Practice question
Question
A second-order reaction has a rate constant of 0.07 L mol⁻¹ s⁻¹ and an initial concentration of 0.25 mol L⁻¹. What is the time for 80% completion?
Explanation
t = (1/0.05 − 1/0.25)/0.07 ≈ 228.57 s.
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