= Solution
Apply <upper-envelope stability for competing mineral dissolution reactions> with $\ell_G=10.53-0.6x$ and $\ell_K=12.228-0.2x$. For each <chlorite> dissolution reaction,
$$
\frac{\Delta_rG_G}{RT\ln10}=5(y-\ell_G),\qquad \frac{\Delta_rG_K}{RT\ln10}=5(y-\ell_K).
$$
Therefore high $y$ suppresses dissolution and stabilizes <chlorite>. Low $y$ favors the alteration products. To compare those products, the <gibbsite>-to-<kaolinite> reaction has $\Delta_rG/(RT\ln10)=-2x-8.49$: <kaolinite> is favored at $x>-4.245$, and <gibbsite> at $x<-4.245$.
The stable fields are consequently \b[<chlorite> above $y=\max(\ell_G,\ell_K)$; <gibbsite> below $\ell_G$ for $x<-4.245$; and <kaolinite> below $\ell_K$ for $x>-4.245$]. The stable <chlorite>–<gibbsite> boundary is the left-hand part $x\leq-4.245$, and the stable <chlorite>–<kaolinite> boundary is the right-hand part $x\geq-4.245$. Their opposite extensions are <metastable> because the third <mineral> is more stable. The vertical <gibbsite>–<kaolinite> boundary is stable below $y=13.077$ and <metastable> above it.
The figure displays the stable upper envelope. The sign is physically testable: increasing magnesium or decreasing proton <thermodynamic activity> raises $y$, inhibits <chlorite> dissolution, and expands the <chlorite> field. This direct <Gibbs energy> comparison avoids inferring stability merely from where three algebraic lines meet.
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