Practice question
Question
A reaction’s rate constant increases from 2.0 × 10⁻⁴ min⁻¹ at 290 K to 6.0 × 10⁻⁴ min⁻¹ at 300 K. What is the activation energy (R = 8.314 J mol⁻¹ K⁻¹)?
Explanation
log 3 = Ea/(2.303R) × (10)/(290×300) → Ea ≈ 79.4 kJ mol⁻¹.