1.2 Vapour Pressure, Raoult's Law and Non-ideal Solutions
Above every liquid mixture sits a vapour. Raoult's law $p_{i}=x_{i}\,p_{i}^{\circ}$ predicts how much each component contributes to it, a non-volatile solute lowers it, and real mixtures that stray from the law can form azeotropes that no distillation can separate.
Open a bottle of nail-polish remover and you smell acetone at once; open a bottle of glycerol and you smell almost nothing. Every liquid sends molecules into the space above it, but some do so far more eagerly than others. That eagerness is measured by the vapour pressure. It decides how fast petrol evaporates from a fuel tank, why a wet towel dries faster in Chennai's hot afternoon than at night, and how a distillery separates alcohol from water.
This section asks what happens to the vapour pressure when two liquids are mixed, or when a solid such as sugar is dissolved in a liquid. For many mixtures the answer is a very simple law, found by François-Marie Raoult in the 1880s: each component contributes in proportion to its mole fraction. Mixtures that obey the law exactly are called ideal. Real mixtures often stray above or below it, and the size and sign of that departure tell us about the forces between the molecules. When the departure is large, the mixture can form an azeotrope, which boils without any change in composition. This is the reason ordinary distillation of fermented sugar stops at about 95.6% ethanol, however many times you repeat it.
Vapour Pressure of a Liquid Mixture: Raoult's Law
Put a little liquid in an evacuated, closed flask at constant temperature. Molecules with enough energy escape from the surface into the empty space. Some of the vapour molecules hit the surface again and are recaptured. At first evaporation is faster than condensation, but as the vapour fills up, condensation speeds up until the two rates are equal. The pressure then stops rising.
The vapour pressure of a liquid at a given temperature is the pressure exerted by its vapour when the vapour is in dynamic equilibrium with the liquid in a closed vessel. We write $p^{\circ}$ for the vapour pressure of a pure liquid, and $p_{i}$ for the partial vapour pressure of component $i$ above a solution.
The vapour pressure depends only on the nature of the liquid and on the temperature, not on how much liquid is present or on the volume of the vapour space. It rises steeply with temperature, and the liquid boils when its vapour pressure equals the external pressure. A liquid with a high vapour pressure at a given temperature is called volatile.
Now mix two volatile liquids, say benzene and methylbenzene (toluene). Both kinds of molecule are at the surface, and both escape. The vapour is a mixture too, and by Dalton's law of partial pressures the total pressure is the sum of the two partial pressures. How big is each one? If the molecules of the two liquids are very similar in size and in the attractions between them, a benzene molecule at the surface "feels" much the same whether its neighbours are benzene or toluene. Its escaping tendency is unchanged, but only a fraction $x_{1}$ of the surface is benzene. So its partial pressure is the fraction $x_{1}$ of the pure-liquid value.
For a solution of volatile liquids, the partial vapour pressure of each component is proportional to its mole fraction in the liquid, and the proportionality constant is the vapour pressure of the pure component at the same temperature: $$p_{1}=x_{1}\,p_{1}^{\circ},\qquad p_{2}=x_{2}\,p_{2}^{\circ}.$$
Conditions. The temperature is constant, the vapour behaves as an ideal gas, and the two components are chemically similar so that the attractions between unlike molecules (1–2) are about the same as those between like molecules (1–1 and 2–2). A mixture that obeys the law at every composition is an ideal solution (Definition 1.6 below).
By Dalton's law the total vapour pressure is $p_{\text{total}}=p_{1}+p_{2}$. Using $x_{1}=1-x_{2}$,
So for an ideal solution $p_{\text{total}}$ is a straight line joining $p_{1}^{\circ}$ (at $x_{2}=0$) to $p_{2}^{\circ}$ (at $x_{2}=1$), and it always lies between the two pure vapour pressures.
