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Physical chemistry
Phase Equilibrium
Dr. R. Usha
Miranda House, Delhi University
Delhi
CONTENTS
Phase equilibrium
Phase
The phase rule
What is a phase diagram?
The phase diagram for water
Equilibrium between solid and vapour (sublimation curve)
Equilibrium between liquid, vapour and solid water (ice)
Equilibrium between solid and liquid (fusion curve)
Metastable equilibrium involving liquid and vapour phases
The phase diagram for carbondioxide
The phase diagram for sulphur
Enantiotropy
Monotropy
Metastable equilibria in the sulphur system
Phase equilibria of two component systems
Thermal analysis
Saturation or solubility method
The bismuth-cadmium system
The Lead-Silver system
The Magnesium-Zinc system
The Sodium chloride-water system
The Ferric chloride - Water system
Efflorescence and deliquescence
Liquid-Liquid mixtures – ideal liquid mixtures
Raoult’s law
Effect of temperature on the solubility of gases
Effect of pressure on the solubility of gases
Konowaloff’s rule
The Duhem-Margules equation
Fractional distillation of non ideal solution
Partially miscible liquids
Phenol-water system
Triethylamine-water system
Nicotine-water system
Steam distillation
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Phase equilibrium
Various heterogeneous equilibria (Box 10.1) have been studied by methods suitable to that type
of equilibrium, such as vaporization by using Raoult’s law and Clausius-Clapeyron equation;
distribution of solutes between phases by using the distribution law etc. A principle called the
phase rule can be applied to all heterogeneous equilibria. The number of variables to which
heterogeneous equilibria are subjected to under different experimental conditions may be defined
using this principle. The phase rule is able to fix only the number of variables involved. The
quantitative relation between the variables is obtained by using the various laws and equations
(Box 10.2).
This rule was first put forward by J. Willard Gibbs, an American Chemist, in the year 1876 (Box
10.3). The full implications of this rule was understood by chemists only when Roozeboom,
Ostwald and van’t Hoff applied it to some well known physical and chemical equilibria in a
language that could be easily followed. After this its value as a fundamental generalization was
fully realized.
Certain terms like phases, components, degrees of freedom, true and metastable equilibrium (Box
10.4) need to be explained in some detail before the phase rule is stated.
Box 10.1 Examples of heterogeneous equilibria:
•
Liquid – vapour (vapourization)
•
Solid – vapour (sublimation)
•
Solid – liquid (fusion)
•
Solid 1 – solid 2 (transition)
•
Solubility of solids, liquids and gases in each other
•
Vapour pressure of solutions
•
Chemical reaction between solids or liquids and gases
•
Distribution of solutes between different phases
Box 10.2 Laws and equations to study heterogeneous equilibria:
•
Raoult’s law
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•
Clapeyron equation
•
Clausius –Clapeyron equation
•
Henry’s law
•
Equilibrium constants
•
Distribution law
Box 10.3 Josiah Willard Gibbs
•
Spent most of his working life at Yale
•
May be regarded as the originator of chemical thermodynamics
•
Reflected for ten years before publishing his conclusions
•
Published his papers in the transactions of the Connecticut Academy of Arts and
Sciences, not a well known journal
His work remained overlooked for 20 years. Roozeboom, Ostwald, van’t Hoff and others
showed how this rule could be utilized in the study of problems in heterogeneous equilibria.
Definitions
Phase
A phase is defined as any homogeneous and physically distinct part of a system which is
separated from other parts of the system by interfaces.
A part of a system is homogeneous if it has identical physical properties and chemical
composition throughout the part.
•
A phase may be gas, liquid or solid.
•
A gas or a gaseous mixture is a single phase.
•
Totally miscible liquids constitute a single phase.
•
In an immiscible liquid system, each layer is counted as a separate phase.
•
Every solid constitutes a single phase except when a solid solution is formed.
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•
A solid solution is considered as a single phase.
•
Each polymorphic form constitutes a separate phase.
The number of phases does not depend on the actual quantities of the phases present. It also does
not depend on the state of subdivision of the phase.
Examples 10.1.1
Counting the number of phases
a)
Liquid water, pieces of ice and water vapour are present together.
The number of phases is 3 as each form is a separate phase. Ice in the system is a
single phase even if it is present as a number of pieces.
b)
Calcium carbonate undergoes thermal decomposition.
The chemical reaction is:
CaCO3(s) CaO(s) + CO2 (g)
Number of phases = 3
This system consists of 2 solid phases, CaCO3 and CaO and one gaseous phase,
that of CO2.
c)
Ammonium chloride undergoes thermal decomposition.
The chemical reaction is:
NH4Cl(s) ' NH3 (g) + HCl (g)
Number of phases = 2
This system has two phases, one solid, NH4Cl and one gaseous, a mixture of NH3
and HCl.
d)
A solution of NaCl in water
Number of phases = 1
e)
A system consisting of monoclinic sulphur, rhombic sulphur and liquid sulphur
Number of phases = 3
This system has 2 solid phases and one liquid. Monoclinic and rhombic sulphur,
polymorphic forms, constitute separate phases.
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Box 10.4 True and metastable equilibrium
•
True equilibrium is obtained when the free energy content of a system is at a
minimum for the given set of variables.
•
A state of true equilibrium is said to exist in a system when the same state can be
obtained by approaching from either direction.
•
An example of such an equilibrium is ice and liquid water at 1 atm pressure and 0oC.
At the given pressure, the temperature at which the two phases are in equilibrium is
the same whether the state is attained by partial freezing of liquid water or by partial
melting of ice.
Liquid water at -4oC is said to be in a state of metastable equilibrium because this state of
water can be obtained by only careful cooling of the liquid and not by fusion of ice. If an ice
crystal is added to this system, then immediately solidification starts rapidly and the
temperature rises to 0oC. A state of metastable equilibrium is one that is obtained only by
careful approach from one direction and may be preserved by taking care not to subject the
system to sudden shock, stirring or “seeding” by solid phase.
Components
The number of components of a system at equilibrium is the smallest number of independently
varying chemical constituents using which the composition of each and every phase in the system
can be expressed. It should be noted that the term “constituents” is different from “components”,
which has a special definition. When no reaction is taking place in a system, the number of
components is the same as the number of constituents. For example, pure water is a one
component system because all the different phases can be expressed in terms of the single
constituent water.
Examples 10.1.2.
Counting the number of components
a)
The sulphur system is a one component system. All the phases, monoclinic, rhombic,
liquid and vapour – can be expressed in terms of the single constituent – sulphur.
b)
A mixture of ethanol and water is an example of a two component system. We need
both ethanol and water to express its composition.
c)
An example of a system in which a reaction occurs and an equilibrium is established
is the thermal decomposition of solid CaCO3. In this system, there are three distinct
phases: solid CaCO3, solid CaO and gaseous CO2. Though there are 3 species
present, the number of components is only two, because of the equilibrium:
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CaCO3 (s) CaO(s) + CO2(g)
Any two of the three constituents may be chosen as the components. If CaO and CO2
are chosen, then the composition of the phase CaCO3 is expressed as one mole of
component CO2 plus one mole of component CaO. If, on the other hand, CaCO3 and
CO2 were chosen, then the composition of the phase CaO would be described as one
mole of CaCO3 minus one mole of CO2.
d)
A system in which ammonium chloride undergoes thermal composition.
NH4Cl (s) NH3(g) + HCl (g)
There are two phases, one solid-NH4Cl and the other gas – a mixture of NH3 and
HCl. There are three constituents. Since NH3 and HCl can be prepared in the correct
stoichiometric proportions by the reaction:
NH4Cl → NH3+HCl
The composition of both the solid and gaseous phase can be expressed in terms of
NH4Cl. Hence the number of components is one.
If additional HCl (or NH3) were added to the system, then the decomposition of
NH4Cl would not give the correct composition of the gas phase. A second
component, HCl (or NH3) would be needed to describe the gas phase.
Degrees of freedom (or variance)
The degrees of freedom or variance of a system is defined as the minimum number of variables
such as temperature, pressure, concentration, which must be arbitrarily fixed in order to define the
system completely.
Examples
Systems of different variance
a)
A gaseous mixture of CO2 and N2. Three variables: pressure, temperature and
composition are required to define this system. This is, hence, a trivariant system.
b)
A system having only liquid water has two degrees of freedom or is bivariant. Both
temperature and pressure need to be mentioned in order to define the system.
c)
If to the system containing liquid water, pieces of ice are added and this system with
2 phases is allowed to come to equilibrium, then it is an univariant system. Only one
variable, either temperature or pressure need to be specified in order to define the
system. If the pressure on the system is maintained at 1 atm, then the temperature of
the system gets automatically fixed at 0oC, the normal melting point of ice.
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d)
If in the system mentioned above, a small quantity of water is allowed to evaporate
and then the system is allowed to come to equilibrium, then the number of phases in
equilibrium will be three. This system has no degrees of freedom or it is invariant.
Three phases, ice, water, vapour can coexist in equilibrium at 0.0075oC and 4.6mm
of Hg pressure only. A change in temperature or pressure will result in one or two
phases disappearing.
Hence the degree of freedom of a system may also be defined as the number of
variables, such as temperature, pressure and concentration that can be varied
independently without altering the number of phases.
The phase rule
The phase rule is the relationship between the number of phases, P, the number of components, C
and the number of degrees of freedom, F of a system at equilibrium at a given P and T. The rule
is P+F = C+2, where 2 stands for the intensive variables pressure, P and temperature, T. This is a
general rule applicable to all types of reactive and nonreactive systems. In a nonreactive system,
the various components are distributed in different phases without any complications, such as
reacting chemically with each other. First let us derive this rule for the nonreactive system and
then show that the same rule applies to the reactive system as well.
Derivation of the phase rule
Before taking up the derivation of the phase rule, let us determine the number of degrees of
freedom of some simple systems without using the phase rule.
a)
Example 1 – A gaseous system having one component.
No. of phases in the system = 1
Every homogeneous phase has an equation of state or phase equation given by f(P,T,C)=0 where
P stands for pressure, T for temperature and C for concentration. This phase equation has three
variables P,T and C. If the values of 2 variables are known, the third can be calculated using this
equation. Hence the number of variables that need to be actually specified is equal to 2.
