# The rate of the chemical reaction doubles for an increase of 10K in absolute temperature from 298K. Calculate Ea.

Given-

Initial temperature, T1 = 298 K

Final temperature, T2 = 298 K + 10 K = 308 K

Knowing that the rate constant of a chemical reaction normally increases with increase in temperature, we assume that,

Initial value of rate constant, k1 = k

Final value of rate constant, k2 = 2k

Using Arrhenius equation, Equation 1

where, R = 8.314 J K-1 mol-1 (gas constant).

Substituting all the values in equation 1, we get, log 2 = Ea = Ea = Ea = 52897 J mol-1

Ea = 52.897 kJ mol-1

The energy of activation, Ea is 52.897 kJ mol-1

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos  Interactive Quiz on Integrated Rate equation58 mins  Interactive quiz on arrhenius equation53 mins  Arrhenius equation & Collision Theory59 mins  Integrated Rate Equation52 mins  Questions based on Chemical Reaction57 mins  Factors influencing Rate of Chemical Reaction45 mins  Questions Based on Factors influencing Rate of Chemical Reaction41 mins  Learn Rate of a Chemical Reaction65 mins  15 Concept Builder Questions of Chemical Bonding - Reminding 11th46 mins  Full Chapter Quiz | Solid State | Check your learning43 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation view all courses 