Chapter 2b

Deviations from Ideality

Raoult's law describes molecules that do not care what they are next to. Real molecules care a great deal — and the consequences run from a mild correction all the way to a separation that is flatly impossible.

Wankat, Ch. 2 2 interactive apps Follows Chapter 2a

By the end of this chapter you should be able to

Explain, at the molecular level, why a mixture shows a positive or a negative deviation from Raoult's law — and predict which, given the two components.
Define the activity coefficient and state what and mean physically.
Use the Margules and van Laar models and know what each assumes.
Explain how an azeotrope arises from deviation, and test whether a given system has one.
Recognise when deviation is severe enough to split the liquid into two phases.
Say what all of this does to a distillation design.

Where Raoult's law breaks

Chapter 2a ended with a construction that works beautifully2a App 3: two Antoine curves, one pressure, and the whole diagram falls out. It rests on — the assumption that a molecule's escaping tendency depends only on how many of its neighbours are of its own kind, not on what the other kind is.

That assumption is excellent for benzene and toluene, which are near enough the same molecule. It is very bad for ethanol and water. And the failure is not a small numerical correction — it can invert the entire behaviour of the system.

The stakes

An ideal binary can always be separated by distillation, given enough stages. A sufficiently non-ideal binary cannot be separated by ordinary distillation at all, at any number of stages, because the equilibrium curve crosses the diagonal. Ethanol and water is the everyday example: no simple column will take you past 89.4 mol % ethanol. That single fact is why this chapter exists.

2b.1 Why real solutions deviate

Consider what has to happen for a molecule of A to escape from a liquid mixture of A and B. It must break its attractions to its neighbours. In pure A, those neighbours are all A. In the mixture, some are B. So the question is simply: are A–B attractions stronger or weaker than the average of A–A and B–B?

A–B weaker than A–A and B–B

The unlike molecules dislike each other. Each is less firmly held than it would be among its own kind, so both escape more readily than Raoult predicts.

— positive deviation
Total pressure lies above the Raoult straight line
Mixing is endothermic; the mixture may even get colder
Extreme case: a minimum-boiling azeotrope, then liquid–liquid splitting
Example: ethanol + hexane. Ethanol hydrogen-bonds to itself; hexane cannot, so adding hexane breaks up the hydrogen-bond network and frees ethanol to escape.

A–B stronger than A–A and B–B

The unlike molecules attract each other specifically. Each is held more firmly in the mixture than among its own kind, so both escape less readily than Raoult predicts.

— negative deviation
Total pressure lies below the Raoult straight line
Mixing is exothermic; the mixture warms up
Extreme case: a maximum-boiling azeotrope
Example: chloroform + acetone. The chloroform hydrogen donates to the acetone carbonyl — a hydrogen bond that exists in the mixture and in neither pure liquid.

A test you can apply without any data

Ask what new interaction the mixture creates or destroys. Does mixing create a hydrogen bond that did not exist before? → negative deviation. Does mixing break up a hydrogen-bonded network? → positive deviation. Neither — both molecules are non-polar and similar in size? → nearly ideal. This heuristic gets the sign right for most of the pairs you will meet, and getting the sign right is most of the battle.

App 1

Predict the deviation

Pick a pair, decide from the molecules alone which way it will deviate, then check. The interaction bars show the reasoning; the sketch shows what it does to the pressure–composition diagram.

System

Your prediction

Which interactions matter

Relative strength of like and unlike neighbour interactions.

Like–like (A–A, B–B)Unlike (A–B)

What it does to

Raoult's straight line, and where the real total pressure actually sits.

Raoult (ideal)Real

?Make a prediction.

γ₁ at infinite dilution

—

γ₂ at infinite dilution

—

Score

0 / 0

2b.2 The activity coefficient

Return to the general statement of vapour–liquid equilibrium2a §2.3. Keep the ideal-gas assumption — reasonable at the pressures in this course — but drop the ideal-liquid one:

(2.18)

and by Dalton's law of partial pressures the total is

(2.19)

Every equation you met in Chapter 2a survives with a inserted. The distribution coefficient becomes

(2.20)

Read equation (2.20) carefully — it is the whole chapter

In an ideal system2a §2.7 is a ratio of vapour pressures, and it can never equal 1 unless the two components boil at the same temperature. With activity coefficients in it, has a second, composition-dependent factor that can pull it towards 1 — and through it. When at some composition, that composition is an azeotrope. Non-ideality does not merely perturb the diagram; it can change its topology.