Number of degrees of freedom = Total number of variables – number of equations connecting the
variables
F = 3-1=2
The above mentioned system is a bivariant one.
b)
Example 2 – a system consisting of water and water vapour in equilibrium with each
other.
This system has 2 phases – liquid water and water vapour. One can write one equation of
state or phase equation for each phase.
fl (T, P, Cl) = 0 for the liquid phase
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fv (T, P, Cv) = 0 for the gaseous phase
Cl and Cv are concentrations in the liquid and gaseous phases respectively. As water and
vapour are in equilibrium at a definite temperature and pressure, there is a chemical potential
relation equating the chemical potential of water in the 2 phases.
µ Hl O (P, T, Cl) = µ Hν O (P, T, Cv)
2
2
This system has four variables, T, P, Cl and Cv, and three equations relating them, two
equations of state and one chemical potential equation. Hence the number of degrees of
freedom, F= number of variables – number of equations relating the variables.
F=4–3=1
This system is thus a univariant one.
c)
Example 3 – a system consisting of ice, liquid water and vapour in equilibrium at
constant temperature and pressure.
There are 3 phases and hence 3 phase equations
For ice f (T, P, Ci) = 0
For water f (T, P, Cl) = 0
For vapour f (T, P, Cv) = 0
Ci, Cl and Cv are the concentrations of ice, liquid water and water vapour
respectively.
When these three phases coexist in equilibrium at a definite temperature and
pressure, the chemical potential of water is the same in each phase. The chemical
potential equations are:
µHi O (T, P, Ci) = µHl O (T, P, Cl)
2
2
µ lH2 O (T, P, Cl) = µ Hv 2 O (T, P, Cv)
Total number of variables = 5. These are T, P, Ci, Cl and Cv.
Total number of equations relating these variables = 5, 3 equations of state and 2
chemical potential equations. The number of degrees of freedom, F=5–5 =0
This is an invariant system.
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d)
Example 4: A homogeneous solution of sugar in water. This system has two
components, sugar and water, and one phase. The phase equation for the system is:
f (T, P, C H O , Csugar) = 0
2
Total number of variables = 4
Number of equations =1
Number of degrees of freedom, F=4-1=3.
These three degrees of freedom are temperature, pressure and concentration of one
component.
e) Example 5: A system made up of 2 phases and 2 components at constant temperature and
pressure. Let the concentration of component 1 in phase 1 be C11 and in phase 2 be C12 . Let the
concentration of component 2 in phase 1 be C12 and in phase 2 be C22 . The phase equations are:
for phase 1: f1 (T, P, C11 , C12 )=0 and
for phase 2: f2 (T, P, C12 , C22 )=0
As the system is at equilibrium, the chemical potential of component 1 in the 2 phases is the
same, as also that of component 2. Let the chemical potential of component 1 in the two phases
be µ11 and µ12 , that of component 2 be µ12 and µ 22 . The chemical potential equations are µ11 = µ12
and µ12 = µ 22 . Total number of variables are 6; P, T, C11 , C12 , C12 and C22 . Total number of equations
are 4, 2 phase equations and 2 chemical potential equations. The number of degrees of freedom,
F=6-4=2.
Derivation of the phase rule for the nonreactive system
Let us consider a system of P phases and C components existing in equilibrium at constant
temperature and pressure.
The number of degrees of freedom or the variance F is equal to the number of intensive variables
required to describe a system, minus the number that cannot be independently varied, which in
turn is given by the number of equations connecting the variables.
We begin by finding the total number of intensive variables that would be needed to describe the
state of the system. Let us assume that all the C components are present in all P phases. The state
of a system is specified at equilibrium if temperature, pressure and the amounts of each
component in each phase are specified. Since the actual amount of material in any phase does not
affect the equilibrium, it is the relative amount and not the absolute amount that is important.
Therefore, mole fractions are used instead of number of moles.
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Number of concentration variables (C mole
fractions to describe one phase;
P×C to describe P phases)
P×C
Temperature, pressure variables
2
Total number of variables
PC + 2
Let us next find the total number of equations connecting the variables.
A phase equation for each phase (For each
phase, the sum of mole fractions equals unity)
X1 + X2 + X3 + ………..+ Xc = 1
P equations for P phases
P
Chemical potential equations (At equilibrium
the chemical potentialof each component is the
same in every phase.) Equations for component
1in P phases
µ11 = µ12 = µ13 = ……………= µ1p
P-1 equations for each component
C(P-1) equations for C components
C(P–1)
Total number of equations
P+C(P–1)
Number of degrees of Freedom, F=Total number of variables – total number of equations
F = P × C + 2 – {P+C(P-1)}
F=PC + 2- P-CP+C
F=C+2-P
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P+F = C+2, which is the Gibb’s phase rule.
This rule gives the number of variables, F that need to be specified in order to define the system
completely and unambiguously.
It was assumed in this derivation that each component is present in every phase. It can be shown
that the phase rule remains unaltered even if all the components are not present in all the phases.
Derivation of the rule taking that one of the components is present only in P-1 phases.
We consider, as in the earlier case, a system consisting of C components and P phases under
equilibrium at constant temperature and pressure. One of the components is missing from one
phase and hence is present in only P-1 phases.
Let us first find out the total number of intensive variables that are needed to describe the state of
the system. As one component is excluded from one phase, the number of concentration variables
will be CP-1.
Number of concentration variables
=
CP-1
Pressure, temperature variables
=
2
Total number of variables
=
CP+1
Let us next find the total number of equations connecting the variables.
Number of phase equations
=
P
Number of chemical potential equations for
C-1 components in P phases
=
(C-1) (P-1)
for one component in P-1 phases
=
P-2
Total number of equations
=
P+(C-1)(P-1)+(P-2) = C(P-1)-1
Number of degrees of freedom, F=total number of variables – total number of equations
F=CP-1-{C(P-1)-1}
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F=C-P+2
P+F = C+2
Thus we see that the effect of one of the components not present in one phase is reduction in both
the number of variables and the number of equations by one.
The difference is thus the same and the phase rule remains unchanged. This shows that the phase
rule is generally valid under any kind of distribution as long as the system is under equilibrium.
Derivation of the phase rule for a reactive system
Let us consider a system of C constituents and P phases under equilibrium at constant
temperature and pressure. Let us assume that four of the constituents are involved in a reaction
given by:
ν1A1+ν2A2 ν3A3 + ν4A4
We next find the total number of variables that are needed to describe the system completely and
the number of equations that are available at equilibrium.
Number of concentration variables (C mole
fractions to describe one phase; P×C to
describe P phases)
P×C
Temperature, pressure variables
2
Total number of variables
PC + 2
Let us find the total number of equations connecting the variables.
A phase equation relating the
mole fractions for each phase
X1 + X2 + X3 + ………..+ Xc = 1
There are P equations for P phases
P
Chemical potential equations:P-1 equations
for each constituent
C(P-1)
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µ11 = µ12 = µ13 = ……………= µ1p
For C constituents in P phases,
there are C(P-1) equations
For a reactive system, there is another
condition that has to be satisfied. At
equilibrium, the reaction potential, ∆rG is
zero. This gives ν3µ3+ ν4µ4 - ν1µ1-v2µ2=0.
Thus we get one more equation.
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Total number of equations available
P+C(P-1)+1
Variance=number of variables – number of equations
F=CP+2-{ P+C(P-1)+1}
F=(C-1)-P+2
If in a system, two independent reactions are possible, then it can be shown that F=(C-2)P+2
Generalizing, we write
F=(C-r)-P+2
Where r is the number of independent reactions that are taking place in a system.
Sometimes a chemical reaction takes place in such a manner that requires additional
equations expressing further restrictions upon the mole fractions to be satisfied. One such
reaction is the thermal decomposition of solid NH4Cl in vacuum.
NH4Cl (s) NH3 (g) + HCl (g)
Additional restriction that exists in the gaseous phase is
XNH = XHCl
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The number of such equations as this one which impose additional restrictions should
also be included in the total number of equations.
A system containing a salt solution is another example in which an additional restricting
equation relating the mole fraction of ions exists.
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AB → A+ + BThe additional restricting equation is
X A + = XB-
In general if there are r independent reactions and Z independent restrictive conditions in
a system, then the total number of equations is given by:
Total number of equations = Number of phase equations + number of chemical potential
equations + number of equations due to chemical reactions + number of equations due to
restricting conditions.
Total number of equations = P+C(P–1)+r+Z
Variance,
F=(CP+2)–{P+C(P–1)+r+Z}
F=(C–r–Z)–P+2
F=C'–P+2
Where C'=C-r-Z and is known as the number of components of the system.
Thus, the number of components of a reactive system is equal to the total number of
constituents present in the system less than the number of independent chemical reactions
and the number of independent restricting equations.
This equation has the same form as that for a nonreactive system with C' in place of C.
Phase rule gives information only about the number of degrees of freedom of a system at
equilibrium. If a variable is altered, and the equilibrium is disturbed, then information
regarding the direction and extent of change that will follow is not provided by the rule.
This is its limitation.
For the application of phase rule to study different heterogeneous systems under
equilibrium, it is convenient to classify all systems according to the number of
components present. We will discuss one and two component systems in that order in the
following sections.
Phase equlibria of one component systems – water, CO2 and S systems
Applying the phase rule to a one component system, we write
F = C-P+2 = 1-P+2 = 3-P
Three different cases are possible with P taking values 1, 2 and 3.
a)
System having only one phase, i.e., P=1
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F=3-P=3-1=2
This is a bivariant system. We need to state the values of 2 variables in order to define the
system completely. These are temperature and pressure. The given component may exist
in any of the three phases, solid, liquid or vapour.
b)
System having 2 phases, i.e. P=2
F=3-P=3-2=1
This is a univariant system and hence the value of either of the 2 variables, temperature
or pressure, would define the system completely. The two phases in equilibrium with
each other may be solid-solid, solid-liquid, liquid-vapour and solid-vapour.
c)
System having 3 phases, i.e. P=3.
