For a certain reaction, the rate constant doubles when the temperature increases from 298 K to 308 K. What is the activation energy?
Q1. For a certain reaction, the rate constant doubles when the temperature increases from 298 K to 308 K. What is the activation energy?
Answer: 53.6 kJ/mol
Explanation: Using Arrhenius equation, Ea = RT1T2 ln(k2/k1)/(T2-T1) = 53.6 kJ/mol.