Vapor-Liquid Equilibrium (VLE) at Low Pressures

Chapter 8

Why study VLE?

Many chemical and environmental processes involve vapor-liquid equilibria Drying Distillation Evaporation

N2 + 3 H2 ⇋ 2NH3

Liquid Ammonia

3 moles H2

1 mole N2

Recycled Product Ammonia and unreacted feed

ChillerCondenses most of the ammonia

Separator

Reactor partially converts H2 and N2

to NH3

Bleed Stream

~15% conversion

Consider the ammonia production process discussed in Chapter 1

Controlled by VLE

+ N2 and H2

NH3 vapor+ N2 and H2

Most of what we’ve discussed so far in the course is VLE

Raoult’s law is a vapor-liquid equilibrium estimating equation

Henry’s law is vapor-liquid equilibrium estimating equation

The biggest use of VLE analysis is in distillation Check out the great picture in the text

book – page 160 Distillation is separation by boiling point

Distillation Columns

Simple VLE Measurement Device

Othmer Still

Some standard conventions used in VLE

The lowest-boiling component (most volatile) is usually called species a.

The next lowest is species b etc. Tables are arranged similarly

Data from Table 8.1Boiling Temp

Mole Fraction Acetone in Liquid

Mole Fraction Acetone in Vapor

100 0 0

74.8 0.05 0.6381

68.53 0.1 0.7301

65.26 0.15 0.7716

63.59 0.2 0.7916

61.87 0.3 0.8124

60.75 0.4 0.8269

59.95 0.5 0.8387

59.12 0.6 0.8532

58.29 0.7 0.8712

57.49 0.8 0.895

56.68 0.9 0.9335

56.3 0.95 0.9627

56.15 1 1

This data is used in examples 8.1 to 8.3, and is plotted on the next slide

Acetone-Water Composition

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.2 0.4 0.6 0.8 1

Acetone fraction in liquid

Ace

ton

e fr

acti

on

in

gas

Lowest boiling point component

If all we ever dealt with were binary systems, and if tables like the one used to create this figure were available for all combinations of species, we wouldn’t need VLE correlations

Equilibrium Curve

Reference Curve

For Multispecies systems, there is no simple graph we can make

The K factor can be used to help solve this problem

Relative volatility is an additional approach

i

ii x

yK

ab

ba

xy

xy

K

K

species volatileLess

species volatileMore

Volatility Measures

0.1

1

10

100

0 0.2 0.4 0.6 0.8 1

Mole Fraction in the Liquid Phase of Acetone

Rel

ativ

e vo

lati

lity

an

d "

K"

fact

ors

Relative Volatility,

K, acetone

K, water

If is greater than 1.5 to 2 over the whole range of composition, then distillation is almost always the cheapest separation technique

Mathematical Treatment of Low-Pressure VLE

The tables and graphs we’ve just looked at are easy to interpret

It would be nice if we could calculate values, instead of reading them off graphs

Our starting point is that the fugacity of the gas phase equals the fugacity of the liquid phase

)2()1(ii ff

0)2()2(0)1()1(iiiiii fxfy

For gases we usually choose the total pressure as the standard fugacity P

For liquids we usually choose the pure component partial pressure as the standard fugacity

0iP

For ideal gases the activity coefficient is one1

0)2(

ii

iii Px

Py

The activity coefficient of the liquid phase

0)2(

ii

iii Px

Py

In example 8.2, we use this equation to find the activity coeffients for water and for acetone at a number of concentrations

Partial pressure of the gas

Example 8.2

0acetoneacetone

Totalacetoneacetone Px

Py

0waterwater

Totalwaterwater Px

Py

The pure component vapor pressures are a function of temperature, and can be calculated from the Antoine equation at the appropriate boiling points.

Calculated activity coefficients

1

10

0 0.2 0.4 0.6 0.8 1

Mole fraction acetone in the liquid phase

acti

vity

co

effi

cien

t

acetone

water

Raoult’s Law

0ii

ii Px

Py If we rearrange this

equation, we find…

0iiii PxPy

Which is the same as Raoult’s law, except for the activity coefficient. Raoult’s law applies to ideal solutions, where the activity coefficient is one.

The acetone water system is not ideal, since the calculated activity coefficients are not one!!

