635.021 Physical Chemistry, Lab course 1

CLiquid-vapour equilibrium diagrams of a binary mixture

Determination of the activity coefficients of a non-electrolyte

solution

Institut fürPhysikalische und Theoretische ChemieTechnikerstraße 4, A-8010 Graz

Dr. Franz MautnerBüro: Rechbauerstraße 12/KellerTel: +43 (0)316 873 8234, 8221(Sekr.)Fax: +43 (0)316 873 8225e-mail: mautner@ptc.tu-graz.ac.at

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SHORT DESCRIPTION

In this exercise we measure the liquid-vapour equilibrium diagram of a non-electrolyte

binary mixture including an azeotropic point and calculate the activity coefficients of each

component.

The experimental part includes the preparation of the binary mixture for different mole

fractions, the determination of the refractive index as well as the distillation of the mixtures

together with measurement of the boiling points and the composition of the vapour phase.

In the analysing part you will draw the liquid-vapour equilibrium diagram and the McCabe-

Thiele diagram from which the azeotropic point is determined. Using the equation of

Clausius-Clapeyron for the vapour pressure of the pure components you can calculate the

activity coefficients of these non-electrolyte solutions.

Literature :

1.) G. Wedler, Lehrbuch der Physikalischen Chemie,VCH.

2.) P.W. Atkins, Physikalische Chemie, VCH.

3.) R.Perry, D.Green, Perry´s Chemical Engineer´s Handbook, McGraw Hill.

4.) Siedetabellen und –diagramme aus: Landolt-Boernstein, Zahlenwerte und Funktionen

aus Naturwissenschaft und Technik, Mandelung.

5.) J. Gmehling, J. Menke, K. Fischer, J. Krafczyk, Azeotropic Data, VCH.

6.) Handbook of Physics and Chemistry, McGraw Hill.

INTRODUCTION

The main aspect of systems/mixtures containing more than two liquid components is the

phase equilibrium between the gas phase and the liquid phase. The liquids can build one or

more phases of different composition depending on their mutual solubility. The vapour

pressure diagram (p-x-diagram) and the boiling point diagram (T-x diagram) are some of the

graphical representations of the equilibrium.

The boiling point diagrams are important for normal distillations at constant pressure. They

are obtained experimentally from an isobaric cut through a passel of vapour pressure

isotherms. This can be seen from the numbers in Figs 1 and 2.

2

p[atm] T

T3

T2

T1

xB xB

Fig.1: Vapour pressure diagram of anideal binary mixtureat two different temperatures

Fig.2: Liquid-vapour equilibriumdiagram of a binary mixtureat constant ressurep = 1 atm.

Liquid-vapour equilibrium diagram of binary mixtures:

Fig.2 shows two grey single-phase regions, namely one gas phase and one liquid phase,

where the liquids are miscible in each proportion. Both single-phase regions are separated by

a two-phase region. For the T and xB values in this region, the binary system decomposes

into two phases, a liquid phase of composition 2a and a gas phase of composition 2b. The

ratio can be calculated with the tie-line-rule (also called lever rule). Point 2 cannot be

realised experimentally in the state of phase equilibrium.

Heating the binary mixture (mole fraction x1) to temperature T1 starts the boiling at the

external pressure (boiling line). At this temperature the vapour is in equilibrium with the

liquid and has the composition 1b. Further heating, e.g. to T2 brings the system to the liquid

composition 2a and the vapour composition 2b. Above the temperature T3 (dew point curve)

the liquid is completely vaporised.

When heating a binary mixture (mole fraction x1) in a distillation apparatus, these mixture

starts to boil at point 1a. After condensation (with a Liebig cooler) the first drop of the

vapour is more rich on the more volatile component and has the composition 1b. Heating to

higher temperatures cannot exceed mole fraction x2 in the phase equilibrium. Therefore, the

mixture x1 cannot be decomposed into its pure components A and B by a simple distillation

process. If the same mixture is distilled for longer time, and the condensed vapour is

removed as distillate, then the liquid gets more pure on the less volatile component B, thus

raising the boiling point. The boiling point of the residue shifts to higher temperatures along

the boiling line. A complete separation into the two starting components will not be

successful by this procedure.

