Suppose we have a sealed container (fixed volume) and we introduce two gases $\ce{A}$ and $\ce{B}$. The two reactants

$$\ce{A + B <=> C}$$

form another gas $\ce{C}$ with equilibrium constant $K$. We know that a closed system at constant $(T,P)$ a system reaches equilibrium where Gibbs free energy $G$ takes its minimum value. Can a reaction reach equilibrium in a closed container (fixed volume)?

Although the system can be coupled to a heat bath (providing the constant temperature condition), by fixing the volume, the necessary condition of constant pressure for $G$ minimization is lost. Can the equilibrium state now be predicted?

I mean if the piston was movable (e.g. coupled to an external pressure of $1$ atm) the system would reach equilibrium and we could find the equilibrium state (e.g. the gas composition) by using the equilibrium constant $K$.

I am asking this question because in many sites I have seen the following Le Chatelier's Principle:

When there is a decrease in volume, the equilibrium will shift to favor the direction that produces fewer moles of gas.

I can't understand why we are free to make such a statement if the pressure is not constant. I am not asking about Le Chatelier's Principle. I just gave that example to justify the motivation behind the question.

Another example is when we want to predict the vapor pressure of a liquid (again sealed container). We put some liquid in the closed container and after a while equilibrium is established where vapor pressure is related to an equilibrium constant $K_ \text{p}$.

In summary I want to understand what let us use equilibrium constants under non-constant pressure conditions.

  • $\begingroup$ Can a reaction reach equilibrium in a closed container (fixed volume)? Sometimes even much faster, like small amount of explosive in a rigid cavity. $\endgroup$
    – Poutnik
    Commented Jul 7, 2021 at 8:21
  • $\begingroup$ The equilibrium constant $K_P$ is just that, constant (at constant T). If the pressure changes then the extent of reaction alters to compensate for this. see chemistry.stackexchange.com/questions/153545/… $\endgroup$
    – porphyrin
    Commented Jul 7, 2021 at 16:46
  • $\begingroup$ @porphyrin But the system is not at constant pressure. $\endgroup$
    – Anton
    Commented Jul 7, 2021 at 18:01
  • $\begingroup$ I checked the link. My question is about if it is proper for a reaction or physical change (like liquid turning into vapor) to use Helmholtz vs Gibbs free energy for the derivation of equilibrium constants. $\endgroup$
    – Anton
    Commented Jul 7, 2021 at 18:30
  • $\begingroup$ Exactly, not at constant pressure, thus the degree of dissociation changes just as in the example I linked to. $\endgroup$
    – porphyrin
    Commented Jul 8, 2021 at 7:47

1 Answer 1


Yes. The universal condition for equilibrium is that, given the current constraints on the system, no further spontaneous change can take place. I.e., that the entropy of the universe (system + surroundings) is maximized.

At constant temperature and pressure, a maximization of the entropy of the universe corresponds to a minimization of the Gibbs free energy (G) of the system, and we have:

$$\Delta G^{\circ} = - R T \ln K_p$$

By contrast, at constant temperature and volume, a maximization of the entropy of the universe corresponds to a minimization of the Helmholtz free energy (A) (sometimes also symbolized using "F") of the system. Here, we have:

$$\Delta A^{\circ} = - R T \ln K_V$$

Indeed, these are why we use the Gibbs and Helmholtz free energies—we can't typically measure the change in the entropy of the universe to determine the point at which it's maximized, so we instead use the minimzation of the Gibbs or Helmholtz free energies of the system (which can be measured) as surrogates for it.

[You're probably going to ask me if that standard state for A includes standard pressure. I'm going to need to think about that.]

  • $\begingroup$ So in the example of the reaction $\ce{A + B <=> C}$ Le Chatelier Principle still applies but the equilibrium constant (and how the equilibrium will shift) will depend on $\Delta A^{\circ} = - R T \ln K_V$? Does the same apply for the vapor liquid equilibrium? $\endgroup$
    – Anton
    Commented Jul 7, 2021 at 18:34
  • 1
    $\begingroup$ Yes to both. For the 2nd, imagine you have a sealed container that (by volume) is half liquid water and half gaseous water. Now suppose you inject some gaeous water into it. By Le Chatelier's principle, that will shift the rxn to towards the liquid side: Gaseous water will be converted to liquid water until the pressure of the gaseous water returns to the equilibrium vapor pressure. Interestingly, in this case, adding gaseous water will result in a net reduction in the total amount of gaseous water in the container, because the volume available for the gaseous water will be reduced! $\endgroup$
    – theorist
    Commented Jul 7, 2021 at 22:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.