F=3-P=3-3=0
The system is invariant and three phases can exist in equilibrium only at definite values
of temperature and pressure. The three phases in equilibrium with each other may be
solid-liquid-vapour, solid-solid-liquid, and solid-solid-vapour.
Thus, we see that the maximum number of phases existing in equilibrium in a one component
system is three. For a one component system, as the maximum number of degrees of freedom is
two, the equilibrium conditions can be represented by a phase diagram in two dimensions
choosing pressure and temperature variables.
What is a phase diagram?
A phase diagram is one in which the relationships between the states or phases of a substance can
be summarized. The diagram shows the various phases present at different temperatures and
pressures. They are also called equilibrium diagrams.
We can see important properties like melting point, boiling point, transition points and triple
points of a substance on the phase diagram. Every point on the phase diagram represents a state
of the system as it described T and p values. The lines on the phase diagram divide it into regions.
These regions may be solid, liquid or gas. If the point that describes a system falls in a region,
then the system exists as a single phase. On the other hand, if the point falls on a line, then the
system exists as two phases in equilibrium. The liquid-gas curve has a definite upper limit at the
critical temperature and pressure as liquid and gas become indistinguishable above this
temperature and pressure.
The phase diagram for water
The simplest case of a one component system is one in which there is only one solid phase. In a
system having more than one solid phase, there are a number of possible equilibria and the phase
diagram gets quite complicated. In the case of “water” system, above –20oC and below 2000
atm pressure, there is only one solid phase, namely, ordinary ice. We will discuss the phase
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diagram for water under moderate pressure (Fig.10.2.2) with only ordinary ice forming the solid
phase (Box 10.2.2.1).
Figure 10.2.2: The phase diagram for water
Equilibrium between solid and vapour (sublimation curve)
At the point B (Fig.10.2.2) ice is in equilibrium with its vapour. The pressure at B is the vapour
pressure of ice at the temperature at B. If this temperature at B is gradually raised keeping the
volume constant, vapour pressure of ice also increases. If the vapour pressure of ice is plotted
against temperature, the curve BO, the sublimation curve is obtained. Along the curve BO, ice
and water vapour are in equilibrium with each other. The slope of the curve at any point as given
by the Clapeyron equation is:
∆H m,Subl
⎛ dp ⎞
⎜ dT ⎟ = T(V − V )
⎝
⎠
m,v
m,s
The variation of sublimation pressure with temperature is given by the Clausius-Clapeyron
equation as:
lnp = –
∆H m,Subl
RT
+I
Where I is the constant of integration
For each temperature of this solid-vapour system, there exists a certain definite pressure of the
vapour given by the curve BO.
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If the system represented by point B is expanded isothermally, then this will decrease the pressure
of the vapour phase. As at a given temperature, the solid-vapour system has a fixed vapour
pressure, some ice will sublime to maintain the pressure. If the isothermal expansion is continued,
more and more ice will sublime till the solid phase disappeared.
If, on the other hand, the system represented by point B is compressed isothermally, then some
vapour will condense to form ice in order to maintain the pressure and prevent its increase. If the
isothermal compression is continued, then the entire vapour phase will disappear leaving only a
solid phase in the system. These show that the regions above and below the curve BO represent
solid and vapour phases, respectively.
Equilibrium between liquid, vapour and solid water (ice)
The system at point B (Fig.10.2.2) is gradually heated keeping the volume constant when the
vapour pressure of ice increases. A temperature is reached at which the vapour pressure of ice
becomes equal to that of liquid water maintained at the same temperature. Then the solid water
starts melting and the system consists of three phases, ice, water and vapour, in equilibrium with
each other. This is an invariant system (F=0) and the temperature and pressure of the system
remain unchanged as long as all the three phases are present together. This is the system at point
O in the figure 10.2.2. and is known as triple point. This point for water lies at 0.0075oC and
4.6mmHg.
Equilibrium between liquid and vapour (vaporization curve)
The system at the triple point is gradually heated at constant volume, the temperature and
pressure do not change till the entire solid melts to give liquid water. There are only 2 phases in
the system – liquid water and vapour. If the heating is continued at constant volume, the
temperature and vapour pressure of the system vary along the curve OA (Fig.10.2.2). The curve
OA is known as the vaporization curve and along the curve OA liquid water and vapour are in
equilibrium with each other. The slope of the curve OA at any point is given by the Clapeyron
equation:
⎛ dp ⎞
⎜ dT ⎟
⎝
⎠l =
v
∆H m,Vap
T(Vm,v − Vm,l )
The Clausius – Clapeyron equation:
lnp =
−∆H m,vap
RT
+I
(I=Integration constant) gives the variation of vapour pressure with temperature, i.e., the curve
OA. This curve OA has an upper limit at the critical pressure and temperature, i.e., the point A.
If a system represented by any point on the curve OA is subjected to isothermal expansion, then
the pressure of the vapour phase decreases, a small quantity of water evaporates to raise the
pressure to a value which is the vapour pressure of liquid water at that temperature. As the
isothermal expansion is continued, more and more liquid water evaporates till the entire liquid
phase disappears and the system is made up of only vapour.
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On the other hand, if the system is subjected to isothermal compression, then vapour condenses to
lower the pressure. This continues till the entire vapour phase disappears and there is only liquid
water in the system. These changes lead us to conclude that the regions above and below the
curve OA represent liquid and vapour phases, respectively.
Equilibrium between solid and liquid (fusion curve)
If solid water (ice) at some high pressure is heated slowly, it starts melting after a certain
temperature is reached. Temperature of the system then remains constant till the entire solid
phase is converted to liquid. The temperature at which the solid melts to give liquid depends on
the pressure on the solid phase. The line OC (Fig.10.2.2) depicts the various conditions of
temperature and pressure at which ice and water are in equilibrium with each other. This line OC
is known as the fusion curve and the slope of this line is given by the Clapeyron equation:
⎛ dp ⎞
⎜ dT ⎟
⎝
⎠s =
l
∆H m,fus
T(Vm,l − Vm,s )
As Vm,l -Vm,s is small, the slope of the line OC is comparatively large and hence the line OC is
almost vertical. For water, Vm,l-Vm,s is negative and hence the line OC is slightly tilted towards
the pressure axis (Box 10.2.2).
Metastable equilibrium involving liquid and vapour phases
If a system represented by a point on the curve AO, liquid water in equilibrium with vapour, is
cooled rapidly, ice may fail to form at the triple point and the vapour pressure of the liquid may
continue along OA'. This represents metastable equilibrium involving liquid and vapour phases.
At the triple point, the slope of the solid-vapour curve, OB is greater than that of the liquidvapour curve, OA. This can be shown by using the Clapeyron equation as follows:
⎛ dp ⎞
⎜ dT ⎟
⎝
⎠s ⎛ dp ⎞
⎜ dT ⎟
⎝
⎠l =
v
=
v
∆H m,sub1
T(Vm,v − Vm,s )
∆H m,vap
T(Vm,v − Vm,l )
Vm,v – Vm,s Vm,v – Vm,l
∆Hm,sub1 = ∆Hm,fus + ∆Hm,vap
since ∆H m,subl > ∆H m,vap
⎛ dp ⎞
⎜ dT ⎟
⎝
⎠s v
⎛ dp ⎞
>⎜
⎟
⎝ dT ⎠l v
19
The continuation of the AO curve, vaporization curve beyond the triple point, i.e., OA' lies above
the OB curve, the curve for the stable phase in that temperature interval. Hence the vapour
pressure of the system in the metastable region is more than that of the stable system, that is, ice
at the same temperature.
Table 10.2.2.Description of the phase diagram for water
1)
BO
Sublimation
curve
Solid vapour
P=2, F=1
T or P
2)
OA
Vaporization
curve
Liquid vapour
P=2, F=1
T or P
3)
OC
Fusion curve
Solid liquid
P=2, F=1
T or P
4)
Area left of BOC
Solid phase
P=1, F=2
T and P
5)
Area AOC
Liquid phase
P=1, F=2
T and P
6)
Area below AOB
Vapour phase
P=1, F=2
T and P
7)
Point O (0.0075oC, 4.6mmHg)
Triple point,
Solid Liquid vapour
P=3, F=0
8)
Point A (374oC, 217.5 atm)
Critical
temperature,
critical pressure
9)
OA'
metastable
equilibrium
Liquid vapour
P=2, F=1
T or P
20
Box 10.2.2.1
In the phase diagram for water under moderate pressure (Fig.10.2.2), there is only one solid
phase, namely ordinary ice. Several crystalline modifications of ice are observed when the
system is studied under very high pressures (of about50,000 atmospheres). Ice I is ordinary
ice. At very high pressures ices II, III, V, VI and VII are stable. Existence of ice IV was
reported but was not confirmed. It was an illusion. It is reported that ice VII melts at about
100oC under a pressure of 25,000 atm. Isn’t the melting of ice hot?
Box 10.2.2.2
We have seen in the water system, the fusion curve OC (Fig.10.2.2) is almost vertical with a
slight tilt towards the pressure axis. This indicates that an increase in pressure decreases the
melting point of ice, a property that contributes to making skating on ice a possibility. The
pressure exerted by the weight of the skater through the knife edge of the skate blade lowers
the melting point of ice. This effect along with the heat developed by friction produces a
lubricating layer of liquid water between the ice and the blade. It is of interest to note that
the skating is not good if the temperature of ice is too low.
The phase diagram for carbondioxide
The system of CO2 (Fig.10.2.3) is very similar to the water system except that the solid – liquid
line OC slopes to the right, away from the pressure axis. This indicates that the melting point of
solid carbon dioxide rises as the pressure increases. The slope of this line follows the clapeyron
equation:
⎛ dp ⎞
⎜ dT ⎟
⎝
⎠s =
l
∆H m,fus
T(Vm,l − Vm,s )
As Vm,l > Vm,s and Vm,l-Vm,s is small, the line OC has a large positive slope.
