Past exam of the year 2 of the earth sciences (geology) course of the University of Oxford / 2025 / Paper 3 / 9

At the one-bar standard state, pure mineral endmembers and water vapour treated as an ideal gas all have unit thermodynamic activity. The Gibbs energy of reaction is ${\Delta }_r{G}^{\circ }={\Delta }_r{H}^{\circ }-T{\Delta }_r{S}^{\circ }$. Setting it to zero givesThe quantities are taken per mole of the reaction as written. The simplified result uses the stipulated temperature-independent enthalpy and entropy differences.

Use the pressure correction to a mineral reaction for the solids, with $P_0=1\mathrm{bar}$. The water contribution must instead use its chemical fugacity. For pure solid phases and water vapour treated as an ideal gas, $f_{{\mathrm{H}}_2\mathrm{O}}=P$, givingThe last term already accounts for the gas-pressure contribution; adding another gas-volume correction would double count it. The pressure ratio inside the logarithm is dimensionless.

With pressure expressed in bars, the solid correction is $-2.28(4500-1)=-10257.72\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$. ThusUsing $4500$ rather than $4499$ for the solid pressure increment would silently shift the specified one-bar reference.

The compositions alone do not specify a unique thermodynamic activity model. Adopt the usual ideal, independent mixing on crystallographic sites approximation: three Mg–Fe positions per formula unit, with the other sites fixed at their stated endmember compositions. Then the mineral endmember activities areThe exponent counts the three mixing cations; treating each whole mineral formula as a single mixing site would be a different model. In particular, the magnesian biotite and garnet components here are phlogopite and pyrope.

For the written reaction, the reaction quotient iswith the other solid activities equal to one. The Gibbs energy of reaction balance therefore givesThe smaller product-endmember activity lowers the reaction chemical potential and lowers the equilibrium temperature in this model. Nonideal activity coefficients or unequal occupancies of crystallographic sites would require further information rather than a revised universal numerical answer.^[ref]

The fugacity coefficient gives $f_{{\mathrm{H}}_2\mathrm{O}}=\varphi P=0.82\times 4500=3690\mathrm{bar}$. Keep the solid pressure correction at the actual $4500\mathrm{bar}$; only the water activity uses the fugacity. ThusThe decrease from part (c) follows because $\varphi <1$ lowers the chemical potential of the water product. All logarithms in these thermodynamic formulae are natural logarithms.