Solvent Extraction and Colligative Properties

Lessons

Senior 6 · Chemistry

Solvent Extraction and Colligative Properties

5 lessons

Lesson 2 of 5

Raoult's Law: Ideal Solutions and Deviations

25 min

What leaves a mixture first?

Spill a little petrol on the forecourt of a filling station and it smells sharp at first, then less so, and it leaves an oily film that evaporates slowly. Petrol is a mixture of many hydrocarbons. The most volatile ones escape first, so the vapour above a liquid mixture does not have the same composition as the liquid. To predict how much of each component is in the vapour, and to design the distillations of the next lesson, we need the vapour pressures of mixtures.

Definition

Saturated vapour pressure, $p^{\circ}$ : the pressure exerted by a vapour in dynamic equilibrium with its pure liquid in a closed container, at a given temperature. It rises with temperature; a liquid boils when its vapour pressure equals the external pressure. Example: water, 3.17 kPa at 25 °C and 101.3 kPa at 100 °C.

Mole fraction, $x_{A}$ : the amount of A divided by the total amount of all components: $x_A = \frac{n_A}{n_A + n_B}$ , so $x_{A} + x_{B} = 1$ .

Raoult's law: the partial vapour pressure of a component in a liquid mixture equals its mole fraction in the liquid times the vapour pressure of the pure component, at the same temperature: $p_A = x_A\,p_A^{\circ}$ .

Ideal solution: a mixture that obeys Raoult's law for every component at all compositions. Example: hexane and heptane.

Raoult's law and Dalton's law together

At the surface of a mixture of A and B, only a fraction $x_{A}$ of the molecules are A, so A escapes at $x_{A}$ times the rate of pure A, and its partial pressure is reduced in the same ratio:

$p_A = x_A\,p_A^{\circ} \qquad p_B = x_B\,p_B^{\circ}$

By Dalton's law of partial pressures, the total vapour pressure is the sum:

$p = p_A + p_B = x_A\,p_A^{\circ} + (1 - x_A)\,p_B^{\circ}$

This is a straight line between $p_B^{\circ}$ (pure B) and $p_A^{\circ}$ (pure A) (Figure 2.1). The mole fraction of A in the vapour, $y_{A}$ , is its share of the total pressure:

$y_A = \frac{p_A}{p}$

Because the more volatile component has the larger $p^{\circ}$ , the vapour is always richer in the more volatile component than the liquid it came from. This single fact is the basis of fractional distillation.

A mixture is ideal when the attractions between unlike molecules (A–B) are the same as those between like molecules (A–A and B–B). Then mixing gives out or takes in no heat ( $\Delta H_{\text{mix}} = 0$ ) and the volumes simply add. Pairs of very similar molecules come closest: hexane + heptane, benzene + methylbenzene, bromoethane + chloroethane.

Figure 2.1 Vapour pressures of hexane–heptane mixtures at 25 °C ( $p^{\circ}$ : hexane 20.2 kPa, heptane 6.09 kPa). Each partial pressure, and so the total, is a straight line. At $x(\text{hexane}) = 0.40$ the total is 11.7 kPa, of which 8.08 kPa is hexane.

Worked example

Problem. Calculate the total vapour pressure at 25 °C of a hexane–heptane mixture in which $x(\text{hexane}) = 0.40$ , and the mole fraction of hexane in the vapour. ( $p^{\circ}$ : hexane 20.2 kPa, heptane 6.09 kPa)

Solution.

$p_{\text{hexane}} = 0.40 \times 20.2 = 8.08\,\mathrm{kPa} \qquad p_{\text{heptane}} = 0.60 \times 6.09 = 3.65\,\mathrm{kPa}$

$p = 8.08 + 3.65 = 11.7\,\mathrm{kPa} \qquad y_{\text{hexane}} = \frac{8.08}{11.73} = 0.689$

Answer: 11.7 kPa; the vapour is 69 % hexane (by moles), although the liquid is only 40 % hexane.

Worked example

Problem. 39.0 g of benzene ( $M = 78.11\,\mathrm{g\,mol^{-1}}$ ) is mixed with 46.0 g of methylbenzene ( $M = 92.14\,\mathrm{g\,mol^{-1}}$ ). Assuming ideal behaviour, find the vapour pressure of the mixture at 25 °C and the composition of the vapour. ( $p^{\circ}$ at 25 °C: benzene 12.7 kPa, methylbenzene 3.79 kPa)

Solution.

