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

Question

If you look at a Table of Thermodynamic Values for chemical substances, most substances have very close values for their standard enthalpy of formation ($\Delta H_{\mathrm{f}}^\circ$) and their standard Gibbs free energy of formation ($\Delta G_{\mathrm{f}}^\circ$). This makes intuitive sense because differences in enthalpy, which in chemical reactions is mostly just a signifier of internal energy because so little work is done by reactions, under normal circumstances correlates with the change in free energy of that system. When heat is lost through a reaction, that system has both less internal energy but also less free energy because the production of heat and then exporting of heat to the surroundings requires the 'use' of free energy.

My question: why do some substances have widely diverging values for $\Delta H_{\mathrm{f}}^\circ$ and ($\Delta G_{\mathrm{f}}^\circ$)? It appears that some substances containing nitrogen have the widest divergence for standard conditions of 25 °C and 1 bar, whereas most substances have a $\Delta H_{\mathrm{f}}^\circ$ and ($\Delta G_{\mathrm{f}}^\circ$) differing by no more than 30%. For example, $H_\mathrm{f}(\ce{AgNO3}) = −124.39$ kJ/mol and $G_\mathrm{f}(\ce{AgNO3}) = -33.41$ kJ/mol, and $H_\mathrm{f}(\ce{N2O4}) = 9.16$ kJ/mol and $G_\mathrm{f}(\ce{N2O4}) = 97.89$ kJ/mol.

A mathematical explanation would be helpful, but I'm most interested understanding this conceptually and why these nitrogen-containing compounds exhibit these properties. Are these properties more common among other substances at extremely high or low temperatures and pressures?

Answer 1

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.