Chapter 2a

Phase Equilibria and the Ideal Binary

Where the diagrams come from. Everything in Chapters 3 to 7 is drawn on a phase diagram — this chapter is about what those diagrams are, and why they look the way they do.

Wankat, Ch. 2 5 interactive apps Leads into Ch. 2b and Ch. 3

By the end of this chapter you should be able to

Apply the Gibbs phase rule and say what its answer means for a real separation.
State the condition for phase equilibrium in terms of chemical potential and of fugacity, and explain why the second form is the useful one.
Derive Raoult's law from the general equilibrium statement, and say exactly which assumptions it needs.
Explain why a mixture boils over a range of temperature while a pure component does not.
Construct a diagram from two vapour-pressure curves and a total pressure.
Define the distribution coefficient and the relative volatility , and judge when treating as constant is safe.
Apply the lever-arm rule on any two-phase diagram.

The two pillars of chemical engineering

Equilibrium	Kinetics
Thermodynamics	Rate
Defines the boundaries of the process.	Defines the speed of the process.
Answers "How much?"	Answers "How fast?"
Chemical potential, fugacity, equilibrium coefficient, equilibrium selectivity, saturation capacity.	Diffusion coefficient, mass transfer coefficient.

This course leans overwhelmingly on the left column. Many unit operations are equilibrium-controlled — mass transfer is fast enough that we can leave it out of a first analysis — so equilibrium data alone tells us what separation is achievable and roughly how large the equipment must be. When mass transfer is slow enough to matter, it shows up as an efficiency correction bolted onto an equilibrium calculation, which is exactly how Chapter 4 handles it.

Chapter 1 argued1 §1.4 that a separation is possible only because the components differ in some property, and that undoing a mixture always costs work. Equilibrium is the language in which "differ" is made precise: everything in this chapter is machinery for saying by how much, and therefore how hard the separation will be.

Why this chapter earns its place

Equilibrium data is expensive. Companies spend serious money measuring it, because a design is only as good as the phase diagram it was drawn on. And small impurities can shift equilibrium out of all proportion to their concentration. A separation engineer who cannot read, question and construct a phase diagram is guessing.

2.1 The phase-equilibrium problem

Figure 2.1 — A typical two-component, two-phase system at equilibrium. is pressure, temperature, the mole fraction of component in the liquid and that in the vapour.

Two phases in contact are at equilibrium when nothing further changes. That single statement has three separate consequences, and it is worth naming all three because students routinely remember only the first two.

Mechanical equilibrium — no net force, so no pressure gradient

(2.1)

Thermal equilibrium — no net heat flow

(2.2)

Chemical equilibrium — no net transfer of any species across the interface

(2.3)

Equation (2.3) is the one that does the work. , the chemical potential of component , plays the role for mass transfer that temperature plays for heat transfer: matter moves from high chemical potential to low, and stops when they are equal. Note carefully that equal chemical potential does not mean equal concentration — which is precisely why separation by phase contacting is possible at all.

2.2 The Gibbs phase rule

Before computing anything, it pays to know how many things you are allowed to specify. Consider components distributed among phases, with no chemical reaction.

Counting variables

To fix the intensive state of one phase you need its independent mole fractions plus and — that is numbers per phase, so

Counting equations

Equilibrium supplies equalities across the phases. Temperature gives equations, pressure another , and each of the chemical potentials a further :

The degrees of freedom are what is left over:

(2.4)

Read the answer correctly

counts intensive variables — pressure, temperature, mole fractions. It says nothing about flow rates or amounts, which are extensive. A flash drum has for a binary two-phase system, yet you still need a feed rate to size it. The degrees-of-freedom table for a flash drum3 §3.1 is a separate, larger count that includes the extensive variables; do not confuse the two.

App 1

Phase-rule counter

Set the number of components and phases and watch the bookkeeping. The diagram on the right highlights where each answer applies on a real phase diagram.

Components C 1

Phases 𝒫 1

Pressure

The count

Every term in equation (2.4), evaluated.

Variables 𝒫(C+1)	—
Equations (C+2)(𝒫−1)	—
Degrees of freedom ℱ = C − 𝒫 + 2	—
…with pressure fixed	—

✓—

Where it applies

—

Figure 2.2 — Phase diagram of a one-component system. An area is two-phase-rule-degrees-of-freedom wide, a coexistence line is one, and the triple point is zero — you cannot choose anything there.

