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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.

At fixed temperature and pressure, liquid-liquid equilibrium is stable coexistence of distinct liquid thermodynamic phases that minimizes total Gibbs free energy for the specified component amounts. Their compositions can differ even though each exchanging chemical species has the same chemical potential in both phases. For neutral, nonreacting components in bulk phases at common temperature $T$ and pressure $p$, neglecting external-field and interface-curvature contributions,To see why, transfer an amount $d{n}_i$ from phase $\alpha$ to phase $\beta$ while holding the total amount of each component fixed. The change in Gibbs free energy is $dG=({\mu }_i^{\beta }-{\mu }_i^{\alpha })d{n}_i$. If the coefficient is nonzero, one transfer direction lowers $G$. Equality removes that driving term; stable equilibrium additionally requires a global minimum over the allowed phase arrangements. Here each chemical potential is the partial derivative of Gibbs free energy with respect to that component amount at fixed temperature, pressure and the other component amounts.^[ref]

Charged species instead require equality of the corresponding electrochemical potential. Equal chemical potentials for neutral species do not require equal mole fractions. With a common pure-component reference, ${\mu }_i={\mu }_i^{\circ }+RT\ln a_i$ and $a_i=x_i{\gamma }_i$, where $a_i$ is thermodynamic activity, ${\gamma }_i$ an activity coefficient, and $R$ the molar gas constant. The activity equality for coexisting liquids is $x_i^{\alpha }{\gamma }_i^{\alpha }=x_i^{\beta }{\gamma }_i^{\beta }$. Different nonideal environments can therefore support different compositions at the same chemical potential.

For a binary mixture, let $x$ be the mole fraction of component B and let $g(x)$ be the homogeneous molar Gibbs free energy at fixed $T,p$. Differentiating $G=(n_A+n_B)g(n_B/(n_A+n_B))$ givesEquality of both component chemical potentials between two compositions becomes the common-tangent construction for phase coexistence:In this molar representation, the tangent slope is ${\mu }_B-{\mu }_A$ and its intercept is ${\mu }_A$. Stable coexistence uses a supporting tangent lying below the homogeneous curve over all admissible compositions. This constructs the relevant convex envelope; two stationary points or equal slopes alone do not establish stability.

An original regular solution calculation makes this distinction visible. Because linear composition bias leaves coexistence compositions unchanged, subtract the linear pure-component reference and writeThe dimensionless parameter $\chi$ penalizes unlike contacts when it is positive. Take $\chi =3$. Symmetry makes the coexistence tangent horizontal, and the two outer solutions of $h^{\prime }(x)=\ln (x/(1-x))+3(1-2x)=0$ give the symmetric regular-solution binodal:These are model compositions, rather than measured hexane–water values. At overall mole fraction $z=0.30$, the liquid-liquid tie line and lever rule give the fraction $\lambda$ of total moles in phase $\beta$:Changing $z$ within the coexistence interval changes phase amounts while leaving the two endpoint compositions fixed. These mole-based phase fractions need not equal volume fractions.

The spinodal answers a different question: local stability of a homogeneous mixture. Here $h^{\prime \prime }(x)=1/[x(1-x)]-2\chi$, so at $\chi =3$ the two local-stability limits are $(1\mp \sqrt{1/3})/2\approx 0.211325,0.788675$. Between them, composition fluctuations lower the bulk free energy. Between a binodal endpoint and the nearer spinodal, the homogeneous state has positive local curvature but a lower-energy phase-separated state exists: this is metastability. Thus the observed coexistence compositions are not obtained merely by setting $h^{\prime \prime }=0$.

A useful experiment measures compositions as well as the presence of layers. One hexane–water record reports overnight shaking, microsyringe sampling of the aqueous solution, and gas chromatography to measure dissolved hydrocarbon. The small dissolved amount is evidence of finite mutual solubility, consistent with liquid immiscibility over other bulk compositions.^[ref] At controlled temperature, repeated sampling of both phases and checks from different initial compositions help test whether the endpoint compositions have stabilized. Critically evaluated reference systems provide a way to validate such instruments and techniques.^[ref]

In a regular solution, ideal combinatorial entropy competes with a nonzero unlike-contact energy. For equal molecular volumes and mole fraction $\varphi$ of species A, write mixing free energy per site aswhereHere $z$ is the coordination number and ${ϵ}_{ij}$ a pair-contact energy counted consistently in the lattice model. A positive $\chi$ penalizes unlike contacts; a negative value favors them. The logarithmic terms lower free energy on mixing. The sign of the enthalpy of mixing alone therefore does not determine miscibility.

At fixed temperature and composition, a homogeneous phase is globally stable only if no division into two compositions lowers its free energy. The common-tangent construction selects coexisting compositions ${\varphi }_a,{\varphi }_b$, equivalent to equality of both species' chemical potentials. Local stability to infinitesimal fluctuations instead requiresZero curvature gives the spinodal, while the coexistence boundary is the binodal. Between them a homogeneous state can be metastable and require nucleation; inside the spinodal its small composition fluctuations grow. The symmetric regular solution hasBelow this contact penalty, the free energy is convex at every composition. Above it a miscibility gap is possible around the central composition. If $\chi =A/T$ with $A>0$, heating promotes mixing, but specific interactions or temperature-dependent noncombinatorial terms can produce other trends.

