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Activity conversion of an equilibrium constant changes the numerical equilibrium constant when a different standard state convention rescales the thermodynamic activities used to define it. The equilibrium composition is unchanged. The essential requirement is to convert the reaction quotient, the equilibrium constant and the standard Gibbs energy of reaction together.^[ref]

At fixed temperature $T$, write a chemical reaction as ${\sum }_i{\nu }_i{B}_i=0$, with positive stoichiometric coefficients for products and negative ones for reactants. Let a standard-state activity conversion give ${\tilde{a}}_i=c_i{a}_i$, where each $c_i$ is positive and independent of composition at the chosen conditions. From the product defining the reaction quotient,At chemical equilibrium, $Q=K$ and $\tilde{Q}=\tilde{K}$, soA scale change for a reactant enters with a negative exponent; multiplying every activity by the same number need not leave $K$ unchanged when the sum of stoichiometric coefficients is nonzero.

The reference chemical potentials must compensate. Since ${\mu }_i={\mu }_i^{\circ }+RT\ln a_i$, the converted standard values are ${\tilde{\mu }}_i^{\circ }={\mu }_i^{\circ }-RT\ln c_i$, where $R$ is the molar gas constant. ConsequentlyThe physical Gibbs energy of reaction and the ratio $Q/K$ are invariant. A change in the numerical equilibrium constant alone therefore does not show that the reaction has become more favorable.

For an original example, consider $2\mathrm{A}\rightleftharpoons \mathrm{B}$, with $K=25$, and choose ${\tilde{a}}_A=10a_A$, ${\tilde{a}}_B=2a_B$. Then $\tilde{K}=25(2/10^2)=0.5$. A composition with $Q=5$ has $\tilde{Q}=0.1$; both conventions give $Q/K=0.2$ and ${\Delta }_{r}G=RT\ln 0.2$. The original standard Gibbs energy of reaction is $-RT\ln 25$, whereas the converted value is $RT\ln 2$. Their opposite signs identify different reference conventions, while the actual reaction driving Gibbs energy at the chosen composition agrees exactly. This calculation assumes a reference rescaling, rather than a composition-dependent alteration of the physical activity model.

Let a chemical reaction have signed stoichiometric coefficients ${\nu }_i$, positive for products. Changes of amount of substance satisfy $d{n}_i={\nu }_{i}d\xi$, with extent of reaction $\xi$. At fixed temperature and pressure, the Gibbs free energy differential isThus the Gibbs energy of reaction is ${\Delta }_{r}G={\sum }_i{\nu }_i{\mu }_i$, using molar chemical potentials when $\xi$ is measured in moles.

An interior chemical equilibrium allows either sign of a small $d\xi$, so it requires ${\Delta }_{r}G=0$. A positive forward change lowers $G$ when ${\Delta }_{r}G<0$; a negative change lowers it when ${\Delta }_{r}G>0$. Stability additionally requires a constrained minimum, rather than stationarity alone.

At a boundary of allowed compositions, only one direction may be feasible. Then the necessary minimum condition is ${\Delta }_{r}Gd\xi \ge 0$ for every allowed small change. The zero-derivative equality must not be imposed on an infeasible reverse change.

For a mineral reaction producing one mole of water, let $H^{\circ },S^{\circ }$ refer to the $P^{\circ }$ standard state and let the solids’ volume change $\Delta V_s$ be constant. If $A_s$ is the product/reactant solid-activity ratio, the Gibbs energy of reaction isFor temperature-independent specified activities and fugacity, equilibrium givesThe gas pressure work is already represented by the fugacity term; adding a second constant water-volume term double counts it. For an ideal gas, $f=P$; a given fugacity coefficient gives $f={\varphi }_{f}P$. If these quantities depend on temperature, the equilibrium equation must instead be solved implicitly. Pressure units in the solid volume correction must match those used for $\Delta V_s$.

For the cell reaction normalized as in 4(b)(i), six moles of electrons are transferred per mole of reaction. ThereforeThis is the Gibbs energy of reaction at the stated composition and gas pressure, not the standard Gibbs energy of reaction. Its negative sign confirms thermodynamic favourability in the written direction under those conditions.

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.

For pure phases at the reference pressure $P_0=1\mathrm{bar}$, the Gibbs energy of reaction is ${\Delta }_{r}G=\Delta H^{\circ }-T\Delta S^{\circ }$. Chemical equilibrium requires ${\Delta }_{r}G=0$, soThe reaction is muscovite dehydration to sanidine, corundum and water; its positive standard entropy of reaction makes the product assemblage favored at high temperature under this constant-property approximation.

At the one-bar standard state, pure phases have unit thermodynamic activities. The Gibbs energy of reaction balance is ${\Delta }_r{H}^{\circ }-T{\Delta }_r{S}^{\circ }=0$, givingThis uses the stipulated temperature-independent reaction enthalpy and entropy differences, taken per mole of the written reaction.

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.

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]

A pairwise equilibrium line restricted to conditions where both minerals are stable against the other allowed phases. Compute the Gibbs energy of reaction for independent conversions to compare them. A pairwise line can continue into a region where a third mineral is favoured; this continuation is a metastable equilibrium relation, rather than a boundary of stable fields.

The Gibbs energy of reaction for specified standard states. It determines the dimensionless equilibrium constant through ${\Delta }_r{G}^{\circ }=-RT\ln K$.

At fixed temperature and pressure, an interior reacting equilibrium minimizes Gibbs free energy under the constraints imposed by stoichiometry. Each independent reversible reaction then has zero actual Gibbs energy of reaction,This is not a requirement that the standard Gibbs energy of reaction vanish.

Using dimensionless thermodynamic activities $a_i$, define the reaction quotient $Q={\prod }_i{a}_i^{{\nu }_i}$. The relationtherefore givesThe equilibrium constant $K$ is dimensionless because each activity is measured relative to its declared standard state. In general $K\ne 1$ even though the actual equilibrium Gibbs energy of reaction is zero. ^[ref]

These equations determine an equilibrium composition within the model. They do not supply the reaction rate or guarantee rapid equilibration; that requires the chemical kinetics.