When we consider a reaction that consists of an equilibrium between reactants and products such as an ionic equilibrium or an acid-base equilibrium we are often concerned about the concentration of the various species in solution. The reaction is said to proceed from reactants toward products and from products toward reactants until no perceptable reaction is taking place. At this point the change in free energy with time has stopped, see Eqn. \eqref{eq:dG} \begin{equation} \frac{{\Delta G}}{{\Delta T}}\left( {\frac{{dG}}{{dT}}} \right) = 0 \label{eq:dG} \end{equation} When this has occurred the reaction is at equilibrium.

How do we find this point?

Let's begin by considering the approach to equilibrium as a reaction where we start with a reactant and proceed towards a product. At any point along the reaction coordinate the weighted average free energy is given by Eqn. \eqref{eq:aDG} \begin{equation} \Delta {G_{ave}} = {f_{react}}\Delta G_{react}^0 + {f_{prod}}\Delta G_{react}^0 \label{eq:aDG} \end{equation} This is the red curve plotted in upper half of Figure 2. This curve would suggest that the lowest/best free energy would always be at the product. This is not the case. What is missing is the entropy of mixing. When two substances (reactants and products in the present case) mix there is an increase in the number of confiugrations, see Figure 1,

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and a favorable configurational entropy given by Eqn. \eqref{eq:Smix} \begin{equation} \frac{{{S_{config}}}}{{{N_0}}} = - R\left[ {\frac{{{N_1}}}{{{N_0}}}\ln \frac{{{N_1}}}{{{N_0}}} + \frac{{{N_2}}}{{{N_0}}}\ln \frac{{{N_2}}}{{{N_0}}}} \right] \label{eq:Smix} \end{equation} where N₀ is the sum of the amount of the first species, N₁ and the second species N₂. For a mixture of reactants and products \eqref{eq:Smix2} \begin{equation} \frac{{{N_1}}}{{{N_0}}} = \left( {1 - x} \right);{\rm{ }}\frac{{{N_2}}}{{{N_0}}} = x \label{eq:Smix2} \end{equation} The configurational entropy is given by \eqref{eq:Smix3} \begin{equation} \frac{{{S_{config}}}}{{{N_0}}} = - R\left( {1 - x} \right)\ln \left( {1 - x} \right) - Rx\ln x\\ \label{eq:Smix3} \end{equation} The entropy associated with the reactant fraction can be called the reactant entropy of mixing Eqn. \eqref{eq:SmixR} and the entropy associated with the product fraction, the product entropy of mixing Eqn. \eqref{eq:SmixP}. \begin{equation} \frac{{S_{mix}^{react}}}{{{N_0}}} = - R\left( {1 - x} \right)\ln \left( {1 - x} \right) \label{eq:SmixR} \end{equation} \begin{equation} \frac{{S_{mix}^{prod}}}{{{N_0}}} = - Rx\ln x\ \label{eq:SmixP} \end{equation} The leads to Eqn. \eqref{eq:Free2} for the free energy of reaction. \begin{equation} \Delta {G_{rxn}} = \left( {1 - x} \right)\Delta G_{react}^0 + x\Delta G_{prod}^0 - T\frac{{S_{mix}^{react}}}{{{N_0}}} - T\frac{{S_{mix}^{prod}}}{{{N_0}}} \label{eq:Free2} \end{equation}

In the upper plot of Figure 2 the green and blue curves are the entropy of mixing for the reactant and product as a function of extent of reaction, respectively. The free energy of reaction is given by the black curve. Equilibrium is achieved when the free energy is minimized (black curve is lowest).

This point, the free energy minimum can be found from the slopes of the reactant and product free energies with respect to amount. This slope is referred to as the chemical potential, \eqref{eq:chmpot}

\begin{equation} {\mu _i} = {\left( {\frac{{\partial G}}{{\partial {N_i}}}} \right)_{T,P,{N_{j \ne i}}}} \label{eq:chmpot} \end{equation}

The chemical potentials for reactant product are given in Eqns. \eqref{eq:chmpotR} and \eqref{eq:chmpotP}, respectively.

\begin{equation} {\mu _{react}} = \Delta G_{react}^0 + RT\ln \left( {1 - x} \right) \label{eq:chmpotR} \end{equation} \begin{equation} {\mu _{prod}} = \Delta G_{prod}^0 + RT\ln x \label{eq:chmpotP} \end{equation}

When the difference of the chemical potentials is zero the free energy is at a minimum. The red curve in the lower plot is the chemical potential for the reactant, the green curve the chemical potential for the product, and the blue curve the difference.

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