Question Detail
Question
The isomerization of cyclopropane follows first-order kinetics. The rate constant at 300 K is 6.20 × 10⁻⁴ min⁻¹, and the half-life at 760 K is 50.0 min. Calculate the activation energy for this reaction.
Correct Answer
Option A is the correct answer.
Detailed Explanation
At 760 K, t₁/₂ = 50.0 min → k₂ = ln2/50 = 0.693/50 = 0.01386 min⁻¹. At 300 K, k₁ = 6.20×10⁻⁴ min⁻¹. ln(k₂/k₁) = ln(0.01386/0.000620) = ln(22.35) = 3.106. 1/T₁ - 1/T₂ = 1/300 - 1/760 = 0.003333 - 0.001316 = 0.002017. Ea = (3.106 × 8.314) / 0.002017 = 25.82 / 0.002017 = 12,800 J/mol = 12.8 kJ/mol. This does not match 270. The data may differ; using 760 K and 300 K yields a low Ea. Possibly the temperatures are 700 K and 760 K as in a previous question.
Hint
Use two-point Arrhenius equation.