Why is the Gibbs free energy (G) considered a spontaneity criterion for phase transformations and chemical reactions? Why are other thermodynamic parameters such as enthalpy (H), entropy (S), and Helmholtz free energy (A) not accepted as spontaneity criteria?

  • 1
    $\begingroup$ That is just convention. It would be confusing to have multiple, sometimes conflicting criteria for a single term. $\endgroup$
    – Karsten
    Commented Aug 2, 2022 at 15:28
  • $\begingroup$ All S, G and A can be used for spontaneity evaluation, depending on system conditions (isolated, closed isobaric, closed isochoric) $\endgroup$
    – Poutnik
    Commented Aug 2, 2022 at 20:13
  • $\begingroup$ 2nd Law of thermodynamics - The Gibbs free energy change of the system is equal to the opposite of the temp times the change in entropy of the universe. Since the universe will change in the way that increases its entropy (second law), that term is positive, as is the temperature (in K), so the Gibbs free energy change must be negative in order to get the opposite. $\endgroup$
    – Andrew
    Commented Aug 3, 2022 at 23:40

2 Answers 2


Spontaneity is simply a system not at equilibrium (this is regardless of direction) with a possible mechanism to reach equilibrium.

Before getting upset note that reactions involving only standard conditions do not reach equilibrium in either direction, so one direction will be spontaneous and the other not. This is incorrect both directions are from the standard states to equilibrium [JG].

For a multistep reaction each elementary step should be considered in turn. At equilibrium forward and reverse reactions are at the same rate and the chemical activities satisfy the equilibrium constant. The equilibrium constant value determines the extent of reaction in each direction.

The energetics of a reaction are determined by the energy change and the entropy change, not by only one of the two. The functions that combine these properly are the Gibbs free energy

$$G = H - TS, \tag{1}$$

especially useful at constant pressure; and the Helmholtz free energy

$$A = E - TS, \tag{2}$$

especially useful at constant volume.

There are cases where either $H$ or $S$ is much larger and seems to dominate the process. But if one thinks it through, that is true when the conditions are far from equilibrium. At equilibrium

$$\Delta G = 0 = \Delta H - T\,\Delta S\quad\implies\quad \Delta H = T\,\Delta S. \tag{3}$$

This means at equilibrium an infinitesimal change in either will perturb the equilibrium in the appropriate direction with an appropriate change in the other to satisfy the equilibrium condition. If energy is added, $\Delta H > 0$; bonds break increasing particles, $\Delta S > 0$; $\Delta G = 0.$

  • $\begingroup$ Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. ALL CAPS shouldn't be used for emphasis on a regular basis (use italics for normal emphasis and bold text for strong emphasis). Square brackets and braces have different meaning and are not universally interchangeable in prose or math. Structuring the text with paragraphs improves readability, too. $\endgroup$
    – andselisk
    Commented Aug 2, 2022 at 19:48
  • $\begingroup$ Why was this answer downvoted? If for formatting give me a break! If for erroneous content please enlighten me. $\endgroup$
    – jimchmst
    Commented Aug 3, 2022 at 20:10
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    $\begingroup$ Most users have a lot to process in real life, and liberating them from putting extra unnecessary work to decipher posts written using own standards is the least an author can do. On SE sites we strive towards highest possible quality and straightforward Q&A experience. If you don't want to improve your communication after you were explicitly guided several times, then you should no longer wonder why the content is downvoted. Sorry, but at the end of working day interpreting a non-formatted wall of text isn't something many are inclined to do, so I won't even tell if this is correct or not. $\endgroup$
    – andselisk
    Commented Aug 3, 2022 at 20:32
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    $\begingroup$ I will correct one error in my post above. Reactions written in the standard states do attain equilibrium so the forward reaction has the delta G for that direction and the reverse the negative delta G. In both cases the energy is the from the standard states to equilibrium. $\endgroup$
    – jimchmst
    Commented Aug 3, 2022 at 21:40
  • $\begingroup$ Thank you in advance. Please feel free to ask if something is unclear, I think the links should give a decent start. Markdown and MathJax are simple, but could be cumbersome. Yet they are ubiquitous markup tools, so I think learning them is worth investing some time. Sorry if I sounded harsh, I could be too straightforward at times. $\endgroup$
    – andselisk
    Commented Aug 3, 2022 at 21:48

Enthalpy $H$ is not a valid criterion for spontaneous reactions. Spontaneous reactions may be exothermic or endothermic. The famous reaction of $\ce{NaHCO3 + HCl}$ is endothermic, but it is spontaneous.

It is the same for entropy. Some reactions produce entropy. Dimerization reactions like

$$\ce{A + A -> A2}$$

lead to a decrease of entropy. So the only valid criterion for spontaneity is the Gibbs free energy. In all cases, a spontaneous reaction occurs if $\Delta G < 0.$


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