William M. White - Geochemistry

Like other properties of state, the Gibbs free energy is additive. Therefore:

(2.127)

In other words, we can use Hess's law to calculate the free energy change of reaction. Values for at the standard state are available in compilations.

The Gibbs free energy is perhaps the single most important thermodynamic variable in geochemistry because it provides this criterion for recognizing equilibrium. This criterion is:

Products and reactants are in equilibrium when their Gibbs free energies are equal.

Another important quality of the Gibbs free energy is closely related:

At fixed temperature and pressure, a chemical reaction will proceed in the direction of lower Gibbs free energy (i.e., ΔG r <0).

The reverse is also true: a reaction will not proceed if it produces an increase in the Gibbs free energy.

On an intuitive level, we can understand the Gibbs free energy as follows. We know that transformations tend to go in the direction of the lowest energy state (e.g., a ball rolls down hill). We have also learned that transformations go in the direction of increased entropy (if you drop a glass it breaks into pieces; if you drop the pieces they don't re-assemble into a glass). We must consider both the tendency for energy to decrease and the tendency for entropy to increase in order to predict the direction of a chemical reaction. This is what the Gibbs free energy does. Example 2.7illustrates how Gibbs free energy of reaction is used to predict equilibrium.

One reason why the Gibbs free energy is useful is that its characteristic variables are temperature and pressure, which are the “external” variables of greatest interest in geochemistry. Since it is a state variable, we can deduce its temperature and pressure dependencies from eqn. 2.124, which are:

(2.128)

(2.129)

Using the thermodynamic data given in Table 2.2, calculate Δ G rfor the reaction:

at 298 K and 0.1 MPa. Which mineral assemblage is more stable under these conditions (i.e., which side of the reaction is favored)? Which assemblage will be favored by increasing pressure? Why? Which side will be favored by increasing temperature? Why?

Answer: We can calculate ΔG rfrom Δ H fand Δ S f, values listed in Table 2.2:

Δ H is calculated as: . Δ S is calculated in a similar manner. Our result is −6.08 kJ/mol. Because Δ G ris negative, the reaction will proceed to the right, so that the assemblage on the right is more stable under the conditions of 298 K and 1 atm.

To find out which side will be favored by increasing pressure and temperature, we use equations 2.128and 2.129to see how Δ G will change. For temperature, . Δ S ris −36.37 J/K-mol, and . The result is positive, so that Δ G will increase with increasing T , favoring the left side. Had we carried out the calculation at 1000°C and 0.1 MPa, a temperature appropriate for crystallization from magma, we would have found that the anorthite–forsterite assemblage is stable. For pressure, . Δ V for the reaction is −20.01 cc/mol (= J/MPa-mol), so will decrease with increasing pressure, favoring the right side. Reassuringly, our thermodynamic result is consistent with geologic observation. The assemblage on the left, which could be called plagioclase peridotite, transforms to the assemblage on the right, spinel peridotite, as pressure increases in the mantle.

Equations 2.128and 2.129allow us to predict how the Gibbs free energy of reaction will change with changing temperature and pressure. Thus, we can predict how the direction of a reaction will change if we change temperature and pressure . To obtain the Δ G rat some temperature T' and pressure P' we integrate:

(2.130)

(See Example 2.8.) For liquids and particularly gases, the effects of pressure and temperature on ΔV are significant and cannot be ignored. The reference pressure is generally 0.1 MPa. For solids, however, we can often ignore the effects of temperature and pressure on Δ V so the first integral reduces to: (see Example 2.9). On the other hand, we cannot ignore the temperature dependence of entropy. Hence we need to express Δ S ras a function of temperature. The temperature dependence of entropy is given by eqn. 2.105. Writing this in integral form, we have:

This is the change in entropy due to increasing the temperature from the reference state to T . The full change in entropy of reaction is then this plus the entropy change at the reference temperature:

(2.131)

Substituting this into 2.130, the second integral becomes:

(2.132)