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.

Answer 2

Step 1: Given data.
Initial concentration of the drug, \( [A_0] = 16 \, \text{mg/mL} \)
Concentration after 12 months, \( [A] = 4 \, \text{mg/mL} \)
Decomposition follows first-order kinetics.
The drug becomes ineffective after 50% decomposition, i.e., when \( [A] = \frac{[A_0]}{2} = 8 \, \text{mg/mL} \).

Step 2: Use the first-order rate equation.
For a first-order reaction:
\[ k = \frac{2.303}{t} \log \left(\frac{[A_0]}{[A]}\right) \] Substitute the given values to find \( k \):
\[ k = \frac{2.303}{12} \log \left(\frac{16}{4}\right) \] \[ k = \frac{2.303}{12} \log (4) \] \[ k = \frac{2.303}{12} \times 0.6021 = 0.1155 \, \text{month}^{-1} \]

Step 3: Find the time for 50% decomposition (expiry time).
When 50% decomposes, \( [A] = [A_0]/2 \).
Using the same formula:
\[ t = \frac{2.303}{k} \log \left(\frac{[A_0]}{[A]}\right) \] \[ t = \frac{2.303}{0.1155} \log(2) \] \[ t = \frac{2.303 \times 0.3010}{0.1155} = \frac{0.693}{0.1155} = 6.0 \, \text{months} \]

Step 4: Final Answer.
The expiry time of the drug is:
\[ \boxed{6 \, \text{months}} \]