The mole fractions in the vapour, written $y_{1}$ and $y_{2}$, follow from Dalton's law, $p_{i}=y_{i}\,p_{\text{total}}$: $$y_{1}=\frac{p_{1}}{p_{\text{total}}}=\frac{x_{1}p_{1}^{\circ}}{x_{1}p_{1}^{\circ}+x_{2}p_{2}^{\circ}},\qquad y_{1}+y_{2}=1.$$ The vapour is always richer in the more volatile component (the one with the larger $p^{\circ}$) than the liquid it came from.
The table gives the pure-liquid vapour pressures used in this section. All values are at $298\ \mathrm{K}$ and rounded to three significant figures.
| Liquid | $p^{\circ}$ / kPa | Liquid | $p^{\circ}$ / kPa |
|---|---|---|---|
| water | 3.17 | propanone (acetone) | 30.8 |
| ethanol | 7.87 | trichloromethane (chloroform) | 26.2 |
| benzene | 12.7 | carbon disulphide | 48.2 |
| methylbenzene (toluene) | 3.80 | hexane | 20.2 |
| heptane | 6.09 | octane | 1.86 |
Compare Raoult's law, $p_{1}=x_{1}\,p_{1}^{\circ}$, with Henry's law from Section 1.1, $p=K_{H}\,x$. Both say that the partial pressure of a component in the vapour is proportional to its mole fraction in the solution. The only difference is the constant. In Raoult's law it is the vapour pressure of the pure component, $p_{1}^{\circ}$. In Henry's law it is $K_{H}$, which is generally quite different from $p^{\circ}$ (a gas such as $\mathrm{O_{2}}$ has no liquid vapour pressure at room temperature at all).
So Raoult's law is Henry's law with $K_{H}=p_{1}^{\circ}$. In any dilute solution the solvent (with $x_{1}$ close to $1$) obeys Raoult's law and the solute (with $x_{2}$ close to $0$) obeys Henry's law. In an ideal solution the two laws coincide for each component over the whole composition range.
Solutions of Non-volatile Solutes
Sugar, salt and urea have practically no vapour pressure at room temperature: their particles do not escape from the liquid. Dissolve one of them in water and the vapour above the solution contains only water. But some of the places at the liquid surface are now taken by solute particles, so fewer water molecules are in a position to escape each second. Condensation is not hindered (any water molecule hitting the surface can still be captured), so the balance is struck at a lower vapour pressure.
If the solute is non-volatile ($p_{2}^{\circ}\approx0$), only the solvent contributes to the vapour, and Raoult's law for the solvent gives the whole vapour pressure of the solution:
Since $x_{1}=1-x_{2}\lt1$, the vapour pressure of the solution is always lower than that of the pure solvent at the same temperature. The lowering is
Conditions. The solution is dilute enough to behave ideally, and the solute neither dissociates nor associates. A plot of $p_{1}$ against $x_{1}$ is a straight line through the origin with slope $p_{1}^{\circ}$. Section 1.3 builds on this result: the lowering depends only on $x_{2}$, the fraction of particles that are solute, and not on what the solute is.
(b) What mass of urea ($60.0\ \mathrm{g\,mol^{-1}}$) must be dissolved in $100\ \mathrm{g}$ of water to lower its vapour pressure by $1.00\%$? ($\mathrm{H_{2}O}=18.0\ \mathrm{g\,mol^{-1}}$.)
Ideal and Non-ideal Solutions
Whether a mixture obeys Raoult's law depends on a competition between attractions. Call the two components A and B. In the pure liquids there are only A–A and B–B attractions. In the mixture some of these are replaced by A–B attractions. If the A–B attractions are just as strong as the ones they replace, nothing changes for a molecule at the surface, and Raoult's law holds.
An ideal solution is one that obeys Raoult's law over the entire range of composition at all temperatures. For such a solution
- the enthalpy of mixing is zero, $\Delta_{mix}H=0$: no heat is absorbed or released when the components are mixed;
- the volume of mixing is zero, $\Delta_{mix}V=0$: the volume of the solution equals the sum of the volumes of the components.
At the molecular level, the A–B attractions are nearly equal to the A–A and B–B attractions. No real solution is perfectly ideal, but mixtures of very similar molecules come close: $n$-hexane and $n$-heptane, bromoethane and chloroethane, benzene and methylbenzene (toluene).