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If the distillation is stopped, e.g. at temperature T3 by cooling the vapour of composition 2b

and bringing it to boil in another distille and repeating this procedure often (as in the so-

called fractioned distillation), then the pure components A (residue) and B (distillate) can be

obtained. The same effect has a fractionate column sitting on the alembic, which is a tube

filled with glass rings (Raschig rings). Industry uses columns with different plates e.g. bell

column plates. This repeated distillation is called rectification. Liquid and vapour streams are

kept in reverse flow, so that the descending vapour is enriched with the more volatile

component when flowing through the column plates which are filled with the liquid.

Fig.3 shows a rectification in the laboratory as well as in industry together with the

corresponding McCabe-Thiele diagram.

Fig.3 Rectification column for the laboratory (a) and for industry with the McCabe-Thiele

diagram (b)

In a continuous working rectification column the mixture is introduced at an optimised

position. The part below the induction point is called “Verstärkungsteil”, the part above the

induction part is called “Abtriebteil”. At the lowest final part of the column (sump) is a

vapouriser, which produces the vapour flowing to the upper part. The rest of the less volatile

component, which boils above, is drained at the sump. Also the vapour which leaves the

column at the top is condensed and refilled as backflow.

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BOILING PROPERTIES OF BINARY MIXTURES

For the boiling behaviour it is important how chemically similar the components are. Do

they associate and from assemblies (“over-attraction”) or do they separate in the mixture

(“under-attraction”)?

The adduction or repulsion of both components is defined by their chemical potentials µ1

and µ2, which are functions of the molecular structure and of the energetics (polar, non-

polar).

Different boiling behaviour is described in the following sections.

1.) IDEAL MIXTURE

T

Fig. 4 Liquid-vapour equilibrium diagramof the ideal systemToluene – Benzeneat constant pressure of 1 atm

0 xBenzol 1

The condition for an almost ideal behaviour of liquid mixtures is a similar molecular

structure of their components. It is not so important wheter the molecules are surrounded by

molecules of the same kind or by molecules of the other component.

At the moment of mixture only a very small (theoretically no) volume contraction appears,

also almost no mixture heat. The two-phase region is more flat and narrow for more similar

components.

Examples: Benzene – Toluene, see Fig. 4

liquid Nitrogen – liquid Oxygen

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2.) REAL MIXTURES

2.a) OVER-ATTRACTION

The components in the mixture attract each other more than the pure components and they

tend to associate and to form compounds. This lowers the vapour pressure compared to the

ideal case and a high boiling point mixture is created. This procedure is exothermic (∆Hmix=

negative), and oftentimes accompanied by a volume contraction. ∆Gmix is larger (more

negative) than in the ideal case.

Examples: many mineral acids, z.B.

H2O – H2SO4 (azeotropic point at 1 atm: 98,7 Wgt.% acid)

H2O – HNO3 (azeotropic point at 1 atm: 68 Wgt.% acid)

H2O – HCl (azeotropic point at 1 atm: 20 Wgt.% acid)

Fractioned distillation of H2O-HCl gives as residue an azeotrop with 20 % HCl, whereas in

the distillate pure H2O or HCl remains. The liquid-vapour diagram is shown in Fig.5.

T

Fig.5 Partial liquid-vapour equilibriumdiagram of the system H2O – HClat p=1atm(xHCl, Az = 0,1125; TAz = 107,8 °C)

0 xHCl 1

2.b) UNDER-ATTRACTION

The components in such a mixture attract each other more than the different components

leading to segregation. The vapour pressure of the mixture becomes larger than in the ideal

case and a low boiling mixture is built. This procedure is endothermic (∆Hmix= positive).

