# 26.1: Equilibrium Results when Gibbs Energy is Minimized

The thermodynamic criterion for phase equilibrium is simple. It is based upon the chemical potentials of the components in a system. For simplicity, consider a system with only one component. For the overall system to be in equilibrium, the chemical potential of the compound in each phase present must be the same. Otherwise, there will be some mass migration from one phase to another, decreasing the total chemical potential of the phase from which material is being removed, and increasing the total chemical potential of the phase into which the material is being deposited. So for each pair of phases present ($$\alpha$$ and $$\beta$$) the following must be true:

$\mu_\alpha = \mu_\beta$

## Gibbs Phase Rule

The Gibbs phase rule describes the number of compositional and phase variables that can be varied freely for a system at equilibrium. For each phase present in a system, the mole fraction of all but one component can be varied independently. However, the relationship

$\sum_i \chi_i =1$

places a constraint on the last mole fraction. As such, there are $$C – 1$$ compositional degrees of freedom for each phase present, where $$C$$ is the number of components in the mixture. Similarly, all but one of the chemical potentials of each phase present must be equal, leaving only one that can be varied independently, leading to $$P – 1$$ thermodynamic constraints placed on each component. Finally, there are two state variables that can be varied (such as pressure and temperature), adding two additional degrees of freedom to the system. The net number of degrees of freedom is determined by adding all of the degrees of freedom and subtracting the number of thermodynamic constraints.

\begin{align} F &= 2+ P(C-1) - C(P-1) \nonumber \\[5pt] &= 2 + PC - P -PC +C \nonumber \\[5pt] F = &= 2+C-P \label{Phase} \end{align}

Equation \ref{Phase} is the Gibbs phase rule.

Example $$\PageIndex{1}$$:

Show that the maximum number of phases that can co-exist at equilibrium for a single component system is $$P = 3$$.

Solution:

The maximum number of components will occur when the number of degrees of freedom is zero.

$0 = 2+1 -P$

$P=3$

Note: This shows that there can never be a “quadruple point” for a single component system!

Because a system at its triple point has no degrees of freedom, the triple point makes a very convenient physical condition at which to define a temperature. For example, the International Practical Temperature Scale of 1990 (IPT-90) uses the triple points of hydrogen, neon, oxygen, argon, mercury, and water to define several low temperatures. (The calibration of a platinum resistance thermometer at the triple point of argon, for example, is described by Strouse (Strouse, 2008)). The advantage to using a triple point is that the compound sets both the temperature and pressure, rather than forcing the researcher to set a pressure and then measure the temperature of a phase change, introducing an extra parameter than can introduce uncertainty into the measurement.

### Contributors

• Patrick E. Fleming (Department of Chemistry and Biochemistry; California State University, East Bay)

Many important chemical reactions -if not most- are performed in solution rather than between solids or gases. Solid state reactions are often very slow and not all chemical species can be put in the vapor phase because they decompose before evaporating.

Often we are not concerned with the temporal aspects of a reaction. That can be technologically very important but it is the domain of kinetics - a different branch of Physical Chemistry - rather than classical thermodynamics.. The latter is more concerned with the endpoint. This is the thermodynamically speaking the (stable) equilibrium, but chemically it can either represent a completed reaction or a chemical equilibrium.

Unfortunately, of the three main aggregation states: gas – liquid – solid, the structure of liquids is least understood and one of the most complex liquids is also one of the most extensively used ones: water. It is vital to many branches of chemistry varying from geochemistry to environmental chemistry to biochemistry. We shall make just a small inroad into its complexity.

### Extent of reaction

To describe the progress of a reaction we define the extent of reaction. It is usually denoted by the Greek letter $$ξ$$.

Consider a simple reaction:

$v_AA + v_BB \rightleftharpoons v_YY + v_ZZ$

Using stoichiometry we can define the extent by considering how the number of moles (or molar amounts) of each species changes during the reaction:

reactants

• $$n_A= n_{A,0} - v_A ξ$$
• $$n_B= n_{B,0} - v_B ξ$$

products

• $$n_Y= n_{Y,0} + v_Y ξ$$
• $$n_Z= n_{Z,0} + v_Z ξ$$

The dimension of ξ is [mol] because the stoichiometric coefficients vi are dimensionless integers. If the reaction goes to completion for one of the reactants -the limiting reactant- nA or B=nlimiting will go to zero. If we start with $$n_{limiting}= v_{limiting}$$ moles, the value of ξ starts at 0 (no products) and goes to 1 at completion (limiting reactant depleted). When approaching an equilibrium $$ξ$$ will not go beyond $$ξ_{eq}$$.

