Liquid–Liquid Phase Diagrams

Shake hexane with water and you get two clear layers, no matter how hard you mix. Shake phenol with water at 60 °C and, surprisingly, you get one layer — yet at 20 °C the same two liquids split apart again. This temperature-dependent stubbornness is the subject of liquid–liquid phase diagrams. They map out exactly which compositions and temperatures give one phase versus two. Understanding these diagrams matters far beyond a lab curiosity. Solvent extraction, polymer processing, oil recovery, and the formulation of paints and foods all hinge on knowing when two liquids mix and when they refuse to. The same thermodynamic machinery — chemical potentials, the lever rule, the phase rule — that describes solid–liquid eutectics governs this behavior too. This article builds the phenol–water-type diagram from a simple free-energy-of-mixing picture. You will see how a single mathematical condition produces a miscibility gap. You will learn to locate the upper and lower critical solution temperatures. Finally, you will apply the lever rule inside the two-phase region, exactly as you would for a eutectic.

Key takeaways

Partially miscible liquids form two liquid phases over a range of compositions bounded by a miscibility gap (binodal curve).
The upper critical solution temperature (UCST) is the highest temperature at which phase splitting occurs; above it, the liquids mix in all proportions.
Some systems show a lower critical solution temperature (LCST), below which mixing is complete but above which phases separate — driven by entropy loss on heating.
Inside the two-phase region, tie-lines connect the compositions of the two coexisting phases; the lever rule gives their relative amounts.
The Gibbs phase rule ( $F = C - P + 2$ ) confirms that at fixed pressure, two components and two phases leave exactly one degree of freedom — a curve, not a point.
Key idea: A miscibility gap exists wherever the Gibbs energy of mixing curve develops a region of negative curvature, allowing the system to lower its total free energy by splitting into two coexisting compositions.

The core insight

A binary liquid mixture separates into two phases whenever the molar Gibbs energy of mixing, $\Delta_{\text{mix}}G$ , curves downward (concave) over some composition range, because the system can then reach a lower overall free energy by resolving into two coexisting compositions joined by a common tangent line, rather than remaining as one homogeneous blend.

The physical picture

Two liquids mix spontaneously when doing so lowers the Gibbs energy of the system. For an ideal solution, $\Delta_{\text{mix}}G$ is always negative and bowl-shaped (convex everywhere), so ideal liquids mix in all proportions at every temperature. Real liquids, however, often have unfavorable interactions — an A–B contact costs more energy than the average of A–A and B–B contacts. This is captured by a positive interaction parameter, most simply through a regular-solution model:

\Delta_{\text{mix}}G = RT\left[x_A \ln x_A + x_B \ln x_B\right] + \chi RT\, x_A x_B

The logarithmic terms (entropy of mixing) always favor blending and dominate at high $T$ . The $\chi RT x_A x_B$ term (enthalpy of mixing, when $\chi>0$ ) penalizes mixing and dominates at low $T$ . When $\chi$ is large enough relative to $RT$ , the $\Delta_{\text{mix}}G$ curve develops a double-well shape: two local minima separated by a hump. A system prepared at an overall composition sitting under the hump minimizes its free energy not by staying uniform, but by splitting into two phases whose compositions lie at the two minima — connected by a common tangent line, exactly the same construction used for solid–liquid eutectics. As temperature rises, the entropy term becomes more dominant, the hump shrinks, and eventually vanishes at a critical temperature. Plotting the two coexistence compositions against $T$ produces the characteristic dome-shaped miscibility gap: the binodal curve. Everything inside the dome is two-phase; everything outside is a single, fully mixed liquid phase.

Figure

Free energy of mixing at three temperatures, showing the emergence of a double well

The core relations

\Delta_{\text{mix}}G = RT\left[x_A \ln x_A + x_B \ln x_B + \chi\, x_A x_B\right]

where: $\chi$ is a dimensionless interaction parameter measuring the enthalpic penalty of A–B contacts relative to $RT$ .

