Fractional Distillation & the Ethanol–Water Azeotrope

Choose a pair of liquids, the composition in the still pot and a number of theoretical stages. The page finds where the pot boils and what its vapour holds, then condenses and re-boils that vapour stage by stage up the column, and shows what reaches the top.

Loading interactive simulation...

Lesson

The theory — Fractional Distillation & the Ethanol–Water Azeotrope

Over a liquid mixture each component has a partial pressure pᵢ = xᵢ γᵢ Pᵢˢᵃᵗ(T). Raoult's law is the ideal case γᵢ = 1, where each liquid evaporates in proportion to how much of the mixture it is. The liquid boils at its bubble point, where the partial pressures add up to the pressure above it, and the first vapour has yᵢ = pᵢ / P. A theoretical stage is one such equilibrium; in a column its vapour, condensed, becomes the liquid of the stage above. Where y = x strictly between the pure liquids, stages stop separating, and that composition is an azeotrope.

What each symbol means

x, y
mole fraction of the more volatile liquid, in the liquid and in the vapour
γ
activity coefficient: 1 in an ideal mixture, above 1 where the two kinds of molecule would rather not be neighbours
Pˢᵃᵗ
vapour pressure of the pure liquid at the same temperature, from its Antoine equation
α
relative volatility, γ₁P₁ˢᵃᵗ / γ₂P₂ˢᵃᵗ: the vapour's ratio of the two liquids divided by the liquid's
Assumes
The pressure is one standard atmosphere, 101.325 kPa, and the vapour is an ideal gas. Every stage reaches equilibrium, which a real tray does not, so a real column needs more trays than theoretical stages. The column runs at total reflux, all its vapour condensed and returned, which gives the fewest stages any separation can need. Vapour pressures are NIST's Antoine fits, and two are stretched past the range they were fitted over: benzene's by 30 K, up to toluene's boiling point, and ethanol's by 6.5 K, up to water's.
Breaks when
Raoult's law breaks for ethanol and water, which is the page's subject. The NRTL parameters that replace it were fitted to measurements rather than derived, and they put the azeotrope at 90.14 mol% where it is measured at 89.4. At lower pressure the measured azeotrope moves toward pure ethanol, 91.5 mol% at 36.78 kPa and 94.1 mol% at 17.52 kPa, and the page, fixed at one atmosphere, cannot show that. Liquids that split into two layers, as butanol and water do, need a model with two liquid phases, which this page does not have.

No number of stages takes ethanol past the azeotrope 🖖

Press Ethanol by Raoult. If the two liquids mixed ideally, twelve stages would lift a 10 mol% wash to 99.954 mol% ethanol, since ethanol boils at 78.3 °C and water at 100 °C and nothing else would matter. Press Ethanol, real: the same twelve stages reach 87.10 mol%. On the right-hand panel the curve crosses the diagonal, and a stage whose vapour has the same composition as its liquid separates nothing. That crossing is the azeotrope, measured at 89.4 mol%, which is 95.57% ethanol by mass, boiling at 78.17 °C, a little below pure ethanol. Forty stages add less than three points more, and stop just under the model's own azeotrope, which sits 0.7 points above the measured one. So ethanol from a still stops near 95.6% by mass, and ethanol sold as absolute, 99.5% and above, has been dried some other way: a third liquid added to break the azeotrope, or a molecular sieve that holds back the water.

Each extra nine of purity costs the same few stages 🖖

Press Benzene–toluene and read the purity table. From an equimolar pot, 99 mol% benzene takes 5 stages, 99.9% takes 8 and 99.99% takes 10. The last nine costs no more than the one before it. At total reflux each stage multiplies the ratio of benzene to toluene by the relative volatility, about 2.5 here, so the ratio grows geometrically and every factor of ten costs the same two or three stages. Open the stage table and multiply its α column: 2.491 × 2.541 × 2.573 × 2.589 × 2.596 = 109.5, and a top at 99.10 mol% benzene is a ratio of 109.5 to 1. These are minimum counts. A column that draws product off the top needs more stages than total reflux does, never fewer.

From a weak wash, the first vapour is three times richer than Raoult's law says 🖖

Press A pot still: one stage, from 5 mol% ethanol. The first vapour is 32.91 mol% ethanol, where Raoult's law gives 10.52. Water holds a few ethanol molecules badly, and an activity coefficient of 4.162 says so in a number. The behaviour that pins the top of the column at the azeotrope makes its lowest stages better than ideal: from a 10 mol% pot the real liquid reaches 50 mol% in 2 stages where Raoult's law needs 3, and falls behind only on the way to 80 mol%, 6 stages against 5. The measurements agree. At 32.86 kPa a liquid of 6.3 mol% ethanol boils off vapour at 35.6 mol%, against 36.8 from the NRTL parameters this page uses and 13.5 from Raoult's law.

Problem solved in full

  1. A column that makes both ends pure 4 steps

    The page steps only upward from its pot. A real column is fed in the middle and has to be pure at both ends. An equimolar benzene–toluene feed is to give 99.9 mol% benzene at the top and 99.9 mol% toluene at the bottom. At total reflux, how many theoretical stages does it need? The page gives the relative volatility as 2.347 at the toluene end and 2.600 at the benzene end: set the pot to 0.1 and then to 99.9 mol% to read them.

    Open this problem in the tool

    1. The vapour's ratio of benzene to toluene is α times the liquid's, and that vapour becomes the next liquid, so every stage multiplies the ratio by its own α.

    2. Check it on the preset's five stages: the product of the stage table's α column is the ratio at the top.

    3. Over the whole column the ratio goes from 1 : 999 at the bottom to 999 : 1 at the top, a factor of 999². α changes slowly along the way, and it is the logarithms that add, so take the geometric mean of its two ends.

    4. The number of stages is the logarithm of the whole factor divided by the logarithm of α. This is Fenske's equation, published in 1932.

    Answer

    15.3, so 16 theoretical stages, counting the reboiler at the bottom as one. Set the pot to 0.1 mol% and read the purity table: stepping up one stage at a time from there, the page also needs 16 to reach 99.9%. The shortcut and the stage-by-stage count agree because α hardly moves along this column, from 2.347 to 2.600. For a pair whose α drifts further, the shortcut drifts with it.

Learning path

Mixtures and solutions

References (3)