A first-order reaction has a rate constant of . What is the time required to reduce of this reactant — Chemical Kinetics Chemistry Question

💡 Solution & Explanation

Reducing from 40g to 10g means the reactant undergoes two half-lives (40 to 20, 20 to 10). The master formula gives $t = \frac{2.303}{k} \log \frac{40}{10} = \frac{2.303}{2.303 \times 10^{-3}} \log 4 = 1000 \times 2(0.3) = 600 \text{ s}$.