How much would we be off if we assumed ideal solution behavior for acetone?

At 1 atm andXacetone=0.05

Experimental values from Table 8.1

Calculated values using =1

Equilibrium boiling temperature

74.8 96.4

Mole fraction acetone in the vapor phase

0.6381 0.1656The calculational details are in Example 8.3

The Four most common types of Low Pressure VLE

Ideal Solution Behavior Positive Deviations from Ideal

Solution Behavior Negative Deviations from Ideal

Solution Behavior Two-Liquid Phase --Heteroazeotropes

1

1

1

Ideal Solution Behavior – Type I behaviorThe activity coefficient of both species is one at all concentrations – Figure 8.7b

Consider a benzene-toluene system

The two species are very similar chemically, and behave as an ideal solution

Benzene has the lower boiling, and therefore the higher vapor pressure Mole fraction of benzene

in the liquid phase, xa

0 0.2 0.4 0.6 0.8 1.0

Act

ivit

y C

oeffi

cient,

0

.1 1

.0 1

0

BenzeneToluene=1

At any P and T

Ideal Solution Behavior Since the activity

coefficients of both species is one, they both follow Raoult’s law

Mole fraction of benzene in the liquid phase, xa

0 0.2 0.4 0.6 0.8 1.0

0iiii PxPy

Pre

ssure

, to

rr

0 4

00

8

00

1

20

00iii PxP

1

1

Ideal Solution Behavior Figure 8.7a

Mole fraction of benzene in the liquid phase, xa

0 0.2 0.4 0.6 0.8 1.0

0iiii PxPy

Pre

ssure

, to

rr –

at

90

C

0 4

00

8

00

1

20

0

0iii PxP

torrxP BenzeneBenzene 1021

torrxP TolueneToluene 074

PBenzene

PToluene

PTotal

TolueneBenzeneTotal PPP

1

Ideal Solution BehaviorFigure 8.7c For an ideal

solution, the mole fraction of the most volatile component, is always higher in the gas than in the liquid

Mole fraction of benzene in the liquid phase, xa

0 0.2 0.4 0.6 0.8 1.0

Mole

fra

ctio

n o

f benze

ne

in t

he liq

uid

phase

0 0

.2 0

.4 0

.6 0

.8 1

.0

Equilibrium curve

Reference curve

Not to Scale

1

Ideal Solution BehaviorFigure 8.7 d

If we heat a mixture of benzene and toluene, the temperature where it starts to boil is called the bubble point

The combined partial pressures equal 1 atm Mole fraction of benzene

in the liquid or vapor phase, xa

0 0.2 0.4 0.6 0.8 1.0

Tem

pera

ture

, C

75

8

5 9

5 1

05

1

15

Not to Scale

Bubble Point

1

Ideal Solution BehaviorFigure 8.7 d

If we cool a mixture of benzene and toluene vapor, the temperature where it starts to condense is called the dew point

The combined partial pressures equal 1 atm Mole fraction of benzene

in the liquid or vapor phase, xa

0 0.2 0.4 0.6 0.8 1.0

Tem

pera

ture

, C

75

8

5 9

5 1

05

1

15

Not to Scale

Bubble Point

Dew Point

1

Mole fraction of benzene in the liquid or vapor phase, xa

Tem

pera

ture

, C

75

8

5 9

5 1

05

1

15

Bubble Point

Dew Point

0 0.2 0.4 0.6 0.8 1.0

Temperature at the bubble point

Vapor Phase composition at the bubble point

Heat a liquid mixture until it starts to boil

Liquid Phase composition at the dew point

Now let’s look at non-ideal solution behavior

First let’s attack positive deviations from ideal solution behavior (Type II behavior)

The activity coefficients of both (or all) species is greater than one.

