Why does the standard enthalpy of formation diverge so far from the standard Gibbs free energy of formation for some substances?
Continuing from the comments above. 

The difference is due to the entropy appearing in $\Delta G$ as the energy $-T\Delta S$. Classical thermodynamics cannot tell us what the entropy is in molecular terms. This is discovered using statistical mechanics via Boltzmann's entropy equation $S=k\ln(\Omega)$ where $\Omega$ is the number of arrangements of particles into their energy levels, sometimes called 'complexions'. 

Different types of molecule have different natures, because they contain different number of atoms, which means that there are different numbers of molecular vibrations, or they contain atoms of different types, which means that vibrations are of different frequencies to one another. Hence molecules of different types have a unique set of energy levels (ignoring for simplicity degenerate vibrations).

As energy is added, the number of ways of filling up the energy levels is clearly different between different types of molecules because the stack of energy levels differs one to another. Single bonds between heavy atoms, for example, have lower frequencies, than those between lighter atoms, and so more levels are populated at the same temperature than in the heavier atom molecule. In this case the entropy is greater than that for a molecule with widely spaced levels. 

The same type of argument can be made for rotational and translational energy levels, and can also be made for phase changes, such as solid-solid, or for melting or vaporisation. Which of these are to be counted depends upon whether the substance is gas, liquid, or solid at the required temperature. 

Notes:

Technically the entropy is calculated as 
$$S=R\ln(Z)+RT\left ( \frac{\partial \ln(Z)}{\partial T}  \right)_V$$

where Z is the partition function and $Z=\Sigma _i g_i \exp(-\epsilon _i/(kT))$ where $\epsilon_i$ is the energy of level i with degeneracy $g_i$ and k is the Boltzmann constant.