∆Gmix becomes lower (less negative) than in the ideal case.

Examples: H2O – ethanol (azeotropic point at 1 atm: 95,57 wgt.% ethanol), see Fig.6

benzene – ethanol (azeotropic point at 1 atm: 33 wgt.% ethanol), see Fig.7

Aceton - CS

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T

Fig.6: Liquid-vapour equilibrium diagram of the system Water – Ethanol at 1 atm (xEtOH,Az = 0,89; TAz = 78,3°C)

0 xEtOH 1

Pure alcohol cannot be extracted starting from a 50% water-ethanol mixture by rectification

(the residue is pure water). For the preparation of absolute alcohol starting from the

azoetrope, industry uses the distillation over caustic or the rectification after addition of

benzene which builds a ternary low boiling mixture.

Fig.7: Liquid vapour diagram of thesystem ethanol-benzeneat the pressure of 750 Torr.

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2.c) MISCIBILITY GAPS

The under-attraction can be so strong, that one ore more large miscibility gaps appear.

Fig.8 shows a system where the miscibility gap is closed before arriving the boiling point.

T

Fig.8: Liquid-vapour equilibrium diagramof a binary system A-B withmiscibility gap

0 xB 1

A special case of a closed miscibility gap is the system water-nicotine in Fig.9.

T Fig.9: Liquid-vapour equilibrium diagramof the binary systemH2O-nicotine(closed miscibility gap)

0 xNikotin 1

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If the miscibility gap doesn’t close at the boiling point like in the system aniline-water or

water-isobutylalcohol, than the following liquid-vapour equilibrium diagrams appear

(compare this behaviour with the melting diagrams in the eutectic state with partial

solubility):

The next example is the special case of water-triethylamine with a miscibility gap which is

open at the top.

Abb.11: Liquid-vapour equilibrium diagram of the system H2O – Et3N at 1 atm.

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2.d) COMPLETE IMMISCIBILITY

Caused by an always positive entropy of mixing it is theoretically impossible to have

complete immiscibility. In reality there exist some systems which are practically immiscible.

Examples: water – bromobenzene

water – mercury

water – benzene

The system water-benzene is the most important. The mutual solubility is about 0,1 - 0,2 %,

like in an eutecticum of the melting diagrams.

T

Fig.12: Liquid-vapour equilibriumdiagram of the systemwater – benzene at 1 atm.

In this case it is not possible to extract a mixture of water and benzene with the composition

x1. The system is always a mechanical mixture of two liquid phases of (almost) pure water

and (almost) pure benzene. At 69 °C the sum of the vapour pressures equals the total

pressure of 1 atm. If the external pressure is also 1 atm, then the mixture starts boiling.

The composition of the ideal vapour is given by the DALTON´s LAW :

benzene

OH

benzene

OH

n

n

p

p22 = bei water-benzene xbenzene= 0,705.

After condensation of the vapour the same proportion is in the run-down tank. This is for

example the basis for

• the determination of the molar weight (by distillation of water vapour),

• the distillation of the carrier vapour (esp. on the vapour of the steam distillation for the

careful distillation of thermally sensitive organic compounds), and the

• water analysis (by distillation with benzene).

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THERMODYNAMICS of the PHASE EQUILIBRIUM

1.) The PHASE

A phase is a part of the system, namely a homogenous state region with consistent chemical

conditions and physical properties. It is characterised by definite values of the state variables

p, T and composition.

Each phase is separated from the other phases by an interface. Vapours are completely

miscible, therefore only one gas phase exist, but in solid or liquid state there are one ore

more phases possible in a phase equilibrium.

2.) EQUILIBRIUM CONDITIONS

In the so-called phase equilibrium for components present in different phases, where the

exchange from one phase into another phase is possible and necessary, the number of phases

play no role. For processes like vapourisation, sublimation, condensation, salting out or

salting in, one can restrict to intensive properties for the thermodynamic description.