The acetone water system is an example of this behavior

So is the isopropanol-water system shown in Figure 8.8

Type II behavior of activity coefficients – always greater than 1

Notice that the behavior of each species approaches ideal (=1) as it’s concentration increases

Mole fraction of isopropanol in the liquid phase, xa

0 0.2 0.4 0.6 0.8 1.0

Act

ivit

y C

oeffi

cient,

1 1

0 1

5 Not to ScaleP=1 atm

isopropanol

water

Pure isopropanol

Pure waterFigure 8.8 b

Type II behavior

Activity coefficients greater than 1 mean that the partial pressure of the vapor is greater than that predicted with Raoult’s law

Mole fraction of isopropanol in the liquid or vapor phase, xa

0 0.2 0.4 0.6 0.8 1.0

Pre

ssure

, to

rr –

at

84

C

1000

800

600

400

200

0

Not to Scale

0iiii PxP

Predicted with =1

Actual isopropanol partial pressure

1i

Type II behavior

Activity coefficients greater than 1 mean that the partial pressure of the vapor is greater than that predicted with Raoult’s law

Mole fraction of isopropanol in the liquid or vapor phase, xa

0 0.2 0.4 0.6 0.8 1.0

Pre

ssure

, to

rr –

at

84

C

1000

800

600

400

200

0

Not to Scale

0iiii PxP

1i

Predicted with =1

Actual water partial pressure

Type II behavior

In this case (but not every type II case) the total pressure curve (at constant temperature) displays a maximum, which produces a minimum boiling azeotrope.

Mole fraction of isopropanol in the liquid or vapor phase, xa

0 0.2 0.4 0.6 0.8 1.0

Pre

ssure

, to

rr –

at

84

C

1000

800

600

400

200

0

Not to Scale

0iiii PxP

1i

water partial pressure

isopropanol partial pressure

Total pressure

Figure 8.8 a

Type II Solution BehaviorFigure 8.8c When the total

pressure displays a maximum, the equilibrium curve crosses the reference curve

Mole fraction of isopropanol in the liquid phase, xa

0 0.2 0.4 0.6 0.8 1.0

Mole

fra

ctio

n o

f is

opro

panol

in t

he v

apor

phase

0 0

.2 0

.4 0

.6 0

.8 1

.0

Equilibrium curve

Reference curve

Not to Scale

Azeotrope

1i

Azeotropes At an azeotrope,

the liquid and vapor have the same concentration

Mole fraction of isopropanol in the liquid phase, xa

0 0.2 0.4 0.6 0.8 1.0

Mole

fra

ctio

n o

f is

opro

panol in

the liq

uid

phase

0 0

.2 0

.4 0

.6 0

.8 1

.0

Not to Scale

Azeotrope

1i

Type II Solution Behavior At mole fractions

below the azeotrope, almost pure water, and the azeotropic composition can be distilled

At mole fractions above the azeotrope, almost pure isopropanol and the azeotropic composition can be distilled Mole fraction of isopropanol

in the liquid phase, xa

0 0.2 0.4 0.6 0.8 1.0

Tem

pera

ture

, C

75

8

0 8

5 9

0 9

5 1

00

Not to Scale

Azeotrope

Bubble Point Curve

Dew Point Curve

1i

Type II Solution Behavior

Type II solutions do not necessarily exhibit an azeotrope

For example, the acetone-water system does not

Two factors contribute to this behavior Difference in boiling point Deviation from ideal behavior

1i

Type III Behavior

for both species is less than 1 Similar to Type II – just insert

maximum for minimum etc Consider the acetone-chloroform

system

Type III behavior of activity coefficients – always less than 1

Notice that the behavior of each species approaches ideal (=1) as it’s concentration increases

Mole fraction of acetone in the liquid phase, xa

0 0.2 0.4 0.6 0.8 1.0

Act

ivit

y C

oeffi

cient,

0.3

0

.4 0

.6 1

.0 1

.5 Not to ScaleP=1 atm

acetone

chloroform

Pure acetone

Pure chloroform

Figure 8.9 b

Type III behavior

Activity coefficients less than 1 mean that the partial pressure of the vapor is less than that predicted with Raoult’s law

Mole fraction of acetone in the liquid or vapor phase, xa

0 0.2 0.4 0.6 0.8 1.0

Pre

ssure

, to

rr –

at

60

C

1000

800

600

400

200

0

Not to Scale

0iiii PxP

Predicted with =1

Actual acetone partial pressure

1i

Type III behavior

Activity coefficients less than 1 mean that the partial pressure of the vapor is less than that predicted with Raoult’s law

Mole fraction of acetone in the liquid or vapor phase, xa

0 0.2 0.4 0.6 0.8 1.0

Pre

ssure

, to

rr –

at6

0 C

1000

800

600

400

200

0

Not to Scale

0iiii PxP

Predicted with =1

Actual chloroform partial pressure

1i

Type III behavior

In this case (but not every type III case) the total pressure curve (at constant temperature) displays a minimum, which produces a maximum boiling azeotrope.

