1. Given data

• Initial concentration: $[X]_0 = 16\ \text{mg mL}^{-1}$
• Concentration after 12 months: $[X] = 4\ \text{mg mL}^{-1}$
• Decomposition follows first-order kinetics.
• Drug becomes ineffective after 50% decomposition.

That means expiry occurs when concentration becomes: $[X]_{\text{expiry}} = 50\%\text{ of }16 = 8\ \text{mg mL}^{-1}$

2. Use first-order integrated rate law

For a first-order reaction, $k = \frac{1}{t}\ln\left(\frac{[X]_0}{[X]}\right)$

Using the data for 12 months: $k = \frac{1}{12}\ln\left(\frac{16}{4}\right) = \frac{1}{12}\ln 4$

Since $\ln 4 = 2\ln 2$, $k = \frac{2\ln 2}{12} = \frac{\ln 2}{6}$

3. Find expiry time

At expiry, concentration falls from $16$ to $8\ \text{mg mL}^{-1}$.

Again using first-order law: $t_{\text{expiry}} = \frac{1}{k}\ln\left(\frac{16}{8}\right)$ $t_{\text{expiry}} = \frac{1}{k}\ln 2$

Now substitute $k = \frac{\ln 2}{6}$: $t_{\text{expiry}} = \frac{\ln 2}{\ln 2/6} = 6\ \text{months}$

4. Check options

• A: 12 ❌
• B: 3 ❌
• C: 6 ✅
• D: 2 ❌

Therefore, the expiry time is 6 months.