The triple point, O (Fig.10.2.3) occurs at -56.4oC and a pressure of about 5 atm. We must note,
that as the triple point lies above 1 atm, the liquid phase cannot exist at normal atmospheric
pressure whatever be the temperature. Solid carbon dioxide hence sublimes when kept in the open
(referred to as “dry ice”). It is necessary to apply a pressure of about 5 atm or higher to obtain
liquid carbon dioxide. Commercial cylinders of CO2 generally contain liquid and gas in
equilibrium, the pressure in the cylinder is about 67 atm if the temperature is 25oC. When this gas
21
comes out through a fine nozzle, it cools and condenses into a finely divided snow-like solid as
the outside pressure is only 1 atm (Box.10.2.3).
Table 10.2.3 Description of the phase diagram for carbon dioxide.
1)
BO
sublimation curve
solid vapour
P=2,
F=1
T or P
2)
OA
Vaporization curve
Liquid vapour
P=2,
F=1
T or P
3)
OC
Fusion curve
Solid Liquid
P=2,F=1 T or P
4)
Area left of
BOC
Solid phase
P=1,
F=2
T&P
5)
Area AOC
Liquid phase
P=1,
F=2
T&P
6)
Area below
AOB
Vapour phase
P=1,
F=2,
T&P
7)
Point O
(-56.4oC,
~5 atm)
Triple point
SolidLiquidvapour
P=3,F=0
8)
Point A
(31.1oC, 73
atm)
Critical
temperature,
critical pressure
22
liquid
solid
p
vapour
T
Figure 10.2.3: The phase diagram for carbondioxide
Box 10.2.3 Super critical carbondioxide
Super critical carbondioxide is obtained by heating compressed carbondioxide to temperatures
above its critical temperature. The critical constants of CO2 are: Tc=304.1K and Pc=73.8 bar
which are not far from ambient conditions. It is inexpensive and easily available in large
quantities. It is non toxic, nonflammable and inert to most materials. It has good dissolving
properties and hence used as a super critical solvent. It is thus an ideal eco friendly substitute for
hazardous and toxic solvents.
It is used for extracting flavours, decaffeination of coffee and tea, recrystallization of
pharmaceuticals etc. It is also used in supercritical fluid chromatography, a form of
chromatography in which the supercritical fluid is used as the mobile phase. This technique can
be used to separate lipids and phospholipids and to separate fuel oil into alkanes, alkenes and
arenes.
The phase diagram for sulphur
Sulphur exists in two solid modifications, the rhombic form stable at ordinary temperatures and
the monoclinic form at higher temperatures. Substances that can exist in more than one crystalline
form, each form having its own characteristics vapour pressure curve, are said to exhibit the
phenomenon of polymorphism. Two types of polymorphism are observed, enantiotropy (Greek:
opposite change) and monotropy (Greek: one change).
23
Enantiotropy
Two crystalline modifications of a substance are said to be enantiotropic (or to exhibit
enantiotropy) when each has a definite range of stability and conversion from one modification to
the other takes place at a definite temperature in either direction. This temperature is the transition
point and it is the only temperature at which the two modifications can coexist in equilibrium at a
given pressure. A change in this temperature results in the complete transformation of one
modification into the other, one being stable above the transition point and the other below it.
We represent, say, the two enantiotropic forms by α and β and we assume that the α form is
stable at lower temperatures while the β form at higher temperatures.
αβ
Figure 10.2.4.1 gives the vapour pressure-temperature curve of this enantiotropic system. Each
form has its own vapour pressure curve, AB is the curve of the α form and BC is that of the β
form. B is the transition point where both α and β forms are at equilibrium with the vapour.
D
E
C
p
B
A
T
Figure 10.2.4.1: Vapour pressure temperature curve of an enantiotropic system
A system represented by a point on the curve AB is heated slowly such that the α form continues
to stay in equilibrium with vapour. The system moves along AB and at B, the α form is
transformed into the β form, the temperature and vapour pressure remain constant. The heat
supplied goes towards converting the α into the β form. After all the α form gets converted into
the β form, the vapour pressure of the system changes along the curve BC. Slopes of the
sublimation curves AB and BC at any point are given by the Clapeyron equation:
⎛ dp ⎞
⎜ dT ⎟
⎝
⎠s =
v
∆H m,subl
T(Vm,v − Vm,s )
Where the variables refer to the α and β forms respectively. At the triple point, we can write
24
∆ Hm,subl (solid α) = ∆ Hm, trans (solid α → solid β) + ∆ Hm, subl (solid β)
∆ Hm,subl (solid α) > ∆ Hm, subl (solid β)
As a result of the above relation, the slope of the vaporization curve of the α form (AB) is greater
than that of the β form (BC) at the triple point B.
The β form melts at C and CD is the vaporization curve. If the α form is heated rapidly, no β
form appears at B, the vapour pressure of the system continues along BE and a metastable
equilibrium exists between the α form and vapour.
Similarly, if a system represented by a point on the curve CD is cooled rapidly, no β form
separates at C, cooling of the liquid continues along CE and a metastable equilibrium exists
between liquid and vapour. The two curves meet at the point E, which represents the melting
point of the α form.
This explains why an enantiotropic substance melts at different temperatures depending upon
whether the solid is heated slowly or rapidly. The two melting points are those of the β and α
forms respectively.
Enantiotropy, a more common form of polymorphism is exhibited by sulphur, tin, ammonium
nitrate, carbon tetrachloride among other substances.
Monotropy
If one crystalline form is stable and the other form metastable over the entire range of their
existence, the substance is said to exhibit monotropy. The transformation from one form into
another takes place in one direction only, that is, from metastable to stable form.
F
p
E
D
B
C
A
T
Figure 10.2.4.2: Vapour pressure temperature curve of a monotropic system
25
Fig.10.2.4.2 gives the vapour pressure – temperature curve of a monotropic system. The stable α
form has the lower vapour pressure curve AB and the metastable β form has the upper vapour
pressure curve CD. BE is the vaporization curve. At the triple point B, α form, liquid and vapour
are in equilibrium with each other. The curve EB is extended to meet the curve CD at D, the
melting point of the β form. The curves CD and AB are extended to meet at a point F, a
hypothetical transition temperature. This has no real existence as it is above the melting points of
both the forms and only one solid form has a stable existence. The points D and F are metastable
triple points.
At point D
β form liquid vapour
At point F
β form α form vapour
As the effect of pressure on the melting and transition points is usually very small, points B, D
and F may be referred to as the melting point of the α form, melting point of the β form and the
transition point respectively.
The metastable β form undergoes transition to the stable α form but the reverse of this process
does not take place. The β form, however, can be obtained by an indirect method from the α
form. The α form is first heated to get a liquid, the liquid is then cooled rapidly when the system
moves along BD instead of BA and the β form separates at D.
Monotropy is exhibited by iodinemonochloride, silica, benzophenone among other substances.
Some substances for example, phosphorous exhibit both enantiotropy and monotropy.
Discussion – the phase diagram of sulphur
Figure 10.2.4.3. gives the phase diagram of sulphur. As mentioned earlier, rhombic and
monoclinic are the two enantiotropic forms of sulphur, rhombic being stable at lower
temperatures. If the temperature of the system, rhombic sulphur in equilibrium with sulphur
vapour, represented by the point A is raised at constant volume, the vapour pressure increases
along the curve AB. The curve AB, sublimation curve of rhombic sulphur, gives the temperatures
and pressures at which rhombic sulphur and its vapour exist in equilibrium.
26
F
E
D
liquid
C
B
A
T
Figure 10.2.4.3: The phase diagram for sulphur
If a system on the curve AB is allowed to expand keeping temperature constant, more of solid
sulphur will sublime to keep the vapour pressure at a constant value. If this process in continued,
then the solid phase will disappear and the system will be all vapour. If on the other hand the
same system is subjected to an isothermal compression, then vapour will condense to keep the
vapour pressure at a constant value. On the continuation of the process, vapour phase will
disappear and the system will be all solid rhombic sulphur. Hence, we conclude that the phases
above and below the curve AB comprise of rhombic sulphur and vapour sulphur respectively.
As the heating of the system, rhombic sulphur in equilibrium with vapour, is continued at
constant volume, the vapour pressure of rhombic sulphur increases along AB and at the point B
becomes equal to that of monoclinic sulphur. Then the rhombic form undergoes transition into
the monoclinic form and the system at the point B has three phases, rhombic, monoclinic and
vapour. This is an invariant system (P=3 and F=0) and the temperature and pressure of the system
remain unchanged as long as the three phases coexist. These variables remain constant as the
heating at constant volume is continued till the entire rhombic sulphur gets converted into the
monoclinic sulphur. The system then has only 2 phases, monoclinic and vapour in equilibrium
27
with each other. If the heating at constant volume is continued, then the monoclinic in
equilibrium with vapour moves along BC, the sublimation curve of monoclinic sulphur. The
slope of the curve AB is larger than that of the curve BC at the triple point B as can be seen by
applying the Clapeyron equation at the point B.
⎛ dp ⎞
⎜ dT ⎟
⎝
⎠ rhom ⎛ dp ⎞
⎜ dT ⎟
⎝
⎠ mono =
v
=
v
∆H m,rhom,subl
T(Vm,v -Vm,rhom )
∆H m,mono,subl
T(Vm,v − Vm,mono )
∆H m,r hom,subl = ∆H m,rhom →mono + ∆H m,mono,subl
∆H m,rhom,subl >∆H m,mono,subl
⎛ dp ⎞
⎜ dT ⎟
⎝
⎠ r hom v
⎛ dp ⎞
>⎜
⎟
⎝ dT ⎠ mono v
It can be shown by taking a system on the curve BC and subjecting to isothermal expansion and
compression that the system above the curve BC is solid monoclinic sulphur and below BC is
vapour (Fig.10.2.4.3).
The vapour pressure of monoclinic sulphur increases as heating at constant volume is continued
and becomes equal to that of liquid sulphur at the point C. At C, a triple point, three phases coexist, monoclinic sulphur, liquid sulphur and vapour. As the heating is continued, all the solid
melts to give liquid, temperature and pressure remaining constant. The vapour pressure of liquid
sulphur in equilibrium with vapour moves along CD with heating and reaches D, the critical
temperature. It can be seen that the slope of the curve BC is larger than that of the curve CD at the
point C in accordance with the Clapeyron equation. It can also be shown by subjecting a system
on the curve CD to isothermal expansion and compression that the phase below CD is vapour and
that above CD is liquid sulphur (Fig 10.2.4.3).
