Page 125 - Mechanism and Theory in Organic Chemistry

Let us assume that we have two isomeric substances A and B in equilibrium
at very low temperature, so that we can say that for practical purposes all mole-
cules are in their lowest energy states. Then we could regard A and B as two
energy states of a system, and since there are only two populated states, the
equilibrium constant K, the ratio of the number of B molecules to the number of A
molecules, would be given by Equation A1.2,
[ - (€0; ;
NB €01)
=
K = - exp I
NA
The quantity E,, is the energy of the lowest state of B, and E,, is the energy of the
lowest state of A. As it is more convenient to have energies on a molar basis, we
multiply the energies and the Boltzmann constant by Avagadro's number, No,
and since kNo is equal to the gas constant R, 1.986 cal OK mole-l, we obtain
Equation A1.3 :

K = exp
RT

E: is the standard state energy at O°K. Equation A1.3 may be written in the
form A1.4, and since at absolute zero AE: = AH: = AG,", this equation is

- AE: = RTln K (A1.4)

indeed the familiar thermodynamic expression for the equilibrium constant.

The Partition Function
The example we have used is of course unrealistic; we are interested in what goes : .
on at temperatures above absolute zero, where many energy levels of the mole-
'
cules are populated. All we need do to correct our equilibrium constant A1.3 is
to find out how many molecules of A and of B are in each energy level at the
temperature of interest. The total number of molecules of A, NA, is the sum of
numbers of molecules in each energy state,

So far we have assumed that each state has a different energy, but this will not
always be true. We have to allow for degeneracies, that is, groups of more than one
state at the same energy. When two states have the same energy, their populations
must be identical; we therefore modify Equation A1.5 to A1.6, where g,, is the

number of states that have energy E~A, and n,, is the number of molecules occupy-
ing a single state of energy €,,. Since the Boltzmann distribution deals with ratios
of numbers in various states, we divide both sides of Equation A1.6 by no, and
.
obtain Equation A1.7, which gives the ratio of the total number of molecules to