PHASE EQUILIBRIUM · STORY 01

About 12 minutesUndergraduate physical chemistryInteractive Alpha

The unreachable
100%

A column can make ethanol richer and richer.
Why does even a taller column lose its direction at the last trace of water?

Scroll down and commit to a prediction

HOW TO READ

The stage at right is not a set of three unrelated illustrations. One thermodynamic state drives them all: the macroscopic view shows what happens, the particulate view helps build an explanation, and the symbolic view supplies a map you can calculate with.

Current state: ethanol mole fraction 0.100 in the liquid and 0.416 in the vapor; bubble point 87.91 degrees Celsius; equilibrium stage 0.

Synchronized state
Liquid x
0.100
Vapor y
0.416
Bubble point
87.91 °C
Still liquid x = 0.100Distillate y = 0.416Equilibrium stage 0
MACROSCOPIC Equilibrium-stage schematic; it does not represent actual tray efficiency
Vapor y = 0.416
Liquid x = 0.100
PARTICULATE Particle proportions are quantitative; positions, collisions, and paths are explanatory only
778389951020.000.250.500.751.00xyEthanol mole fraction · x (liquid) / y (vapor)Temperature / °CVaporLiquidLiquid + vapor
Solid model curve · liquid x (bubble curve) Dashed model curve · vapor y (dew curve) 1.013 bar
MODEL Three-suffix Margules teaching model (λ=1.00), not a fit to the experimental data

00 · THE HOOK

One more stage—will it reach 100%?

Distill once and the ethanol gets richer. Distill again and it gets richer again. Intuition says that repeating the same step often enough must end at pure ethanol.

Do not accept the answer yet. Add stages to this “theoretical-stage column” — every stage truly reaches vapor–liquid equilibrium — and watch where the top composition chooses to stop.

PREDICT FIRST
Stage 0: x = 0.100

01 · A COMMON LANGUAGE

What exactly does “10% alcohol” mean?

Volume fraction, mass fraction, and mole fraction may all be called “concentration,” but they are not the same scale. The horizontal axis of this phase diagram is the ethanol mole fraction.

Orange denotes ethanol and blue-green denotes water. Both branches of the diagram record ethanol composition, so solid versus dashed lines and circles versus triangles distinguish liquid x from vapor y.

zE=nEnE+nWz_E=\frac{n_E}{n_E+n_W}

02 · THE FIRST BUBBLE

Boiling begins when the sum catches the external pressure.

Water and ethanol both contribute vapor pressure at all times. The first bubble can survive inside the liquid only when their partial pressures add up to the external pressure.

More volatile means that ethanol is enriched in the vapor; it does not mean that water stops evaporating. We begin with an ideal solution because it is calculable. Scene 4 will test it against experimental data.

xEpE∗+(1−xE)pW∗=Px_Ep_E^*+(1-x_E)p_W^*=P
THE FIRST BUBBLE WILL BE

03 · ONE TIE LINE

At one temperature, why are there two compositions?

Call the liquid composition in the still x and the equilibrium vapor composition y. They share the same temperature and pressure, yet their compositions can differ.

Condense that vapor completely. This changes the phase but not the overall composition of the sample, so the new liquid droplet has exactly the former vapor composition y.

yE=xEpE∗Py_E=\frac{x_Ep_E^*}{P}

04 · BUILD THE MAP

Turn an experimental table back into a map.

Lai and co-workers reported 16 ethanol–water x–T–y equilibrium measurements at 101.3 kPa. Each row gives a liquid composition, an equilibrium temperature, and a vapor composition. These points were measured; they were not drawn by an equation.

Choose rows from the table. Place the circle (x,T) on the liquid, or bubble-point, branch and the triangle (y,T) on the vapor, or dew-point, branch. Then use dashed lines to sketch your proposed phase boundaries.

The circle and triangle from one row are joined by a horizontal tie line. Its ends give the two phase compositions; where the overall composition z lies between them determines how much of each phase is present.

β=zE−xEyE−xE\beta=\frac{z_E-x_E}{y_E-x_E}
Pick experimental compositions directly
Sampling coverage low compositionmiddle compositionhigh composition

You currently cover 0 / 3 composition regions. Do not choose only neighbouring measurements.

0 / 16 samples selected

Pick different compositions directly above. The circle and triangle are liquid x and vapor y from the same equilibrium state. Select at least 5 samples across the composition axis before comparing your sketch with the complete table of 16 measurements.

05 · EQUILIBRIUM-STAGE CASCADE

The ideal model builds a composition ladder.

Keep the experimental points as a reference and ask what ideal Raoult’s law predicts. Bring a liquid to equilibrium, remove its vapor, and condense it completely; then repeat with the new liquid.

In the ideal model, the ladder moves to the right. Each equilibrium stage brings a smaller gain, but the direction always points toward pure ethanol.

xn+1=y(xn)x_{n+1}=y(x_n)

06 · WHERE IDEALITY FAILS

Mixing changes how strongly molecules “want to escape.”

