I have read in my textbook that for a reaction $\Delta G=\Delta G^o-RT\ln Q $, where $\Delta G^o$ is the Gibbs free energy change when the initial concentration of products and reactants are unity. But when I searched for the definition of $\Delta G^o$ and it is only mentioned that the reactants and products must be in their standard states, it doesn't mention that initial concentration of products and reactants should be unity.
So if we carry out a reaction (with no gaseous reactants or product) at standard condition, then shouldn't $\Delta G=\Delta G^o $ instead of $\Delta G=\Delta G^o -RT\ln Q$ ?

  • $\begingroup$ See chemistry.stackexchange.com/a/139640 $\endgroup$
    – Karsten
    Commented Mar 14, 2022 at 19:00
  • $\begingroup$ Standard states are: activities equal one for pure solids or liquids, fugacities equal one for gases and concentrations equal one Molar for solutions. So in the standard states Q equals one and lnQ is zero and DeltaG = DeltaG[0]. At equilibrium Delta G=0 so DeltaG[0] = RTlnKeq. This is almost what you are stating, think it through. The standard free energy change is the free energy difference from reactants and products both in their standard states to the equilibrium concentrations. $\endgroup$
    – jimchmst
    Commented Mar 16, 2022 at 4:13

1 Answer 1


First note, I believe you mean this equation: $$\Delta G= \Delta G^\circ + RT \ln{Q}$$

The Gibbs free energy for reactions in their standard state is defined relative to their stoichiometric relations (what I believe you mean by "unity" here.) So for a sample reaction, $$2A + B \ce{<=>} 3C$$ we would define the standard Gibbs free energy as the free energy (non-expansion work, specifically) associated with one mole of this reaction happening. One mole of this reaction consumes two moles of $A$, one mole of $B$, and produces three moles of $C$. You can assume this reaction occurs in one liter of solvent in order to make the generalization to concentration easier.

The standard Gibbs free energy $\Delta G^\circ$ is associated to how this system comes to equilibrium; you can express the standard Gibbs free energy for a reversible reaction the following way: $$\Delta G^\circ = -RT \ln{K_{eq}} $$

This equation carries the implicit statement that our standard Gibbs free energy is only suitable when the system has accomplished equilibrium. We often run into systems that have not reached equilibrium yet—if we want to quantify Gibbs free energy for these systems too, then we will need to transform our equation a bit.

Here is where it is helpful to think of equilibrium as a sort of "home base." The equilibrium of a reaction is where a reaction will naturally tend if perturbed in some way. Adding more product can be a perturbation of the system, adding more reagent perturbs the reaction, et cetera. These perturbations take the system away from equilibrium, much like stretching or compressing a spring, and then over time they relax back to the equilibrium concentrations. We represent the perturbed system with $Q$. Any time you see $Q$, you know that it is a general statement about your chemical system not necessarily in equilibrium. If your system is already at equilibrium, then there is no change in the Gibbs free energy ($\Delta G=0$) which you can confirm for yourself by combining the previous equations listed with your $Q$ set equal to $K_{eq}$. If your system is a little bit away from equilibrium, then as it reestablishes equilibrium, it will do some work in some "direction" (chemically speaking.)


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