This is the premise of Chapter 11 §1.2 in its purest failure mode. A separation needs a property difference; at the azeotrope the property difference we are exploiting has gone to zero, and no amount of equipment will recover it. The escape is to change the property you exploit4b §4b.6 — which is why ethanol is finished on a molecular sieve rather than in a taller column. On a McCabe–Thiele diagram4a §4.6 this is the moment the equilibrium curve meets the diagonal and the staircase has nowhere left to step.

γ_i > 1: Positive deviation. Component is less comfortable in the mixture than in its own pure liquid, so it escapes more readily and exerts a higher partial pressure than Raoult predicts.
γ_i = 1: Ideal — Raoult's law exactly. Always true in the limit , where a molecule's neighbours are all its own kind.
γ_i < 1: Negative deviation. Component is more comfortable in the mixture, so it is held back and exerts a lower partial pressure than Raoult predicts.

Two limits are worth committing to memory. As , : a nearly pure component always obeys Raoult's law. As , , the infinite-dilution activity coefficient — the single most useful number for characterising a non-ideal pair, because it measures the deviation where it is largest.

2b.3 Positive deviations

Figure 2.11 — Ethanol–benzene at 67.77 °C. The total pressure bulges above the Raoult straight line, and both activity coefficients exceed 1 — each rising sharply towards its infinite-dilution value at its own dilute end.

Figure 2.12 — The same system. Top: at 67.77 °C. Bottom: at 1 atm. Note the curve crossing the diagonal, and the corresponding minimum in the boiling temperature.

Look at the diagram, and compare it with the well-behaved lens you built in the construction app2a App 3. Both pure components boil around 78–80 °C, yet the mixture boils at 67 °C — lower than either pure component. That is not a mistake. Positive deviation means both components escape more easily from the mixture than from their own pure liquids, so the mixture develops a higher vapour pressure and therefore a lower boiling point. Push that far enough and you get a minimum-boiling azeotrope.

2b.4 Negative deviations

Figure 2.13 — Chloroform–acetone at 64 °C. The total pressure sags below the Raoult line and both activity coefficients are less than 1, because the mixture forms a hydrogen bond that neither pure liquid has.

Figure 2.14 — The same system. The curve crosses the diagonal from the other side, and the diagram shows a maximum in the boiling temperature.

The mirror image. Chloroform boils at 61.7 °C and acetone at 56.1 °C, yet their mixture boils at 64.5 °C — higher than either. The specific C–H···O=C hydrogen bond between the two holds both molecules in the liquid, lowering the vapour pressure and raising the boiling point. A maximum-boiling azeotrope.

Which end does the azeotrope come out of?

At a minimum-boiling azeotrope the azeotropic mixture is the most volatile thing in the system, so it leaves at the top of a column and the pure components can only be recovered at the bottom. At a maximum-boiling azeotrope it is the least volatile, so it accumulates at the bottom. Minimum-boiling azeotropes are much the more common of the two.

2b.5 Models for the activity coefficient

is a function of composition (and, more weakly, of temperature). Rather than tabulate it, we correlate it with a model containing a small number of adjustable constants fitted to experimental data. Two two-parameter models are worth knowing; your slides also name Wilson, and NRTL and UNIQUAC are the industrial standards.

Two-constant Margules

(2.21)

(2.22)

van Laar

(2.23)

Both models are written so that the constants have the same meaning: setting in either gives , and gives . So the two constants are just the two infinite-dilution activity coefficients, and the models differ only in how they interpolate between them. Setting recovers Raoult's law in both.

These are correlations, not truth

Nothing derives Margules or van Laar from first principles for a real mixture. They are curve fits with a thermodynamically consistent form. Two consequences follow. First, fitted constants are only valid over the temperature and composition range of the data they were fitted to. Second, two models fitted to the same data will agree where the data is and can disagree noticeably where it is not — which is exactly the region where you are most likely to need them. Toggle between the two in App 2 and watch the gap.

App 2

Activity-model explorer — watch an azeotrope appear

Start at : Raoult's law, a straight bubble line, an equilibrium curve bowed safely above the diagonal. Now raise the constants together and watch the curve flatten, touch the diagonal, and cross it. The moment it crosses, an azeotrope exists and the separation is capped.

Preset system

Model

A₁₂ = ln γ₁^∞ 0.00

A₂₁ = ln γ₂^∞ 0.00

Activity coefficients

Both → 1 at their own pure end. Ideal is the flat line at 1.

γ₁γ₂ideal

at 1 atm

A minimum or maximum in the boiling curve is an azeotrope.

