of one reactant.
- Rate = k[A]
- Integrated rate law: ln[A]t = -kt + ln[A]0
- Half-life (t1/2) = ln(2)/k; half-life is independent of initial concentration.
Memory trick: First-order is a Lonely Logarithm, Zero-order is a Straight Line.