A first-order reaction has a rate constant of 0.0346 min⁻¹. What percentage of the reactant decomposes in 40 minutes?
log([R]₀/[R]) = kt/2.303 = (0.0346×40)/2.303 ≈ 0.602 → [R]₀/[R] ≈ 4. Fraction remaining = 0.25 → % decomposed = 75%.
Ref: NCERT Class 12 Chemistry > Chapter 3: Chemical Kinetics > Topic: Rate of Reaction - Average and Instantaneous Rate and Rate Law