What is the difference between $\Delta G$ and $\Delta G^\circ$? I couldn't figure it out.


  1. $\Delta G^\circ = -RT\ln K$
  2. $\Delta G = \Delta G^\circ + RT \ln Q$

What is the difference, is it that $\Delta G^\circ$ is the value predicted by the standard free energies and that $\Delta G$ is just the actual value?

  • $\begingroup$ Related questions have come up a lot around here lately: Here's an example with G and here is one with enthalpy. $\endgroup$
    – Curt F.
    Commented Apr 13, 2015 at 23:46

2 Answers 2


A simple explanation would be that the "o on top" denotes standard state and the one without the "o on top" denotes conditions that are not standard state. However, that may not allow you to "get it" so let's look at the equations.

  1. $\Delta G^{\circ} = -RT\ln K$
  2. $\Delta G = \Delta G^{\circ} + RT \ln Q$

The first one allows us to find the Gibbs free energy at the standard state, as was mentioned. However, it would be useful to review what exactly the standard state is:

For the standard state of formation:

The partial pressure of any gas involved in the reaction is 0.1 MPa. The concentrations of all aqueous solutions are 1 M. Measurements are generally taken at a temperature of 25 °C (298 K). The standard-state free energy of formation is the change in free energy that occurs when a compound is formed from its elements in their most thermodynamically stable states at standard-state conditions. In other words, it is the difference between the free energy of a substance and the free energies of its constituent elements at standard-state conditions.

Or, very simply:

  • Partial Pressure of gases = 0.1 MPa
  • Concentration of Aqueous solutions are = 1 M (Molar)
  • Temperature usually 298 K

So if it's so nice and well, why would we need the one without "o on top"? Well, if we meet a problem or situation without the concentrations or pressures as stated above, we need to use our handy equation to relate the standard state which may have been given to the actual situation, with perhaps 0.1 M or some other concentration. This is powerful because it allows you to view different states,perhaps with more or less reactants or products, and realize how one can affect equilibrium.

  • 1
    $\begingroup$ In equation 2, "K" should be "Q". Also, link isn't working. $\endgroup$
    – DavePhD
    Commented Dec 8, 2015 at 16:03

An add : the Gibbs energy is at its standard state only when the extent of reaction ( greek letter Xi ) is equal to zero. That counts also for the standard state of enthalpy.

  • $\begingroup$ That's incorrect. The condition for them to be equal is when the reaction quotient $Q = 1$ - you can see that from the equation $\Delta G = \Delta G^\circ + RT \ln Q$. In general, that does not correspond to the point at which $\xi = 0$. $\endgroup$ Commented Sep 28, 2015 at 5:19

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