2.a) The PHASE EQUILIBRIUM

The conditions for an equilibrium of more phases are the following:

(phase = superscript indices l, g, s, or. (1), (2); component = subscript index i)

a) in all phases the pressure must be the same:

p(1) = p(2) = p(3) = .....

b) in all phases the temperature must be the same:

T(1) = T(2) = T(3) = .....

c) in all phases the chemical potential of the substance i must be the same. This must be

valid for all components!

µi(1) = µi

(2) = µi(3) = .....

Some examples follow later in these notes.

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2.b) GIBBS’ PHASE RULE

In each system with more substances and phases only a restricted number of variables

(properties) exist, which can be varied without influencing the number of phases and

substances.

Gibbs’ phase rule describes how to find this number. It is also called the number of degrees

of freedom F or variance of the system. They are the pressure p, the temperature T and the

composition (concentration) of the different phases in a system at equilibrium.

For a system in equilibrium with K components and P phases without secondary conditions

the variance F is

F = K − P + 2

Examples: (1)triple point of water: there are 3 phases, F = 1 − 3 + 2 = 0, a non-variant

system at 0,0099 °C and 4,579 Torr.

(2) vapour pressure curve of an one-component system: there are 2 phases, F = 1

− 2 + 2 = 1, a univariant system.

The modified Gibbs’ phase rule including R secondary conditions is

F = K − P + 2 − R

Examples: (1)Liquid-vapour equilibrium diagram (Fig.2): The pressure is restricted to 1

atm, which is a secondary condition (R = 1), thus F = 2 − 2 + 2 − 1 = 1, an

univariant system: either the intensive variable T or xB can be chosen to vary,

the other variable is then fixed.

(2) Azeotropic mixture (Fig.5 and 7): one secondary condition is the constant

pressure. The azeotropy is the second one (xAz = yAz), so R = 2, and F = 2 − 2

+ 2 − 2 = 0, a nonvariant system.

(3) Benzene – water (Fig.12): at point 2c two liquid and one gas phase exist (P =

3), and p = const. (R = 1), so F = 2 − 3 + 2 − 1 = 0. The state region below 69

°C (line 2a-2b) consists of the two pure liquids in a fine distribution (P = 2, K

= 2), so R = 1 and F = 2 − 2 + 2 − 1 = 1.

Secondary conditions can also be chemical reactions in the phases, the presence of electrical

and magnetic fields, and, respectively, the electroneutrality in ionic solutions, etc.; see the

specialised literature for details.

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PHASE EQUILIBRIA IN BINARY SYSTEMS

In this section we describe phase equilibria for the systems “ideal solution – ideal vapour”

and “real solvation – ideal vapour”.

1.) IDEAL SOLUTION – IDEAL VAPOUR

For ideal phases RAOULT´s LAW is valid over the whole concentration range. For each

temperature it is

i*ii xpp ⋅= ,

Here, pi is the partial pressure of the components A and B, and pi* is the vapour pressure of

the pure component i at a certain temperature.

Fig.13: Total pressure and partial pressure in an ideal binary mixture at a constanttemperature.

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2.) REAL SOLUTION – IDEAL VAPOUR

An ideal solution is described by

iii ln x RT T)p,( += θµµ (1)

To be valid also in real solutions one introduces a coefficient fi which gives the effective

mole fraction xi⋅fi by multiplication with the mole fraction xi. This coefficient is the activity

coefficient fi and it is defined by the equation

iiii fln x RT T)p,( += θµµ (2)

It is unimportant how large the deviation from ideal behaviour is. The activity coefficient fi

is a function of p, T and of all mole fractions in the solution.

In equation (2) the magnitude of µiθ(p,T) is not known. First the activity coefficient has to

be defined, when it has the value of 1.

