Azeotropic, extractive and steam distillation
Minimum- and maximum-boiling azeotropes, azeotropic distillation with an entrainer, extractive distillation with a high-boiling solvent, pressure-swing distillation, and steam distillation with steam-consumption calculations.
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Why it matters
Ordinary distillation fails when relative volatility is close to 1 or equals 1 at an azeotrope (ethanol–water, isopropanol–water, acetone–methanol), or when the material would decompose at its normal boiling point (essential oils, fatty acids, high-boiling organics). Azeotropic, extractive and steam distillation are the standard industrial answers; knowing how each works, and how to estimate steam consumption, is expected in GATE and in plant interviews.
Key ideas
Azeotropes. An azeotrope is a liquid mixture whose equilibrium vapour has the same composition as the liquid (y = x, α = 1) at a given pressure. It boils at constant temperature like a pure substance, so ordinary distillation cannot cross it.
- Minimum-boiling (positive-deviation) azeotrope: activity coefficients > 1; the azeotrope boils below either pure component and goes overhead (ethanol–water, 95.6 wt % ethanol, about 78.2 °C at 1 atm).
- Maximum-boiling (negative-deviation) azeotrope: activity coefficients < 1; it boils above both components and leaves as bottoms (HCl–water, nitric acid–water).
- At the azeotrope, modified Raoult's law gives γ₁·P₁^sat = γ₂·P₂^sat = P.
- Azeotrope composition shifts with pressure, which is exploited in pressure-swing distillation (two columns at different pressures).
Azeotropic distillation. A third component, the entrainer, is added that forms a new, usually minimum-boiling and often heterogeneous (two-liquid-phase) azeotrope with one or both original components. The new azeotrope goes overhead, is condensed and split in a decanter; one liquid layer is refluxed and the other sent to recovery. Classic case: dehydration of ethanol with benzene (now cyclohexane or another hydrocarbon) – anhydrous ethanol leaves as bottoms, water is carried overhead in the ternary azeotrope.
Extractive distillation. A high-boiling, relatively non-volatile solvent is fed near the top of the column. It interacts more strongly with one component, changing activity coefficients so that the relative volatility of the original pair moves well away from 1. It does not form an azeotrope. The less-attracted component goes overhead; the other leaves with the solvent at the bottom and is separated in a second column, the solvent being recycled. Examples: ethylene glycol for ethanol–water, furfural or N-methylpyrrolidone for butadiene–butenes, phenol for toluene from paraffins. Solvent criteria: high selectivity, low volatility, easy recovery, thermal stability, low cost.
Steam distillation. Used for high-boiling, water-immiscible, heat-sensitive organics. Immiscible liquids each exert their own full vapour pressure regardless of the amounts present, so the mixture boils when p_water + p_organic = P. This happens below 100 °C at 1 atm and below the organic's own boiling point – the boiling point is not "reduced" for the organic alone; the mixture boils because the two vapour pressures add. Distillate composition is set by the ratio of vapour pressures. Live (open) steam can also be used without a separate liquid-water phase; then incomplete saturation of the steam is allowed for with a vaporisation efficiency E (typically 0.6–0.9). Vacuum lowers steam consumption further. Steam is also used to strip volatile organics from non-volatile oils.
Formulas
Azeotrope condition (modified Raoult): y_i = x_i and γ₁·P₁^sat = γ₂·P₂^sat = P
Relative volatility (non-ideal): α₁₂ = γ₁·P₁^sat / (γ₂·P₂^sat); azeotrope where α₁₂ = 1
Steam distillation with an immiscible water phase:
P = p_w^sat(T) + p_o^sat(T) (fixes the boiling temperature)
n_w/n_o = p_w/p_o, m_w/m_o = (p_w·18)/(p_o·M_o)
Live steam with vaporisation efficiency E and a non-volatile diluent (organic mole fraction x_o in the still liquid, ideal solution):
p_o = E·x_o·p_o^sat, p_w = P − p_o
m_w/m_o = (P − E·x_o·p_o^sat)·18 / (E·x_o·p_o^sat·M_o)
- P – total pressure, kPa; p – partial pressure, kPa; M_o – molar mass of organic, kg/kmol; m – masses, kg; n – moles; E – dimensionless.