A non-ideal solution does not obey Raoult's law over the whole composition range. Its vapour pressure is either higher or lower than Raoult's law predicts.
- Positive deviation: each $p_{i}\gt x_{i}p_{i}^{\circ}$, so $p_{\text{total}}$ lies above the Raoult line. The A–B attractions are weaker than the A–A and B–B attractions, so molecules escape more easily from the mixture. Mixing absorbs heat and the liquid expands slightly: $\Delta_{mix}H\gt0$, $\Delta_{mix}V\gt0$. Examples: ethanol + propanone (acetone), carbon disulphide + propanone, ethanol + water.
- Negative deviation: each $p_{i}\lt x_{i}p_{i}^{\circ}$, so $p_{\text{total}}$ lies below the Raoult line. The A–B attractions are stronger than the A–A and B–B attractions, so molecules are held more tightly in the mixture. Mixing releases heat and the volume shrinks slightly: $\Delta_{mix}H\lt0$, $\Delta_{mix}V\lt0$. Examples: trichloromethane (chloroform) + propanone, phenol + aniline, nitric acid + water, hydrochloric acid + water.
The molecular reasons are worth knowing for every example. Ethanol molecules are held together by hydrogen bonds. Propanone molecules slip in between them and break some of those hydrogen bonds without forming equally strong ones in return, so the escaping tendency of both rises: a positive deviation. In carbon disulphide + propanone, the dipole–dipole attractions of propanone are weakened when the non-polar $\mathrm{CS_{2}}$ dilutes them. In trichloromethane + propanone, the opposite happens: the $\mathrm{C{-}H}$ hydrogen of $\mathrm{CHCl_{3}}$ (made slightly acidic by the three chlorine atoms) forms a hydrogen bond with the oxygen of propanone, $\mathrm{Cl_{3}C{-}H\cdots O{=}C(CH_{3})_{2}}$. That new attraction holds both molecules back, so the vapour pressure falls: a negative deviation. In phenol + aniline, a hydrogen bond forms between the phenolic $\mathrm{O{-}H}$ and the nitrogen of aniline, which is stronger than those in either pure liquid.
1. Mixing up the direction. Weaker A–B attraction means molecules escape more easily: higher vapour pressure, a positive deviation, and heat absorbed ($\Delta_{mix}H\gt0$). Every word flips for a negative deviation.
2. Thinking a deviation means the solution is "wrong". Raoult's law is a reference line. Deviations from it are real and useful: they tell us about molecular attractions.
3. Assuming volumes always add. They add only for an ideal solution. That is why recipes say "make up to $100\ \mathrm{mL}$" (Section 1.1).
4. Treating the $\Delta H$ and $\Delta V$ signs as laws. They follow from the simple picture of attractions and are what exams expect. A few real mixtures do not fit every part of it: ethanol + water shows a positive deviation in vapour pressure, yet it shrinks slightly on mixing (Section 1.1) and warms slightly when water-rich. The vapour-pressure deviation is the measured fact; the signs of $\Delta_{mix}H$ and $\Delta_{mix}V$ are the usual, but not universal, companions.
Azeotropes
When the deviation from Raoult's law is large, the total vapour pressure curve stops being a gentle bulge and develops a maximum (strong positive deviation) or a minimum (strong negative deviation) at some intermediate composition. At that special composition the liquid and the vapour have the same composition, so boiling does not change the composition at all.
An azeotrope is a binary mixture that has the same composition in the liquid and the vapour phase and boils at a constant temperature. Its components cannot be separated by fractional distillation.
- A minimum-boiling azeotrope forms in solutions with a large positive deviation. Its vapour pressure is a maximum, so it boils at a temperature below that of either component. Example: ethanol + water, about $95.6\%$ ethanol by mass, boiling at $351.3\ \mathrm{K}$ ($78.2\ ^{\circ}\mathrm{C}$) at $1\ \mathrm{atm}$.