BubbleDew

— the one that matters

Where this curve crosses the diagonal, distillation stops.

van LaarMargules

γ₁^∞

—

γ₂^∞

—

α at x→0

—

α at x→1

—

Azeotrope at x =

—

…boiling at

—°C

Liquid splits?

—

✓—

2b.6 Azeotropes

An azeotrope is a composition at which the vapour and the liquid have the same composition:

(2.24)

Boiling such a mixture produces vapour identical to the liquid, so nothing separates. It behaves, for that one composition, exactly like a pure substance — including boiling at a fixed temperature, which is why azeotropes were once mistaken for compounds.

Does a given system have one?

Because as and as , the relative volatility at the two ends of the composition range takes a particularly simple form:

(2.25)

The azeotrope test

If one of these two values is greater than 1 and the other less than 1, then passes through 1 somewhere in between — and there is an azeotrope. If both are on the same side of 1, there is none. That is the whole test, and it needs only the two infinite-dilution activity coefficients and the vapour-pressure ratio. Watch these two numbers in App 2 as you move the sliders; the azeotrope appears at the exact moment one of them crosses 1.

One subtlety about evaluating (2.25)

The two limits sit at opposite ends of the composition range, and at fixed pressure those two ends boil at different temperatures — the two pure boiling points. Strictly, each limit should use the vapour-pressure ratio at its own temperature. Hand calculations usually evaluate both at a single convenient temperature, which is fine for deciding whether an azeotrope exists, because the sign of the test is robust; it is not good enough for locating one precisely. App 2 evaluates each limit at its own boiling temperature, so its numbers will differ a little from a single-temperature hand calculation — that difference is this subtlety, not an error in either.

Think!

Two components have very different boiling points, so is large — say 8. How large would have to be before an azeotrope appeared? What does that tell you about which systems are at risk?

Reveal

From (2.25) an azeotrope needs , i.e. . That is a very large activity coefficient, achievable only for strongly non-ideal pairs.

So the systems at risk are those where the vapour-pressure ratio is small — components of similar volatility — and the non-ideality is large. Ethanol and water is exactly that combination: boiling points 78.3 and 100 °C, so the vapour-pressure ratio near the azeotrope is only about 1.2, while . It takes very little non-ideality to overcome a ratio of 1.2.

Conversely, a wide-boiling pair is nearly immune. This is a useful design instinct: close boiling points plus chemical dissimilarity is the danger zone. Test it in App 2 — set a large and see how the azeotrope moves as you change the preset system.

Living with an azeotrope

An azeotrope caps ordinary distillation, but it does not end the story. The standard industrial responses are:

Pressure-swing distillation. Azeotropic composition usually shifts with pressure. Two columns at different pressures can walk the mixture around the azeotrope. Works only if the shift is large enough to be economic.
Extractive distillation. Add a high-boiling solvent that changes enough to destroy the azeotrope. The solvent leaves with the bottoms and is recovered separately.
Heterogeneous azeotropic distillation. Add an entrainer that forms a heteroazeotrope; the condensate splits into two liquid layers and a decanter does the separation that the column could not.
Change technology entirely. Membranes — pervaporation is the standard route for the last few per cent of water out of ethanol — or adsorption on molecular sieves, which is how fuel-grade ethanol is actually dried. Chapters 8 and 9.

2b.7 When the liquid splits in two

Push positive deviation further still and the molecules stop tolerating each other altogether: the liquid separates into two phases. This is not a new phenomenon requiring new physics — it is the same activity coefficients continuing along the same trend.

The criterion is thermodynamic stability. A single liquid phase is stable only if the Gibbs energy of mixing curves upward:

(2.26)

The first two terms are the ideal entropy of mixing, which always favours a single phase. The excess term is where non-ideality lives. When grows large enough to overcome the entropy term, the second derivative goes negative over a range of composition and the mixture splits. App 2 reports this; push both constants above about 2 and watch it trigger.

That is the bridge to Chapter 7. Liquid–liquid extraction is not an unrelated unit operation — it is the deliberate exploitation of exactly this instability, and the lever-arm rule2a §2.9 you learned on a tie line will be waiting for you there on a ternary diagram.

2b.8 What non-ideality costs a design

If the system is…	then…
Nearly ideal	Antoine constants and Raoult's law2a §2.4 are enough. Hand calculations are reliable; a constant is usually safe — check how safe2a App 4. Chapters 3–4 methods apply directly.
Moderately non-ideal, no azeotrope	You need experimental VLE data or a fitted activity model. The curve is no longer symmetric, so constant- shortcuts become unreliable — use the real curve for stage-stepping. The separation is still achievable.
Azeotropic	Ordinary distillation is capped at the azeotropic composition — you can watch a cascade stall against it3 App 3. A different process — pressure swing, extractive, entrainer, membrane, adsorption — is mandatory for higher purity, and it will dominate the cost of the plant.
Phase-splitting	Liquid–liquid extraction becomes available and is often the cheapest route. The heteroazeotrope plus a decanter is a classic, and very effective, combination.