$n(\text{benzene}) = \frac{39.0}{78.11} = 0.4993\,\mathrm{mol}$

$n(\text{methylbenzene}) = \frac{46.0}{92.14} = 0.4992\,\mathrm{mol}$

So $x(\text{benzene}) = 0.500$ . Then $p_{\text{benzene}} = 0.500 \times 12.7 = 6.35\,\mathrm{kPa}$ , $p_{\text{methylbenzene}} = 0.500 \times 3.79 = 1.90\,\mathrm{kPa}$ , and

$p = 6.35 + 1.90 = 8.25\,\mathrm{kPa} \qquad y_{\text{benzene}} = \frac{6.35}{8.25} = 0.770$

Answer: 8.25 kPa; the vapour is 77.0 % benzene by moles. Note that equal masses are not equal amounts: always convert to moles first.

Real solutions: deviations from Raoult's law

Most mixtures are not ideal, because the A–B attractions differ from the A–A and B–B attractions.

Positive deviation. If A–B attractions are weaker than the average of A–A and B–B, molecules escape more easily from the mixture, and the vapour pressure is higher than Raoult's law predicts. Energy is needed to separate the like molecules and less is given back, so mixing is usually endothermic (the mixture cools) and the volume increases slightly. Example: ethanol + hexane. Hexane molecules push between ethanol molecules and break many of their hydrogen bonds, but can offer only weak London forces in return.
Negative deviation. If A–B attractions are stronger, molecules are held back, and the vapour pressure is lower than predicted. Mixing is exothermic (the mixture warms) and the volume decreases slightly. Example: trichloromethane + propanone. Neither pure liquid is hydrogen-bonded, but in the mixture the H of $\mathrm{CHCl_3}$ , made δ+ by three electronegative Cl atoms, forms a hydrogen bond with the O of propanone (Figure 2.3). Nitric acid + water and hydrogen chloride + water also show large negative deviations.

Figure 2.2 Total vapour pressure for non-ideal mixtures. When the deviation is large, the curve passes through a maximum (positive deviation) or a minimum (negative deviation); these mixtures form azeotropes (Lesson 3).

Figure 2.3 In a mixture of trichloromethane and propanone, a new hydrogen bond, C–H···O, forms between unlike molecules. It holds molecules in the liquid and lowers the vapour pressure below the Raoult's law value.

Table 2.1: Ideal and non-ideal mixtures compared

Property	Ideal	Positive deviation	Negative deviation
A–B attractions compared with A–A and B–B	the same	weaker	stronger
vapour pressure	as Raoult's law	higher	lower
$\Delta H_{\text{mix}}$	zero	usually positive (cools)	negative (warms)
volume on mixing	additive	usually increases slightly	decreases slightly
boiling point of mixtures	between those of A and B	lower than expected; may show a minimum	higher than expected; may show a maximum
examples	hexane + heptane; benzene + methylbenzene	ethanol + hexane; ethanol + water	$\mathrm{CHCl_3}$ + propanone; $\mathrm{HNO_3}$ + water

Worked example

Problem. At 25 °C, $p^{\circ}$ (propanone) = 30.8 kPa and $p^{\circ}$ (trichloromethane) = 26.2 kPa. An equimolar mixture has a measured vapour pressure of 23.3 kPa (illustrative value). Classify the mixture and explain the result in terms of intermolecular forces. Predict what happens to the temperature when the liquids are mixed.

Solution. Raoult's law predicts $p = 0.500 \times 30.8 + 0.500 \times 26.2 = 28.5\,\mathrm{kPa}$ . The measured 23.3 kPa is lower, so the mixture shows a negative deviation. Hydrogen bonds form between the δ+ H of $\mathrm{CHCl_3}$ and the δ− O of propanone, so the unlike molecules attract each other more strongly than like molecules do, and fewer molecules escape into the vapour. Because new, stronger attractions form, energy is released: the mixture warms up ( $\Delta H_{\text{mix}} < 0$ ).

Warning

Common mistakes. (1) Using mass fractions or volume fractions in Raoult's law: it needs mole fractions. (2) Using the liquid mole fraction for the vapour: $y_A = \frac{p_A}{p}$ , which is larger than $x_{A}$ for the more volatile component. (3) Reversing the link: weaker A–B attractions give a higher vapour pressure (positive deviation). (4) Comparing $p^{\circ}$ values at different temperatures.

