Chemical Equilibrium

The Gibbs free energy allows us to define equilibrium conditions for a system at constant TT and PP, which means three conditions need to be satisfied for a system to be at equilibrium:

  1. There is no net flow of energy or matter within the system.

  2. There is no net exchange of energy or matter with the outside world.

  3. There are no unbalanced driving forces or thermodynamic potentials.

When we say there is no “net” flow/exchange of energy and matter, it doesn’t mean there isn’t any flow or exchange of energy and matter, but such flow/exchange should be balanced out when summed in all available directions. More formally, GG is minimized with respect to perturbations in the composition of the system at constant TT and PP.

dGT,P=iμidni=0\dd G\big|_{T, P}=\sum_i \mu_i \dd n_i = 0

In other words, an infinitesimal change in the composition of the system yields no change in the Gibbs free energy. For example, in a chemical reaction at constant TT and PP, the number of moles of each species (A, B, C, D) does not change independent of each other as seen in the following reaction formula

aA+bBcC+dDa\text{A} + b\text{B} \Leftrightarrow c\text{C} + d\text{D}

where aa, bb, cc, and dd are the stoichiometric coefficients. To make this more concrete, consider the reaction 2H2+O22H2O2\text{H}_2 + \text{O}_2 \Leftrightarrow 2\text{H}_2\text{O}. The convention is to consider the forward reaction, where the stoichiometric coefficients are -2, -1, and 2 for H2\text{H}_2, O2\text{O}_2, and H2O\text{H}_2\text{O}, respectively. Now, let’s write out the infinitesimal change in the Gibbs free energy for the above equation.

dGT,P=iμidni=μAdnA+μBdnB+μCdnC+μDdnD\begin{aligned} \dd G \big|_{T, P} &= \sum_i \mu_i\dd n_i \\ &= \mu_{\text{A}} \dd n_{\text{A}} + \mu_{\text{B}} \dd n_{\text{B}} +\mu_{\text{C}} \dd n_{\text{C}} +\mu_{\text{D}} \dd n_{\text{D}} \end{aligned}

As written in Equation 2.4.3, it may seem like dG\dd G has 4 degrees of freedom, namely the change in the number of moles for each of the four species (dnA,dnB,dnC,dnD\dd n_{\text{A}}, \dd n_{\text{B}}, \dd n_{\text{C}}, \dd n_{\text{D}}). However, since we are dealing with a chemical reaction with fixed stoichiometric coefficients, in order to observe conservation of mass, the dni\dd n_i variables do not change independently from each other. More specifically, for every aa moles of species A consumed, bb moles of species B must also be consumed, while cc moles of species C and dd moles of species D must be produced.

As a result, since the four dni\dd n_i variables need to change in concert, there is only 1 true degree of freedom. We name this degree of freedom as extend of the reaction, denoted by dξ\dd \xi. If we consider the forward reaction from reactants to products, we can write

dnA=adξdnB=bdξdnC=cdξdnD=ddξ\begin{aligned} \dd n_{\text{A}} &= -a\,\dd \xi \\ \dd n_{\text{B}} &= -b\,\dd \xi \\ \dd n_{\text{C}} &= c\,\dd \xi \\ \dd n_{\text{D}} &= d\,\dd \xi \end{aligned}

Substituting these expressions into Equation 2.4.3, we then get

dGT,P=aμAdξbμBdξ+cμCdξ+dμDdξ=(aμAbμB+cμC+dμD)dξ(Gξ)T,P=aμAbμB+cμC+dμD=0\begin{aligned} \dd G \big|_{T, P}&= -a\mu_{\text{A}}\,\dd \xi - b\mu_{\text{B}}\,\dd \xi + c\mu_{\text{C}}\,\dd \xi + d\mu_{\text{D}}\,\dd \xi \\ &= \left(-a\mu_{\text{A}}-b\mu_{\text{B}}+c\mu_{\text{C}}+d\mu_{\text{D}}\right)\dd \xi\\ \left(\frac{\partial G}{\partial \xi}\right)_{T, P} &= -a\mu_{\text{A}}-b\mu_{\text{B}}+c\mu_{\text{C}}+d\mu_{\text{D}}=0 \end{aligned}

We can see that at constant TT and PP, the Gibbs free energy is minimized with respect to changes in the extent of the chemical reaction. Equation 2.4.5 can be generalized to describe chemical equilibria with the following expression

(Gξ)T,P=iviμi=0\left(\frac{\partial G}{\partial \xi}\right)_{T, P} = \sum_i v_i \mu_i = 0

where viv_i is the stoichiometric coefficient of species ii in the chemical reaction (negative for reactants and positive for products). Finally, we can express the condition of chemical equilibrium for the water production reaction (2H2+O22H2O2\text{H}_2 + \text{O}_2 \Leftrightarrow 2\text{H}_2\text{O}) as

(Gξ)T,P=2μH2μO2+2μH2O=0\left(\frac{\partial G}{\partial \xi}\right)_{T, P} = -2 \mu_{\text{H}_2} - \mu_{\text{O}_2} + 2 \mu_{\text{H}_2\text{O}}= 0
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