Figure 2.3 — Phase diagram of a two-component system at constant pressure. With , ; fixing pressure spends one, so the two-phase region needs exactly one more number — a temperature, or a composition — to be pinned down.

2.3 Fugacity — making equation (2.3) usable

Chemical potential is the right variable physically but a poor one computationally: it runs to as concentration goes to zero, and it has no natural units to build intuition with. So we define a new property, the partial fugacity of component :

(2.5)

Fugacity has units of pressure and can be read as an "escaping tendency" — the effective pressure a component exerts in its attempt to leave a phase. Because the exponential is monotonic, equality of chemical potentials is exactly equivalent to equality of fugacities:

(2.6)

Written out for a vapour–liquid system in the form actually used for calculation:

(2.7)

γ_i: activity coefficient — corrects the liquid for non-ideality. Obtained from Margules, van Laar, Wilson, NRTL, UNIQUAC. This is the subject of Chapter 2b.
φ̂_i^V: partial fugacity coefficient — corrects the vapour for non-ideality. Obtained from an equation of state such as Peng–Robinson.
f_i^0L: pure-component fugacity at the standard state.
p: total pressure.

The whole of VLE in one line

Equation (2.7) is the general statement. Every specific model you will meet — Raoult's law, Henry's law6 §6.2, K values, relative volatility — is (2.7) with particular choices of and . When a correlation surprises you, come back to this equation and ask which of its two correction factors has been assumed away. Chapter 2b relaxes the first of them: see what happens when 2b §2b.2.

2.4 Raoult's law

Now make the two simplest possible assumptions.

Assumption	Consequence	When it is reasonable
Ideal liquid	and , the pure-component vapour pressure	Chemically similar molecules — benzene/toluene, adjacent n-alkanes, xylene isomers
Ideal gas		Low pressure, well away from the critical point

Substituting both into (2.7) collapses it to Raoult's law:

(2.8)

and summing over the components, since the mole fractions of the vapour must add to one, the total pressure is a simple mole-fraction-weighted average of the two pure vapour pressures:

(2.9)

Figure 2.4 — Raoult's law at constant temperature. Each partial pressure is a straight line from zero to its own pure vapour pressure, and the total is their sum — hence a straight line from to . Straightness is the signature of ideality, and it is the first thing to go once real molecules are involved2b §2b.1.

Note the vapour pressure notation. Your thermodynamics course probably wrote this as or ; this course writes . They are the same quantity. It is obtained from the Antoine equation,

(2.10)

with tabulated constants , , — and a stated temperature range and pressure unit that you must check before using them, because different sources tabulate mmHg, kPa and bar.

2.5 Boiling a mixture

Here is the first place your intuition will mislead you, because your intuition was trained on water.

Figure 2.5 — Heating a mixture at constant pressure. All liquid at composition ; the first bubble of vapour ; roughly half vaporised; the last drop of liquid ; all vapour at .

Follow the compositions across that sequence. The first bubble is much richer in the volatile component than the liquid it came from — that is the separation we exploit. But that bubble has to come from somewhere, so the liquid left behind is depleted, and a depleted liquid boils hotter. The temperature therefore climbs continuously throughout boiling. It only stops climbing when the last drop of liquid vanishes.

The single most important sentence in this chapter

A pure component boils at one temperature. A mixture boils over a range — from its bubble point to its dew point. The width of that range is a direct measure of how separable the mixture is, and a mixture whose range has collapsed to zero cannot be separated by distillation at all. That is exactly what an azeotrope2b §2b.6 is, and it is the subject of Chapter 2b.

Figure 2.6 — The generic diagram at constant pressure. Below the lower (bubble) curve everything is liquid; above the upper (dew) curve everything is vapour; between them the two coexist, and a horizontal tie line joins the compositions that are in equilibrium at that temperature.

App 2

The heating curve — why a mixture has no boiling point

Heat a fixed charge at constant pressure and plot temperature against heat added. Compare a mixture with a pure component. Then move the feed composition towards either pure end and watch the mixture behaviour turn back into the plateau you expected.

Feed composition z (benzene) 0.50

Heat added 0 %

compare with pure benzene

Heating curve

Temperature against cumulative heat added, at 1 atm. The sloped section is the boiling range.

