Ideal and Non-ideal Solutions

Mix two volatile liquids and pour their vapour pressures onto a graph, and one question decides everything: do the unlike molecules get along with each other exactly as well as they get along with their own kind? If A–B attractions match A–A and B–B attractions, the mixture obeys Raoult's law at every composition — an ideal solution. If the unlike molecules attract each other more weakly or more strongly, the observed vapour pressure drifts away from the Raoult's-law prediction, and the mixture is non-ideal, showing either a positive or a negative deviation. This single idea — compare A–B forces to A–A/B–B forces — is the key that unlocks vapour-pressure behaviour, mixing enthalpy and volume changes, and even why some mixtures (azeotropes) refuse to separate by ordinary distillation, all from one comparison.

The core insight

A solution is ideal when unlike-molecule forces equal like-molecule forces (Raoult's law holds exactly, $\Delta H_{mix}=\Delta V_{mix}=0$ ); weaker unlike forces push vapour pressure above the Raoult's-law line (positive deviation, endothermic mixing), while stronger unlike forces pull it below (negative deviation, exothermic mixing).

The core idea

Ideal solutions: the baseline. An ideal solution obeys Raoult's law across the whole concentration range, at every temperature:

p_1 = x_1p_1^0, \qquad p_2 = x_2p_2^0, \qquad p_{total}=p_1+p_2

where: $p_1, p_2$ are the partial vapour pressures of components 1 and 2 above the mixture, $x_1, x_2$ are their mole fractions in the liquid, $p_1^0, p_2^0$ are the vapour pressures of the pure liquids at that temperature, and $p_{total}$ is the total vapour pressure over the solution. Because A–B forces match A–A and B–B forces in an ideal solution, mixing changes neither the internal energy nor the packing of molecules:

\Delta H_{mix} = 0, \qquad \Delta V_{mix} = 0

Benzene–toluene, n-hexane–n-heptane, and ethyl bromide–ethyl iodide are the standard textbook examples — pairs of structurally similar molecules with near-identical intermolecular forces.

Non-ideal solutions: two directions of drift. Real mixtures usually deviate from this baseline in one of two ways, and the direction tells you everything about the underlying forces. Positive deviation occurs when A–B attractions are weaker than A–A/B–B attractions. Molecules escape the liquid more easily than the ideal formula predicts, so $p_{total}$ is higher than $x_1p_1^0+x_2p_2^0$ . This costs energy to set up, so $\Delta H_{mix}>0$ (endothermic) and $\Delta V_{mix}>0$ (slight expansion). Ethanol–acetone, carbon disulfide–acetone, and water–ethanol are classic cases. Negative deviation occurs when A–B attractions are stronger than A–A/B–B attractions (frequently hydrogen bonding between unlike molecules). Molecules are held back from escaping, so $p_{total}$ falls below the Raoult's-law prediction, mixing releases energy ( $\Delta H_{mix}<0$ , exothermic), and the solution contracts slightly ( $\Delta V_{mix}<0$ ). Chloroform–acetone, phenol–aniline, and nitric acid–water illustrate this.

Azeotropes: when deviation blocks separation. At certain compositions in a strongly non-ideal mixture, the vapour and liquid end up with identical composition, so the mixture boils at a fixed temperature and distils over unchanged — an azeotrope. Strong positive deviation produces a minimum-boiling azeotrope (95% ethanol–water boils below either pure component); strong negative deviation produces a maximum-boiling azeotrope (about 20% HCl in water boils above either pure component). Past the azeotropic point, ordinary fractional distillation cannot purify further, because there is no longer any difference between the vapour and liquid compositions to exploit.

Figure

Three vapour pressure versus composition curves showing the straight ideal line, a positive deviation curve bulging above it, and a negative deviation curve dipping below it

Derivation ATotal vapour pressure and vapour composition of an ideal binary solution

The goal is to start from Raoult's law for each component of an ideal binary solution and derive both the total vapour pressure above the liquid and the mole fraction composition of the vapour phase, using only the two partial-pressure equations and Dalton's law of partial pressures — no calculus, just algebra with mole fractions that add to one.

Step 1. Write Raoult's law for each component. For an ideal solution of components 1 and 2 with liquid mole fractions $x_1$ and $x_2$ , the partial vapour pressures are $p_1 = x_1p_1^0$ and $p_2 = x_2p_2^0$ , where $p_1^0$ and $p_2^0$ are the pure-liquid vapour pressures at the working temperature.

Step 2. Apply Dalton's law of partial pressures. The total pressure of vapour above the liquid is the sum of the partial pressures: $p_{total} = p_1 + p_2 = x_1p_1^0 + x_2p_2^0$ .

Step 3. Eliminate one mole fraction using $x_1 + x_2 = 1$ , so $x_2 = 1-x_1$ . Substituting, $p_{total} = x_1p_1^0 + (1-x_1)p_2^0 = p_2^0 + x_1(p_1^0-p_2^0)$ . This shows $p_{total}$ varies linearly with liquid composition $x_1$ — the straight line in an ideal solution's vapour-pressure diagram, running from $p_2^0$ at $x_1=0$ to $p_1^0$ at $x_1=1$ .