F = C - P + 2

where: $F$ is the number of degrees of freedom, $C=2$ components, $P$ the number of liquid phases present (phase rule, at variable $T$ and $p$ ).

\frac{n^{\alpha}}{n^{\beta}} = \frac{l^{\beta}}{l^{\alpha}}

where: the lever rule; $l^\alpha$ and $l^\beta$ are the horizontal distances from the overall composition to phases $\alpha$ and $\beta$ on a tie-line.

T_c = \frac{\chi_c \, RT_c}{2R} \quad\Longrightarrow\quad \chi_c = 2

where: within the symmetric regular-solution model, phase separation first appears when $\chi = 2$ , defining the critical temperature $T_c = \chi \cdot \dfrac{w}{2R}$ with interaction energy $w$ .

The core relations continued: locating the binodal

\ln\!\left(\frac{x_B}{x_A}\right) + \chi\,(1-2x_B) = 0

where: this is the condition that the chemical potential of each component is equal in both coexisting phases, solved simultaneously at the two roots $x_B'$ and $x_B''$ .

Derivation ALocating the critical solution temperature from the regular-solution model

Goal. Find the composition and temperature at which the two-phase region first pinches to a single point — the upper critical solution temperature (UCST).

Step 1. Write the molar Gibbs energy of mixing. For a regular solution,

\Delta_{\text{mix}}G = RT\left[x\ln x + (1-x)\ln(1-x)\right] + w\, x(1-x)

where $x \equiv x_B$ and $w$ is the (temperature-independent) interaction energy per mole, related to the earlier dimensionless parameter by $\chi = w/RT$ .

Step 2. Demand a stationary inflection point. Phase separation begins exactly where the curvature of $\Delta_{\text{mix}}G$ first touches zero at a single composition — the point where the double well merges into one. This requires both

\frac{d^2(\Delta_{\text{mix}}G)}{dx^2} = 0 \qquad\text{and}\qquad \frac{d^3(\Delta_{\text{mix}}G)}{dx^3} = 0

simultaneously (a stationary inflection, not just any inflection).

Step 3. Differentiate. The second derivative is

\frac{d^2(\Delta_{\text{mix}}G)}{dx^2} = RT\left[\frac{1}{x} + \frac{1}{1-x}\right] - 2w

and the third derivative is

\frac{d^3(\Delta_{\text{mix}}G)}{dx^3} = RT\left[-\frac{1}{x^2} + \frac{1}{(1-x)^2}\right]

Step 4. Solve the third-derivative condition. Setting it to zero gives $x^2 = (1-x)^2$ , whose physically meaningful root is

x_c = \tfrac{1}{2}

By symmetry of the regular-solution model, the critical composition sits exactly at equimolar mixing.

Step 5. Substitute back to get $T_c$ . At $x=\tfrac12$ , the second-derivative condition becomes

RT_c\left[2 + 2\right] = 2w \quad\Longrightarrow\quad T_c = \frac{w}{2R}

Boxed result.

\boxed{x_c = \tfrac{1}{2}, \qquad T_c = \dfrac{w}{2R}}

Below $T_c$ the system phase-separates over a composition range straddling $x=\tfrac12$ ; above $T_c$ a single homogeneous phase is stable at all compositions. This is the regular-solution prediction of the UCST.

Worked examples

Example 1 — Reading a phenol–water phase diagram with the lever rule. The classic phenol–water system has a UCST near 66 °C at a critical mass fraction of phenol of about 34%. Suppose at 50 °C the two coexisting liquid layers have compositions (from tabulated data close to the real system) of $w_{\text{phenol}}^{\alpha} = 0.08$ (water-rich phase) and $w_{\text{phenol}}^{\beta} = 0.63$ (phenol-rich phase), by mass fraction. A sample is prepared by mixing 40.0 g phenol with 60.0 g water, total mass 100.0 g, giving an overall mass fraction $w_{\text{phenol}} = 0.40$ , and equilibrated at 50 °C. Step 1 — confirm two phases form. Since $0.08 < 0.40 < 0.63$ , the overall composition lies inside the miscibility gap at 50 °C: two liquid layers must be present. Step 2 — apply the lever rule. Treating the tie-line as a mass-fraction axis, the lever arms are

