# What exactly is minimized when we say that the Gibbs Free Energy of Reaction is minimized at equilibrium?

I often see the statement- "gibbs free energy of a reaction is minimized at equilibrium"

The reasoning given is that $$\Delta_rG<0$$ before equilibrium and $$\Delta_r G=0$$ at equilibrium. Thus, it is monotonically decreasing until equilibrium where it stabilizes and therefore, the value of $$\Delta_rG$$ at equilibrium is the minimum value.

This makes sense mathematically, but considering the definition of $$\Delta_r G$$ i.e. $$\Delta_r G$$=$$\Sigma G_{products}-\Sigma G_{reactants}$$=o, I don't understand what quantity is being "minimized". $$\Delta_r G$$ doesn't seem to denote the change of the same function but rather the difference between gibbs free energies of products and reactants. In other words, the minimum value of a quantity $$x$$ can occur when function $$\Delta x$$=0, but the quantity $$x$$ needs to have meaning in isolation. The quantity "$$_r G$$" is meaningless so what exactly is being minimized here?

• You have asked two very similar questions, this and physics.stackexchange.com/questions/719273/… (and also posted on the Chemistry stack). Rather than asking another separate question, it is preferred that to clarify your first question; it may help to take into account the comments that other users posted on your first question.
– Buzz
Jul 20, 2022 at 20:49
• @Buzz I think I will delete my old post as this is more clear. Thanks for the feedback :) Jul 20, 2022 at 20:51
• The Gibbs free energy of the reaction mixture is being minimized. Jul 21, 2022 at 10:16
• @ChetMiller Also, do you mean $G_A+G_B+G_C+G_D$ is minimized? Jul 21, 2022 at 11:32
• $G=\mu_AN_A+\mu_BN_B+\mu_CN_C+\mu_DN_D$ is minimized, where the $\mu"s$ are the chemical potentials. This is subject to the stoichiometric constraints. Jul 22, 2022 at 12:07

There are often confusions in the notations. By definition, $$∆rG=∂G/∂ξ$$ is the derivative with respect to the extend of reaction $$\xi$$ at constant temperature and pressure.
To find equilibrium, the method consists in having an expression of the Gibbs free energy $$G(T,P,\xi)$$ and to seek for which value of the reaction progress $$G(T,P,\xi)$$ is minimal. Except in the case of a break of equilibrium, this minimum is obtained by cancelling the derivative of the Gibbs free energy. So At equilibrium, $$∆rG$$ is zero.
For multiple reactions occurring simultaneously, one would minimize $$G=\sum{\mu_iN_i}$$where $$\mu_i$$ is the chemical potential of the i'th species (a function of the species mole fractions) and $$N_i$$ is the number of moles of that component. This would be constrained by conservation of mass considerations on the number of moles of each atomic species. The constrained minimization would typically be carried out using Lagrange Multipliers. See Smith and van Ness, Introduction to Chemical Engineering Thermodynamics.
Building on Vincent's answer, there is a $$G$$ of a reacting mixture and a $$\Delta G$$ of reaction. Let's see how the two are different using as an example the reaction $$A \leftrightharpoons 2 B\tag{1}$$ If we have a mixture with $$n_A$$ and $$n_B$$ moles of the species the Gibbs enrgy of the mixture is $$G = n_A\mu_A+n_B \mu_B \tag{2}$$ Suppose we react $$\delta\xi$$ moles of $$A$$ to produce $$2\delta \xi$$ moles of B. The change in $$G$$ is $$\delta G = -\delta\xi\, \mu_A + 2\delta\xi\, \mu_B = \delta \xi(-\mu_A+2\mu_B) \tag{3}$$ At equilibrium $$G$$ is minimized, which means $$\delta G/\delta \xi = 0$$: $$\boxed{\frac{dG}{d\xi} = 0 \Rightarrow \underbrace{-\mu_A+2\mu_B}_{\Delta_r G} = 0} \tag{4}$$ This expresses the minimization condition on $$G$$. The quantity $$-\mu_A+2\mu_B$$ is what we call $$\Delta G_r$$ and represents the change in $$G$$ under stoichiometric reaction, i.e., all species change by their stoichiometric coefficients.
Therefore, the minimization of $$G$$ is equivalent to $$\Delta_r G = 0$$.