A first-order reaction has a specific rate constant of exactly . How much absolute time will it take — Chemical Kinetics Chemistry Question
Question
A first-order reaction has a specific rate constant of exactly $10^{-2} \text{ s}^{-1}$. How much absolute time will it take for 20 g of the reacting substance to rigorously reduce to 5 g? (Assume $\ln 2 \approx 0.693$)
Answer: B
💡 Solution & Explanation
Reducing the mass from 20 g to 5 g represents exactly two half-lives ($20 \rightarrow 10 \rightarrow 5$). For a first-order reaction, $t_{1/2} = 0.693 / k = 0.693 / 10^{-2} = 69.3 \text{ s}$. The total time is $2 \times 69.3 = 138.6 \text{ s}$.
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