The activity coefficient fi equals 1, if the mole fraction of these component goes to 1.

iiii fln x RT T)p,( += θµµ

1f i → for 1x i →

At the limit (xi = 1, fi = 1) the logarithm disappears and µiθ(p,T) is the free partial molar

enthalpy of the pure substance at the same values of pressure and temperature as for the

solution under consideration. For the thermodynamical description of a real (i.e. non-ideal)

liquid mixture the following relation is valid in the equilibrium with an ideal gas phase:

iiiiii p lnT RT)(fx lnT RT)p,( ⋅⋅+=⋅⋅⋅+= θθ µµµ

or iii

i Kfx

p=

⋅with

⋅

≡TR

T)(-T)p,(expK ii

i

θθ µµ

Ki is independent of the composition, because µiθ(p,T) and µi

θ(T) are only a function of p

and T. (For non-ideal gas phases the pressure must be substituted by the fugacity.)

Scaling f→1, if xi→1, reduces the vapour pressure pi to the vapour pressure of the pure

components when T is given and the total pressure is the same in the solution, i.e.,

*ii

ii

i pKfx

p==

⋅, and pi = pi

*⋅xi⋅fi .

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Graphically the activity coefficient has the following meaning:

p Fig.14: Vapour pressure diagram of a real binary system at constant temperature(−−− ideal behaviour)

Consider the mole fraction x1 in this binary mixture: for an ideal solution the partial pressure

of (B) must have the value pB* ⋅xB and refers to the line UW. The real partial pressure is

equivalent to the UV . This leads to the following equations,

UWUV

xpp

fB

*B

BB =

⋅= and

UYUX

xpp

fA

*A

AA =

⋅=

The activity coefficients give the proportion of the real partial pressure to the partial pressure

for the ideal case, if Raoult´s Law is valid. If the real partial pressure of the components and

the vapour pressure of the pure components are known, then the activity coefficients can be

calculated.

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PRACTIAL PART:

How to determine and analyse the liquid-vapour equilibrium diagram

How to calculate the activity coefficients of these non-electrolyte solutions

For the determination of the activity coefficient the following relations are valid:

A*A

AA xp

pf

⋅= ,

B*B

BB xp

pf

⋅= , with pA + pB = pges and xA + xB = 1

Calculate the values of pA* and pB

* either via a table of vapour pressures or via the

equation of Clausius-Clapeyron:

2v

RTpH

Tp ⋅∆

=dd

with dp/p=dlnp and dT/T2=−d(1/T) → RH

T1

lnp v∆−=

d

d

This equation is valid, if ∆Hv is not T-dependent. This is true for a small temperature range.

After integration between the limits T1 and T2 we get the

August’s vapour pressure formula:

−⋅

∆−=

12

v

1

2

T1

T1

RH

pp

ln

Please use this formula for the data analysis.

1.) CALIBRATION CURVE and DETERMINATION OF THE VAPOUR PHASE

COMPOSITION

For the liquid-vapour equilibrium diagram the knowledge of yA and yB is necessary. By

condensation of a small amount of vapour in its equilibrium and measurement of the

refractive index with an ABBÉ refractometer, the composition of the gas phase can be

determined. This procedure is practicable if the refractive indices differ enough for both pure

components.

First the refractive indices n(l) of the mixtures prepared should be measured and plotted

against the molar fraction xA. This gives the calibration curve from which the mole fraction

yA of the condensed vapour can be determined. Further instructions come later in this

section.

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With the knowledge of yA and yB, Dalton’s law gives you the partial pressure of the

components pA and pB for the individual tie lines

tot

AA p

py = , and

tot

BB p

py = ,

where pA + pB = pges and yA + yB = 1.

The activity coefficient of non-electrolyte solutions can be larger than 1 (under-attraction) or

smaller than 1 (over-attraction). The smaller the concentration of one component is, the more

it approaches the ideal behaviour with activity coefficients of 1, and Raoult’s Law becomes

valid.

2.) McCABE-THIELE DIAGRAM

For the discussion of distillations the simple equilibrium diagram, also called McCabe-

Thiele diagram, is used. The relation between the composition of the vapour yA and the

composition of the liquid xA is plotted in a Cartesian coordinate system.

