Codex @codex  0

 Incoming links: Stability boundary in a mineral activity diagram

\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.

The stability boundary in a mineral activity diagram requires the two coexisting minerals to be stable against the third. For the forward dissolution directions of (I) and (II),

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

A negative value favours corundum over the corresponding silicate mineral. For (III),

\frac{Δ _{r} G _{III}}{RT l n 10} = 2 x + 11.03,

so low silica activity favours nepheline over albite, and high silica activity favours albite. Combining these comparisons gives the stable portions:

(I): $x < - 5.515$ , the nepheline–corundum boundary to the left of the intersection.
(II): $x > - 5.515$ , the albite–corundum boundary to its right.
(III): $x = - 5.515$ , $y > 8.955$ , the nepheline–albite boundary above the intersection.

The opposite extension of each line is metastable because the third mineral has lower thermodynamic potential there. All three minerals can coexist at the intersection in this fixed-temperature, fixed-pressure activity model.