T15: K from Dose, AUC and V

50. Pharmacokinetics 50.1) PK Basic parameters 50.1.3) Area Under Curve (AUC) 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 145.34 L after administering 280.51 mg of the drug. The area under the plasma drug concentration over time (AUC) was found to be 12.45 mg × h/L.

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lb equals 0.16 h-1 kg

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

    `⇒ K=(dose)/(V × AUC)`

    Plugging the values in this equation we get:
    ` K = (280.51 \quad mg)/(145.34 \quad L × 12.45 (mg × h)/L)`

    ` ∴ K=0.155 \quad h^{-1}`. Ans.

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