Question Detail
Question
The isomerization of cyclopropane follows first order kinetics. The rate constant at 700 K is 6.20 x 10⁻⁴ min⁻¹, and the half-life at 760 K is 29.0 min. Calculate the activation energy for this reaction.
Correct Answer
Option C is the correct answer.
Detailed Explanation
At 760 K, t₁/₂ = 29.0 min → k₂ = ln2 / 29.0 = 0.693/29.0 = 0.0239 min⁻¹. At 700 K, k₁ = 6.20×10⁻⁴ min⁻¹. Use ln(k₂/k₁) = (Ea/R)(1/T₁ - 1/T₂). ln(0.0239/0.000620) = ln(38.55) = 3.652. 1/T₁ - 1/T₂ = 1/700 - 1/760 = 0.0014286 - 0.0013158 = 0.0001128. Ea = (3.652 × 8.314) / 0.0001128 = 30.36 / 0.0001128 = 269,000 J/mol = 269 kJ/mol ≈ 270 kJ/mol.
Hint
Use two-point Arrhenius equation.