- A maximum-boiling azeotrope forms in solutions with a large negative deviation. Its vapour pressure is a minimum, so it boils at a temperature above that of either component. Example: nitric acid + water, about $68\%$ $\mathrm{HNO_{3}}$ by mass, boiling at $393.5\ \mathrm{K}$ ($120.4\ ^{\circ}\mathrm{C}$) at $1\ \mathrm{atm}$.
What fractional distillation does. Distillation always moves the vapour towards the composition with the highest vapour pressure (the lowest boiling point), and the residue towards the composition with the lowest vapour pressure. For an ideal mixture these are the two pure liquids, so they can be separated. For a mixture with a minimum-boiling azeotrope the azeotrope itself has the highest vapour pressure: the distillate approaches the azeotrope and only the excess component can be left pure in the flask. Fermented sugar (about $10\%$ ethanol) gives at best $95.6\%$ ethanol at the top of the column. For a maximum-boiling azeotrope the roles reverse: the azeotrope collects in the flask and the excess component distils off.
Exercises
- 1 $1.00\ \mathrm{mol}$ of hexane is mixed with $3.00\ \mathrm{mol}$ of heptane. Find the partial pressure of each component and the total vapour pressure.
- 2 For the mixture in part 1, find the mole fraction of hexane in the vapour.
- 3 $50.0\ \mathrm{g}$ of benzene is mixed with $50.0\ \mathrm{g}$ of toluene. Find the total vapour pressure.
- 4 A benzene–toluene mixture has a total vapour pressure of $5.00\ \mathrm{kPa}$. Find the mole fraction of benzene in the liquid and in the vapour.
- 5 Show that in an ideal binary solution with $p_{1}^{\circ}\gt p_{2}^{\circ}$, the vapour is always richer in component 1 than the liquid ($y_{1}\gt x_{1}$ for $0\lt x_{1}\lt1$).
- 6 For heptane + octane: (a) find the mole fraction of heptane in the vapour above an equimolar liquid; (b) find the liquid composition whose vapour contains equal amounts of the two.
- 7 Fuel vapour. A simple model of a petrol is an ideal mixture with mole fractions hexane $0.20$, heptane $0.30$ and octane $0.50$. Find its total vapour pressure at $298\ \mathrm{K}$ and the mole fraction of hexane in the vapour. Why does the petrol left in an open can become "heavier" over time?
- 8 A little propanone is dissolved in a large amount of water. Which law describes the partial pressure of propanone, and which describes that of water? What are the two proportionality constants?
- 1 $34.2\ \mathrm{g}$ of sucrose is dissolved in $90.0\ \mathrm{g}$ of water. Find the vapour pressure of the solution.
- 2 Find the vapour pressure of water above a $10.0\%$ (by mass) aqueous urea solution.
- 3 A solution of a non-volatile solute in benzene has a vapour pressure of $12.4\ \mathrm{kPa}$. Find the mole fraction of the solute.
- 4 An aqueous glucose solution has a vapour pressure of $3.10\ \mathrm{kPa}$. Find its molality.
- 5 Solution P is glucose in water and solution Q is naphthalene in benzene, both with $x_{2}=0.0200$. Which has the larger lowering $\Delta p$? Which has the larger fractional lowering $\Delta p/p^{\circ}$?
- 6 To water ($x_{1}=0.90$) is added either (a) ethanol or (b) glucose, with $x_{2}=0.10$ in each case. Treating both as ideal, find the vapour pressure of each solution. Explain why one is above and one below $3.17\ \mathrm{kPa}$.
- 7 What mass of glucose must be dissolved in $500\ \mathrm{g}$ of water to lower its vapour pressure by $0.50\%$?
- 1 Classify each mixture as nearly ideal, positive deviation or negative deviation: (a) $n$-hexane + $n$-heptane; (b) ethanol + propanone; (c) trichloromethane + propanone; (d) phenol + aniline; (e) bromoethane + chloroethane; (f) carbon disulphide + propanone.
