Original concentration, $$a = 16$$ mg/mL

Concentration at time

t = 12 months, $$a - x = 4$$ mg/mL

For first order kinetics, the equation for late constant

$$k = {{2.303} \over t}\log {a \over {a - x}}$$

a $\to$ Initial concentration

$a-x\to$ Concentration at time, t

$$ = {{2.303} \over {12\,months}}\log {{16\,mg/mL} \over {4\,mg/mL}}$$

$$ = {{2.303} \over {12}}\log 4$$

$$ = 0.1155\,month{s^{ - 1}}$$

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.

$${t_{1/2}}$$ formula is

$${t_{1/2}} = {{0.693} \over k}$$

Substitute k = 0.1155 months$^{-1}$,

$${t_{1/2}} = {{0.613} \over {0.1155}}$$ months$^{-1}$

= 6 months

Expiry time can also be calculate as,

$$t = {{2.303} \over k}\log {a \over {a - x}}$$

Initial is taken as 100%

The percent at expiry time is 50%

$$ = {{2.303} \over {0.1155\,month{s^ - }}} \times \log {{100\% } \over {50\% }}$$

$$ = {{2.303} \over {0.1155}} \times \log {{100} \over {50}}$$

$$ = {{2.303} \over {0.1155}} \times 0.301$$

$$ = 6.002$$

= 6 months

Correct answer is option (3) 6