The steps are as follows:

1. Build the molecule

2. Optimize the geometry using PM3

3. Compute the frequencies for the optimized geometry to

a. verify that you found a minimum

b. compute the sum of the translational, rotational, and vibrational free energies at the PM3 level.

4. Compute the B3LYP/6-31G(d) energy for the PM3 optimized geometry

Some of the keywords I used have been described in a previous post, and I actually copy the keywords from this post to save typing.

The electronic energy for the water dimer is -152.7533 atomic units (au) while the sum of the translational, rotational, and vibrational free energies is 13.820 kcal/mol. If you repeat these calculations for the water molecule you get -76.3715 au and 2.271 kcal/mol (1 au = 627.51 kcal/mol).

So, the change in electronic energy on going from two water molecules to the water dimer is

ΔE = (-152.7533 - 2(-76.3715))*627.51 = -6.4 kcal/mol

The corresponding free energy change is

ΔG(298K) = -6.4 + 13.820 - 2(2.271) = -6.4 + 9.3 = 2.9 kcal/mol

Here I make the usual assumption that the electronic free energy is the electronic ground state energy.

Notice that the free energy change is positive meaning the water dimer is not predicted to form at this temperature and pressure (1 bar) under equilibrium conditions (the relative probability of observing the water dimer can be computed with this equation). This is because of the entropy of loss on going from 2 particles to 1.