Mixture Pure benzene Current state

What is in the vessel

Liquid below, vapour above; the numbers are the two compositions.

…and where that is on the diagram

Benzene–toluene at 1 atm, computed from Raoult's law and Antoine constants.

Bubble (liquid) Dew (vapour) Tie line

Readings

Everything the state of the vessel is described by.

Temperature

—°C

Fraction vaporised

—

x (liquid)

—

y (vapour)

—

Bubble point

—°C

Dew point

—°C

Boiling range ΔT

—°C

✓—

Do it now

Set the composition slider to and then to . What happens to the boiling range, and to the shape of the heating curve? Now try . Is a 2 % impurity enough to destroy the plateau?

Reveal

At or the bubble and dew points coincide, ΔT collapses to zero, and the sloped section becomes a vertical jump — the flat plateau of a pure component. The plateau is not a different phenomenon from the mixture behaviour; it is the limiting case of it.

At the range is a couple of degrees rather than zero. Small, but not nothing — and this is why boiling point is used as a purity test in the laboratory. A "sharp" boiling point means a pure substance; a substance that boils over several degrees is telling you it is a mixture. It is also why the boiling range, not the boiling point, is the meaningful specification for a petroleum cut.

2.6 Building the diagram from vapour-pressure curves

The diagram is not a primitive object. For an ideal system it is fully determined by two Antoine curves and one number — the total pressure. Here is the construction.

Pick a temperature between the two pure boiling points. Evaluate both vapour pressures at that temperature. Raoult's law for the total pressure, equation (2.9) written for a binary, is

which contains only one unknown, so solve it for the liquid composition that boils at exactly this temperature and pressure:

(2.11)

and then get the vapour in equilibrium with it straight from Raoult's law:

(2.12)

That is one tie line — one horizontal slice of the diagram. Repeat for every temperature between the two boiling points and the whole diagram appears. Equation (2.11) is sometimes called the flash equation for temperature, and it is worth noticing that it needs no iteration at all for an ideal system — a convenience you will exploit again in binary flash calculations3 §3.2.

App 3

From vapour-pressure curves to a diagram

Drag the temperature line on the left panel — or press Sweep — and watch the diagram draw itself, one tie line at a time. The middle panel shows the isothermal picture at that temperature; the right panel accumulates the result.

Component pair

System pressure 101.3 kPa

Temperature —

1 · The two Antoine curves

Vapour pressure of each pure component against temperature.

Light Heavy System p

2 · at that temperature

Bubble line straight (Raoult); the system pressure cuts it at , and the dew line at .

Bubble Dew

3 · …which builds the

Each temperature contributes one point to each curve.

Bubble Dew

Temperature

—°C

(VP)_light

—kPa

(VP)_heavy

—kPa

—

α at this T

—

Boiling pts

—°C

✓—

Think!

Drop the system pressure to 20 kPa and sweep again. Three things change: the whole diagram moves, its width changes, and changes. Which of the three is the reason vacuum distillation is used for heat-sensitive materials, and which is a bonus?

Reveal

The reason is that the whole diagram moves down in temperature: both boiling points fall, so a material that would decompose at its atmospheric boiling point can be distilled intact. That is the point of vacuum distillation.

The bonus is that generally increases as pressure falls, because the two vapour-pressure curves diverge in ratio at lower temperature. A wider two-phase lens means an easier separation and fewer stages. You pay for both in vacuum equipment, larger column diameter for the same molar flow, and condenser duty at a lower temperature.

Try p-xylene / m-xylene at any pressure and you will see the third case: a lens so thin the two curves are almost superimposed. That is the , 1000-plate separation in the table below.

2.7 The distribution coefficient and the relative volatility

Rearranging Raoult's law (2.8) puts it in the form used everywhere downstream:

(2.13)

is the distribution coefficient (or K value, or equilibrium ratio) of component : the factor by which that component is concentrated in the vapour relative to the liquid. means the component prefers the vapour. Two routes to a value:

Graphical correlations — the DePriester chart for light hydrocarbons, read at a given and . See the reference box below.
Mathematical correlations — the Antoine equation (2.10), or Clausius–Clapeyron. The Rachford–Rice solver3 App 2 will compute this way for you.

