Question

Drugs ‘A’, ‘B’ and ‘C’ follow zero order, first order and second order degradation kinetics, respectively, but have the same rate constant. Which of the following statement is true in this respect:

• A. Drug ‘A’ will be the first to degrade by 50%
• B. Drug ‘B’ will be the first to degrade by 50%
• C. Drug ‘C’ will be the first to degrade by 50%
• D. Drugs ‘B’ and ‘C’ will degrade by 50% at the same time

Answer 1

The Correct Option is C

To determine which drug will be the first to degrade by 50%, we need to understand the degradation kinetics of zero-order, first-order, and second-order reactions and how these relate to the time required for a substance to reduce by 50% (half-life). 

• Zero-order kinetics (Drug A): The rate of degradation is constant and does not depend on the concentration of the drug. The formula for zero-order reactions is: \([\text{A}] = [\text{A}_0] - kt\), where \([\text{A}]\) is the concentration at time \(t\), \([\text{A}_0]\) is the initial concentration, \(k\) is the rate constant. The time to degrade by 50% is: \(t_{50\%} = \frac{[\text{A}_0]}{2k}\).
• First-order kinetics (Drug B): The rate of degradation is directly proportional to the concentration of the drug. The formula is: \([\text{A}] = [\text{A}_0]e^{-kt}\). The time to degrade by 50% (half-life) is independent of the initial concentration: \(t_{50\%} = \frac{\ln(2)}{k}\).
• Second-order kinetics (Drug C): The rate depends on the square of the concentration of the drug. The formula is: \(\frac{1}{[\text{A}]} = \frac{1}{[\text{A}_0]} + kt\). The time to degrade by 50% is: \(t_{50\%} = \frac{1}{k[\text{A}_0]}\).

Given that all drugs have the same rate constant \((k)\), we can compare the times for the concentration of each drug to reduce by 50%:

1. For Drug A (zero-order), the time depends on the initial concentration.
2. For Drug B (first-order), the half-time is a constant value \(t_{50\%} = \frac{\ln(2)}{k}\).
3. For Drug C (second-order), the time to degrade by 50% is inversely proportional to the initial concentration. If the concentration is high, this can be very quick.

From the formulas, \(t_{50\%}\)for second-order (Drug C) is smallest compared to the zero-order and first-order when given the same initial concentration and rate constant.

Thus, the correct statement is: Drug ‘C’ will be the first to degrade by 50%.