First-Order Elimination
Drug concentration decreases exponentially — a constant fraction is eliminated per unit time.
Formula
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Example
C₀ = 100 mg/L, kₑ = 0.1/hr → after 7 hr: C = 50 mg/L
How it works
This equation models drugs that are removed at a rate proportional to how much is currently in the body: the more drug present, the faster the absolute amount falls, but the same fraction is eliminated in each equal time interval. That proportional-loss rule leads to an exponential decay curve, with the elimination rate constant controlling how steeply the concentration drops. It is the standard model when metabolism and excretion are not saturated and the pharmacokinetics are approximately linear.
When to use it
Use this formula for first-order pharmacokinetics, such as estimating a drug concentration at a later time after an IV bolus or during the elimination phase after absorption is complete. It is also useful for comparing half-life behavior, dosing intervals, and how long a drug remains above a target concentration.
Worked example, step by step
Given: C_0 = 80 mg/L, k_e = 0.25 hr^-1, t = 6 hr Step 1: Substitute into the formula C(t) = C_0 × e^(-k_e t) Step 2: C(6) = 80 × e^(-(0.25)(6)) Step 3: C(6) = 80 × e^(-1.5) Step 4: e^(-1.5) ≈ 0.2231 Step 5: C(6) = 80 × 0.2231 ≈ 17.8 mg/L Final answer: The concentration after 6 hours is about 17.8 mg/L.
Common mistakes to avoid
- using the wrong time units for k_e and t, which makes the exponent incorrect
- adding k_e t instead of multiplying by e to the negative product
- assuming the same amount is eliminated each hour rather than the same fraction
Did You Know?
Most drugs follow first-order kinetics, but alcohol is a famous exception — it's eliminated at a constant rate (zero-order) regardless of concentration.
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