Drug X becomes ineffective after 50 % decompo | vedclass

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(D) For a first-order reaction,the rate constant $k$ is given by $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.Given $[A]_0 = 16 \ mg/mL$,$[A]_t = 4

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$6$

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(D) For a first-order reaction,the rate constant $k$ is given by $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.Given $[A]_0 = 16 \ mg/mL$,$[A]_t = 4 \ mg/mL$,and $t = 12 \ months$.$k = \frac{2.303}{12} \log \frac{16}{4} = \frac{2.303}{12} \log 4 = \frac{2.303 	imes 0.602}{12} \approx 0.1155 \ month^{-1}$.The drug becomes ineffective after $50 \%$ decomposition,meaning $50 \%$ remains. Thus,$[A]_t = 0.5 	imes [A]_0$.The time taken for $50 \%$ decomposition is the half-life $t_{1/2} = \frac{0.693}{k}$.$t_{1/2} = \frac{0.693}{0.1155} = 6 \ months$.

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.

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