Reference chart — not reproduced here

The DePriester chart gives values for light hydrocarbons directly, as a nomograph in pressure and temperature. It is a copyrighted figure, so rather than reproduce it we point you to it:

Original source: C. L. DePriester, "Light-hydrocarbon vapor–liquid distribution coefficients", Chemical Engineering Progress Symposium Series 49(7), 1–43 (1953).
In your textbook: Wankat, Separation Process Engineering, §2.5 — the chart is reproduced there in both low- and high-temperature versions, and that is the copy to use for assignments.
Also in: Perry's Chemical Engineers' Handbook, section on vapour–liquid equilibrium. Available through the University of Alberta Library.
A high-resolution scan is posted on the course Canvas page under Chapter 2 resources.

If you would rather compute than read a nomograph, the underlying vapour-pressure data is in the NIST Chemistry WebBook, and App 2 in Chapter 3 will compute from Antoine constants for you.

The trouble with is that it depends strongly on temperature — and in a distillation column the temperature changes from tray to tray, so changes everywhere. The fix is to work with a ratio of K values, because the strong temperature dependences largely cancel. For a binary, the relative volatility is defined as

(2.14)

and for an ideal system this is simply the ratio of the two vapour pressures, — the pressure cancels entirely.

If can be treated as constant across the whole composition range, equation (2.14) can be solved explicitly for :

(2.15)

This is the single most useful equation in the course. It turns a table of experimental data into one number, and it is the reason so many problems can be worked analytically — the constant-α option in the binary flash explorer3 App 1 uses exactly this. It is not recommended for wide-boiling mixtures, nor for non-ideal systems, where activity coefficients2b §2b.2 enter the definition of and can drive it to 1.

Figure 2.8 — Equation (2.15) plotted for several constant relative volatilities. The further the curve bows above the diagonal, the easier the separation; at it collapses onto the diagonal and separation by distillation becomes impossible.

The whole difficulty of a separation, in one number
System	, kPa		Plates needed
Water (A) – glycerol (B)	101.3	191 – 683	1
Methanol (A) – water (B)	101.3	2.45 – 7.58	30
p-Xylene (A) – m-xylene (B)	101.3	1.002 – 1.018	1000

App 4

How constant is "constant "?

and vary enormously with temperature; their ratio much less. This app quantifies "much less" for a real pair, and then shows what the constant- shortcut costs you on the diagram.

Component pair

System pressure 101.3 kPa

Compare using

, and across the column's temperature range

Between the two pure boiling points at this pressure. Note the log scale on K.

(light) (heavy)

What the shortcut costs

True Raoult equilibrium curve against equation (2.15) using a single .

True (Raoult + Antoine) Constant α

T range

—°C

α at low T

—

α at high T

—

α used

—

Variation in α

—%

Max error in y

—

✓—

2.8 The same data, three ways

Equilibrium data for one binary at one pressure can be drawn as a , an , or an diagram. They contain overlapping information and each is convenient for a different job, so you must be fluent in all three.

Figure 2.9 — to . Each tie line contributes one point: its two endpoints become the coordinates . Temperature is discarded, which is exactly why the diagram is the natural stage-counting diagram in Chapter 4.

Figure 2.10 — to . Replacing temperature with enthalpy makes energy balances graphical, which is how the heat duties in Chapter 3 problems are read off.

2.9 The lever-arm rule

You know the composition of a two-phase mixture and the two compositions in equilibrium with each other at that condition. How much of each phase do you have? This comes up so relentlessly that it is worth deriving once and then never again.

Take moles of overall composition , splitting into moles of liquid at and moles of vapour at .

(2.16)

Substituting the first into the second and rearranging:

(2.17)

Look at what that says geometrically. On the tie line, is the distance from the liquid end to the feed point, and is the distance from the feed point to the vapour end. So the ratio of the two amounts is the inverse ratio of the two distances — a lever, balanced at .

Remember it as a seesaw, not a formula

Put the fulcrum at . Hang the liquid at and the vapour at . The lever balances: . The phase you are nearer to on the tie line is the phase you have more of — a heavy weight sits on a short arm. Get that picture straight and you will never invert the ratio by accident, which is the single most common arithmetic error in this course.

App 5

The lever

Drag the feed point along the tie line, or drag the tie line itself. The beam below always balances — that is the rule. Switch the diagram to see that the same construction works on , and alike.

