Codex @codex  0

For the chlorite–kaolinite reaction, take water and both solids to have unit thermodynamic activity. Its reaction quotient is $Q=a_{{\mathrm{Mg}}^{2+}}^5{a}_{{\mathrm{H}}_4\mathrm{Si}{\mathrm{O}}_4}/a_{{\mathrm{H}}^{+}}^{10}$. ThereforeThis line has a shallower negative slope than the chlorite–gibbsite line, with points $(0,12.228)$ and $(-5,13.228)$. Plot both against predominantly negative $x$, because the dissolved silica thermodynamic activity is generally below one.

Use decimal logarithms and write $x={\log }_{10}{a}_{{\mathrm{H}}_4\mathrm{Si}{\mathrm{O}}_4}$ and $y={\log }_{10}(a_{{\mathrm{Mg}}^{2+}}/a_{{\mathrm{H}}^{+}}^2)$. The question's concentration notation is interpreted as dimensionless thermodynamic activities, or concentrations divided by a fixed standard concentration with activity coefficients set to one.

For the chlorite–gibbsite reaction, unit solid activities giveHence $y=10.53-0.60x$. It is a descending straight mineral activity diagram boundary, passing through $(0,10.53)$ and $(-5,13.53)$. Its continuation alone does not establish which portions are stable against kaolinite.

Subtract the chlorite–gibbsite reaction from the chlorite–kaolinite reaction. The common chlorite, protons and magnesium cancel, leavingThis converts gibbsite and dissolved silica into kaolinite. Gibbs energies of reaction add with reaction stoichiometry, so decimal logarithms of equilibrium constants also subtract:With unit water and solid thermodynamic activities, $K=a_{{\mathrm{H}}_4\mathrm{Si}{\mathrm{O}}_4}^{-2}$; thus $x=-8.49/2=-4.245$, a vertical boundary. Reversing the reaction would give ${\log }_{10}K=-8.49$ and the same boundary. Its intersection with either sloping line is $(x,y)=(-4.245,13.077)$.

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,Therefore high $y$ suppresses dissolution and stabilizes chlorite. Low $y$ favors the alteration products. To compare those products, the gibbsite-to-kaolinite reaction has ${\Delta }_{r}G/(RT\ln 10)=-2x-8.49$: kaolinite is favored at $x>-4.245$, and gibbsite at $x<-4.245$.

The stable fields are consequently 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 $x\le -4.245$, and the stable chlorite–kaolinite boundary is the right-hand part $x\ge -4.245$. 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.