For two polymers, let $N_A,N_B$ count equal-volume segments per chain, and now let $\varphi$ be the segment volume fraction of A. Connecting segments into chains reduces their independent translational entropy. The binary Flory-Huggins polymer blend hasThe energetic contact term remains extensive in the number of segments, while the favorable entropy per segment is reduced by $1/N_A$ and $1/N_B$. This is the key long-chain limit of polymer-blend miscibility: a small positive incompatibility can overcome the mixing entropy. The lattice treatment is a controlled model for this argument, rather than a claim that every real blend has composition-independent $\chi$.^[ref]

The local stability condition and critical interaction parameter becomeFor equal chain lengths, ${\chi }_c=2/N$. Similar segment chemistry and matched cohesive interactions can reduce the unlike-contact penalty; favorable specific interactions, such as complementary hydrogen bonding, can make mixing favorable even for long chains. Lower molar mass increases mixing entropy. temperature can help when it lowers the effective $\chi$, but heating is not universally sufficient. A suitable common solvent can facilitate processing or ternary mixing, and a block-copolymer compatibilizer can stabilize fine domains; neither guarantees that the final solvent-free binary material is one equilibrium molecular phase.

An oil–water emulsion illustrates emulsion stability versus molecular miscibility. A surfactant adsorbs with its water-affine head toward water and its oil-affine tail toward oil, lowering interfacial tension and the work $\gamma dA$ needed to make additional interface. Breaking a fixed oil volume $V_o$ into spherical droplets of radius $R$ gives total area $3V_o/R$, so creating small droplets still requires substantial area and usually mechanical work.

Adsorbed surfactants can oppose droplet coalescence through charged double-layer repulsion, hydrated-head interactions or steric repulsion from sufficiently extended groups. Adequate coverage, adsorption time, ionic conditions and film drainage affect the lifetime. A lower interfacial tension does not itself remove the bulk miscibility gap, and a protected emulsion is commonly kinetically stabilized. A thermodynamically stable microemulsion is a distinct situation requiring an appropriate composition and free-energy balance. The two liquids must not be called molecularly miscible simply because droplets disperse visibly.

For the capillary, balance upward contact-line force against hydrostatic column weight, neglecting meniscus volume and gas density. For inner radius $r$,The contact angle is measured through the liquid. complete wetting means ${\theta }_2=0$, so the contact-angle change of capillary rise givesThis follows from the stipulated wetting change, holding radius, bulk liquid surface tension and density fixed. One should not replace that premise with a general claim that grease always improves glass wetting; if contamination also changes the liquid surface tension, the original height and new angle alone would not determine the new height.

For two wide vertical plates separated by gap $d$, use unit width. There are two upward contact-line contributions and liquid cross-sectional area $d$, giving the capillary rise between parallel plates:Only one principal meniscus curvature contributes between wide plates, whereas a circular tube has two. For the same fluid and angle,If the plate gap equals the tube diameter, $d=2r$, the rise is half as large: the initial and completely wetting heights would be $2.50$ cm and $3.05$ cm. If instead $d=r$, the heights are equal. In either geometry, changing only the angle from $35^{\circ }$ to zero multiplies the height by $1/\cos 35^{\circ }$. Since the question specifies no plate spacing, the general gap-dependent relation is the unambiguous result; the half-height answer requires the equal-gap-to-diameter convention.

For a symmetric binary regular solution, define the dimensionless molar mixing Gibbs free energywhere all logarithms are natural, $x$ is one component's mole fraction and $\chi$ the dimensionless interaction parameter. The symmetric regular-solution binodal gives the compositions of two coexisting phases, rather than the overall composition of their mixture. The common-tangent construction supplies the equilibrium condition.^[ref]

Symmetry gives $g(x)=g(1-x)$. For distinct coexistence endpoints $x_{\alpha }<1/2$ and $x_{\beta }=1-x_{\alpha }$, their joining line is horizontal. Tangency at both ends therefore requiresThis derives the displayed binodal equation. The ever-present stationary solution $x=1/2$ is not a distinct two-phase pair.

To see when the nontrivial pair exists, differentiate:For $\chi \le 2$, this is nonnegative throughout the interval, so the mixing free energy is convex. For $\chi >2$, let $f=g^{\prime }$. On $(0,1/2)$, $f^{\prime }$ decreases from infinity to $4-2\chi <0$. Consequently $f$ first increases and then decreases. It starts at negative infinity, is positive just below1/2, and ends at0. There is exactly one zero in the open interval before its maximum. That zero has positive curvature and gives the lower coexistence composition; symmetry gives the other.

For an original numerical example, $\chi =3$ yieldsThe spinodal endpoints instead solve $g^{\prime \prime }=0$:which give approximately 0.211325 and 0.788675 at $\chi =3$. The binodal marks equilibrium phase compositions; the spinodal marks loss of local curvature stability. Their intervening metastable ranges are therefore not interchangeable boundaries. At $\chi =2$ the endpoints meet at the symmetric critical composition 1/2.