Step 4. Find the vapour-phase composition. The vapour is itself a gas mixture, so its mole fractions follow from partial pressure over total pressure: $y_1 = p_1/p_{total} = x_1p_1^0/p_{total}$ and $y_2 = p_2/p_{total} = x_2p_2^0/p_{total}$ , with $y_1+y_2=1$ . Because $p_1^0 \ne p_2^0$ in general, $y_1 \ne x_1$ — the vapour is always richer in the more volatile component (the one with the larger $p^0$ ) than the liquid it came from, which is the entire basis of fractional distillation.

Worked example

Problem: At a certain temperature, pure benzene has a vapour pressure $p_1^0 = 150\ \text{mmHg}$ and pure toluene has $p_2^0 = 50\ \text{mmHg}$ . An ideal solution is made with liquid mole fractions $x_{benzene}=0.6$ and $x_{toluene}=0.4$ . Find the partial vapour pressures, the total vapour pressure, and the vapour-phase mole fractions.

Step 1 — Partial pressures from Raoult's law. $p_{benzene} = x_{benzene}\,p_1^0 = 0.6\times150 = 90\ \text{mmHg}$ , and $p_{toluene} = x_{toluene}\,p_2^0 = 0.4\times50 = 20\ \text{mmHg}$ .

Step 2 — Total vapour pressure. By Dalton's law, $p_{total} = p_{benzene}+p_{toluene} = 90+20 = 110\ \text{mmHg}$ . This lies between $p_2^0=50$ and $p_1^0=150$ , exactly as the linear relation from Step 3 of the derivation requires.

Step 3 — Vapour-phase mole fractions. $y_{benzene} = p_{benzene}/p_{total} = 90/110 = 0.818$ , and $y_{toluene} = p_{toluene}/p_{total} = 20/110 = 0.182$ .

Step 4 — Interpret. The liquid was 60% benzene by mole fraction, but the vapour above it is 81.8% benzene — richer in the more volatile component (benzene, with the higher pure vapour pressure). Answer: $p_{benzene}=90\ \text{mmHg}$ , $p_{toluene}=20\ \text{mmHg}$ , $p_{total}=110\ \text{mmHg}$ , $y_{benzene}=0.818$ , $y_{toluene}=0.182$ .

Competitive stretch

If this same benzene–toluene pair instead showed a positive deviation from ideality at $x_{benzene}=0.6$ , the measured $p_{total}$ would exceed $110\ \text{mmHg}$ (the Raoult's-law value computed above), because weaker benzene–toluene forces would let both components escape more readily than the ideal calculation predicts — the 110 mmHg figure is the ideal reference line any real measurement is compared against.

Common traps

Confusing "positive deviation" with "more ideal." Positive deviation means vapour pressure is higher than the Raoult's-law prediction — it is a departure from ideality, not a step toward it.
Mixing up the sign of $\Delta H_{mix}$ with the deviation type. Weaker new (A–B) interactions in positive deviation require energy to form, so mixing is endothermic ( $\Delta H_{mix}>0$ ); stronger new interactions in negative deviation release energy ( $\Delta H_{mix}<0$ ).
Treating $x_1$ (liquid mole fraction) and $y_1$ (vapour mole fraction) as the same number. They are equal only in the trivial case $p_1^0=p_2^0$ ; otherwise the vapour is always enriched in the more volatile component, as the worked example shows.
Assuming an azeotrope can be pushed further apart by more distillation. By definition, an azeotrope's vapour has the same composition as its liquid, so no amount of additional fractional distillation can separate it further.

Ask the tutor: unsure whether a given liquid pair will show positive, negative, or no deviation from Raoult's law, or want to work through another Dalton's-law vapour-composition problem — ask the AI tutor to walk through the intermolecular forces and the algebra with you.

Why does a stronger A-B attraction lower the vapour pressure instead of raising it?

Vapour pressure comes from molecules escaping the liquid surface into the gas phase. If A and B attract each other more strongly than A attracts A or B attracts B (often through hydrogen bonding, as in chloroform–acetone), each molecule is held back more tightly inside the mixture than it would be in its own pure liquid. Fewer molecules have enough energy to escape, so the total vapour pressure measured above the mixture falls below the Raoult's-law prediction — a negative deviation.

Is benzene-toluene the only pair that behaves ideally, or are there others?

Benzene–toluene is the standard textbook example because the two molecules are similar in size, shape and polarity, so A–B forces closely match A–A and B–B forces. Other near-ideal pairs given in the syllabus are n-hexane and n-heptane (adjacent straight-chain alkanes) and ethyl bromide and ethyl iodide (same alkyl group, similar halogens). The common thread is structural and electronic similarity between the two molecules — true ideality is the exception, not the rule, among real liquid pairs.

Why can fractional distillation not separate an azeotrope completely?

Fractional distillation works because, away from the azeotropic composition, the vapour above a boiling mixture is richer in the more volatile component than the liquid — repeated vaporisation-condensation cycles gradually concentrate that component. At the azeotropic composition itself, however, the vapour and liquid have exactly the same composition by definition, so no cycle of boiling and condensing changes the composition at all. The mixture simply boils at a fixed temperature and distils over unchanged, so ordinary fractional distillation reaches the azeotrope and then stalls.

Ideal and Non-ideal Solutions

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads