 Solution

ID: past-exam-of-the-year-2-of-the-earth-sciences-geology-course-of-the-university-of-oxford/2024/paper-3/10/iv/solution

\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) .

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.

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