Gibbs free energy is defined as the total non-expansion work (i.e. electrical work) that a system can perform. However, it also indicates whether or not a reaction is spontaneous. I am therefore confused on the interpretation of the Gibbs free energy when it comes to reactions such as the combustion of methane, which is evidently spontaneous - meaning that its $\Delta G$ is negative. By the above logic, the combustion of methane should therefore be capable of producing non-expansion work, but I can't see how this is the case (i.e. what types of non-expansion work can it do?).

Can someone explain, or if I'm misunderstanding something?

  • 2
    $\begingroup$ You can make an engine based on methane burning. $\endgroup$
    – Poutnik
    Aug 21, 2023 at 6:14
  • $\begingroup$ Thank you for your comment. I am assuming that you are talking about rocket boosters? If so, could you please clarify what type of work this would be? $\endgroup$ Aug 21, 2023 at 7:55
  • $\begingroup$ I mean a classical thermal engine, similar to one in petrol or LPG powered cars, converting (part of) chemical energy to mechanical work via the torsion of rotating crank. But rocket boosters would be possible too, if liquid methane would be used. Another way could be fuel cells. $\endgroup$
    – Poutnik
    Aug 21, 2023 at 8:06
  • $\begingroup$ The best non-expansion work that a system can perform is the electric work (or electric energy). This is what you wrote in the first line of the text, To get an electric work, the reaction must be carried out in a galvanic cell. If the reagents (methane and oxygen for example) are used in an electric cell, the reaction proceeds if the reaction is spontaneous (if Delta G is negative). If Delta G is negative, the reaction will produce a tension and a current, whatever the value of Delta H. In a spontaneous reaction, Delta G is negative, and Delta H can be either positive or negative. $\endgroup$
    – Maurice
    Aug 21, 2023 at 8:40
  • $\begingroup$ @Poutnik please correct me if I am wrong, but aren't heat/thermal engines devices converting heat energy into work? The heat given off by the combustion of methane itself is not work. Furthermore, dG correlates to the maximum non-expansion work that can be performed, while a heat engine, doesn't it produce mechanical/expansion work anyway? $\endgroup$ Aug 21, 2023 at 9:38

2 Answers 2


The (standard) reaction Gibbs energy change is the top limit of nonexpansion work, that can be provided by a closed system at isothermal and isobaric conditions, where 1 mol of the referred chemical reactions happens (at standard conditions).

It does not say that the limit work can be practically achieved, nor it say by what way such a work could be achieved.

Note that the expansion work is already involved in the enthalpy change $ΔH = ΔU + pΔV$.

Heat engines transform the Gibbs energy of fuel to Gibbs energy of heat reservoirs via creation of temperature difference, which is then used for nonexpansion work $\delta W = T \cdot \omega \cdot \mathrm{d}t$, where $T$ is torsion and $\omega$ is angular speed.

Similarly, human metabolism transforms part of the Gibbs energy of food and oxygen to nonexpansion work of body muscles.

The schema generally is not

$$\text{Reaction} \rightarrow \text{nonexpansion work}$$

but rather

$$\text{Reaction} \rightarrow \overset{ΔG \rightarrow W}{\text{"black box"}} \rightarrow \text{nonexpansion work}$$

  • $\begingroup$ So if a reaction happens and the number of moles changes, would that be PV work or non-PV work? P∆V = ∆nRT $\endgroup$ Aug 25, 2023 at 6:32
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    $\begingroup$ @Proscionexium See the 3rd paragraph. $\endgroup$
    – Poutnik
    Aug 25, 2023 at 7:02

I would say you have your definitions backwards, in a sense. The Gibbs free energy is the Legendre transform of the enthalpy. When you do this and derive the differential form of $G$ from the fundamental thermodynamic relation, you find that $G$ is a function of $P$, $T$, and the number of moles of each type of species in your system. The special thing about $G$ is that, because of the second law, you will always have $dG<0$ for a spontaneous process $\textit{under constant temperature and pressure}$. That conditional is critical. Without it, there is nothing special about $G$ at all, and it tells you nothing about spontaneity. It just so happens that chemists often consider processes where this happens to hold, and so $G$ is a very useful function. This is sort of like how $H$ is quite useful under conditions of constant pressure, as it gives you the amount of heat transferred in a process. But in general the enthalpy is just a Legendre transform of the internal energy: no more, no less.

Through a different set of arguments, you get that $G$ allows you to determine the maximum amount of non $PV$ work obtainable from your system. It is not obvious at all how you would actually obtain this work, but you know that it is theoretically possible to get it, as suggested in another answer. Unfortunately, thermodynamics has a nasty habit of telling us interesting things about nature without giving us a guidebook on how to actually capitalize on them.

  • 1
    $\begingroup$ Chemists are not usually aware of the Legendre transform (its TD application). I have met the orthogonal Legendre polynomials few times, but not the transform. $\endgroup$
    – Poutnik
    Aug 25, 2023 at 7:17
  • $\begingroup$ That’s fair. I’m actually teaching physical chemistry this fall, and my students will be very familiar with them! $\endgroup$ Aug 25, 2023 at 11:31

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