Mole fraction of acetone in the liquid or vapor phase, xa

0 0.2 0.4 0.6 0.8 1.0

Pre

ssure

, to

rr –

at

60

C

1000

800

600

400

200

0

Not to Scale

0iiii PxP

1i

Total pressure

Figure 8.9 a

Actual chloroform partial pressure

Actual acetone partial pressure

Type III Solution BehaviorFigure 8.9c When the total

pressure displays a minimum, the equilibrium curve crosses the reference curve

Mole fraction of acetone in the liquid phase, xa

0 0.2 0.4 0.6 0.8 1.0

Mole

fra

ctio

n o

f ace

tone iin

th

e v

apor

phase

0 0

.2 0

.4 0

.6 0

.8 1

.0

Equilibrium curve

Reference curve

Not to Scale

Azeotrope

1i

Azeotropes At an azeotrope,

the liquid and vapor have the same concentration

0 0

.2 0

.4 0

.6 0

.8 1

.0

Not to Scale

1i

Mole fraction of acetone in the liquid phase, xa

0 0.2 0.4 0.6 0.8 1.0

Mole

fra

ctio

n o

f ace

tone iin

th

e v

apor

phase

Equilibrium curve

Azeotrope

Type II Solution Behavior At mole fractions

below the azeotrope, almost pure chloroform, and the azeotropic composition can be distilled

At mole fractions above the azeotrope, almost pure chloroform and the azeotropic composition can be distilled Mole fraction of acetone in the

liquid phase, xa

0 0.2 0.4 0.6 0.8 1.0

Tem

pera

ture

, C

50

5

5 6

0 6

5 7

0 7

5

Not to Scale

Azeotrope

Bubble Point Curve

Dew Point Curve

1i

What controls whether the activity coefficients are greater or less than one?

When >1 it indicates that the solution species are repelled by each other, or at least are more attracted to themselves than to the other species

When <1 it indicates that the solution species are more attracted to each other than to their own kind

More about azeotropes For a given degree of mutual

attraction or repulsion, an azeotrope is more likely for species with a small difference in boiling point

For pairs of compounds with the same difference in boiling point, an azeotrope is more likely for the pair whose activity coefficients deviate from one by the largest amounts

Binary azeotropes between compounds with wide differences in boiling point are rare

Most azeotropes are of the minimum boiling type (over 90%)

Type IV Behavior Two Liquid Phase --Heteroazeotropes

Type II behavior occurs when two species repel each other

For a moderately strong repulsion an azeotrope forms

If the repulsion is strong enough, the two liquids actually separate into two separate phases

For example

Consider the water and butanol system

Between 65% water and 98% water, two phases occur

At less than 65% water only a single phase exists

At greater than 98% water only a single phase occurs

0 0.2 0.4 0.6 0.8 1.0

Mole fraction water

Water-butanol system From 0 to 65%

water one phase exists

From 98% to 100% water one phase exists

Between these values, two phases exist – one of 65% water and one of 98% water

1 phase

1 phase

2 phases

In the single phase region…

The solution exhibits Type II behavior

Mole fraction of water in the liquid phase, xa

0 0.2 0.4 0.6 0.8 1.0

Act

ivit

y C

oeffi

cient,

1.0

2

.0 5

.0 1

0. 2

0. 3

01butanol

1

1water1

1butanol2

1water2

Type IV BehaviorPressure behavior

In the two phase region the total pressure is the sum of the pressure from each phase

However, both phases must be exposed to the gas phase

Mole fraction of water in the liquid phase, xa

0 0.2 0.4 0.6 0.8 1.0

Pre

ssure

, to

rr –

at

90

C

0 4

00

8

00

1

20

0

Partial Pressure of n-butanol

Partial Pressure of water

Total Pressure

Type IVVapor-Liquid Equilibrium

The composition of the gas phase stays constant when two liquid phases are present

