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Co-distillation is the vaporization and collection of multiple components together during distillation. A component can enter the collected vapor alongside another volatile substance instead of being isolated as a pure vapor. Steam distillation is the case in which water supplies a volatile carrier; co-distillation is not restricted to water.

The result depends on vapor-liquid equilibrium. In an ideal, fully mixed binary liquid at a specified temperature, Raoult's law gives $p_A=x_A{p}_A^{\ast }$ and $p_B=x_B{p}_B^{\ast }$. The vapor fraction is $y_A=p_A/(p_A+p_B)$. The McMaster distillation manual uses liquid/vapor composition diagrams to explain why the vapor can be enriched in one component while still containing both.^[ref]

For original hypothetical data, let $x_A=0.25$, $p_A^{\ast }=80\mathrm{kPa}$ and $p_B^{\ast }=20\mathrm{kPa}$. Then $p_A=20$ and $p_B=15\mathrm{kPa}$, giving $y_A=20/35\approx 0.571$. Both components co-distil, but the vapor is richer in $A$ than the starting liquid. These are equilibrium pressures at the stated temperature; boiling additionally requires their sum to match the external pressure.

With two persistent immiscible liquid phases, a different limiting model uses the pure-component vapor pressures directly, rather than multiplying each by its fraction of a homogeneous mixture. An azeotrope is another special equilibrium situation. Specify the phase behavior before assigning a distillate composition, and use a distillation material balance to distinguish enrichment from total recovery. Collecting a component together with a carrier does not itself establish complete separation from every impurity.

Condensation is conversion of a vapor into a liquid. Cooling vapor-bearing air below its dew point is a familiar example: droplets form on a cold surface as the saturation vapor pressure falls. In a mixture, the relevant comparison uses the condensable component's partial pressure, rather than the mixture's total pressure alone.

Creating a new droplet costs interfacial free energy, while conversion to the stable liquid lowers bulk free energy. A nucleation barrier can therefore permit supersaturated vapor before droplets appear. A pre-existing surface can lower that barrier through heterogeneous nucleation. Condensation also releases latent heat, so a cold plate or condenser must remove heat as liquid accumulates. At vapor-liquid equilibrium, evaporation and condensation continue microscopically with equal average opposing rates.

Distillation separates a mixture by producing vapor from a liquid and collecting the condensed liquid, called the distillate. The more volatile component can have a higher mole fraction in the vapor. Simple distillation collects this condensate directly; fractional distillation adds repeated contact in a fractionating column.^[ref]

Two questions must be kept separate. Vapor-liquid equilibrium describes which compositions can coexist. A distillation material balance accounts for how much of each component leaves in each product. A product can be enriched without containing all the original component, and even a feasible distillation material balance does not establish that an apparatus can make those products.

An equilibrium stage models a contact whose outgoing liquid and vapor satisfy vapor-liquid equilibrium at the stage's temperature and pressure. It is a composition model, not a claim that each piece of packing reaches thermodynamic equilibrium.^[ref]

A fractionating column provides contact between rising vapor and returning liquid during fractional distillation. Packing or trays provide contact area; the resulting separation depends on both vapor-liquid equilibrium and the actual flow and exchange between the thermodynamic phases.^[ref]

For the stipulated A/B ideal solution, put $x=x_A$. At the fixed temperature $300\mathrm{K}$, Raoult's law reduces toThe relative volatility is $\alpha =p_A^{\ast }/p_B^{\ast }=4$. For $0<x<1$,Hence every interior vapor-liquid equilibrium pair in this chosen model has A-enriched vapor; neither pure endpoint is enriched. Conversely,For example, $y_A=1/2$ corresponds to $x_A=1/5$, $p_A=p_B=16\mathrm{kPa}$ and $P=32\mathrm{kPa}$. Equal vapor mole fractions therefore need not mean equal liquid mole fractions.

These are fixed-temperature predictions. During boiling at a specified external pressure, the temperature instead adjusts until ${\sum }_i{x}_i{p}_i^{\ast }(T)=P_{\mathrm{external}}$. The two stipulated vapor pressures specify only one temperature; they provide no boiling-temperature curve.

An ideal solution is a chemical solution whose component chemical potentials depend on composition throughHere $n_i$ is an amount of substance, $x_i$ its mole fraction, $R$ the molar gas constant, and ${\mu }_i^{\ast }$ the pure-component liquid reference at the same temperature $T$ and pressure $P$. This model applies to every component across the composition range; it is stronger than an approximation valid only near a pure solvent.^[ref]

With this pure-liquid standard state convention, the thermodynamic activity satisfies ${\mu }_i={\mu }_i^{\ast }+RT\ln a_i$, so an ideal solution has $a_i=x_i$, or activity coefficient ${\gamma }_i=a_i/x_i=1$.^[ref] This statement concerns the liquid. Assuming that a coexisting vapor is an ideal gas mixture is an additional approximation. Its mole fractions $y_i$ generally differ from the liquid mole fractions $x_i$.

For an original two-component example, stipulate $T=300\mathrm{K}$ and pure-component vapor pressures $p_A^{\ast }=80\mathrm{kPa}$ and $p_B^{\ast }=20\mathrm{kPa}$. A and B are invented distinguishable components; these numbers are chosen model inputs, not material data. Combine an ideal solution with an ideal gas vapor, neglect the pure-liquid reference's pressure correction, and exclude other gases. At liquid mole fractions $x_A=0.4$ and $x_B=0.6$, Raoult's law and addition of partial pressures giveThus the coexisting vapor contains about $72.73\%$ A by amount of substance, although the liquid contains only $40\%$. The total pressure is neither pure-component vapor pressure, and $x_A$ cannot be substituted for $y_A$. This enrichment is the basis of distillation.

