Department of Energy Argonne National Laboratory Office of Science NEWTON's Homepage NEWTON's Homepage
NEWTON, Ask A Scientist!
NEWTON Home Page NEWTON Teachers Visit Our Archives Ask A Question How To Ask A Question Question of the Week Our Expert Scientists Volunteer at NEWTON! Frequently Asked Questions Referencing NEWTON About NEWTON About Ask A Scientist Education At Argonne Activation Energy
Name: shifaan
Status: student
Age: 18
Location: N/A
Country: N/A
Date: 2000

I'm really confused about a concept. In chemical kinetics according to the maxwell-boltzmann distribution of energy, the activation energy for a reaction remain the same irrespective of the temperature. But on the other hand, according to the Arrhenius equation the activation energy of a reaction is dependant on temperature. Why do both of them say the totally opposite when talking of the same thing>?

You are referring to approximations that focus on different aspects of the process. The Maxwell-Boltzmann distribution of energy does not require that activation energies be temperature-independent. It just tells you the occupancies of states at thermal equilibrium given the temperature and the energies of the states. (Even what I just wrote is an approximation - I'm neglecting degeneracies.) The discussion you encountered no doubt explained how the temperature accelerates the rate of a reaction by making the activation energy more accessible. If you allow the activation energy itself to change with temperature, the Maxwell-Boltzmann distribution of states still holds. The rate, however, becomes more complicated to calculate, because you need to also take the new activation energy into account.

Actually, the Arrhenius equation, k = A exp(-Ea/RT), does not require that the activation energy Ea changes with temperature. The rate certainly is temperature-dependent, but not tyhe activation energy. In fact, rates of a reaction at different temperatures are sometimes used, in concert with this equation, to determine the temperature-independent Ea. All you do is plot ln k against 1/T, giving a (hopefully) straight line with a slope of Ea/R.

Now, not all chemical reactions will give a straight line to such a plot. This means that the Ea changes with temperature. One way for this to occur is if Ea is actually a Gibbs free energy of activation, which can be broken down into an enthalpy and entropy of activation, G = H - TS. Then, even if H and S are temperature dependent (which they may not be), G (= Ea) will depend upon the temperature. In any event, as long as you know the Ea at a given temperature, you can calculate the rate (given some reasonable assumptions).

Richard E. Barrans Jr., Ph.D.
Assistant Director
PG Research Foundation, Darien, Illinois

Click here to return to the Chemistry Archives

NEWTON is an electronic community for Science, Math, and Computer Science K-12 Educators, sponsored and operated by Argonne National Laboratory's Educational Programs, Andrew Skipor, Ph.D., Head of Educational Programs.

For assistance with NEWTON contact a System Operator (, or at Argonne's Educational Programs

Educational Programs
Building 360
9700 S. Cass Ave.
Argonne, Illinois
60439-4845, USA
Update: June 2012
Weclome To Newton

Argonne National Laboratory