Tip

Think of vapour pressure as "escaping tendency". Anything that holds molecules in the liquid lowers it; anything that loosens them raises it. Stronger A–B attraction: holds them, lower $p$ , negative deviation, heat given out.

Khan Academy India: Ideal and non-ideal solutions. Watch how the partial and total vapour pressure lines compare with Raoult's law for ideal mixtures and for mixtures with positive and negative deviations.

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Practice

State Raoult's law, and give two properties of an ideal solution.
A benzene–methylbenzene mixture has $x(\text{benzene}) = 0.25$ . Calculate its vapour pressure at 25 °C and the mole fraction of benzene in the vapour. ( $p^{\circ}$ : benzene 12.7 kPa, methylbenzene 3.79 kPa)
Using the same data, which liquid composition gives a vapour containing equal amounts of the two components?
A benzene–methylbenzene mixture has a total vapour pressure of 8.00 kPa at 25 °C. Find $x(\text{benzene})$ .
Predict whether a mixture of ethanol and hexane shows a positive or a negative deviation from Raoult's law, and explain.
When trichloromethane and propanone are mixed, the temperature rises. Explain what this tells you about the vapour pressure of the mixture.
Sketch and label the vapour pressure–composition diagram for an ideal mixture of A ( $p^{\circ} = 30\,\mathrm{kPa}$ ) and B ( $p^{\circ} = 10\,\mathrm{kPa}$ ), showing both partial pressures and the total.
A student says: "An ideal mixture gives a vapour with the same composition as the liquid." Evaluate this statement.

Show answersHide answers

The partial vapour pressure of a component equals its mole fraction in the liquid times the vapour pressure of the pure component: $p_A = x_A p_A^{\circ}$ . Ideal solutions: A–B forces equal A–A and B–B forces; $\Delta H_{\text{mix}} = 0$ ; volumes add; Raoult's law obeyed at all compositions.
$p_{\text{benzene}} = 0.25 \times 12.7 = 3.18\,\mathrm{kPa}$ ; $p_{\text{methylbenzene}} = 0.75 \times 3.79 = 2.84\,\mathrm{kPa}$ ; $p = 6.02\,\mathrm{kPa}$ ; $y = \frac{3.18}{6.02} = 0.528$ .
Equal partial pressures: $12.7 x = 3.79 (1 - x)$ , so $x = \frac{3.79}{16.49} = 0.230$ .
$12.7x + 3.79(1 - x) = 8.00 \Rightarrow 8.91x = 4.21 \Rightarrow x = 0.473$ .
Positive: hexane breaks up the hydrogen bonds between ethanol molecules but provides only weak London forces, so A–B attractions are weaker; molecules escape more easily and the vapour pressure is higher than predicted.
Heat is given out, so the new A–B attractions (hydrogen bonds) are stronger than those broken; molecules are held more strongly, so the vapour pressure is lower than Raoult's law predicts (negative deviation).
Three straight lines: $p_{A}$ from 0 (at $x_{A} = 0$ ) to 30 kPa (at $x_{A} = 1$ ); $p_{B}$ from 10 kPa to 0; total from 10 kPa to 30 kPa. All axes labelled: vapour pressure / kPa against $x_{A}$ .
Incorrect. Raoult's law gives $y_A = \frac{x_A p_A^{\circ}}{p}$ , which equals $x_{A}$ only if $p_A^{\circ} = p_B^{\circ}$ . Normally the vapour is richer in the more volatile component, e.g. 0.689 hexane in the vapour above 0.40 hexane in the liquid. (This is what makes fractional distillation work.)

Sources and credits

Vapour pressure data: standard values at 25 °C as tabulated in the CRC Handbook of Chemistry and Physics; the measured value for propanone + trichloromethane in the third example is illustrative.
Video: Ideal and Non-Ideal Solutions, Khan Academy India, embedded from YouTube.
The text, graphs, diagram, table, worked examples and Practice items were written and drawn by Ganzaa.

1What leaves a mixture first?

2Raoult's law and Dalton's law together

3Real solutions: deviations from Raoult's law

What leaves a mixture first?

Raoult's law and Dalton's law together

Real solutions: deviations from Raoult's law