Past exam of the year 2 of the earth sciences (geology) course of the University of Oxford / 2023 / Paper 1 / 8 / c / 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,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.