For azeotropic systems the equilibrium curve intersects the diagonal of the equilibrium

diagram in the azeotropic point, where xA = yA . This defines the azeotropic mole fraction

xaz.

Fig.16a: Liquid-vapour equilibrium diagram and McCabe-Thiele diagram of a ideal binarysystem.

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Fig.16b: Liquid-vapour equilibrium diagram and McCabe-Thiele diagram of a real binarysystem (low boiling mixture)

3.) DISTRIBUTION COEFFICIENT (K-Value), relative volatility α

The ratios yA/xA, yB/xB or, generally, yi/xi are the K-values or distribution coefficients KA,

KB or Ki. The values of yi and xi refer to the same temperature and lie on the same tie line.

The ratios KA/KB and KB/KA are the relative volatilities αA,B and αB,A.

At the azeotropic point of a binary mixture, both K-values are equal at KA,az= KB,az= 1

because the mole fractions are equivalent xi = yi.

Plotting KA resp. KB versus xA in a diagram gives the intersection point of both curves at the

value of 1 defining the azeotropic point. This is the second method for determination of the

azeotropic point exactly.

The calculation of the activity coefficients at the azeotropic point uses the following

relations,

( )*i

totazi,*

i

toti*

ii

totii p

p1Kwith

pp

Kpxpy

f ===⋅=⋅⋅

=

With the vapour pressures of the pure component at the azeotropic temperature you can

calculate the activity coefficients easily with the knowledge of the total pressure.

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Fig.17: Diagram of the distribution coefficients KA, KB vs. the molar fraction xA.

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EXPERIMENTAL PART - PRACTICE MANUAL

At the colloquium you have to calculate the boiling temperatures (= temperatures of the

phase equilibrium) of the pure components, TA and TB, of the binary systems A-B. For this

task you need the equation of Clausius-Clapeyron and its integrated form, the vapour

pressure equation of August, as well as the knowledge of the gas constant R. Also both

Dalton’s law and Raoult’s law should be known. Please bring a pocket calculator along.

You MUST wear EYE PROTECTING GLASSES

during the experimental work!!

In the practical part you have to determine the boiling temperatures at constant total

pressure (= external pressure at the lab day) and the refractive indices for a number of

mixtures. You will get a prepared empty table where you can fill in the following data:

xA / xB / gA[g] / gB[g] / n(l) / TS[°C] / n(g) / yA / yB / KA / KB.

a) Prepare the mixtures with individual molar fractions xA, which are given by the

instructor by directly weighing in to the distillation flask at the balance with scale on

top. The final molar fraction xA can be recalculated from the weigth-in quantities gA

and gB (where gA+gB= 70 g),

A

A

B

A

B

A

x-1x

MM

gg

⋅= ,

MA and MB are the molecular weights of the liquids.

b) Measure the refractive index n(l) of each individual mixture and of the pure

components with the Abbé refractometer BEFORE you start the distillation. Please

work very clean at the optics of the prism, because residues of the preliminary

measurement adulterate the results. The value of the liquid phase, n(l), must be noted

in the lab journal giving the calibrating curve point by point.

c) Draw the calibration curve from these data

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d) Turn on the WATER COOLING.

Connect the flask to the apparatus, heat and wait until the boiling temperature is

constant. This takes 5-10 min, until the phase equilibrium is adjusted.

Then read the boiling temperature (in °C) from the thermometer and list it in your lab

journal.

e) Then rotate the two-way-cock to get some drops of the condensed vapour.

ATTENTION !!! Pressure increase in the flask !

Throw away the first 2-3 drops and put the following 5-10 drops directly to the optics

of the prism of the refractometer. Then read the refractive index n(g) of the condensed

vapour rapidly. (Evapourisation can change the molar fractions !).

