William M. White - Geochemistry

, 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′ .

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. 2.130for P such that .

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

(2.133)

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

(2.134)

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

We found in Example 2.7. Computing Δ a as , we find . Computing Δ b and Δ c similarly, they are −0.01732 J/(K-mol) and 1.66 × 10 6J-K 2/mol, respectively. Substituting values into eqn. 2.136, we find .

Since we may assume the phases are incompressible, the solution to the pressure integral is:

(2.135)

Equation 2.130may now be written as:

Let . (calculated from values in Table 2.2), so . is positive, meaning that the left side of the reaction is favored at 1000°C and atmospheric pressure, consistent with our prediction based on ∂G/∂T .

Solving for pressure, we have

(2.136)

With , we obtain a value of 1.49 GPa (14.9 kbar). Thus, assemblages on the right and left will be in equilibrium at 1.49 GPa and 1000°C. Below that pressure, the left is stable, and above that pressure, the right side is the stable assemblage, according to our calculation.

The transformation from “plagioclase peridotite” to “spinel peridotite” actually occurs around 1.0 GPa in the mantle. The difference between our result and the real world primarily reflects differences in mineral composition: mantle forsterite, enstatite and diopside are solid solutions containing Fe and other elements. The difference does not reflect our assumption that the volume change is independent of pressure. When available data for pressure and temperature dependence of the volume change are included in the solution, the pressure obtained is only marginally different: 1.54 GPa.

The compressibility ( β ) of forsterite (Mg 2SiO 4) is 8.33 × 10 –6MPa −1. Using this and the data given in Table 2.2, what is the change in molar volume and Gibbs free energy of forsterite at 100 MPa and 298 K?

Answer: Let's deal with volume first. We want to know how the molar volume (43.79 cc/mol) changes as the pressure increases from the reference value (0.1 MPa) to 1 GPa. The compressibility is defined as:

(2.12)

So the change in volume for an incremental increase in pressure is given by:

(2.137)

To find the change in volume over a finite pressure interval, we rearrange and integrate:

Performing the integral, we have:

(2.138)

This may be rewritten as:

(2.139)