A first-order reaction has a rate constant of . What is the time required to reduce of this reactant — Chemical Kinetics Chemistry Question
Question
A first-order reaction has a rate constant of $2.303 \times 10^{-3} \text{ s}^{-1}$. What is the time required to reduce $40 \text{ g}$ of this reactant to $10 \text{ g}$? ($\log 2 = 0.3$)
Answer: C
💡 Solution & Explanation
Reducing from 40g to 10g means the reactant undergoes two half-lives (40 to 20, 20 to 10). The master formula gives $t = \frac{2.303}{k} \log \frac{40}{10} = \frac{2.303}{2.303 \times 10^{-3}} \log 4 = 1000 \times 2(0.3) = 600 \text{ s}$.
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