Ideal and non-ideal solutions

Mix two liquids and you might expect the total vapour pressure to sit neatly between the two pure-liquid values, changing in step with composition. Most real mixtures refuse to cooperate. Ethanol and water together escape into vapour far more eagerly than a simple average would suggest; chloroform and acetone hold each other back instead. The reason lies entirely in intermolecular forces — whether A-B attractions in the mixture are weaker, equal to, or stronger than the A-A and B-B attractions in the separate pure liquids. This single comparison sorts every binary liquid mixture into ideal solutions, positive-deviation solutions, and negative-deviation solutions, and it explains why some mixtures can never be fully separated by ordinary distillation — they form azeotropes, constant-boiling mixtures that distil over unchanged.

The core insight

A solution's vapour pressure equals Raoult's-law prediction only when solute-solvent forces match solute-solute and solvent-solvent forces exactly; any mismatch produces positive or negative deviation and, at extreme compositions, an azeotrope that resists further separation by fractional distillation.

The core idea

Ideal solutions obey Raoult's law at every composition and temperature, with $\Delta H_{mix} = 0$ and $\Delta V_{mix} = 0$ . This requires A-B intermolecular forces to be essentially identical in strength to A-A and B-B forces, so a molecule at the surface feels no net change in environment on mixing — nothing is gained or lost by swapping neighbours. Benzene + toluene and n-hexane + n-heptane are the standard textbook pairs because both components are structurally similar non-polar molecules held together mainly by comparable dispersion forces. For any ideal binary mixture of components A and B, the total vapour pressure above the liquid is the sum of each component's own partial contribution:

p_{total} = p_A + p_B = x_A\, p_A^{\circ} + x_B\, p_B^{\circ}

where: $p_A^{\circ}$ and $p_B^{\circ}$ are the vapour pressures of pure A and pure B at that temperature, $x_A$ and $x_B$ are their mole fractions in the liquid ( $x_A + x_B = 1$ ), and $p_{total}$ is the vapour pressure actually measured above the mixture. Plotted against $x_A$ , this traces a dead-straight line joining $p_B^{\circ}$ (at $x_A = 0$ ) to $p_A^{\circ}$ (at $x_A = 1$ ) — the ideal reference line every real mixture is judged against.

Non-ideal (real) solutions deviate from this straight line in one of two directions, depending on how A-B forces compare with the pure-component forces.

Positive deviation occurs when A-B attraction is weaker than A-A/B-B attraction. Molecules escape more easily than the ideal line predicts, so the measured curve bulges above the straight line: $p_{total} >$ Raoult's-law value, $\Delta H_{mix} > 0$ (mixing absorbs heat, since breaking the stronger A-A/B-B bonds costs more than the weaker A-B bonds return), and $\Delta V_{mix} > 0$ (the liquid expands slightly). Ethanol + acetone and $CS_2$ + acetone are classic examples.

Negative deviation occurs when A-B attraction is stronger than A-A/B-B attraction, often because of new hydrogen bonds between unlike molecules. Molecules are held back from escaping, so the curve dips below the straight line: $p_{total} <$ Raoult's-law value, $\Delta H_{mix} < 0$ (exothermic — new, stronger bonds release energy), and $\Delta V_{mix} < 0$ (the liquid contracts). Chloroform + acetone and phenol + aniline show this behaviour.

Figure

Vapour pressure against composition for an ideal line and for curves bulging above it for positive deviation and dipping below it for negative deviation

Azeotropes. At certain compositions a deviating mixture's vapour-pressure curve reaches an extremum, and at that exact composition the vapour and liquid have identical composition. Such a mixture boils at a constant temperature and distils over unchanged — a constant-boiling mixture or azeotrope. A minimum-boiling azeotrope sits at the vapour-pressure maximum of a positive-deviation system: ethanol-water, 95.6% ethanol by mass, boils at 78.2°C, lower than either pure component. A maximum-boiling azeotrope sits at the vapour-pressure minimum of a negative-deviation system: HCl-water boils higher than either pure component. Fractional distillation of an azeotropic mixture proceeds only until the azeotropic composition is reached; beyond that point vapour and liquid compositions match, so no further enrichment is possible.

Derivation ATotal vapour pressure of a binary liquid mixture from Raoult's law

The goal is to show, from Raoult's law applied separately to each component, why an ideal mixture gives a straight-line vapour-pressure-versus-composition plot, and how the same reasoning explains why real mixtures bulge above or dip below that line depending on whether intermolecular attraction is weaker or stronger across the A-B interface than it is within pure A or pure B.

Step 1. Apply Raoult's law to component A alone. In an ideal solution, A's partial vapour pressure above the mixture is proportional to its own mole fraction in the liquid: $p_A = x_A\, p_A^{\circ}$ . The same statement holds for B: $p_B = x_B\, p_B^{\circ}$ . Each component behaves as if the other were not there, because A-B forces equal A-A and B-B forces — there is no extra push or pull to distort the simple proportionality.

Step 2. Add the two partial pressures (Dalton's law of partial pressures for the vapour phase) to get the total: $p_{total} = p_A + p_B = x_A p_A^{\circ} + x_B p_B^{\circ}$ . Since $x_B = 1 - x_A$ , this rewrites as $p_{total} = p_B^{\circ} + x_A(p_A^{\circ} - p_B^{\circ})$ , which is linear in $x_A$ — a straight line running from $p_B^{\circ}$ at $x_A = 0$ to $p_A^{\circ}$ at $x_A = 1$ .

