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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.

The standard Gibbs energy of reaction givesThe positive standard Gibbs energy of reaction means that dissociation is unfavourable when every thermodynamic activity equals one; it does not forbid partial dissociation. The corresponding $\mathrm{p}{K}_a=-{\log }_{10}{K}_a$ is about $7.21$.

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 the nepheline–corundum reaction from the albite–corundum reaction and divide by two. In the albite-to-nepheline direction the resulting net reaction isAdditivity of the standard Gibbs energy of reaction makes the logarithms of the equilibrium constants combine with the same coefficients:This normalization is per one albite formula unit. The doubled reaction has ${\log }_{10}K=-22.06$, and reversing the reaction changes its sign.

The signed amount of species $i$ in a written chemical reaction, positive for products and negative for reactants. Multiplying all coefficients by a factor multiplies the standard Gibbs energy of reaction by that factor and raises the equilibrium constant to that power.

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.