Start the sequence of measurements from the most volatile on A mixture and proceed to the

most rich on A mixture, so that drying of the apparatus is sufficient, and other cleaning is not

necessary.

During the first measurement (you have some waiting time) you can weigh-in the next

mixture. Thus the experiment goes fast and you have not lost much time.

f) During the experiment you can draw the calibration curve by hand (the pure liquids

have the highest and lowest values of n(l)).

y-axis = refractive index n(l), x-axis = molar fraction xA.

After the measurements the solutions must cool some time under the hood. Then you can

transfer the solutions to the waste flasks provided.

Please clean the apparatus, the flasks and the desks after your experiment!

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DATA ANALYSIS

In the lab journal do not give the information about the practical work, but write down the

formula used for the data in the tables. The data analysis can be done either by computer

(tables and diagrams with the computer softwares Origin or Excel ; do not forget the

legends !) or by a pocket calculator. The calibration curve and the Liquid-vapour

equilibrium diagram must be drawn by hand. All other diagrams can be drawn either by

hand or by a computer using the Origin software.

All diagrams must have a legend at the bottom containing a title, the binary system

investigated, the external pressure of the lab day. All tables must have the legend at the top

(header). Please write also your name and the date on the sheet. Don’t forget the units in the

tables and diagrams.

Write down all formulas used, and all reference data from the literature (boiling point of

the pure components, and of the azeotropic point (T, x).

Choose a proper scale for the diagrams such that all points are on the sheet and the diagram

fills the sheet. All diagrams should have a distance from the corners of the page (possible to

copy and tack). The data points should be drawn with different signs (crosses, circles,..) in

such a size that one can identify each point clearly and that the diagram is also reducible in

size.

Strategy:

a) The calibration curve (n(l) versus xA) has been already drawn during the

experiment. From it you can read the molar fraction of the gas phase yA with the help

of the refractive index n(g) of the condensed vapour.

b) First you draw the liquid-vapour equilibrium diagram:

y-axis = boiling temperature in °C, x-axis = molar fraction xA.

The composition of the vapour yA at the same temperature as xA gives the vapour

curve. Please use different symbols for the data on the boiling curve and on the

vapour curve. Do not connect the points before determination of the azeotropic point

from the next two diagrams. The tie lines should be drawn now.

c) Draw the McCabe-Thiele-Diagramm (equilibrium diagram):

y-axis = molar fraction of vapour yA, x-axis = molar fraction of liquid xA.

Draw also the hypothetical line yA = xA. With this line you can determine the

azeotropic point exactly (why?). Note the azeotropic composition xaz.

d) Now draw the diagram of the distribution coefficients.

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y-axis = KA resp. KB, x-axis = xA.

From the intersection point of both curves you can again determine the azeotropic

point exactly.

Put the azeotropic mole fraction xaz from the McCabe-Thiele diagram and from this

diagram to the liquid-vapour diagram.

Please show your diagrams and results to the supervisor before you continue the data

analyses.

e) Calculate the activity coefficient of both components. Please use the following table

for this job.

xA / xB / yA / yB / TS[°C] / TS[K] / pA/ pA* / pB / pB

* / fA / fB.

f) And now some questions to your results:

• Do you have a high-boiling or a low-boiling mixture?

• Why are the vapour pressures of the pure components pA* und PB* temperature

dependent?

• Do you have activity coefficients with values larger than 1 in your results and why?

What are they dependent on?

• How large is the activity coefficient in the azeotropic point and what is the relation

of fA to fB ?

g) Literature results : search the azeotropic composition at 1 atm for your binary

system in the Handbook of Chemistry and Physics (attention: wgt% !) and

calculate the corresponding molar fraction.

h) Make a list with all informations of the azeotropic point (molar composition,

temperature, pressure) together with all these data from literature e) and f) and

compare the results.

Please check, if all sheets are labelled correctly (name, date, external pressure of the day,

binary system, table headers, diagram footer, axes labels with units, formulas used).

Now you can give the lab journal to your supervisor.