Step 3. Now relax the "ideal" assumption. If A-B forces are weaker than A-A/B-B forces, molecules at the surface escape more readily than the proportional formula allows, so the true $p_A$ and $p_B$ each exceed their Raoult's-law value at every intermediate $x_A$ , and the summed $p_{total}$ curve lies above the straight line — positive deviation. If A-B forces are stronger, molecules are held back more than the formula predicts, both partial pressures fall below their Raoult's-law value, and $p_{total}$ lies below the line — negative deviation.

Step 4. Read the extremes. Because the curve must still meet the axis endpoints at the pure-component values ( $p_B^{\circ}$ and $p_A^{\circ}$ , where there is no A-B interface at all to deviate), a bulging or dipping curve joining the same two endpoints must pass through a maximum or minimum somewhere in between. That extremum is exactly the azeotropic composition — the one point where the deviation curve's slope is zero and vapour and liquid compositions coincide.

Worked example

Benzene ( $p_A^{\circ} = 150\ mmHg$ ) and toluene ( $p_B^{\circ} = 50\ mmHg$ ) form an ideal solution at a given temperature. Find the total vapour pressure of a mixture with mole fraction of benzene $x_A = 0.6$ , and the mole fraction of benzene in the vapour above it.

Step 1. Apply Raoult's law to each component. $p_A = x_A p_A^{\circ} = 0.6 \times 150 = 90\ mmHg$ . $p_B = x_B p_B^{\circ} = 0.4 \times 50 = 20\ mmHg$ (using $x_B = 1 - 0.6 = 0.4$ ).

Step 2. Sum for the total vapour pressure. $p_{total} = p_A + p_B = 90 + 20 = 110\ mmHg$ . This lies exactly on the straight ideal line between $p_B^{\circ} = 50$ and $p_A^{\circ} = 150$ , as expected for an ideal solution.

Step 3. Find the vapour composition using Dalton's law: each component's mole fraction in the vapour equals its partial pressure divided by the total. $y_A = p_A / p_{total} = 90/110 = 0.818$ .

Step 4. Compare liquid and vapour composition. Answer: $p_{total} = 110\ mmHg$ , and the vapour is richer in benzene ( $y_A = 0.82$ ) than the liquid ( $x_A = 0.6$ ) — the more volatile component (higher $p_A^{\circ}$ ) always concentrates preferentially in the vapour, which is exactly the enrichment principle fractional distillation relies on, and exactly what fails once a mixture reaches its azeotropic composition.

Competitive stretch

Extend Worked example 2 from the syllabus's chloroform-acetone case: if the negative deviation were strong enough to create a vapour-pressure minimum at some intermediate composition, what would happen to fractional distillation started from a dilute chloroform-in-acetone mixture? Distillation would enrich the vapour in whichever component is more volatile at that composition, driving the liquid composition toward the vapour-pressure-minimum point from either side — the maximum-boiling azeotrope. Once reached, vapour and liquid compositions coincide and no further separation occurs, exactly as with HCl-water; the still bottoms would be left at the azeotropic composition rather than pure acetone or pure chloroform.

Common traps

Assuming all real solutions show positive deviation — chloroform-acetone, HNO₃-water and phenol-aniline all show negative deviation.
Mixing up which azeotrope goes with which deviation: minimum-boiling pairs with positive deviation (vapour-pressure maximum), maximum-boiling pairs with negative deviation (vapour-pressure minimum) — the boiling-point and vapour-pressure trends are inverses of each other.
Treating an azeotrope as a pure compound with a fixed formula — it is a mixture at one specific composition that happens to boil without any further change in composition; it can still be separated by non-distillation methods (e.g. chemical drying agents, pervaporation).
Reading the vapour-composition formula $y_A = p_A/p_{total}$ as if it used the liquid mole fraction directly — always compute the partial pressures from Raoult's law first, then divide.

Ask the tutor: want to work through why the vapour-pressure curve must pass through an extremum between two fixed endpoints, or see the ethanol-water and HCl-water phase diagrams compared side by side?

Why does an azeotrope stop fractional distillation from giving 100 percent pure ethanol?

Fractional distillation works by repeatedly vaporising and recondensing, each cycle enriching the vapour in the more volatile component because Raoult's law and Dalton's law together make $y_A$ differ from $x_A$ at every ordinary composition. At the azeotropic composition, though, the vapour-pressure curve hits an extremum where vapour and liquid compositions become numerically equal — there is no longer any enrichment step to exploit, so successive vaporisations leave the composition unchanged and distillation stalls at 95.6% ethanol.

Is a negative deviation solution always going to form a maximum-boiling azeotrope?

Only if the deviation is strong enough to push the vapour-pressure curve down to an actual minimum somewhere between the pure components; mild negative deviation just gives a dipped curve with no turning point and no azeotrope at all. Chloroform-acetone and phenol-aniline show clear negative deviation without necessarily reaching that minimum at accessible compositions in every textbook treatment, while HCl-water is the standard example that does form a true maximum-boiling azeotrope. The syllabus expects you to know the direction of deviation and the general azeotrope idea, not to calculate whether a given pair crosses the threshold.

How is deltaH_mix connected to positive or negative deviation, physically?

Mixing breaks the pure-component A-A and B-B contacts and replaces them with A-B contacts. If A-B attraction is weaker than what was broken, the process costs net energy, so heat is absorbed from the surroundings and $\Delta H_{mix} > 0$ — this is exactly the positive-deviation case, where the weaker A-B pull also lets molecules escape into vapour more easily. If A-B attraction is stronger, new bonds form that release more energy than was needed to break the old ones, so $\Delta H_{mix} < 0$ and molecules are held back from escaping — negative deviation. The sign of $\Delta H_{mix}$ and the direction of vapour-pressure deviation are two views of the same intermolecular-force comparison.

Ideal and non-ideal solutions

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads