Practice question
Question
A first-order reaction has a rate constant of 0.0346 min⁻¹. What percentage of the reactant decomposes in 40 minutes?
Explanation
log([R]₀/[R]) = kt/2.303 = (0.0346×40)/2.303 ≈ 0.602 → [R]₀/[R] ≈ 4. Fraction remaining = 0.25 → % decomposed = 75%.
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