T15: K from Dose, AUC and V

50. Pharmacokinetics 50.1) PK Basic parameters 50.1.3) Area Under Curve (AUC)

50.1.3.1) AUC basic concepts Normal 1


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Determine the elimination rate constant of a drug from the provided information. The volume of distribution is 75.48 L after administering 107.94 mg of the drug. The area under the plasma drug concentration over time (AUC) was found to be 5.3 mg × h/L.


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lb equals 0.27 h-1 kg

    We know that, `AUC=(dose)/(K × V)`.

    `⇒ K=(dose)/(V × AUC)`

    Plugging the values in this equation we get:
    ` K = (107.94 \quad mg)/(75.48 \quad L × 5.3 (mg × h)/L)`

    ` ∴ K=0.27 \quad h^{-1}`. Ans.