Practice question
Question
A first-order reaction has a rate constant of 0.0115 min⁻¹. What percentage of the reactant remains after 60 minutes?
Explanation
log([R]₀/[R]) = kt/2.303 = (0.0115×60)/2.303 ≈ 0.301 → [R]₀/[R] ≈ 2 → % remaining = 50%.
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