The usefulness of the ideal solution is that one composition law connects chemical potentials, vapor-liquid equilibrium, entropy of mixing and Gibbs free energy of mixing. Its liquid components can interact strongly: the defining condition is their composition dependence, rather than the absence of all interactions as in an ideal gas model.

A laboratory condenser removes heat from vapor so that it becomes liquid. In distillation, the downstream laboratory condenser makes collectable distillate. Complete condensation, without loss or reaction, preserves the entering vapor's overall component mole fractions; cooling alone does not give another enrichment step. Partial condensation can instead leave two streams of different composition, requiring both vapor-liquid equilibrium and a distillation material balance.

A concrete motivating experiment is the acetone–chloroform mixture in Truman State University's Liquid-Vapor Equilibrium of a Binary System, Procedure and Figure 1. A two-neck flask feeds a distillation head with a water-cooled laboratory condenser, condensate sampling arm and temperature probe. Starting with one pure liquid, the procedure adds the other component successively and waits for stable temperature with returning condensate. Small paired samples from the flask and condensate are read on an Abbé refractometer; a tabulated refractive index–composition relation gives the two compositions. Controls include three-point thermometer calibration, corrected atmospheric pressure readings, and agreement of the initial pure-liquid and condensate refractive indices with the reference value. Small withdrawals and little retained condensate reduce disturbance of composition. Sample evaporation can bias the readout. The inferred paired mole fractions, temperature and pressure describe vapor-liquid equilibrium, subject to the conversion from mass fraction to mole fraction and to representative sampling. This teaching protocol reports no measured results of a particular student run.^[ref]

How does such a measurement expose activity coefficients? Let $x_i$ and $y_i$ be the mole fractions in the liquid and gas thermodynamic phases, respectively, and let $p_i^{\ast }(T)$ be the pure-component vapor pressure. Assume the gas thermodynamic phase is an ideal gas mixture and neglect the change of the pure-liquid reference chemical potential between its saturation pressure and the experimental pressure. Equality of each component's chemical potential between the two thermodynamic phases then givesThis turns the paired composition readings into estimates of the activity coefficients. The approximation needs refinement when the gas departs appreciably from ideal gas behavior or the neglected liquid pressure effect matters.

The apparatus measures a boiling-composition series in which temperature changes as the mixture changes. Its readings therefore cannot be inserted directly into the fixed-temperature, fixed-pressure activity coefficient constraint. They motivate the more general task: test whether the measured mixture properties can arise from a common Gibbs free energy while accounting for changing state variables. A failure can reveal incompatible property fits or experimental bias; a pass supplies a consistency check, not proof that every reading is correct.

In the ideal solution and ideal gas vapor approximation, each component's partial pressure equals its liquid mole fraction times its pure-component vapor pressure at the same temperature:These pressure formulas neglect the change in the pure-liquid reference between its saturation pressure and $P$; appreciable real gas behavior also requires a more general treatment. The ideal solution definition in terms of chemical potentials does not require either approximation.^[ref]

A short derivation explains the reference choice. At vapor-liquid equilibrium, each component has the same chemical potential in the liquid and vapor. For an ideal gas mixture, chemical potential of an ideal gas giveswhere $p^{\circ }$ is a reference pressure. Approximate ${\mu }_i^{\ast }(T,P)$ by ${\mu }_i^{\ast }(T,p_i^{\ast })$ and subtract the pure-component equation from the mixture equation:Taking the exponential function gives Raoult's law. Summing partial pressures and normalizing gives $P$ and $y_i$; this last step uses the ideal gas vapor assumption separately from the liquid model.

The law is named after François-Marie Raoult, who measured vapor pressures and freezing behavior of chemical solutions.^[ref]

For a two-component mixture, let $x$ be component A's mole fraction in the liquid and $y$ its mole fraction in the coexisting vapor. At interior compositions, define the relative volatility of A with respect to B by the ratio of composition odds:Thus $\alpha >1$ means A is enriched in the vapor, while $\alpha =1$ means $y=x$. Solving this definition givesHere a constant relative volatility is a stipulated model of vapor-liquid equilibrium, not a measured property of a named chemical solution. Real relative volatility can vary with composition, temperature, and pressure. The relation specifies no absolute temperature scale: even fixing $\alpha =2$ does not determine a boiling-temperature curve or a heat requirement.

Stipulate a constant relative volatility $\alpha =2$. The vapor-liquid equilibrium model then becomesAn initial liquid mole fraction $x_0=3/10$ gives an equilibrium vapor mole fraction $y_0=6/13\approx 0.461538$. Now imagine completely condensing a small sample of that vapor and using this condensate as a fresh liquid for another equilibration. Repeating this formal operation givesThese are original model calculations of ideal-stage enrichment. They are not measured compositions, a reflux design, or successive compositions of one pot undergoing simple distillation. That pot loses the more volatile component and becomes depleted; each step here instead starts from the previous enriched condensate. The displayed sequence specifies compositions, not the recoverable amount at each step.

The algebra makes the enrichment especially transparent. With composition odds $r_n=x_n/(1-x_n)$, every formal step doubles the odds, soFor every finite $n$, $x_n<1$, although $x_n\to 1$ as $n\to \infty$. This explains why repeated ideal equilibration can improve composition without making finite-step perfect separation a consequence of the model.

In simple distillation, pot vapor passes directly to the laboratory condenser. An initial small condensate sample can approximate vapor-liquid equilibrium with the current pot liquid. Removing enriched distillate depletes the pot's more volatile component. A collected fraction averages the changing condensate composition, rather than preserving the first-drop mole fraction.^[ref]