Drug $X$ becomes ineffective after $50 \%$ decomposition. The original concentration of drug in a bottle was $16 \mathrm{mg} / \mathrm{mL}$ which becomes $4 \mathrm{mg} / \mathrm{mL}$ in 12 months. The expiry time of the drug in months is _________.

Assume that the decomposition of the drug follows first order kinetics.

<p>Original concentration, $a = 16$ mg/mL</p> <p>Concentration at time</p> <p>t = 12 months, $a - x = 4$ mg/mL</p> <p>For first order kinetics, the equation for late constant</p> <p>$k = {{2.303} \over t}\log {a \over {a - x}}$</p> <p>a $\to$ Initial concentration</p> <p>$a-x\to$ Concentration at time, t</p> <p>$= {{2.303} \over {12\,months}}\log {{16\,mg/mL} \over {4\,mg/mL}}$</p> <p>$= {{2.303} \over {12}}\log 4$</p> <p>$= 0.1155\,month{s^{ - 1}}$</p> <p>Drug is ineffective after 50% decomposition. So, the expiry time is after 50% decomposition. The time at which 50% decomposition occurs in a reaction is known as half life $({t_{1/2}})$. So, ${t_{1/2}}$ is the expiry time of the drug.</p> <p>${t_{1/2}}$ formula is</p> <p>${t_{1/2}} = {{0.693} \over k}$</p> <p>Substitute k = 0.1155 months$^{-1}$,</p> <p>${t_{1/2}} = {{0.613} \over {0.1155}}$ months$^{-1}$</p> <p>= 6 months</p> <p>Expiry time can also be calculate as,</p> <p>$t = {{2.303} \over k}\log {a \over {a - x}}$</p> <p>Initial is taken as 100%</p> <p>The percent at expiry time is 50%</p> <p>$= {{2.303} \over {0.1155\,month{s^ - }}} \times \log {{100\% } \over {50\% }}$</p> <p>$= {{2.303} \over {0.1155}} \times \log {{100} \over {50}}$</p> <p>$= {{2.303} \over {0.1155}} \times 0.301$</p> <p>$= 6.002$</p> <p>= 6 months</p> <p>Correct answer is option (3) 6</p>