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?

  • 1
    $\begingroup$ Welcome to Chemistry Stack Exchange. The derived quantities H and G measure different things; have a look at their definitions. The difference is the energy $-T\Delta S$ , when this is large so is the difference between enthalpy and Gibbs energy. Nothing particular to nitrogen. $\endgroup$
    – porphyrin
    May 5, 2017 at 15:49
  • $\begingroup$ Thanks. However, saying that H and G measure different quantities doesn't explain why they are so extremely similar for most substances on the standard data tables. Thinking about it more myself, this is what I have: given that the system in question is at constant temperature, where does the difference in $\Delta S$ come from? Mostly, the work done by a chemical reaction is minimal, but some nitrates are used in explosives, which do work by expanding rapidly. This is what separates some nitrogen compounds from other substances and leads to differences in H and G on thermodynamic tables. $\endgroup$ May 7, 2017 at 2:04
  • $\begingroup$ I think I didn't say that right, but is the overall idea correct? $\endgroup$ May 7, 2017 at 2:18
  • $\begingroup$ It is not due to any reactivity a molecule may have but just the entropy. The absolute entropy $S^{\mathrm O}$ is normally obtained from calorimetry as the sum of the entropy contributions from $0$ K to $298.16$ K and this means integrating as $\int C_P/T$ between phases plus $\Delta H/T$ for any solid phase transitions as well as melting and vaporisation transitions as necessary. $\endgroup$
    – porphyrin
    May 7, 2017 at 9:29
  • $\begingroup$ I realize I'm being rather stubborn, but I'd like a conceptual explanation for why some substances have widely varying degrees of ΔH∘f and ΔG∘f. Since the difference reduces to −TΔS, what is it about some substances that makes the −TΔS special as compared to 95+% of other substances? Merely citing equations is one thing, but bereft of any conceptual explanation to illuminate these anomalies and give flesh to the mathematical skeleton, I still feel the need to understand why. If there truly is no conceptual explanation that exists, I guess it's my loss. Thanks for your patience. $\endgroup$ May 7, 2017 at 13:34

1 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.


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.


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