- 2 $50\ \mathrm{mL}$ of propanone is mixed with $50\ \mathrm{mL}$ of trichloromethane. Is the final volume more or less than $100\ \mathrm{mL}$? Does the mixture warm up or cool down? Explain.
- 3 When ethanol is mixed with propanone the temperature of the mixture falls. Predict the type of deviation from Raoult's law and the sign of $\Delta_{mix}V$.
- 4 An equimolar mixture of carbon disulphide and propanone at $298\ \mathrm{K}$ has a measured total vapour pressure of about $59\ \mathrm{kPa}$. Compare this with Raoult's law and give the percentage deviation.
- 5 The total vapour pressure of a binary mixture, plotted against composition at constant temperature, has a minimum. What kind of azeotrope does the mixture form? How does the boiling point of the azeotrope compare with those of the pure components?
- 6 Explain why a $10\%$ ethanol solution cannot be turned into pure ethanol by fractional distillation. What would you obtain at the top of the column and in the flask if you distilled a $98\%$ (by mass) ethanol–water mixture?
- 7 Hydrochloric acid forms an azeotrope with water at $20.2\%$ $\mathrm{HCl}$ by mass, boiling at $381.7\ \mathrm{K}$ at $1\ \mathrm{atm}$. Find the mole fraction of $\mathrm{HCl}$ in it, and state the type of azeotrope and of deviation.
- 1 Starting from Raoult's law and Dalton's law, show that for an ideal binary solution $$\frac{1}{p_{\text{total}}}=\frac{y_{1}}{p_{1}^{\circ}}+\frac{y_{2}}{p_{2}^{\circ}}.$$
- 2 Use part 1 to find the total pressure and the liquid composition when the vapour above a benzene–toluene mixture at $298\ \mathrm{K}$ is equimolar ($y_{\text{benzene}}=0.500$).
- 3 Liquids A and B form an ideal solution. At $300\ \mathrm{K}$ the total vapour pressure is $55\ \mathrm{kPa}$ when $x_{\mathrm{A}}=0.25$ and $60\ \mathrm{kPa}$ when $x_{\mathrm{A}}=0.50$. Find $p_{\mathrm{A}}^{\circ}$ and $p_{\mathrm{B}}^{\circ}$.
- 4 Laboratory safety. An equimolar benzene–toluene mixture (Example 1) stands in a closed cupboard at $298\ \mathrm{K}$ until the air inside is saturated. The air pressure is $101.3\ \mathrm{kPa}$. Express the benzene concentration in the air in ppm by volume, and compare it with a typical workplace limit of $1\ \text{ppm}$.
- 5 Liquid A ($p^{\circ}=40\ \mathrm{kPa}$) and liquid B ($p^{\circ}=20\ \mathrm{kPa}$) give a mixture with $x_{\mathrm{A}}=0.50$ whose total vapour pressure is $34\ \mathrm{kPa}$. What is the type of deviation and the expected sign of $\Delta_{mix}H$? Can you tell from this one measurement whether the pair forms an azeotrope?
E1. Board · 3 marks (a) State Raoult's law for a solution of two volatile liquids. (b) What is an ideal solution? (c) Give one example each of a solution showing positive and negative deviation from Raoult's law, with a reason for each.
E2. JEE Main Numerical value. Liquids A ($p_{\mathrm{A}}^{\circ}=80\ \mathrm{kPa}$) and B ($p_{\mathrm{B}}^{\circ}=40\ \mathrm{kPa}$) form an ideal solution. The vapour in equilibrium with a certain liquid mixture contains $60\ \text{mol}\ \%$ of A. The mole percentage of A in the liquid, to the nearest integer, is ______.
E3. NEET Which of the following mixtures shows a negative deviation from Raoult's law? (A) ethanol + propanone (B) carbon disulphide + propanone (C) trichloromethane + propanone (D) benzene + methylbenzene
E4. JEE Advanced One or more correct. At constant temperature, the total vapour pressure of a binary liquid mixture of A and B passes through a maximum as the composition is changed. Which statements are correct? (A) The mixture shows a negative deviation from Raoult's law. (B) At the maximum, the liquid and its vapour have the same composition. (C) The mixture of that composition boils at a lower temperature than either pure liquid at the same external pressure. (D) A–B attractions are weaker than the average of the A–A and B–B attractions.
