A zero-order decay reaction operates seamlessly with a strict half-life given an initial starting co — Chemical Kinetics Chemistry Question
Question
A zero-order decay reaction $A \rightarrow 2B$ operates seamlessly with a strict half-life $t_{1/2} = 6 \text{ hours}$ given an initial starting concentration of precisely $0.2 \text{ M}$. Exactly how many elapsed hours will it require for the total concentration of A to steadily drop from $0.5 \text{ M}$ strictly down to $0.2 \text{ M}$?
Answer: C
💡 Solution & Explanation
For a zero-order mechanism, $t_{1/2} = [A]_0 / 2k$. Plugging in data: $6 = 0.2 / 2k \Rightarrow k = 0.2 / 12 = 1/60 \text{ M h}^{-1}$. To drop precisely from $0.5 \text{ M}$ to $0.2 \text{ M}$, the total consumed mass $x = 0.3 \text{ M}$. Because $x = kt$, the total elapsed time $t = x / k = 0.3 / (1/60) = 18 \text{ hours}$.
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