Activity equality for coexisting liquids ($a_i^{\alpha }=a_i^{\beta }$)

For exchanging neutral components at a common temperature and pressure, activity equality for coexisting liquids expresses liquid-liquid equilibrium as $a_i^{\alpha }=a_i^{\beta }$, provided both thermodynamic activities use the same standard state. The basic equilibrium condition is equality of chemical potentials; equality of numerical activities follows only after their references have been matched.

Write the definition of thermodynamic activity aswhere $R$ is the molar gas constant and $T$ is temperature. Subtraction at equilibrium givesThe ratio is one when the two standard chemical potentials are the same. IUPAC defines activity relative to the specified standard chemical potential and distinguishes pure-phase and solute reference conventions.^[ref] Mixing conventions can therefore produce unequal activities for physically coexisting phases. For instance, if one reference exceeds the other by $RT\ln 2$, its corresponding activity is half as large at the same chemical potential. Standard-state activity conversion changes the numerical coordinate, not the physical driving force.

Equal thermodynamic activities also need not mean equal mole fractions. With a common pure-liquid reference, $a_i=x_i{\gamma }_i$, where ${\gamma }_i$ is an activity coefficient. An original symmetric regular solution example makes this quantitative. Set the dimensionless interaction parameter to $\chi =2\ln 3$. The model has coexisting component-A mole fractions $x_A^{\alpha }=1/4$ and $x_A^{\beta }=3/4$, withThus ${\gamma }_A^{\alpha }=3^{9/8}\approx 3.44161$ and ${\gamma }_A^{\beta }=3^{1/8}\approx 1.14720$, whileComponent B satisfies the analogous equality by symmetry. The different compositions are compensated by different nonideal environments. These are chosen model values, not experimental compositions. As in liquid-liquid equilibrium, satisfying equalities must be supplemented by stability against competing phase arrangements.

Standard-state activity conversion changes the numerical thermodynamic activity assigned to a fixed physical state when its reference chemical potential changes. The observable equilibrium and physical chemical potential stay the same. The activity and its standard chemical potential must therefore be converted together.

At fixed temperature $T$, writeIf ${\delta }_i={\tilde{\mu }}_i^{\circ }-{\mu }_i^{\circ }$, subtraction givesHere $R$ is the molar gas constant. This follows from the IUPAC definition of relative activity and its specified standard state.^[ref] Changing just the activity in the first equation would instead change the calculated physical chemical potential.

For an invented conversion ${\tilde{a}}_i=10a_i$, the compensating reference shift is ${\delta }_i=-RT\ln 10$. An old activity $0.03$ becomes $0.30$; the two expressions for ${\mu }_i$ are identical. An activity of one consequently identifies a reference condition under the chosen convention, not an invariant amount of material.

An activity conversion of an equilibrium constant follows the same logic. For the reaction A $\rightleftharpoons$ B, take $K=a_B/a_A$ at equilibrium and convert by ${\tilde{a}}_A=10a_A$, ${\tilde{a}}_B=2a_B$. Then $\tilde{K}=K/5$. If $K=5$, the converted value is $1$, whilebecause originally ${\Delta }_r{G}^{\circ }=-RT\ln 5$. For any physical composition, the corresponding reaction quotient also changes by $1/5$, so ${\Delta }_r{G}^{\circ }+RT\ln Q$ remains unchanged. These are reference changes, not altered reaction chemistry.

The same requirement governs activity equality for coexisting liquids: compare activities only after identifying their standard states, composition scales and reference conventions. Equal numerical values expressed under incompatible conventions do not, by themselves, establish liquid-liquid equilibrium.

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.