Open circles and triangles are the Lai 2014 measurements; smooth solid and dashed curves are model results. λ=0 is the ideal Raoult-law baseline. λ=1 is a Margules teaching model calibrated to the atmospheric azeotrope.

A model does not replace experiment. Its value is to explain trends with a few parameters, and its misses matter too. Values 0<λ<1 are a teaching morph—not a measurable property and not a point-by-point regression.

pi=xiγipi∗p_i=x_i\gamma_i p_i^*

07 · THE FIXED POINT

The wall is not a line. It is the moment y = x.

When the equilibrium vapor and liquid have exactly the same composition, vaporization followed by condensation no longer changes composition. This azeotropic composition is a fixed point of the distillation map.

From the low-composition side, y−x is positive; from the high-composition side, y−x is negative. Cascades from both sides approach the same point and do not cross it. Experiment and the teaching model give nearby values, but not the same result.

Δ(x)=y(x)−x=0\Delta(x)=y(x)-x=0

08 · CHANGE THE OPERATION

Do not push the same step harder. Choose a different step.

The azeotrope limits repeated vapor–liquid equilibrium separation under fixed conditions. It does not forbid anhydrous ethanol from existing.

Selective dehydration removes water directly and changes the material-balance path. It no longer follows the distillation ladder x→y(x), so it can cross the former fixed point.

xE(q)=nEnE+(1−q)nWx_E(q)=\frac{n_E}{n_E+(1-q)n_W}

CHANGE THE OPERATION

Remove the water
instead of distilling once more.

Starting at the azeotropic composition, a 3A molecular sieve or another dehydration process selectively removes water. It changes the process constraint; it does not make vapor–liquid equilibrium “stop applying.”

Material-balance schematic for the selective removal of water by a molecular sieve
Azeotropic distillate 89.1 mol%
3A
molecular sieve
After dehydration 89.10 mol%

CONCEPT CHECK · PREDICT, THEN TEST

Why does molecular-sieve dehydration not contradict the statement that ordinary distillation cannot cross the azeotrope?

FREE PLAY · LEAVE THE STORY WITH A QUESTION

Now the controls are yours.

The author-controlled camera ends here. Change the initial composition, nonideality strength, and number of theoretical stages. Watch which quantities move together and which relationships remain invariant.

TRY THIS

Start at x₀ = 0.95 and repeat the distillation step. Does the ladder move left or right? What composition does it approach?

Current state: ethanol mole fraction 0.777 in the liquid and 0.806 in the vapor; bubble point 78.33 degrees Celsius; equilibrium stage 4.

Synchronized state
Liquid x
0.777
Vapor y
0.806
Bubble point
78.33 °C
Experimental azeotrope
≈95.4 wt%
Still liquid x = 0.777Distillate y = 0.806Equilibrium stage 4
MACROSCOPIC Equilibrium-stage schematic; it does not represent actual tray efficiency
Vapor y = 0.806
Liquid x = 0.777
PARTICULATE Particle proportions are quantitative; positions, collisions, and paths are explanatory only
778389951020.000.250.500.751.00xyModel fixed pointLai experimental azeotropeEthanol mole fraction · x (liquid) / y (vapor)Temperature / °CVaporLiquidLiquid + vapor
Solid model curve · liquid x (bubble curve) Dashed model curve · vapor y (dew curve) ○ x / △ y · experimental data 1.013 bar
MODEL Comparison · open symbols are Lai 2014 measurements; smooth curves are the Margules teaching model at λ=1.00

MODEL CARD · V0.1

What this diagram knows—
and what it does not.

Evidence layer: 16 public measurements

The open circles and triangles are x–T–y data reported by Lai and co-workers in 2014 at 101.3 kPa. Temperature and vapor-composition uncertainties are the 95% expanded uncertainties published through NIST ThermoML. Lines between points are visual guides only; they do not create new measurements.

Model layer: interpretable, but not itself a measurement

The vapor phase is treated as an ideal gas, pure-component vapor pressures use Antoine equations, and the liquid phase uses a three-suffix Margules activity-coefficient model. Parameters are calibrated to the atmospheric azeotrope at xE≈0.895 and T≈78.15°C (the model gives 95.61 wt% ethanol). Against the 16 Lai points, the current model has a temperature RMS error of about 0.57 K and a vapor-composition RMS error of about 0.016. Experiments in the literature give about 95.4–95.6 wt%, so the story says “about 95.5 wt%.”

What you cannot infer from it

This is not a full temperature-dependent NRTL regression of the experimental data. It does not simulate tray efficiency, reflux ratio, mass-transfer rates, or batch-distillation time. Do not use these teaching parameters for pressure extrapolation or industrial design.

Evidence boundary for the particulate animation

Ethanol/water proportions in each phase come from the equilibrium calculation. Dot positions, collisions, bubble shapes, and animation timing are explanatory metaphors—not real-time molecular dynamics.

You did more than “read a phase diagram.”

You rebuilt a map from an experimental table,
then used a model to explain its wall.