l^{\alpha} = w_{\text{overall}} - w^{\alpha} = 0.40 - 0.08 = 0.32

l^{\beta} = w^{\beta} - w_{\text{overall}} = 0.63 - 0.40 = 0.23

Step 3 — mass ratio. The lever rule states

\frac{m^{\alpha}}{m^{\beta}} = \frac{l^{\beta}}{l^{\alpha}} = \frac{0.23}{0.32} = 0.72

Step 4 — solve for actual masses. With $m^\alpha + m^\beta = 100.0\text{ g}$ and $m^\alpha = 0.72\, m^\beta$ :

0.72\,m^{\beta} + m^{\beta} = 100.0 \text{ g} \implies m^{\beta} = 58.1\text{ g}, \quad m^{\alpha} = 41.9\text{ g}

Sanity check. The overall sample is much richer in water (60%) than phenol, and it lies closer to the phenol-poor phase composition (0.08) than the phenol-rich one (0.63) only moderately — the lever rule correctly assigns somewhat more mass to the phenol-rich layer ( $\beta$ , 58.1 g) because 0.40 sits nearer the midpoint, weighted by how far it is from each boundary. Checking phenol mass balance: $0.08(41.9) + 0.63(58.1) = 3.35 + 36.6 = 40.0$ g phenol. ✓ Matches the 40.0 g phenol charged.

Example 2 — Estimating the interaction energy $w$ and comparing to real phenol–water data. Using the regular-solution result $T_c = w/(2R)$ , estimate $w$ for the phenol–water system, taking $T_c \approx 339$ K (66 °C, the experimental UCST). Step 1 — rearrange for $w$ .

w = 2RT_c = 2 \times 8.314\ \text{J K}^{-1}\text{mol}^{-1} \times 339\ \text{K}

w = 5.64 \times 10^{3}\ \text{J mol}^{-1} = 5.64\ \text{kJ mol}^{-1}

Step 2 — check the order of magnitude. Typical hydrogen-bond enthalpies are 15–25 kJ mol⁻¹; a partial disruption of the water hydrogen-bond network on introducing phenol, worth a few kJ mol⁻¹ per mole of contacts, is physically reasonable — the regular-solution model is a crude mean-field approximation, so agreement to within a factor of a few is expected, not exact quantitative agreement. Step 3 — predict $\chi$ at 50 °C (323 K) and compare with the fitted compositions.

\chi = \frac{w}{RT} = \frac{5.64\times 10^{3}}{8.314 \times 323} = 2.10

Since $\chi > 2$ at 323 K (below $T_c$ ), the model correctly predicts that phase separation persists at 50 °C — consistent with Example 1, where two layers were indeed observed. This self-consistency check confirms the model's qualitative prediction, even though real phenol–water shows an asymmetric (non-regular-solution) binodal, with the critical composition nearer 34 mass % phenol rather than the symmetric 50 mass % predicted by the simplest model.

Limiting cases & common traps

Situation	What happens	Common mistake
$T \gg T_c$	Single phase at all compositions; entropy wins	Assuming immiscibility is permanent, ignoring $T$ -dependence
$T$ just below UCST	Narrow miscibility gap near $x_c$	Forgetting the gap width shrinks continuously to zero at $T_c$ , not abruptly
Nicotine–water, some polymer solutions	Both a UCST and a lower LCST exist (closed-loop diagram)	Assuming all systems only have an upper critical point
Overall composition outside the dome	Fully mixed, single phase — no tie-line applies	Applying the lever rule outside the two-phase region
At the critical point itself	$P=1$ marginally; $F$ jumps as the tie-line length goes to zero	Treating the critical point as an ordinary two-phase point

Common trapThe lower critical solution temperature is not "just the UCST in reverse"

An LCST arises when heating a system causes phase separation — the opposite of ordinary intuition, where heating usually promotes mixing. This happens when strong, orientation-specific interactions (often hydrogen bonds, as in water–triethylamine) lower the enthalpy of mixing so much that mixing is favored at low $T$ , but on heating those bonds break and the entropy gained by disrupting an ordered mixed state (not the entropy of mixing itself) drives phase separation. The regular-solution model with a constant $w>0$ cannot produce an LCST at all — it requires temperature-dependent interactions or an entropic mismatch (as captured in Flory–Huggins-type treatments), so do not expect the simple derivation above to explain closed-loop or LCST-only diagrams.

