Boltzmann Distribution Law

B.T. Luke & Associates, Inc


It is well known from the kinetic theory of gasses that the average kinetic energy of a noninteracting (ideal) gas molecule is given by ½kT for each degree of translational freedom. In this equation, k is the Boltzmann constant and T is the absolute temperature in Kelvin. Boltzmann used this to derive a relationship which states that the natural logrithm of the ratio of the number of particles in two different energy states is proportional to the negative of their energy separation.

The diagram on the right shows two energy levels that are populated by different numbers of molecules. The lower energy state contains N0 molecules at an energy of , and the upper state contains N1 molecules at an energy of . The Boltzmann distribution law is then given by the expression

This expression states that the natural logrithm of the ratio of the number of particles in the upper state to the number in the lower state is equal to the negative of their energy separation divided by kT.

If there are a total of N molecules in the system that can reside in many different energy states, the above expression can be rearranged and used to determine the fraction of the molecules in the jth state (Nj/N). This fraction, which is the more common form of the Boltzmann distribution law, is given by the equation

The sum in the denominator is over all energy states and is known as the molecular partition function.

By looking at the NIST Physical Reference Data, one can find that the current value for the Boltzmann constant, k, is 1.38065×10-23 J K-1. By using Avagadros number and realizing that 1 calorie is exactly 4.184J, the Boltzmann constant is also 1.9872 cal mol-1 K-1, which is just the gas constant, R.

Assume that there is a molecule with three low-lying conformations. The lowest, 0th, state has an energy of -18.0 kcal/mol, while the 1st and 2nd have energies of -17.0 and -15.5 kcal/mol. respectively. We can now determine the fraction of molecules in each state at a given temperature.

Before we run off and do this, we should first realize that we are only interested in energy differences, not the absolute value of the actual energies. Therefore, we can first shift the energies so that these three states have relative energies of 0.0, 1.0 and 2.5 kcal/mol, respectively. In addition, we can assume that we have a total of 1 mol of these molecules so that Ni is the fraction of molecules (moles) that are in the ith state.

The following table shows the calculation of Ni at 300K.

0 0.0 1.0 .8320
1 1.0 0.18686 .1555
2 2.5 0.01509 .0125
Sum   1.20195  

This table shows that at 300K, 83.20% of the molecules are in the 0th state, 15.55% are in the 1st state, and 1.25% are in the 2nd.

If the temperature is increased to 1000K, the following results are obtained.

0 0.0 1.0 .5294
1 1.0 0.60458 .3201
2 2.5 0.28421 .1505
Sum   1.88879  

As expected, as the temperature increases, the fraction of molecules in the higher-energy states increases. Increasing the temperature from 300K to 1000K causes the fraction in the 2nd state to increase from 1.25% to 15.05%.

Calculating the fraction of molecules in each state at a given temperature can also be used to calculate the Boltzmann average of physical properties at a given temperature. This is done by using the following formula

where <A> is the average property you wish to calculate. This equation simply states that the average of a property is just the weighted average taken over all of the states that the molecule can occupy.

To continue with our example, assume that we want to calculate the average of the dipole moment, <D>. If the dipole moment of the molecule in the 0th state, D0, is 1.872 debye, and D1 and D2 are 2.031 and 2.218 for the 1st and 2nd states, respectively, the terms in the above equation are given in the following table for the system at 1000K.

0 1.872 0.0 1.0 1.872
1 2.031 1.0 0.60458 1.228
2 2.218 2.5 0.28421 0.630
Sum     1.88879 3.730

Therefore, the average dipole moment at 1000K is

<D> = 3.730/1.88879 = 1.975 debye