E5. NEET Board $9.0\ \mathrm{g}$ of glucose ($180\ \mathrm{g\,mol^{-1}}$) is dissolved in $90\ \mathrm{g}$ of water at $298\ \mathrm{K}$ ($p^{\circ}=3.17\ \mathrm{kPa}$). The vapour pressure of the solution is (A) $3.14\ \mathrm{kPa}$ (B) $3.11\ \mathrm{kPa}$ (C) $3.17\ \mathrm{kPa}$ (D) $2.85\ \mathrm{kPa}$
E6. Board · Case study A sugar-cane distillery ferments molasses to a liquid containing about $10\%$ ethanol and distils it through a tall column. The product leaving the top never contains more than about $95.6\%$ ethanol by mass, and it boils at $351.3\ \mathrm{K}$, slightly below pure ethanol ($351.5\ \mathrm{K}$) and well below water ($373.2\ \mathrm{K}$). (a) Name the type of mixture obtained at the top, and state the type of deviation from Raoult's law that ethanol + water shows. (2 marks) (b) Why can't further distillation of this product give pure ethanol? (1 mark) (c) Would you expect $\Delta_{mix}H$ for this kind of deviation to be positive or negative according to the simple model of attractions? (1 mark)
Research question. Ethanol blended into petrol (such as India's E20 fuel, 20% ethanol) must be almost free of water. If distillation stops at the azeotrope, how much water is left, and how do fuel plants remove the rest?
Why it matters. Water in ethanol–petrol blends can separate out as a second layer at the bottom of a fuel tank and damage engines. Fuel-grade ethanol is therefore specified as "anhydrous", typically at least $99.5\%$ ethanol by volume. Brazil, the United States and India together produce tens of billions of litres of fuel ethanol each year, and the drying step costs real energy.
What to do.
- Use Example 5 to write the azeotrope ($95.6\%$ ethanol by mass) as a mole fraction. Then estimate its volume percentage, using densities $0.789\ \mathrm{g\,mL^{-1}}$ (ethanol) and $0.997\ \mathrm{g\,mL^{-1}}$ (water) and ignoring the small volume change on mixing.
- How many grams of water must be removed from $1.00\ \mathrm{kg}$ of the azeotrope to reach $99.5\%$ ethanol by mass?
- In Interactive 2, choose the positive pair and drag the deviation slider. Record how the azeotrope composition moves as the deviation becomes stronger or weaker, and the setting at which it disappears.
- Find out how industry gets past the azeotrope. Compare at least two methods: molecular sieves (zeolites that trap water molecules), distillation with a third "entrainer" liquid, and changing the operating pressure.
Tools: Calculator or spreadsheet; Interactive 2; any reliable engineering or government source on fuel-ethanol production.
Hand in: A half-page note with your three numbers, a sketch of the vapour-pressure curve with the azeotrope marked, and a short recommendation of one drying method with one advantage and one disadvantage.
Going further. The position of an azeotrope depends on pressure. Look up how the ethanol–water azeotrope composition changes when distillation is run under reduced pressure, and why that makes "pressure-swing" distillation possible for some mixtures.
🧪 Self-check quiz
Q1. According to Raoult's law, the partial vapour pressure of component 1 in an ideal solution is
Q2. In the vapour above an ideal mixture of benzene ($p^{\circ}=12.7\ \mathrm{kPa}$) and toluene ($p^{\circ}=3.80\ \mathrm{kPa}$), compared with the liquid,
Q3. A solution that shows a negative deviation from Raoult's law has
Q4. A minimum-boiling azeotrope is formed by a mixture that shows
Q5. An ideal benzene–toluene liquid contains $20\ \text{mol}\ \%$ benzene. At $298\ \mathrm{K}$ its total vapour pressure is