Q. What is the activation energy for a reaction if its rate doubles when the temperature is raised from 20°C to 35°C? (R = 8.314 J mol⁻¹ K⁻¹)
A) 342 kJ mol⁻¹
B) 269 kJ mol⁻¹
C) 34.7 kJ mol⁻¹
D) 15.1 kJ mol⁻¹
Explanation
[Solution from source] log(k₂/k₁) = (Ea/2.303R)(T₂ − T₁)/(T₁T₂) with T₁ = 293 K, T₂ = 308 K. Ea = 2.303 × 8.314 × 0.3010 × (293 × 308)/15 = 5.763 × 6016 ≈ 3.47 × 10⁴ J mol⁻¹ = 34.7 kJ mol⁻¹. Answer (3).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.