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For the reaction direction in part ii, pure solids and unit water activity give $K_c=1/a_{{\mathrm{H}}_4\mathrm{Si}{\mathrm{O}}_4^0}^2$. ThereforePlot a vertical line through the common intersection, independent of $y$. Increasing dissolved silica favors albite, consistently with its larger Si/Al ratio. The vertical line’s stable and metastable portions are selected by comparison with corundum, not by its own mass-action equation alone.

Subtract reaction (b) from reaction (a) and divide by two. In the direction chosen here, the resulting balanced conversion isThus nepheline plus dissolved silica converts to albite. Additivity of the Gibbs energy of reaction, equivalently Hess's law, givesFor the reverse reaction the answer is $-11.03$; the reaction direction must accompany the equilibrium constant.

Let ${\ell }_N(x)=3.44-x$ and ${\ell }_A(x)=-7.59-3x$. The common-potential test for a three-mineral activity diagram gives relative potentials, in one positive energy unit,The stable phase minimizes these potentials. Consequently the stable nepheline–corundum branch has $x\le -5.515$, the stable albite–corundum branch has $x\ge -5.515$, and the stable nepheline–albite branch has $x=-5.515$, $y\ge 8.955$. Draw them solid.

The other halves of each line are metastable: a third mineral has lower potential there. In particular, below the triple point the vertical equality describes two phases both less stable than corundum. Extend those halves as dashed lines. This thermodynamic comparison prevents the incorrect use of all three complete pairwise lines as stable field boundaries.

The stable fields areThus corundum occupies the low-$y$ field, nepheline the upper-left field and albite the upper-right field. Pairwise coexistence applies on the indicated stable boundaries, and all three coexist at $(-5.515,8.955)$. These are stability statements within the three-mineral chemical system, not a claim that other possible minerals have been excluded in nature.

Subtract the nepheline–corundum reaction from the albite–corundum reaction and divide by two. In the albite-to-nepheline direction the resulting net reaction isAdditivity of the standard Gibbs energy of reaction makes the logarithms of the equilibrium constants combine with the same coefficients:This normalization is per one albite formula unit. The doubled reaction has ${\log }_{10}K=-22.06$, and reversing the reaction changes its sign.

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,Negative values favour corundum. For the albite-to-nepheline conversion, ${\Delta }_{r}G/(RT\ln 10)=2x+11.03$, so low silica activity favours nepheline and high silica activity favours albite. These comparisons give:

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

Subtract reaction (I) from reaction (II), cancelling the common aqueous ions and corundum, and divide by two. Choose the albite-to-nepheline direction:This is albite plus water converting to nepheline and dissolved silica. Standard Gibbs energies of reaction add with their stoichiometric coefficients, so logarithms of equilibrium constants do too. For this single-formula normalization,The doubled reaction has ${\log }_{10}K=-22.06$, while reversing either reaction changes the sign. Specifying the direction and normalization removes that apparent ambiguity.

For pure albite, pure nepheline and unit water activity, $Q_{III}=a_{{\mathrm{H}}_4\mathrm{Si}{\mathrm{O}}_4}^2$. Consequentlya vertical line through the intersection of (I) and (II). It does not depend on $y$ because the net reaction has no hydrogen or sodium ions.

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),A negative value favours corundum over the corresponding silicate mineral. For (III),so low silica activity favours nepheline over albite, and high silica activity favours albite. Combining these comparisons gives the stable portions:

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.