Worked examples
Example 1 – steam requirement with a water layer (standard). Given: an immiscible organic (M = 130 kg/kmol) is steam-distilled at 101.3 kPa. The mixture boils at the temperature where p_w^sat = 86.3 kPa and p_o^sat = 15.0 kPa (sum 101.3 kPa). Find the kg of steam carried over per kg of organic.
m_w/m_o = (p_w × 18)/(p_o × M_o).- = (86.3 × 18)/(15.0 × 130) = 1553.4/1950 = 0.797.
0.797 kg steam per kg organic distilled, at a temperature well below 100 °C (p_w^sat = 86.3 kPa corresponds to about 95.5 °C).
Example 2 – live steam with incomplete saturation (GATE level). Given: the same organic is stripped with superheated live steam (no liquid water in the still) at 101.3 kPa and at a still temperature where p_o^sat = 15.0 kPa. The organic is practically pure (x_o ≈ 1) and the vaporisation efficiency is 0.75. Find steam needed to distil 100 kg of organic.
- p_o = E·p_o^sat = 0.75 × 15.0 = 11.25 kPa; p_w = 101.3 − 11.25 = 90.05 kPa.
- m_w/m_o = (90.05 × 18)/(11.25 × 130) = 1620.9/1462.5 = 1.108.
- Steam for 100 kg = 110.8 kg.
About 111 kg of steam (vs. 80 kg at E = 1) – poor contacting raises steam use by about 39 %.
Common mistakes
- Saying steam "lowers the boiling point of the component"; it is the sum of independent vapour pressures of immiscible liquids that reaches P at a lower temperature.
- Using mole ratios when mass of steam is asked (multiply by molar masses).
- Treating miscible mixtures as if each component exerted its full vapour pressure – that is only for immiscible liquids.
- Confusing the entrainer (forms a new azeotrope, goes overhead) with the extractive solvent (non-volatile, no azeotrope, leaves at the bottom).
- Thinking an azeotrope composition is fixed – it changes with pressure.
- Feeding the extractive solvent with the feed instead of near the top of the column.
For GATE CH
Expect conceptual questions comparing azeotropic, extractive and pressure-swing distillation and identifying minimum- and maximum-boiling azeotropes from T–x–y or x–y diagrams. Numericals cover steam consumption per kg of organic, boiling temperature of an immiscible mixture from vapour-pressure data, and the azeotrope condition with activity coefficients. Watch for vaporisation efficiency in steam problems.
Quick check
- What is the relative volatility at an azeotrope?
- Does an extractive solvent leave with the distillate or the bottoms?
- Why does an immiscible water–organic mixture boil below 100 °C at 1 atm?
- Name a method that exploits the pressure dependence of azeotrope composition.
Answers: 1. 1; 2. with the bottoms; 3. the two liquids exert their full vapour pressures independently, and their sum reaches 101.3 kPa below 100 °C; 4. pressure-swing distillation.
Interview questions
All Mass Transfer interview questionsTry answering each one aloud before you open it.
1.What is azeotropic distillation?Concept
An azeotrope is a mixture whose equilibrium vapour has the same composition as the liquid at a given pressure, so ordinary distillation cannot cross it. In azeotropic distillation a third component, the entrainer, is added that forms a new, usually minimum-boiling and often heterogeneous azeotrope with one of the components. That azeotrope goes overhead, separates into two liquid layers in a decanter, one layer is refluxed and the other recovered, while the desired component leaves pure as bottoms. The classic example is dehydrating ethanol with benzene or cyclohexane.
