This has confused me for a long time now. Why do we compute Gibbs free energy when we figure out the spontaneity of a reaction involving gases? In my mind, I envision a reaction involving gases to be always in a closed container, which would clearly change the pressure of the gas, but it seems to be that constant pressure is a requirement for using Gibbs free energy.

Now if it is that we are supposed to assume that the reaction involving gases does not occur in a closed container, why do we use molar concentration of gases when computing equilibrium constants? In this situation, it seems to be that volume of gases will change.

I'm suspecting that most of my confusion is on what constant pressure actually means in a reaction, so as a third question, what does constant pressure mean when we are referring to a reaction?

  • $\begingroup$ Hello, could you describe a little more specifically which problem you think would arise in this situation, or which kind of calculation you think that is not feasible to do with Gibbs energy? Just to be sure to adress your question with my answer. $\endgroup$ Mar 6, 2016 at 11:44
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    $\begingroup$ Who says that the pressure of the gas has change in a closed container? Certainly, if there is a piston, the volume of the gas can be caused to increase such that the pressure is held constant. $\endgroup$ Mar 6, 2016 at 12:55
  • $\begingroup$ Gibbs is for constant pressure, Helmholtz is for constant volume. $\endgroup$
    – A.K.
    Mar 6, 2016 at 13:47
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    $\begingroup$ The change in Gibbs Free Energy can be readily determined if pressure is not constant, and the change in Helmholtz Free Energy can be readily determined if volume is not constant. $\endgroup$ Mar 6, 2016 at 14:26

3 Answers 3


The Gibbs Free Energy has been found to be the most convenient thermodynamic function to work with when determining the equilibrium constant for a reaction or in establishing the chemical equilibrium state for a chemically reacting system involving multiple reactions. The condition for equilibrium of these is that, at a given temperature and pressure, the Gibbs Free Energy is minimized with respect to changes in extent of reaction. This applies to whatever temperature and pressure are currently present in the reactor, irrespective of whether they have been changing. For a mixture of ideal gases, it is possible to show that this Free Energy minimization is equivalent to requiring that the sum of the free energies of the pure reactants at pure gas pressures identical to their partial pressures in the reactor are equal to the sum of the free energies of the pure products at pure gas pressures identical to their partial pressures in the reactor.

  • $\begingroup$ Sir can you explain why Gibbs Free Energy is minimized w.r.t changes in extent of the reaction even if pressure and temperature is changing? Everywhere I've seen they derive dG<0 for reactions only at constant pressure and temperature. $\endgroup$
    – Shivansh J
    Apr 19, 2020 at 16:26

Imagine conducting a reaction in a piston that is filled with the gases. (As an example, think of a syringe with the "needle" port closed off; on the other end of the syringe there is a plunger that is gas tight but can move up and down.) The atmosphere presses against the other side of the piston/plunger.

If the reaction causes the pressure to rise, even a tiny bit, the piston (e.g. plunger) will move up, expanding the volume of the gas inside the reaction, which reduces the pressure back to atmosphdoing work against the atmosphere. If the reaction causes the pressure to fall, even infintessimally, the piston (i.e. plunger) will fall, as the atmosphere presses the piston down to maintain the pressure in the chamber to be one atmosphere.

The key is that reactions at constant pressure do not necessarily correspond to reactions at constant volume. The container the gaseous reactants are held in needs to be able to expand or contract in order to maintain constant pressure, which of course means it changes in volume. Such containers can still be closed in the sense that no atoms pass from the inside to the outside.


I think it makes no sense if you computes Gibbs free energy in a closed container.As @A.K. and @Chester Miller said,Gibbs is for constant pressure, the change in Gibbs Free Energy can be readily determined if pressure is not constant.We can computes Helmholtz free energy

The Helmholtz energy is defined as: A=U—TS where

  • A is the Helmholtz free energy
  • U is the internal energy of the system
  • T is the absolute temperature of the surroundings, modelled as a heat bath
  • S is the entropy of the system

The Helmholtz energy is the Legendre transform of the internal energy, U, in which temperature replaces entropy as the independent variable.

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    $\begingroup$ You've copied and pasted a section directly from Wikipedia here. You might want to elaborate on your answer to the OP's question. $\endgroup$ Mar 6, 2016 at 15:36

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