A medication is eliminated from the body with a constant half-life of 6 hours. A patient takes a 400 mg dose. How much remains after 18 hours, and what does the constant half-life reveal about the elimination kinetics?
A200 mg; a constant half-life indicates zero-order elimination at a fixed rate
B50 mg; a half-life that is constant regardless of dose indicates first-order elimination kinetics
C0 mg; all drug is eliminated exactly by the third half-life
D100 mg; successive half-lives are additive, so 18 hours leaves 25% of the dose
After 18 hours (3 half-lives): 400 → 200 → 100 → 50 mg. A half-life that remains constant regardless of how much drug remains is the definitive signature of first-order kinetics. In first-order elimination, rate is proportional to concentration (rate = k[drug]), so the same fraction (50%) is eliminated per half-life regardless of starting amount. Zero-order elimination proceeds at a fixed mg/hour regardless of concentration, so its half-life depends on starting dose and shortens over time. Option C is wrong — exponential decay approaches but never mathematically reaches zero.
Question 2 Multiple Choice
A chemist measures the concentration of reactant A over time and plots [A] vs. t, ln[A] vs. t, and 1/[A] vs. t. Only the 1/[A] vs. t plot is a straight line. What can she conclude?
AThe reaction is first-order; k equals the negative of the slope of the 1/[A] plot
BThe reaction is zero-order; k equals the slope of the 1/[A] plot
CThe reaction is second-order; k equals the slope of the 1/[A] vs. t line
DNo conclusion is possible from graphical methods alone — a kinetic model must be assumed first
This is exactly the graphical method in action. The integrated rate law for second-order kinetics is 1/[A] = 1/[A]₀ + kt — a linear equation in 1/[A] with slope +k (positive). If [A] vs. t were linear, the reaction would be zero-order (slope = −k). If ln[A] vs. t were linear, it would be first-order (slope = −k). Whichever linearization gives a straight line reveals the order, and the slope directly gives k. Option D is wrong — this graphical method IS the standard experimental approach to determining reaction order, requiring no prior assumption.
Question 3 True / False
For a zero-order reaction, successive half-lives are most equal, because the constant reaction rate ensures that the same fraction of reactant is consumed in each time interval.
TTrue
FFalse
Answer: False
Constant half-lives are the exclusive property of first-order reactions, not zero-order. For a zero-order reaction, rate = k (constant, independent of concentration), so a fixed amount of reactant depletes per unit time — not a fixed fraction. The half-life is t₁/₂ = [A]₀/2k, which depends on starting concentration, so each successive half-life is shorter (each 'half' starts with less material, and a constant rate depletes it faster). Only first-order kinetics produce a concentration-independent half-life.
Question 4 True / False
The first-order integrated rate law predicts that reactant concentration decays exponentially and theoretically never reaches exactly zero at any finite time.
TTrue
FFalse
Answer: True
First-order integrated rate law: [A] = [A]₀e⁻ᵏᵗ. Since e⁻ᵏᵗ > 0 for all finite t, the concentration asymptotically approaches zero without ever reaching it — this is the mathematical nature of exponential decay. Each half-life halves the remaining amount, but halving a positive number always leaves a positive number. In practice, once concentrations fall below detection limits we say the reaction is 'complete,' but mathematically the curve approaches the axis asymptotically. Radioactive decay follows the same mathematics for the same reason.
Question 5 Short Answer
Explain why the constant half-life property is unique to first-order reactions and does not hold for zero-order or second-order reactions.
Think about your answer, then reveal below.
Model answer: For a first-order reaction, rate = k[A], so the rate decreases proportionally as concentration decreases. The fraction consumed per unit time remains constant, and the half-life t₁/₂ = 0.693/k contains no concentration term — it is independent of how much reactant remains. For zero-order reactions, rate = k (constant amount depleted per time), so the time to consume half the remaining material shortens as less material is present — half-lives decrease. For second-order reactions, rate = k[A]², so the rate drops dramatically as [A] falls and successive half-lives grow longer.
The mathematical signature is in the integrated forms. Only first-order [A] = [A]₀e⁻ᵏᵗ produces a t₁/₂ formula (0.693/k) with no [A]₀ term. Zero-order: t₁/₂ = [A]₀/2k (shorter each half-life). Second-order: t₁/₂ = 1/(k[A]₀) (longer each half-life). This distinction has clinical relevance: drugs following first-order kinetics (most common) reach steady-state in a predictable number of half-lives regardless of dose, while zero-order drugs (like alcohol at saturation of alcohol dehydrogenase) have dose-dependent kinetics that can lead to dangerous accumulation.