When a system consisting of rhombic sulphur at some high pressure is gradually heated, a
temperature is reached when rhombic gets converted to monoclinic sulphur. This temperature,
known as the transition point, remains constant till all the rhombic form gets converted to the
monoclinic form. The transition temperature depends on the pressure of the system and the
transition line BE (Fig.10.2.4.3) gives the dependence. The line BE has a positive slope because
rhombic sulphur is more dense than monoclinic. Rhombic sulphur exists to the left of the line BE
and monoclinic sulphur to the right.
As the monoclinic form is heated, a temperature is reached when it starts melting and the system,
monoclinic S liquid S, is represented by a point on the line CE. The temperature remains
constant till the change of phase is completed. Along the line CE, the equilibrium between
monoclinic and liquid sulphur exists whereas only solid exists to the left of the line and only the
liquid to the right of the line. The two lines BE and CE meet at E, a triple point where rhombic,
monoclinic and liquid sulphur coexist in equilibrium. If the pressure of the system is higher than
28
the triple point pressure, then the rhombic form gets converted directly to the liquid along the line
EF.
Metastable equilibria in the sulphur system
Heating a system on the curve AB rapidly may not result in the conversion of rhombic to
monoclinic at B and the vapour pressure curve may continue along BG. There exists a metastable
equilibrium between rhombic and vapour along BG. Similarly cooling rapidly a system on DC
may not result in the formation of solid monoclinic form at C and the system may continue along
CG. Liquid and vapour sulphur coexist in a state of metastable equilibrium along CG. The point
G where the curves BG and CG meet is a triple point (metastable) where rhombic, liquid and
vapour sulphur coexist in equilibrium.
If a system consisting of rhombic sulphur at some high pressure is heated rapidly, then transition
to monoclinic form may not occur on the line BE. Rhombic form may continue until the system
meets the dotted line GE when it would melt to give liquid sulphur. Along the line GE, rhombic
sulphur would exist in a state of metstable equilibrium with liquid sulphur. In the area BGEB,
rhombic sulphur exists in a metastable state. Similarly in the area CGEC, liquid sulphur exists in
a metastable state. These metastable states are formed only if rhombic form fails to undergo
transition to monoclinic form on the line BE and liquid sulphur does not pass over to monoclinic
form on the line CE. As the monoclinic form is the stable form in this region BCEB, any other
form has a metastable existence and has a tendency to spontaneously change over to the
monoclinic form. Table 10.2.4 describes, in brief, the phase diagram of sulphur.
Table10.2.4 Description of the phase diagram for sulphur
AB
sublimation curve of rhombic sulphur
rv
P=2
F=1
BC
sublimation curve of monoclinic sulphur
mv
P=2
F=1
CD
vaporization curve of liquid sulphur
lv
P=2
F=1
BE
Transition line of rhombic to monoclinic
rm
P=2
F=1
CE
Fusion line of monoclinic sulphur
ml
P=2
F=1
29
EF
Fusion line of rhombic sulphur
rl
P=2
F=1
BG
Metastable sublimation of
rhombic S
rv
P=2
F=1
CG
Metastable vaporization curve of liquid
S
lv
P=2
F=1
GE
Metastable fusion line of rhombic to
liquid
rl
P=2
F=1
B
Triple point (95.5oC, 0.01mmHg)
rmv
P=3
F=0
C
Triple point (119.2oC, 0.025mmHg)
mlv
P=3
F=0
E
Triple point (151oC, 1290atm)
rml
P=3
F=0
G
Metastable triple point (114.5oC,
0.03mmHg)
rlv
P=3
F=0
Area to the left Rhombic sulphur
of ABF
P=1
F=2
Area
above Liquid sulphur
CD and right
of CEF
P=1
F=2
Area BCEB
Monoclinic sulphur
P=1
F=2
Area below
ABCD
Vapour sulphur
P=1
F=2
30
Area BGEB
Metastable rhombic S
P=1
F=2
Area CGEC
Metastable liquid S
P=1
F=2
Phase equilibria of two component systems
Applying the phase rule to two component systems we have the degrees of freedom, F=C–
P+2=4–P. when a single phase is present in a two component system, the number of degrees of
freedom, F=3, means that three variables must be specified to describe the phase and these are
temperature, pressure and composition of the phase. When two phases are present, the number of
degrees of freedom, 4–P, is reduced to 2, temperature and composition of the liquid phase. The
values of the other variables get automatically fixed. If there are three phases present, then F=4–
3=1 which means the value of only one variable needs to be stated to describe the phases.
The maximum number of degrees of freedom for a two component system we see from the
preceding discussion is three. In order to represent the variation in three variables graphically, we
require a three dimensional diagram, a space model, which is difficult to construct on paper. To
overcome this difficulty, a common practice that is adopted is to keep one of the variables
constant. There are various types of equilibria, that are generally studied at constant external
pressure. Thus, out of the three variables (F=3), one is already stated and the variation in the other
two can be represented on a two dimensional diagram. Equilibria such as solid-liquid equilibria
are such systems in which the gas phase is absent and hence are hardly affected by small changes
in pressure. Systems in which the gas phase is absent are called condensed systems.
Measurements in these systems are generally carried out at atmospheric pressure. As these
systems are relatively insensitive to small variation in pressure, the pressure may be considered
constant. The phase rule takes the form
P+F=C+1
For such systems and in this form it is known as the reduced phase rule. For a two component
system, this equation becomes F=3-P where the only remaining variables are temperature and
composition. Hence solid-liquid equilibria are represented on temperature – composition
diagrams. In the following sections we will be discussing systems involving only solid-liquid
equilibria.
Determination of solid-liquid equilibria
Many experimental methods are used for the determination of equilibrium conditions between
solid and liquid phases. The two most widely used are the thermal analysis and saturation or
solubility methods. Whenever required additional data are obtained by investigating the nature of
the solid phases occurring in a system.
Thermal analysis
Thermal analysis method involves a study of cooling curves (temperature time plots) of various
compositions of a system. A system of known composition is prepared, heated to get a melt,
31
allowed to cool on its own and then its temperature noted at regular time intervals (say, half a
minute). A cooling curve is obtained by plotting temperature versus time.
Information regarding the initial and final solidification temperatures is obtained from the breaks
and halts in the cooling curves. This analysis method is applicable under all temperature
conditions but is especially suitable for investigations at temperatures quite above and below
room temperature.
Saturation or solubility method
In this method the solubilities of one substance in another are determined at different constant
temperatures, then the solubilities are plotted against temperature.
To determine the solubility of A in B, excess of A is added to B (molten B, if B is a solid), kept at
a desired constant temperature, stirred well until equilibrium is reached. The undissolved solid A
is filtered off and the saturated solution of A in B is analysed for both constituents. Similarly
where possible saturated solutions of B in A at different temperatures are analyzed.
This is the principal method of analysis of systems containing water and similar solvents as one
of the constitutents. This method is very difficult to carry out at temperatures below -50oC and
also above 200oC. Then the thermal analysis method is preferred.
Determination of the nature of solid phases
It is important to know the nature and composition of the solid phases which appear during
crystallization and in the final solid for the complete interpretation of a phase diagram. These
solid phases may be pure components, compounds formed by reaction between pure constituents,
solid solutions and mixture of solids.
It is possible to arrive at the nature of the solid phases from the shape of the phase diagram quite
often. However, in some cases, a more careful study may be required. To do this, the solid mass
may be inspected under a microscope or may be studied using X rays.
Classification of two component solid-liquid equilibria
Phases diagrams of some two component solid liquid equilibria are simple while those of others
are quite complicated. The complex ones may be considered to be made up of a number of simple
types of diagrams. The classification of these equlibria is done according to the miscibility of the
liquid phases. These are further divided into various types based on the nature of the solid phases
crystallizing from the solution.
The classification is as follows:
Class A
The two components are completely miscible in the liquid phase
Type 1.
Only the pure components crystallize from the solution
32
Type 2
A solid compound stable upto its melting point is formed by the two constituents
Type 3
A solid compound which decomposes before it reaches its melting point is
formed by the two constituents
Type 4
The two constituents are completely miscible in the solid phase. This results in
the formation of a series of solid solutions
Type 5
In the solid state, the two constituents are partially miscible and they form stable
solid solutions
Type 6
Solid solutions formed by the two constitutents are stable only upto a transition
temperature
Class B
In the liquid phase, the two components are partially miscible
Type 1
Only pure components crystallize from the solution
Class C
In the liquid phase, the two components are immiscible
Type 1
Only pure components crystallize from the solution
Cooling curve of a pure component
A pure component, say A, is taken and heated to get a melt. The liquid A is allowed to cool on its
own and the temperature of the melt is noted at, say, every half a minute. A cooling curve is
obtained by plotting temperature versus time as shown in the fig. 10.3.3.1.
33
a
Temperature →
liquid cooling
liquid
c
d
solid cooling
e
T
Figure 10.3.3.1: Cooling curve of a pure component
Cooling of liquid A takes place along ac, when at c solidification starts and the system has two
phases in equilibrium becoming an invariant system (F=C+1-P=1+1-2=0). The temperature
remains constant till the entire liquid A solidifies along cd. The system is cooling, losing heat to
the surroundings, yet the temperature remains constant during solidification. This is due to the
fact that heat is released during solidification. Cooling of solid A takes place along de. The
system represented by any point on either ac or de has only one phase and hence is univariant.
Cooling curve of a mixture of two components with only pure components crystallizing on
cooling the system.
Solid B is added to solid A to get a mixture of known composition. This mixture is heated to get
it in the liquid phase. The cooling curve of this liquid is depicted in Fig. 10.3.3.2.