The chemist behind it

The systematic study of partially miscible liquids and their critical solution temperatures grew out of 19th-century investigations into how solubility depends on temperature, feeding directly into the framework Gibbs had just supplied. Alexander Naumann and later researchers cataloged UCST and LCST behavior across many organic–water pairs in the late 1800s, but it was the marriage of this experimental cataloging with Gibbs's phase rule and van der Waals's theory of critical phenomena that turned liquid–liquid phase diagrams from a curiosity into a quantitative branch of physical chemistry.

Origins · 1 minJosiah Willard Gibbs and how this was discovered

Josiah Willard Gibbs

1839–1903 · American physicist-chemist whose phase rule and Gibbs energy underpin every phase diagram, including the miscibility gaps discussed here

Ask the tutor: How would this miscibility-gap picture change if we added a third component, turning it into a ternary liquid–liquid system?

Why do some liquid pairs never show a miscibility gap at any temperature?

If the interaction parameter $\chi$ (or $w/RT$ ) never exceeds the critical value of 2 predicted by the regular-solution model, the $\Delta_{\text{mix}}G$ curve stays convex everywhere at all accessible temperatures, so no double well ever forms. This happens when A and B have similar polarities and comparable intermolecular forces — for example, benzene and toluene mix in all proportions at essentially all liquid temperatures because their mutual interactions barely differ from their self-interactions.

How is a liquid–liquid phase diagram different from a solid–liquid eutectic diagram?

Both use the same underlying tools: a common-tangent construction on Gibbs energy curves, tie-lines, and the lever rule. The key difference is that eutectic diagrams involve two distinct phases (solid and liquid) with very different entropies and often nearly straight solidus/liquidus boundaries, while liquid–liquid diagrams involve two liquid phases of the same substances, so the miscibility gap is a smooth dome that must close at a critical point where the two phases become identical — a feature solid–liquid systems rarely show except at special congruent points.

Can a single binary system show both an upper and a lower critical solution temperature?

Yes — such systems produce a closed-loop phase diagram, with two-phase behavior confined between an LCST and a UCST, and single-phase behavior both below the LCST and above the UCST. Nicotine–water is the textbook example, phase-separating only between roughly 61 °C and 210 °C at atmospheric pressure. This requires a delicate balance where hydrogen bonding favors mixing at low temperature but a change in effective interactions at higher temperature promotes phase separation, and demixing then reverses again as thermal agitation overwhelms all specific interactions at very high $T$ .

Does pressure affect liquid–liquid miscibility gaps?

Yes, though the effect is usually much weaker than for temperature, because liquids are nearly incompressible. Increasing pressure generally shifts the UCST because it changes the balance between the (small) volume of mixing and the enthalpic and entropic terms; near-critical mixtures and systems with large volume changes on mixing show more pronounced pressure sensitivity. In most everyday laboratory diagrams reported at 1 atm, pressure effects are neglected, and $T$ and composition alone define the diagram, consistent with the phase rule reducing to two variables once pressure is fixed.

Why does the critical composition in a real system rarely sit exactly at x = 0.5?

The symmetric regular-solution model assumes A and B molecules are the same size and that the interaction energy $w$ is composition-independent. Real molecules differ in size and shape (phenol and water, for instance), and $\chi$ often depends on composition itself. These asymmetries skew the binodal curve so the critical point sits off-center — for phenol–water the experimental critical mass fraction is near 34%, not 50%, showing the limits of the simplest mean-field treatment.