The engineering habit to build

Before you draw a single stage, ask: is this system ideal? If you cannot answer from the chemistry — similar molecules, no hydrogen bonding, no strong polarity difference — then assume it is not, and go and find the VLE data. Designing a column on Raoult's law for a system that turns out to be azeotropic is not a small error; it is a plant that cannot make its product.

2b.9 Worked examples

Worked example 2b.1Testing for an azeotrope

Ethanol (1) and water (2) at 1 atm. van Laar constants fitted to experimental data are and . Near 78 °C the vapour pressures are kPa and kPa. Determine whether an azeotrope exists, and if so, roughly where.

Solution

Infinite-dilution activity coefficients. For van Laar the constants are the logarithms of these:

Both exceed 1, so this is a positive-deviation system — as expected, since adding water disrupts the ethanol hydrogen-bond network and vice versa.

Apply the test, equation (2.25). The vapour-pressure ratio is .

The two values straddle 1, so passes through 1 somewhere in between: an azeotrope exists.

Where? Since is only just below 1 while is far above it, the crossing must be very close to the pure-ethanol end. Solving numerically puts it at and about 78.2 °C. The accepted experimental values are 0.894 mole fraction and 78.15 °C — a two-parameter model fitted to bulk data reproduces the azeotrope to about 2 %.

The engineering consequence. No ordinary distillation column, at any reflux and any number of stages, will produce ethanol purer than about 89 mol % from a dilute feed. Fuel-grade ethanol is dried on molecular sieves; absolute ethanol for laboratory use is made by azeotropic distillation with an entrainer or by pervaporation.

Check it in App 2. Select the ethanol–water preset — the constants load automatically and the azeotrope readout should show .

Worked example 2b.2A negative-deviation system

Chloroform (1) boils at 61.7 °C and acetone (2) at 56.1 °C, yet the mixture forms an azeotrope boiling at 64.5 °C. Explain, and estimate the activity coefficients at the azeotrope given that at 64.5 °C the vapour pressures are kPa and kPa.

Solution

Why the boiling point is raised. Chloroform has a mildly acidic C–H; acetone has a carbonyl oxygen with lone pairs. In the mixture they form a hydrogen bond C–H···O=C which exists in neither pure liquid. Both molecules are therefore held more tightly in the mixture than in their own pure liquids, so both vapour pressures fall below the Raoult prediction — and — and the mixture is harder to boil than either component. Hence a maximum-boiling azeotrope. Consistently, mixing chloroform and acetone is exothermic.

Activity coefficients at the azeotrope. At an azeotrope , so equation (2.18) gives for each component independently. With kPa:

Both below 1, confirming negative deviation.

Note the trick. An azeotrope is the one composition at which you can read activity coefficients straight off a boiling point, with no model and no fitting — because collapses equation (2.18) to . Azeotropic data is therefore disproportionately valuable for fitting activity models, which is why it is so often the anchor point in a data set.

2b.10 Check your understanding

Five multiple-choice questions, two short problems and two long problems, all graded in your browser and stored only on this device.

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Short problems

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Long problems

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Where this chapter connects

Back to: Raoult's law2a §2.4 and relative volatility2a §2.7 — every equation here is one of those with a inserted.
Compare directly: set both constants to zero in the model explorer2b App 2 and you are back to the ideal binary of Chapter 2a2a §2.6.
Consequences you can see: the ethanol–water azeotrope is why the flash cascade3 App 3 stops enriching at , no matter how many stages you add.
Ahead: phase splitting is the physical basis of liquid–liquid extraction in Chapter 7, and azeotropes are why Chapter 4 ends with extractive and pressure-swing distillation.

Summary & key equations

Non-ideal VLE

Equilibrium (ideal gas)

Total pressure

K value

Relative volatility

Reading the sign

positive deviation · above Raoult · endothermic mixing · minimum-boiling azeotrope

negative deviation · below Raoult · exothermic mixing · maximum-boiling azeotrope

Limits as ; as

Models

Margules (2 constant)

van Laar

Constants mean,

Azeotropes

Definition, ,

Existence test and on opposite sides of 1

At an azeotrope — read directly, no model needed

Liquid splits when

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