Question:medium

The unit of zero order elimination rate constant is:

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Keep reactions order units straight: - Zero-order ($K_0$): $\text{Concentration/Time}$ (e.g., $\text{mg/mL}\cdot\text{hr}$) - First-order ($K_1$): $\text{1/Time}$ (e.g., $\text{hr}^{-1}$) - Second-order ($K_2$): $\text{1}/(\text{Concentration}\cdot\text{Time})$
Updated On: Jul 4, 2026
  • \(\text{mg/ml*hour}\)
  • \(\text{1/hour}\)
  • \(\text{mg/hour}\)
  • \(\text{1/ml*hour}\)
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The Correct Option is A

Solution and Explanation

Understanding the Concept: The rate of a chemical or pharmacokinetic reaction is mathematically represented by the change in drug concentration ($C$) per unit time ($t$): \[ \text{Rate} = -\frac{dC}{dt} = K_0 \cdot C^n \] Where $K_0$ is the rate constant and $n$ defines the order of the reaction. For a zero-order process, the rate of elimination is completely independent of the concentration of the drug remaining because the operating systems (such as metabolic enzymes or transport carriers) are fully saturated ($n = 0$).

Step 1:
Set up the zero-order differential expression.
Substituting $n = 0$ into the rate equation: \[ -\frac{dC}{dt} = K_0 \cdot C^0 \implies -\frac{dC}{dt} = K_0 \]

Step 2:
Derive the units of the constant.
From the simplified relation, the units of the zero-order rate constant $K_0$ must match the units of the change in concentration over time: \[ \text{Units of } K_0 = \frac{\text{Units of Concentration}}{\text{Units of Time}} \] Concentration is expressed as mass per unit volume (e.g., $\text{mg/ml}$), and time is expressed in hours ($\text{hour}$). Substituting these units gives: \[ \text{Units of } K_0 = \frac{\text{mg/ml}}{\text{hour}} = \text{mg} \cdot \text{ml}^{-1} \cdot \text{hour}^{-1} = \text{mg/ml*hour} \]
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