2.Explain extractive distillation and how it differs from azeotropic distillation.Concept
Extractive distillation is a process where a solvent is added to a mixture to change the relative volatility of the components, allowing for their separation. Unlike azeotropic distillation, extractive distillation does not rely on forming or breaking azeotropes. Instead, it uses a solvent that selectively interacts with one or more components, enhancing their separation.
3.What is steam distillation and when is it used?Concept
Steam distillation is used for high-boiling, water-immiscible, heat-sensitive organics such as essential oils or fatty materials. Because immiscible liquids each exert their full vapour pressure independently, the mixture boils when p_water + p_organic equals the total pressure, which happens below 100 °C at 1 atm and far below the organic's own boiling point. The distillate ratio is fixed by the vapour pressures: kg steam per kg organic = (p_w × 18)/(p_o × M_o). Vacuum further reduces temperature and steam consumption.
4.Why is a solvent used in extractive distillation?Application
A solvent is used in extractive distillation to alter the relative volatility of the components in the mixture. By selectively interacting with one or more components, the solvent enhances the separation process, making it possible to separate components that are otherwise difficult to separate by conventional distillation methods.
5.What happens if the wrong solvent is chosen for extractive distillation?Application
If the wrong solvent is chosen for extractive distillation, it may not effectively alter the relative volatility of the components, leading to poor separation. Additionally, the solvent might react with the components or introduce impurities, complicating the separation process and potentially damaging the equipment.
6.How does the presence of an azeotrope affect the distillation process?Application
The presence of an azeotrope in a mixture means that the components have a constant boiling point and composition during distillation, making it impossible to separate them by simple distillation. Special techniques like azeotropic or extractive distillation are required to break the azeotrope and achieve separation.
7.An immiscible organic of molar mass 130 kg/kmol is steam-distilled at 101.3 kPa; at the boiling temperature p_water = 86.3 kPa and p_organic = 15.0 kPa. How much steam is needed to distil 100 kg of the organic?Numerical
With a liquid water layer present, the vapour contains water and organic in the ratio of their vapour pressures. Mass ratio = (86.3 × 18)/(15.0 × 130) = 0.797 kg steam per kg organic, so about 80 kg of steam leaves with 100 kg of organic. Extra steam is needed to heat the charge and cover losses, and if the steam is not fully saturated with organic (vaporisation efficiency below 1) consumption rises further.
8.What are the advantages of using steam distillation over simple distillation for essential oils?Application
The oil distils at below 100 °C instead of at its normal boiling point, often above 200 °C, which prevents thermal degradation, polymerisation and loss of aroma. No vacuum equipment is needed, the steam also supplies heat, and the condensate separates easily into an oil layer and a water layer because the two are immiscible. The price is a large steam consumption when the oil's vapour pressure is low.
9.Explain how the addition of a third component can break an azeotrope in azeotropic distillation.Concept
The entrainer is chosen so that it forms a new azeotrope, typically a minimum-boiling ternary or binary one, that boils below the original azeotrope. That new azeotrope becomes the overhead product and removes one of the original components (for example, water in ethanol dehydration), so the other component can be drawn off pure at the bottom. If the new azeotrope is heterogeneous, the condensed overhead splits into two liquid phases in a decanter, which lets the entrainer be recycled and the removed component be withdrawn.
10.A mixture of ethanol and water forms an azeotrope at about 95.6 % ethanol by weight. How would you separate this mixture using azeotropic distillation?Numerical
First concentrate the dilute feed to near the azeotrope in an ordinary column. Feed this to a second column with an entrainer such as cyclohexane (historically benzene), which forms a minimum-boiling heterogeneous ternary azeotrope with ethanol and water. The ternary azeotrope goes overhead, condenses and splits in a decanter into an entrainer-rich layer, refluxed to the column, and a water-rich layer, sent to a stripper to recover entrainer and ethanol. Anhydrous ethanol leaves as the bottoms product.
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