Diagram

Temperature 225 °F

Feed z 0.400

F, lbmol

Hexane–octane, at 101 kPa

Tabulated course data. Drag anywhere on the plot to move the feed point.

The lever

Fulcrum at ; the beam is level because .

x (liquid)

—

y (vapour)

—

—lbmol

L/V

—

Left arm z−x

—

Right arm y−z

—

Torque check

—

Predict before you lookMove the feed point somewhere new, decide which phase you have more of, then check.

✓—

You will meet this again

The lever-arm rule is not a distillation trick. It follows from a mass balance and nothing else, so it applies on any diagram where a mixture point lies between two phase points on a straight line: the diagram for energy balances, the ternary diagram in Chapter 7 for extraction, and the mixing constructions throughout. You meet it immediately in the binary flash operating line3 §3.2, where the same ratio sets how much vapour and liquid leave the drum. Learn it once, properly, here.

2.10 Worked examples

Worked example 2.1Lever-arm rule on a diagram

Using the hexane–octane data at 101 kPa: at with an overall composition , find , , , and the compositions of the two phases. Then for lbmol find the amounts of liquid and vapour.

Solution

Step 1 — read the tie line. Draw a horizontal line at 225 °F. It cuts the bubble curve at and the dew curve at . Interpolating the tabulated data between 223.22 °F (, ) and 227.72 °F (, ) gives those values.

Step 2 — check you are actually in the two-phase region. : ✓. If this fails, the mixture is single-phase and the lever-arm rule does not apply.

Step 3 — lever arm.

Since ,

Step 4 — amounts. For lbmol: lbmol and lbmol.

Sanity check. The feed point at 0.400 sits closer to the dew end (0.526) than to the bubble end (0.178)? No — it is 0.222 from the liquid and 0.126 from the vapour, so it is nearer the vapour, and indeed we have more vapour than liquid. The lever picture and the arithmetic agree.

Check it in App 5. This is the default state — F, .

Worked example 2.2Constructing one tie line from Antoine constants

For benzene (1) and toluene (2), Antoine constants in mmHg and °C are , , and , , . At a total pressure of 101.3 kPa (760 mmHg) and a temperature of 95 °C, find , and .

Solution

Vapour pressures at 95 °C.

so mmHg kPa. Similarly

giving mmHg kPa.

Liquid composition from equation (2.11):

Vapour composition from equation (2.12):

Relative volatility, which for an ideal system is just the vapour-pressure ratio:

Cross-check with equation (2.15). ✓ — agreement to three figures, because at a single temperature the constant- form is not an approximation at all.

Check it in App 3. Select benzene/toluene, leave the pressure at 101 kPa and drag the temperature to 95 °C.

2.11 Check your understanding

Five multiple-choice questions, two short problems and two long problems, all graded in your browser. Nothing is submitted or recorded; your answers stay on this device so a refresh does not lose them.

Marks earned

0 / 0

Multiple choice

Short problems

Enter numbers only. Answers are marked correct within the tolerance shown after you check them.

Long problems

Multi-part. Each part is marked separately.

Where this chapter connects

Straight to work: every diagram you construct here is the diagram a flash calculation3 §3.2 is drawn on, and the lever-arm rule is the arithmetic behind its phase split3 App 1.
Where it breaks: Raoult's law assumes an ideal liquid. Chapter 2b2b §2b.1 removes that assumption and shows what it costs — including mixtures that cannot be distilled at all2b §2b.6.
K values defined here are the input to the Rachford–Rice equation3 §3.3 for multicomponent flash.
Relative volatility is the number that sets how hard every separation in the course will be; it reappears in the McCabe–Thiele construction of Chapter 4 and the Fenske and Underwood equations of Chapter 5.

Summary & key equations

Equilibrium

Conditions for equilibrium, ,

In terms of fugacity

General VLE statement

Gibbs phase rule (intensive variables only)

The ideal binary

Raoult's law

Total pressure

Antoine equation

Tie line at a given , ,

K values and relative volatility

Distribution coefficient

Relative volatility (ideal)

Constant-α equilibrium curve

Separation impossible when

Two-phase bookkeeping

Mass balances,

Lever-arm rule

In wordsThe phase you are nearer to is the one you have more of

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1 · Introduction

2b · Deviations from Ideality