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My intuition says no, but I am having trouble coming up with concrete examples.

I know that minimizing Gibbs free energy predicts a state of equilibrium, while minimizing kinetic or internal energy does not correspond to equilibrium. However, I am not so sure about potential energy--- when PE is at a minimum, is the system also at equilibrium?

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    $\begingroup$ If you write a wavefunction of H$_2^+$ molecule concentrating at nuclei, e.g. $\delta(r-R)$, the potential energy is minimized. By the uncertainty principle, the kinetic energy is quite large. This wavefunction neither corresponds to the ground state of H$_2^+$ nor the ground state/equilibrium of an ensemble H$_2^+$ molecules. $\endgroup$
    – user26143
    Dec 8, 2013 at 7:07
  • $\begingroup$ If you minimize Gibbs free energy, than you have equilibrium at constant pressure and temperature, etc... so the type of energy you choose corresponds to given ensemble. Roughly speaking. $\endgroup$
    – ssavec
    Dec 9, 2013 at 12:53
  • $\begingroup$ No. None of the thermodynamic potentials is at its minimum at the equilibrium point of ANY system. This includes G. $\endgroup$
    – sencer
    Dec 10, 2013 at 2:49

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When you find a minimum in the thermodynamic potential energy of a system, you will have found an equilibrium point.

The key is the thermodynamic potential part. Depending on the constraints of the system (i.e. which thermodynamic state variables are held constant) the thermodynamic potential energy function will be different.

For example, in a system of a fixed number of particles at constant temperature and pressure (we would call it NPT), the Gibbs free energy is the thermodynamic potential, and it will be minimized at equilibrium.

For NVT (constant volume and temperature) the Helmholtz free energy is the thermodynamic potential, and that is what will be minimized at equilibrium.

In a system consisting of a rock dropped from a height above the ground, gravitational potential energy is the thermodynamic potential (I am abusing the language a little bit here) and it will be minimized at equilibrium (when the rock hits the ground).

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