Question

Drug X becomes ineffective after 50% decomposition. The original concentration of drug in a bottle was 16 mg/mL which becomes 4 mg/mL in 12 months. The expiry time of the drug in months is ____ . Assume that the decomposition of the drug follows first order kinetics.

• A. 12
• B. 2
• C. 3
• D. 6

Hint:

For first-order reactions, the time for 50% decomposition can be calculated using the equation: \( \ln \left( \frac{[A]_0}{[A]_t} \right) = kt \). The rate constant can be used to calculate the time for any given concentration change.

Answer 1

The Correct Option is D

For a first-order reaction, the equation for the concentration at time \( t \) is given by: \[ \ln \left( \frac{[A]_0}{[A]_t} \right) = kt \] where \( [A]_0 \) is the initial concentration, \( [A]_t \) is the concentration at time \( t \), \( k \) is the rate constant, and \( t \) is the time.
Given that the drug becomes ineffective after 50% decomposition, we know that \( [A]_0 = 16 \, \text{mg/mL} \) and \( [A]_t = 4 \, \text{mg/mL} \). Substitute the values into the equation: \[ \ln \left( \frac{16}{4} \right) = k(12) \] \[ \ln(4) = 12k \quad \Rightarrow \quad 1.386 = 12k \quad \Rightarrow \quad k = 0.1155 \, \text{month}^{-1} \] Now, to find the expiry time, we use the same equation for 50% decomposition (i.e., \( [A]_t = \frac{[A]_0}{2} \)): \[ \ln \left( \frac{[A]_0}{\frac{[A]_0}{2}} \right) = k \cdot t_{\text{expiry}} \] \[ \ln(2) = 0.1155 \cdot t_{\text{expiry}} \quad \Rightarrow \quad 0.693 = 0.1155 \cdot t_{\text{expiry}} \quad \Rightarrow \quad t_{\text{expiry}} = 6 \, \text{months} \] Thus, the expiry time of the drug is 6 months.