### Measuring ξ

The extent of reaction is what is the central subject of reaction kinetics. Its value is typically measured as a function of time indirectly by measuring a quantity q that is linearly dependent on ξ(t):

$q( ξ ) = aξ +b$

Consider the situation at the extremes $$ξ=0$$ and $$ξ=1$$:

$q_0= a.0+b= b$

$q_1= a.1+b= a+b$

$q_1-q_0= a$

Thus, $$ξ$$ can be found from

$\dfrac{q(t)-q0}{q_1-q_0}=\dfrac{q(t)-b}{a}$

The nature of $$q$$ can vary widely from UV/Vis absorption, conductivity, gravimetric to caloric data.

In practice, $$q_0$$ at $$ξ=0$$ is often hard to observe because it takes time to mix the reactants, particularly in solutions, and q1 at ξ=1 may never be reached if the reaction goes to equilibrium. Nevertheless the values of a and b can often be found from the available data by fitting techniques.

In (equilibrium, static) thermodynamics we are only concerned with the endpoint:

• $$ξ=1$$: the reaction runs to completion
• $$ξ=ξ_{eq}$$: the react ion goes to a state of chemical equilibrium

### Thermodynamic Potentials

As we have seen we can write any change in the Gibbs free energy due to changes in the molar amounts of the species involved in the reaction (at T,P constant) as:

$dG =\sum \dfrac{∂G}{∂n_i} dn_i = \sum μ_idn_i$

where μ is the thermodynamic potential, often called chemical potential when dealing with reactions. From the definition of ξ we can see by differentiation that

• $$d n_A=- v_Adξ$$
• $$d n_B=- v_Adξ$$
• $$d n_Y= v_Ydξ$$
• $$d n_Z= v_Zdξ$$

This allows us to unify the changes in the molar amount of all the species into one single variable dξ.

We get

$dG = \left[ \sum -v_{i,reactants} μ_{i,reactants} + \sum+v_{i,products} μ_{j,products} \right]dξ$

or

$\left (\dfrac{∂G}{∂ξ} \right)_{T,P} = -\sum v_{i,r}μ_{i,r}+ \sum v_{i,p}μ_{j,p}$

This quantity is also written as:

$\left( \dfrac{∂G}{∂ξ} \right)_{T,P} =Δ_rG$

This quantity gives the change in Gibbs free energy for the reaction (as written!!) for Δξ=1 mole. (Units are [J/mol] therefore).

### Gas Reactions

Let us assume that our reaction is entirely between gas species and that the gas is sufficiently dilute that we can use the ideal gas law. Then we can write for each species:

$μ_i= μ_i^o+RT \ln \dfrac{P_i}{P_i^o}$

We can then split up the ΔrG expression in two parts:

$Δ_rG = Δ_rG^o + RT\ln Q$

The standard potentials:

$Δ_rG^o = -\sum v_i,_r μ^o_{i,r} + \sum v_{i,p} μ^o_{j,p}$

and the logarithmic terms:

$RT \ln Q= - v_A RT \ln \left( \dfrac{P_A}{P_A^o} \right)- v_B RT\ln \left( \dfrac{P_B}{P_B^o} \right) + v_YRT \ln \left( \dfrac{P_Y}{P_Y^o} \right) + v_ZRT \ln \left( \dfrac{P_Z}{P_Z^o} \right)$

We can combine all the logarithmic terms into Q, called the reaction quotient. The stoichiometric coefficients become exponents and the reactants' factors will be 'upside down' compared to the products, because of the properties of logarithms:

$a \ln x = \ln x^a$

$- a \ln x = \ln \left( \dfrac{1}{x^a} \right)$

We have kept the standard pressures $$P_i^o$$ in the expression, but often they are omitted. They are usually all 1 bar, but in principle we could choose 1 bar for A 1 Torr for B an 1 psi for the products. It creates a valid (though ridiculous) definition of what o stands for. (Of course the value of $$Δ_rG^o$$ does depend on that choice!).

We could write

$RT \ln Q = RT \ln \dfrac{Q_P}{Q^o}$

$$Q^o$$ is typically unity in magnitude but it cancels the dimensions of $$Q_P$$. That means that $$Q$$ and $$Q_P$$ are equal in magnitude and we can get $$Q$$ from $$Q_P$$ by simply dropping the dimensions. $$Q$$ is dimensionless but $$Q_P$$ usually is not. Often this fine distinction is simply not made and $$Q^o$$ is omitted, we get:

$Δ_rG = Δ_rG^o + RT\ln \dfrac{P_Y^{v_Y}P_Z^{v_Z} }{P_A^{v_A}P_B^{v_B}}$

Notice the difference between $$Δ_rG$$ which denotes the conditions (e.g. pressures) of your reaction and $$Δ_rG^o$$ denotes standard conditions.

### ΔrG versus ΔfG

If Gibbs free energies of formation are known for all the species involved in our reaction we can calculate the Gibbs free energy of the reaction.

$Δ_rG^o = v_YΔ_fG^o(Y)+ v_ZΔ_fG^o(Z)- v_AΔ_fG^o(A)- v_BΔ_fG^o(B)$

If Gibbs values are not available we can calculate them from enthalpies and entropies:

$Δ_rG^o =Δ_rH^o -TΔ_rS^o$