34
a
Temperature→
liquid cooling
b
second solid
starts separating
solid starts separating
c
d
solid mixture cooling
e
Time →
Figure 10.3.3.2: Cooling curve of a liquid mixture of two components
The liquid cools along ab and at b solid A starts solidifying. Temperature and composition of
liquid phase have to be stated to define the liquid phase completely (P=1, F=3-P, F=2) when A
starts solidifying, the system becomes univariant (P=2, F=3-P=1). Thus, the temperature at which
solid A starts solidifying from the liquid mixture of known composition will have a definite
value given by the point b. This temperature is expected to be a little lower than the freezing point
of pure A as the addition of B to A lowers the freezing point of A. Solidification of a small
quantity of A changes the composition of the liquid phase and hence the temperature at which A
solidifies from this liquid will take place at another fixed value but lower than the previous
temperature as the molality of B in the liquid increases with the solidification of A. Thus as more
and more of A separates from the liquid, the temperature of the system falls along bc. The rate of
cooling is affected by the heat evolved due to the solidification of A and hence a break is
observed in the curve at b. The break point indicates the temperature at which A just starts
solidifying. Along the curve bc, there are two phases, liquid and solid A, hence the system is
univariant (P=2, F=3-P=1). As cooling continues along bc, more and more of solid A separates
and the liquid gets richer in B. At the point C, the liquid becomes saturated in B and hence B
starts separating along with A. Along cd both the solids A and B separate and the system becomes
invariant (P=3, F=3-P=0). Solidification from a solution of fixed composition, one corresponding
to the saturation solubility of B in A takes place at constant temperature. This results in a halt or
complete arrest of the cooling curve (cd). As the saturation solubility of B in A has to be
maintained at this temperature, the composition of the solid phase that separates will be the same
as that of the liquid phase. The temperature of the systems will remain unchanged till the whole
of the liquid phase solidifies. The cooling of the solid phase is represented by de and the system is
univariant (P=2, F=3-P=1).
35
a
liquid cooling
Temperature
b
b'
solid starts separating
f
c
d
solid mixture cooling
second solid starts separating
e
Time →
Fig 10.3.3.3: Cooling curve of a liquid mixture of two components in which supercooling has
occurred
36
In some cases, the separation of the solid phase does not occur readily at b and the liquid continues to cool along bf (super cooling occurs). As this
represents an unstable state, there is a sudden increase in temperature which is followed by a normal cooling curve along b'c. The correct freezing
point is then obtained by extrapolation to the point b as shown in the figure 10.3.3.3.
Various mixtures, B in A and A in B show the type of cooling curve given in Fig.10.3.3.2 except one particular composition. With this
composition, on cooling the liquid, both components start solidifying at the same time. In this case the first break does not occur and the cooling
curve resembles the one given in fig. 10.3.3.1.
Systems in which solid solutions and not pure components separate on cooling, the cooling curves are similar to that given in Fig.10.3.3.2 with the
horizontal portion (the halt) replaced by a break.
The bismuth-cadmium system
A number of mixtures of the two metals ranging in overall composition of 100% bismuth to 100% cadmium are prepared. They may be spaced at
10% intervals and should preferably be of equal mass. Crucibles made of inert material, say fire-clay or graphite are taken, each mixture of
bismuth and cadmium is placed in a crucible and heated in an electric furnace to get the melt. An inert or a reducing atmosphere is maintained by
passing hydrogen, nitrogen or carbon dioxide through the furnace, to prevent oxidation of the metals. As an additional precaution, a molten flux
such as borax or a layer of powdered graphite is used to cover the mixture in the crucibles. After melting the mixture and mixing thoroughly, the
furnace and contents are allowed to cool slowly. Temperature and time readings are taken until the mixtures are completely solidified. Then
cooling curves are plotted and break and halt temperatures are noted. If one desires to check on the composition of the mixtures prepared, then the
solids are removed from the crucibles and carefully analyzed.
The equilibrium or phase diagram for this system is then constructed by plotting break and halt temperatures from the cooling curves of the
mixtures on a temperature composition diagram. Smooth curves are drawn to yield the diagram shown in the figure 10.3.4
Temperature
37
Figure 10.3.4: The phase diagram for bismuth-cadmium system
38
Bismuth-cadmium system is one in which pure components only separate (crystallize) as the
liquid (Bi and Cd are completely miscible in the liquid state) cools. This system exhibits a simple
eutectic diagram. Points A and C are the freezing points of pure bismuth (271oC) and pure
cadmium (321oC) respectively. The figure 10.3.4. shows how the initial solidification
temperatures (ti, break points) and final solidification temperatures (tf, halts) are taken off the
cooling curves for the various compositions and plotted on a temperature – composition diagram.
Curve AB indicates the temperatures at which bismuth begins to solidify from various
compositions of melt while BC provides the same information for initial separation of cadmium.
Temperature at which all mixtures become completely solid is indicated by the line DE. Curve
AB may be considered not only as the initial freezing point curve for bismuth, but also as the
solubility curve of bismuth in molten cadmium. Hence points on this curve give the solubilities
of bismuth in molten cadmium at various temperatures. Similarly, solubilities of cadmium in
molten bismuth at various temperatures is given by the curve BC. At the point B, where the two
curves meet, the solution is saturated with respect to both solids.
The curves AB and BC represent monovariant two phase equilibria (P=2, F=3-P=1). At B, three
phases are in equilibrium, the solution is saturated with both bismuth and cadmium and hence B
is an invariant point (P=3, F=3–P=0). At this point the temperature D and the composition G of
the solution must remain constant as long as three phases coexist. The temperature can be brought
below B only when one of the phases disappears and on cooling this must be the saturated
solution. Hence at temperature D, solution B must solidify completely. D is the lowest
temperature at which a liquid phase may exist in the system, bismuth-cadmium. Below this
temperature, the system is completely solid. Temperature D is called the eutectic (Greek: easily
melted) temperature, composition B the eutectic composition and point G the eutectic point in the
system.
Above the curves AB and BC is the area, 1 in which unsaturated solution or melt exists. There is
only one phase in this area and the system is divariant. Temperature and composition, both, must
be specified to describe any point representing a system in this area. To understand the phase
diagram better, we will consider the behaviour of some mixtures of bismuth and cadmium on
cooling.
Let us take first a mixture of overall composition given by a. This mixture is heated to a point a'''
(isopleth a'''-a''-a'-a) when an unsaturated solution is obtained. On cooling this solution, a drop in
temperature occurs until a'', corresponding to temperature x'', is reached. At this point the solution
becomes saturated in bismuth. Temperature x'' is the freezing point of this solution. On further
cooling, solid bismuth separates and the composition of the saturated solution changes along a'' B.
At a temperature such as x', solid bismuth is in equilibrium with saturated solution of composition
y' and so on. Thus, we can see that for any over-all composition falling in area ADB, solid
bismuth is in equilibrium with various solutions of compositions given by AB at each
temperature.
As cooling continues, at temperature D cadmium also separates and the system becomes
invariant. Both solids, bismuth and cadmium, separate from the saturated solution in a fixed ratio
given by G until the solution has been completely solidified. The system, a mixture of solid A
and solid B, becomes univariant. The cooling continues below the temperature D into the region
DHGB, the solid in this region consists of large crystals of bismuth, called primary crystals
39
(primary because these appear first) an intimate mixture of finer crystals of bismuth and cadmium
in the ratio given by C (eutectic mixture).
Applying similar considerations to systems whose overall compositions lie between G and I, say
b, we see that in the area BEC solid cadmium is in equilibrium with saturated solutions along the
curve BC. At temperature D, solid bismuth also separates. The system becomes invariant and
stays so till the solution at B solidifies. When the solidification is complete, the mixture moves
into the area BGIE where primary cadmium and eutectic mixture of composition given by G
separate.
On cooling a mixture of bismuth and cadmium having an overall composition given by B, the
temperature of the melt decreases and no solid separates until point B is reached. At this point,
both bismuth and cadmium separate and the system solidifies to yield the eutectic mixture,
temperature remaining constant. System having composition B behaves like a pure substance on
freezing but the solid separating out is an eutectic mixture of bismuth and cadmium.
If a system depicted by a''' is cooled to a' along the isopleth a'''-a, the system will consist of solid
bismuth in equilibrium with liquid phase of composition given by the point y'. The point y' is the
intersection point of the tie-line drawn from the point a' with the curve AB. The tie-line is defined
as a line that connects different phases in equilibrium with one another. The phases in equilibrium
in this case are solid bismuth (x') and liquid (y'). The relative amounts of the two phases is
determined by using the lever rule, such that
Amount of solid bismuth
a'y'
=
Amount of liquid phase of composition y' a'x'
As the system cools from a'' and moves towards the line DBE, the ratio a'y'/a'x' increases
indicating thereby that more and more of solid bismuth separates when the eutectic temperature is
reached, solid cadmium also separates. Hence in order to get the maximum amount of pure solid
bismuth, the system is cooled to a temperature slightly above the point B. Similarly in order to get
the maximum amount of pure solid cadmium, a system represented by, say, the point b' is cooled
to a temperature slightly above the eutectic temperature. Point B represents the lowest
temperature at which any melt of bismuth and cadmium will freeze out and hence is also the
lowest temperature at which any mixture of solids bismuth and cadmium will melt. The eutectic
mixture melts sharply at the eutectic temperature D, to form a liquid of the same composition
while other mixtures melt over a range of temperature. Microscopic examination of the eutectic
under high magnification shows its heterogeneous character. Eutectics are, hence mixtures. Table
10.3.4.1 describes the phase diagram (Figure 10.3.4) of bismuth-cadmium system.