0 0

.2 0

.4 0

.6 0

.8 1

.0

Mole fraction of water in the liquid phase(s), xa

0 0.2 0.4 0.6 0.8 1.0

Mole

fra

ctio

n o

f w

ate

r in

the

vapor

phase

Reference Line

Equilibrium Curve

Type IVBubble Points and Dew Points

Call the n-butanol rich phase L1

Call the water rich phase L2

Mole fraction of water in the liquid phase, xa

0 0.2 0.4 0.6 0.8 1.0

Tem

pera

ture

, C

80

9

0 1

00

1

10

1

20

1

30

Vapor Phase

L1

L2

L1 + L2

Vapor + L1

Vapor + L2

Dew Point Curve

Bubble Point Curve

Boiling is necessary for this behavior The predicted behavior

will only occur if both phases are in contact with the gas phase

Realistically this only occurs during boiling

This type of behavior is common in petroleum refining

Liquid Phase I

Liquid Phase II

Gas Phase

Steam Distillation

If we take type IV behavior to the extreme, we consider two liquids that are essentially completely immiscible

For example water and mercury In this case we are always in the two

phase region

ii PPyP Mercurywater PP 00

The total pressure is the sum of the pure component vapor pressures

Consider example 8.7 If n-butanol and water were

completely insoluable, what would the boiling point be at one atm?

What would the composition of the vapor be?

ii PPyP butanoln00

PP water

Estimate the partial pressures of each component with the Antoine Equation

TTC

BAPwater

02.227

462.165794917.7log

TP

8812.196

190.1558838.7log butanol-n

mmHg 760butanol-n waterPP

Solve for T iteratively

T=89 0C

At the boiling point of 89 C

PH20 = 0.67 atm Pn-butanol= 0.33 atm Compare these results to Figure 8.12

Remember that butanol and water actually do exhibit considerable solubility

Distillation of the Four Types of Behavior

For systems that do not have an azeotrope, distillation columns can produce practically pure products The highest vapor pressure component is

separated into the overhead component The lower vapor pressure component is

separated into the bottoms

Gas-Liquid Equilibrium – Henry’s Law

In the previous discussions, both components could exist as a pure liquid at the temperatures of interest

In other words we were interested in vapor-liquid equilibria

How can we extend this discussion to include gas-liquid equilibria, for species that do not condense

Use Henry’s Law

Our discussions so far have been about systems at low pressure

At low pressures, the gas phase obeys the ideal gas law

At higher pressures we’ll need to consider the fugacity coefficient in our calculations

At both low and moderate pressures (up to ~1/2 the critical pressure) we won’t need to adjust our liquid phase calculations

Low Pressure VLE Calculations

Graphs are great to get a general idea of how systems behave, but they aren’t very accurate

We need to develop a standard approach to calculate vapor-liquid equilibrium properties

At low pressures the fugacity coefficients are 1 – but the liquid phase activity coefficients aren’t

Estimate them using the Van Laar equation

2

2

)log(

ba

ba

xxBA

Ax 2

2

)log(

ba

ab

xAB

x

Bx

There are other estimating equations – this one is just easy to use -- the basis for the Van Laar equation is developed in chapter 9

The 6 Most Common VLE Calculations

Find the dew point, for a known temperature

Find the dew point, for a known pressure Find the bubble point for a known

temperature Find the bubble point for a known pressure Isothermal flash calculations Adiabatic flash calculations

These are all examples problems that can be formulated as equilibrium flash calculations

Flash Calculations

Liquid

Vapour

F

V

L

T, P

F can be a liquid, a gas or a two phase mixture

This is called a flash calculation, because if the pressure is reduced enough, the liquid changes to vapor in a “flash”

To solve any of these problems we need to identify our equations and unknowns -- Table 8.6

Material Balances

Summation of mole fractions

LVF LxVyFx iii

Remember from Process Engineering that you can write n independent material balances, if you have n components

1 ix 1 iy

More equations to be solved

Equilibrium

Vapor Pressure equations Antoine equation is probably the most

accurate of the simple equations

P

P

x

yK ii

i

ii

Simplified version assuming ideal gas

More equations to be solved

Calculate the activity coefficients Van Laar equation is simplest analytical

approach

Energy balance Adiabatic flashes

LvF HHH

We also need the inlet conditions

Feed specification xi

Temperature Pressure

Let’s do some example calculationsDew Point – Example 8.9

Estimate the boiling pressure and the composition of the vapor in equilibrium with a liquid that is:

