Past exam of the year 2 of the earth sciences (geology) course of the University of Oxford / 2024 / Paper 3 / 10

Subtract the nepheline–corundum reaction from the albite–corundum reaction and divide by two. In the albite-to-nepheline direction the resulting net reaction is

NaAlS i_{3} O_{8} + 4 H_{2} O ⇌ NaAlSi O_{4} + 2 H_{4} Si O_{4} .

(1)

Additivity of the standard Gibbs energy of reaction makes the logarithms of the equilibrium constants combine with the same coefficients:

lo g_{10} K = \frac{- 15.18 - 6.88}{2} = - 11.03.

(2)

This normalization is per one albite formula unit. The doubled reaction has

lo g_{10} K = - 22.06

, and reversing the reaction changes its sign.

Solution

 0  0

The net reaction quotient is

a_{H_{4} Si O_{4}}^{2}

when the solid and water activities equal one. Therefore its equilibrium line is

x = \frac{1}{2} lo g_{10} K = - 5.515,

(1)

a vertical line through the intersection of the other two. There is no dependence on

y

, because the net reaction contains neither hydrogen ions nor sodium ions.

Figure 1.
Stable and metastable mineral boundaries for the common albite-nepheline-corundum equilibrium example
. Original diagram computed from the supplied equilibrium constants. Solid lines show stable coexistence; dashed continuations are metastable. Logarithms are base ten. **Details**.

Solution

 0  0

A stability boundary in a mineral activity diagram must also be stable against the third mineral. Use the common-potential test for a three-mineral activity diagram. For the two corundum-producing directions,

\frac{Δ _{r} G _{a}}{2 RT l n 10} = y - (3.44 - x), \frac{Δ _{r} G _{b}}{2 RT l n 10} = y - (- 7.59 - 3 x) .

(1)

Negative values favour corundum. For the albite-to-nepheline conversion,

Δ_{r} G / (RT ln 10) = 2 x + 11.03

, so low silica activity favours nepheline and high silica activity favours albite. These comparisons give:

Nepheline–corundum: stable for $x < - 5.515$ , metastable on the opposite extension.
Albite–corundum: stable for $x > - 5.515$ , metastable on the opposite extension.
Nepheline–albite: stable for $y > 8.955$ along $x = - 5.515$ , metastable below the intersection.

All three coexist at the intersection. Pairwise mass-action equations alone do not distinguish the stable segments from their metastable extensions.

The bounding equalities give the stable two-mineral coexistences. In particular, the stable corundum region lies below the lower of the two oblique lines; treating every line extension as stable would mislabel parts of the mineral activity diagram.