Table 10.3.4.1. Description of the phase diagram for bismuth-cadmium system
A (271oC)
Freezing point of bismuth
C=1, P=2, F=0
Fixed T
C (321oC)
Freezing point of cadmium
C=1, P=2, F=0
Fixed T
40
B (144oC, 60% Bi)
Eutectic point
C=2, P=3, F=0
Fixed T and
composition
AB
Crystallization of Bi
begins
C=2, P=2, F=1
T or
composition
BC
Crystallization of Cd
begins
C=2, P=2, F=1
T or
composition
Area above ABC
Liquid phase
C=2, P=1, F=2
T and
composition
Area below DBE
Solid mixture
C=2, P=2,F=1
T or
composition
Area ADBA
Solid bismuth in
equilibrium with liquid
having composition given
by the curve AB
C=2, P=2, F=1
T or
composition
Area CEBC
Solid cadmium in
equilibrium with liquid
having composition given
by the curve BC
C=2, P=2, F=1
T or
composition
DBE
Both Bi and Cd separate
from liquid of composition
B
C=2, P=3, F=0
Fixed T and
composition
Table 10.3.4.2: A few examples of systems exhibiting simple eutectic phase diagram
Eutectic
temperature/oC
System
A
m.pt./oC
B
m.pt./oC
Eutectic
composition
41
Antimony
631
Lead
327
246
87 mass% B
Sodium
sulphate
884
Sodium chloride
800
628
48.2 mol%A
Naphthalene
80
Benzoic acid
120
68
30 mol%B
Acetanilide
112
Benzoic acid
121
76
42.4mol%B
Benzoic acid
121
Cinnamic acid
133
81
43.5 mol%B
Resorcinol
115
Cinnamic acid
133
87
41 mass% B
Lead
327
Silver
961
308
2.4 mass% B
The Lead-Silver system
\The metals lead and silver are completely miscible in the liquid state and do not form any
compound. Hence the phase diagram of this system is similar to that of the bismuth-cadmium
system we discussed in the last section.
42
C
Temperature →
liquid
a''
liquid + solid Ag
A
a'
1
Da
B
2
H
Pure Pb
solid silver + eutectic
G
Composition →
E
I
Pure Ag
Figure 10.3.5: The phase diagram for lead-silver system. 1. Liquid+solid Pb and 2. Solid
Pb+eutectic
Figure 10.3.5 gives the phase diagram of the Pb-Ag system. In the last section, the Bi-Cd system
has been discussed extensively and the same arguments hold good for the Pb-Ag system as well.
Hence we give only a brief description here.
In the figure 10.3.5 A and C are the freezing points of pure lead and silver respectively. Curve
AB indicates the temperatures at which lead begins to separate from various compositions of melt
while BC indicates initial separation of silver. D is the eutectic temperature of the system and the
eutectic composition is given by B. Curve ABC is the liquidus curve as it gives the composition
of the liquid phase that is in equilibrium with the solid phase. ADBEC is the solidus curve; AD
represents solid lead, DBE mixture of lead and Ag in equilibrium with liquid phase of
composition B and EC solid silver. Table 10.3.5 describes the phase diagram (Fig.10.3.5) of lead
and silver.
Table 10.3.5
Description of the phase diagram for lead-silver system.
A (327oC)
Freezing point of lead
C=1, P=2, F=0
Fixed T
C (961oC)
Freezing point of silver
C=1, P=2, F=0
Fixed T
43
B (303oC, 2.5 mass Eutectic point
% Ag)
C=2, P=3, F=0
Fixed T and
composition
AB
Crystallization of lead
begins
C=2, P=2, F=1
T or
composition
BC
Crystallization of silver
begins
C=2, P=2, F=1
T or
composition
Area above ABC
Liquid phase
C=2, P=1, F=2
T and
composition
Area below DBE
Solid mixture
C=2, P=2, F=1
T or
composition
Area ADBA
Solid lead in equilibrium
with liquid having
composition given by the
curve AB
C=2, P=2, F=1
T or
composition
Area CEBC
Solid silver in equilibrium
with liquid having
composition given by the
curve BC
C=2, P=2, F=1
T or
composition
DBE
Both lead and silver
separate from liquid of
composition B
C=2, P=3, F=0
Fixed T
When a phase diagram is available for a system, we are able to know from the diagram the
conditions under which particular solid phases may be obtained. We are also able to describe how
a mixture of a given overall composition behaves on cooling. It can be seen that pure solid lead
may be separated only from mixtures falling in the area ADBA and only between temperatures A
and D. Similarly pure solid silver may be obtained in area CEBC from overall compositions
between G and I and only between temperatures C and E. The proportion of solid to saturated
solution at each temperature can be obtained by drawing a tie line and using the lever rule.
44
Pattinson’s process for the desilverisation of argentiferous lead
The process of heating argentiferous lead containing a very small quantity of silver (~0.1 mass%)
and cooling to get pure lead and liquid richer in silver is known as the Pattinson’s process. This
process can be understood by following the phase diagram of the lead-silver system.
The argentiferous lead is melted and heated to a temperature above the melting point of pure lead.
Let the point a'' represent this system on the diagram (Fig.10.3.5). This system is then allowed to
cool slowly and the temperature of the melt decreases along a''-a'. At a', solid lead starts
separating. As the system further cools, more and more lead separates and the liquid in
equilibrium with the solid lead gets richer in silver. The lead that separates floats and is
continuously removed by ladles. When the temperature of the liquid reaches ‘a’ on the line DBE,
the eutectic temperature, solid lead is in equilibrium with the liquid having the composition B.
After removing the lead that separates, the liquid is cooled further when it solidifies to give a
mixture of lead and silver having the eutectic composition of 2.5 mass % of silver. This solid
mixture of lead and silver is subjected to other processes for the recovery of silver.
The Magnesium-Zinc system
The magnesium-zinc system is an example of a two component system forming a solid compound
stable upto its melting point. Such a compound has its own characteristic melting point which
may be greater or smaller than the melting points of the two pure components. The compound on
heating remains in the solid phase upto its melting point and then melts sharply to give a liquid
having the same composition as the solid. The temperature remains constant till the entire solid
compound melts. Such a melting point where both solid and liquid of the same composition can
co-exist is known as the congruent melting point. Magnesium and zinc form a compound, an
alloy having the formula Mg (Zn)2, with a congruent melting point of 590oC. Zinc and
magnesium are completely miscible in the liquid state. The phase diagram of the Mg-Zn system is
given in figure 10.3.6.
45
E
liquid
C
Temperature →
Zn+Melt
Mg(Zn)2+Melt
A
F
B
Mg
+
Melt
G
H
D
Zn+Mg(Zn)2
I
Mg+Mg(Zn)2
J
Pure Zn
Composition →
K
Pure Mg
C'
Figure 10.3.6: The phase diagram for magnesium – zinc system
The melting points of pure zinc and pure magnesium are represented by points A and E
respectively. AB is the freezing point curve of zinc. In the area ABFA, solid zinc is in equilibrium
with liquid containing zinc and magnesium, the composition of which is given by the curve AB.
Similarly, ED is the freezing point curve of magnesium. Solid magnesium is in equilibrium with
liquid containing magnesium and zinc, composition of the liquid phase lying on the curve DE.
If a liquid having the composition c' is cooled, the liquid merely cools till it reaches the point C.
At the point C, a solid compound, having the same composition as the liquid starts separating.
This temperature, the congruent melting point of the compound, remains constant till the entire
liquid phase freezes. On further cooling, the temperature of the solid decreases. Thus, the cooling
pattern of the liquid of composition c' is similar to that of a pure component. The formula of the
compound formed is Mg (Zn)2 which corresponds to the composition c'.
CB and CD give the freezing point curves of Mg (Zn)2. Addition of zinc depresses the freezing
point of Mg(Zn)2 and the temperatures at which the solid compound Mg (Zn)2 will begin to freeze
(separate) from various liquids, composition lying between B and G, fall on the curve CB.
46
Similarly CD gives the temperatures at which the solid compound Mg(Zn)2 starts freezing from
liquids having their composition lying between H and D. The curve CD gives the depression in
the freezing point of Mg (Zn)2 due to the addition of Mg. B is an eutectic point at which solid
zinc and solid Mg (Zn)2 are in equilibrium with liquid of composition B. Similarly at D, another
eutectic point, solid magnesium and solid Mg(Zn)2 are in equilibrium with liquid of
composition D.
Phase diagrams, such as that of Mg-Zn system (Figure 10.3.6), can be considered to be made up
of two simple eutectic diagrams placed side by side. There is the simple eutectic phase diagram
of zinc and Mg(Zn)2 on the left of the line cc' and that of Mg and Mg(Zn)2 on the right.
Theoretically, the curves BC and DC should meet to give a sharp point at C. But normally a
rounded maximum is observed as shown in the phase diagram (Figure 10.3.6). This is because the
compound formed is usually not very stable and dissociates partly. The dissociation products in
the liquid phase depress the actual melting of the compound resulting in a rounded melting point.
Table 10.3.6.1 gives a description of the phase diagram (Figure 10.3.6) of the magnesium – zinc
system.
Table 10.3.6.1Description of the phase diagram for magnesium – zinc system
A (420oC)
Freezing point of Zinc
C=1, P=2, F=0
Fixed T
E (651oC)
Freezing point of Mg
C=1, P=2, F=0
Fixed T
C (590oC)
Freezing point of Mg(Zn)2
C=2, r=1, P=2,F=0
Fixed T and
composition
B
Eutectic point (Zn,
Mg(Zn)2, liq. Of
composition B)
C=2, P=3, F=0
Fixed T and
composition
D
Eutectic point (Mg,
Mg(Zn)2, liq of
composition D)
C=2, P=3, F=0
Fixed T and
composition
AB
Freezing point curve of Zn,
crystallization of Zn begins
C=2, P=2, F=1
T or
composition
BC and CD
Crystallization of Mg(Zn)2
begins
C=2, P=2, F=1
T or
composition
47
ED
Crystallization of Mg
begins
C=2, P=2, F=1
T or
composition
Area ABFA
Zn liquid
(composition given by AB)
C=2, P=2, F=1
T or
composition
Area CBGC
Mg(Zn)2 liquid
(composition given by CB)
C=2, P=2, F=1
T or
composition
Area CHDC
Mg(Zn)2 liquid
(composition given by CD)
C=2, P=2, F=1
T or
composition
Area EDIE
Mg liquid
(composition given by ED)
C=2, P=2, F=1
T or
composition
Area below FBG
Solid mixtures of Zn and
Mg (Zn)2
C=P, P=2, F=1
T or
composition
Area below HDI
Solid mixtures of Mg and
Mg (Zn)2
C=2, P=2, F=1
T or
composition
Area above ABCDE
Liquid containing Zn and
Mg
C=2, P=1, F=2
T and
composition
In many systems, the two components combine to form more than one compound. In such cases,
the phase diagram has a curve similar to BCD for each and every compound. A few examples of
systems forming more than one congruent melting compound are included in the table 10.3.6.2.
