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
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
Richard E. Barrans Jr., Ph.D.
PG Research Foundation, Darien, Illinois
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Update: June 2012