Geochemistry. William M. White

      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_r from ΔH_f and ΔS_f, values listed in Table 2.2:

      ΔH is calculated as: . ΔS is calculated in a similar manner. Our result is −6.08 kJ/mol. Because ΔG_r is 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 2.128 and 2.129 to see how ΔG will change. For temperature, . ΔS_r is −36.37 J/K-mol, and . The result is positive, so that ΔG 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, . ΔV 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 2.128 and 2.129 allow 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 ΔG_r at 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 ΔV 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 ΔS_r as 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)

      , as we have defined it here, is the change in free energy of reaction as a result of increasing temperature from the reference state to T′.

Example 2.8 Predicting the equilibrium pressure of a mineral assemblage

      Using the thermodynamic reaction and data as in Example 2.7:

      determine the pressure at which these two assemblages will be in equilibrium at 1000°C. Assume that the volume change of the reaction is independent of pressure and temperature (i.e., α and β = 0).

      Answer: These two assemblages will be in equilibrium if and only if the Gibbs free energy of reaction is 0. Mathematically, our problem is to solve eqn. 2.130 for P such that .

      Our first step is to find ΔG_r for this reaction at 1000°C (1273 K) using eqn. 2.130. Heat capacity data in Table 2.2 is in the form: . Substituting for _, we have:

      (2.133)

      Performing the double integral and collecting terms, and letting , this becomes:

(2.134)

      equation 2.134 is a general solution to eqn. 2.130 when the Maier-Kelley heat capacity is used.

We found in Example 2.7. Computing Δa as , we find . Computing Δb and Δc similarly, they are −0.01732 J/(K-mol) and 1.66 × 10⁶ J-K²/mol, respectively. Substituting values into eqn. 2.136, we find .