0.1238 mol fraction ethanol remainder water

85.3 oC

First – Use Antoine’s equation to find the pure component partial pressures at this temperature

TC

BAPi

log

TPEtOH

65.222

3.155404494.8log

TPwater

02.227

462.165794917.7log

At 85.3 C, Pwater=0.5772 atm PEtOH=1.3088

SecondFind Activity Coefficients

Use the Van Laar Equation

2

2

)log(

EtOHwater

EtOHwater

xxBA

Ax

2

2

)log(

waterEtOH

waterEtOH

xxBA

Ax

Find the values of A and B for water and ethanol in Table A.7

water=1.0388EtOH=2.9235

NextCalculate the partial pressures

0waterwaterwaterwater PxP

5772.0*9235.2*8762.0waterP

atm 0.5254waterP

atm 0.4737EtOHP

Similarly for Ethanol…

atm 0.9991TotalPand…

FinallyFind the vapor phase mole fractions

Total

ii P

Py

5259.09991.0

5254.0

Total

waterwater P

Py

4741.09991.0

4737.0

EtOH

EtOHEtOH P

Py

Pressure Specified Calculations are Similar

They are harder, because the Temperature appears in the Antoine Equation

They need to be solved interatively

Calculational Steps

Guess a Temperature Calculate the Pure Component Vapor

Pressures Calculate the activity coefficients Calculate the partial pressures Calculate the total pressure, and

compare to the given pressure

Graphical Solution

Add Figure 8.17

Bubble Point Calculations

Mirror image of Dew Point Calculations Temperature specified calculations are

easier Pressure specified calculations require an

iterative approach

Isothermal Flash Calculations Both T and P are specified The division of mass between liquid and

vapor is unknown Consider Example 8.13

An ethanol-water mixture xaFeed=0.126 is brought to equilibrium at

1 atm 91.8 C

Estimate the vapor fraction and the mole fraction of each species in the vapor phase

Let’s Solve it Graphically First

Add figure 8.19

Analytical Solution Find the pure component vapor

pressures Assume a value for V/F Estimate the activity coefficients Calculate the K factors Use a material balance to find the

mole fractions in the vapor and liquid phase

LxVyFx aaaF

LxVyFx aaaF Divide by F

F

Vx

F

Vy

F

Lx

F

Vyx

aa

aaaF

1 Equation 8.12

Adiabatic Flash Calculations In addition to the equations from the

previous example, you need an energy balance

You must “guess” a temperature, then perform the calculations, and finally check to see if the energy balance requirements are met

LVF HHH

Solutions using K factor approximating tools

Activity coefficients are functions of T,P and x (the liquid composition)

Thus, K is also a function of T,P and the liquid composition

By ignoring the contribution of composition, DePreister created Figure 8.20 – an estimating tool for finding K

K factor chart

Colligative Properties – Another application of Raoult’s Law

Boiling Point elevation Freezing Point depression Osmosis (Chapter 14)

Predictable using Raoult’s Law

What happens as the concentration of the solvent approaches 1?

The solvent activity coefficient, , approaches 1

Thus for dilute solutions of anything the solution obeys Raoult’s law

The identity of the solute doesn’t matter, as long as it’s a dilute solution and it’s not volatile!!

Boiling Point Elevation

When the solute has a very high boiling point, the solution vapor pressure only depends on the solvent

solutesolventT PPP 0

solventsolventT PxP 0

Boiling Occurs when the vapor pressure equals the pressure of the surroundings

Example 8.15 One mole of sucrose (MW=342.3

g/mole) is dissolved in 1000 g of water

What is the vapor pressure of this solution at 100 0C?

atm 0.982atm 1*982.0 TP

Boiling won’t occur, unless you’re at an altitude above sea level

What temperature will cause boiling?

First – determine what pure component vapor pressure is required

solventsolventT PxP 0

solventPatm 0982.01

atmP solvent 018.10

From the steam tables we find T=100.51 0C – or a boiling point elevation of 0.51 0C

We can approximate the behavior of this solution at low concentrations of solute

The change in boiling point is directly proportional to the molality of the solution

mKT bboiling

Figure 8.22

Freezing Point Depression

Just as adding a non-volatile solute increases the boiling point, it also decreases the freezing point

See Example 8.16

mKT ffreezing