48
Table 10.3.6.2 A few examples of systems exhibiting compound formation with congruent
melting point.
System
Compound
m.pt./oC
A
m.pt./oC
B
m.pt./oC
Gold
1064
Tin
232
AB
425
CaCl2
777
KCl
776
AB
754
Urea
133
Resorcinol
110
AB
~103
Benzamide
125
Resorcinol
109
AB
~89
Acetamide
79
Phenol
41
AB2
43
Silver
961
Strontium
757
A4B
781
A5B3
760
AB
680
A2B2
665
Aluminium
657
Magnesium
650
A3B4
463
Potassium
63.6
Antimony
613
A3B
812
AB
605
49
The Sodium chloride-water system
Sodium chloride-water is a system in which the two components form a solid compound which is
not stable up to its true melting point and hence decomposes before reaching it. The compound
decomposes to give another solid and a solution. The composition of the solid formed is different
from that of the original compound. Infact, the compound does not have a true melting point. The
dehydrate of sodium chloride, NaCl.2H2O, the compound formed in this system decomposes to
give anhydrous sodium chloride and a solution. The temperature at which this decomposition
reaction takes place is known as the incongruent melting point of the compound. It is also known
as the peritectic (Greek; melting around) temperature or transition temperature.
The compound is said to undergo a transition or peritectic reaction or incongruent fusion.
NaCl.2H2O NaCl + Solution
At the peritectic temperature, the number of degrees of freedom of the system is zero (C=2, P=3,
F=C+1-P, F=2+1-3=0), the system, thus, is invariant. The peritectic reaction takes place at a
definite temperature and is reversible. The melt (or the solution) also has a definite composition.
In such a peritectic reaction, the new solid formed may be a pure component as in this system, or
may be a congruent or incongruent melting new compound. The melting point of this solid is
always more than that of the compound from which it is formed. The phase diagram of the NaClH2O system is given in Figure 10.3.7.
50
D
d
C'
Unsaturated solution
d'
NaCl + solution
a
b
A
e
C
d''
G
e'
J
Temperature
a'
NaCl.2H2O
+
solution
ice + solution
E
a''
B
c'''
d'''
F
e''
ice + NaCl.2H2O
a'''
H
O
b'
NaCl
+
NaCl.2H2O
c'''
G'
Mass % NaCl
Figure 10.3.7: The phase diagram for sodium chloride – water system
I
40
51
The freezing point of water is represented by A in the phase diagram. Addition of sodium
chloride lowers the freezing point of water along AB. The curve AB, hence, is the freezing point
curve of water. In the area ABEA, ice is in equilibrium with a solution of sodium chloride in
water, the composition of which is given by the curve AB. The point G' corresponds to the
composition of the compound NaCl.2H2O. On heating, this compound remains in the solid state
until the point G at temperature tp (peritectic temperature) is reached. At the point G, the
following reaction occurs:
NaCl.2H2O
NaCl + Solution
The point G, therefore, represents the incongruent melting point of the system. The system is
invariant at this point G as NaCl.2H2O is in equilibrium with anhydrous NaCl and a solution of
composition given by the point C (F=C+1-P, F=2+1-3=0). As NaCl separates from NaCl.2H2O at
the peritectic temperature G, the solution with which it is in equilibrium should be less rich in
NaCl and hence its composition should lie to the left of point G, say, C. Line CG is a part of the
tie line drawn at point G. A system anywhere on this line has NaCl.2H2O, NaCl and solution of
composition C in equilibrium.
If the heating of NaCl.2H2O at G is continued, the decomposition to give NaCl continues and the
temperature remains constant till the reaction is over. Further heating causes an increase in
temperature of the system which is NaCl in equilibrium with solution. The system becomes
univarant (F=C+1-P=2+1-2=1). At temperatures more than the peritectic temperature, only NaCl
exists in equilibrium with solution. CD is a part of the freezing point curve of solid NaCl as it
extends beyond D but ends at C.
Curve BC can also be called the solubility curve of NaCl.2H2O and CD is a part of the solubility
curve of anhydrous NaCl. Solutions of NaCl in water having mass % NaCl more than that
represented by point C show a halt on cooling when they reach the temperature at C.
The freezing point curves, AB, of water and BC, of NaCl.2H2O meet at B, the eutectic point of
the system. There are three phases, ice, NaCl.2H2O and solution, in equilibrium at this point and
hence it is an invariant point (F=C+1-P=2+1-3=0). The area CBFGC has NaCl.2H2O in
equilibrium with solution whose composition is given by the curve CB whereas NaCl and
NaCl.2H2O exist in equilibrium in the area GG'IJG. This can be explained on the basis of the
peritectic reaction:
NaCl + Solution
NaCl.2H2O
The reaction between NaCl and solution takes place at a constant temperature till either all the
NaCl or all the solution has been consumed. If NaCl is present in a quantity more than that
required to convert the whole of the solution into NaCl.2H2O, then the system will have after the
reaction NaCl.2H2O, and the excess NaCl in equilibrium. If on the other hand, NaCl is present in
a quantity less than that required, then the system will have after the reaction NaCl.2H2O and the
excess solution in equilibrium. A point on the right of the line GG' represents excess of NaCl
whereas between C and G deficiency of NaCl.
If a solution of NaCl in water represented by a is cooled, then at a' ice starts separating and the
solution with which it remains in equilibrium is given by the curve AB. The lever rule gives the
ratio of the quantity of ice to that of solution. As it cools to a'', NaCl.2H2O, also starts separating.
52
At a'', three phases are in equilibrium, ice, the dihydrate and solution of composition B and the
system is invariant. On further cooling to, say, a''', the temperature of ice and the eutectice, a
mixture of ice and NaCl.2H2O, decreases to a'''. The cooling curve of the system at a, thus,
exhibits a break at the temperature a' and a halt at a''.
On the other hand, if a system at b, an unsaturated solution of NaCl in water, is cooled, the
temperature of the solution decreases along bB and at B both ice and NaCl.2H2O start separating.
The temperature remains constant till the entire solution solidifies to give the eutectic mixture of
ice and NaCl.2H2O. The cooling curve of the system at b(eutectic composition) has only a halt at
temperature B followed by the cooling of the eutectic mixture.
A slightly more concentrated solution of NaCl in water represented by c' cools till the point C is
reached, when NaCl.2H2O starts separating. As cooling continues, more and more NaCl.2H2O
separates and the dihydrate is in equilibrium with solution whose composition is given by the
curve BC. The ratio of the quantity of the dihydrate to the solution is given by the lever rule.
When the temperature c'' is reached, ice also separates alongwith the dihydrate forming the
eutectic mixture. The temperature remains constant till the entire solution, composition given by
B, freezes to form the eutectic mixture which then cools to c'''.
When a system at d cools to d', anhydrous NaCl starts separating and continues to separate on
further cooling, the solid remaining in equilibrium with solution, its composition given by the
curve CD. At d'', the peritectic reaction starts and continues at constant temperature till the solid
NaCl disappears. The system has now NaCl.2H2O in equilibrium with the solution, whose
composition is given by C. Further cooling results in the separation of more and more
NaCl.2H2O with the composition of the solution moving along the curve CB. When d''' is
reached, solution has composition B and eutectic mixture of ice and NaCl.2H2O separates.
Temperature remains constant till the entire solution freezes to give the eutectic mixture, the
temperature of which decreases on further cooling. Thus the cooling curve of a solution at d
shows a break at d', a halt at d'' and another halt at d'''.
Let us next takes the system e and cool it. The system at e is made up solid NaCl and solution,
composition given by CD. As it cools, more and more NaCl separates and at e', solid NaCl and
the solution of composition C undergo peritectic reaction to give NaCl.2H2O. Temperature
remains constant at e' till the completion of the reaction. Further cooling lowers the temperature
of NaCl.2H2O. Table 10.3.7.1 describes the phase diagram 10.3.7.
Table 10.3.7.1Description of the phase diagram for sodium chloride – water system
A(0oC)
Freezing point of H2O
B
Eutectic point (ice, NaCl.2H2O, solution B)
-21.1oC, 23.3 mass% of NaCl.
C(0.15oC)
Peritectic point (NaCl, NaCl.2H2O, solution C)
53
AB
Freezing point curve of ice
BC
Freezing point curve of NaCl.2H2O
CD
Freezing point curve of NaCl.
HA
Solid water (ice). Liquid water beyond A on the line HA.
GG'
Solid NaCl.2H2O
CGJ
A point on this line represents NaCl and NaCl.2H2O in
equilibrium with solution C.
EBF
A point on this line represents ice and NaCl.2H2O in
equilibrium with solution B.
Area AEBA
Ice Area to the right of DCGJ
NaCl Area of BCGFB
NaCl.2H2O Area below GJ
NaCl and NaCl.2H2O
Area above ABD
Unsaturated solution
solution (composition given by AB)
solution (composition given by CD)
solution (composition given by BC)
Cooling produced by freezing mixtures
Freezing mixtures are obtained by adding salt to ice, usually sodium chloride to ice. The low
temperatures obtained in these freezing mixtures can be explained by using the NaCl.2H2O phase
diagram.
54
When solid sodium chloride is added to ice at 0oC, the freezing point of ice gets lowered to
below 0oC and some ice melts. If this system is maintained under adiabatic conditions, then the
melting of ice reduces the temperature of the ice, sodium chloride mixture. If sufficient quantity
of the salt is added, then the temperature of the system drops to the eutectic temperature of 21.1oC. At this temperature ice, solid salt and solution of composition B are at equilibrium. The
temperature remains constant as long as all three phases coexist in equilibrium. The invariance of
the system at the eutectic point enables one to use eutectic mixtures as constant temperature
baths. Table10.3.7.2 gives the eutectic temperature and composition of some ice salt systems.
Table 10.3.7.2 Eutectic temperatures of a few ice-salt systems
Salt
~Eutectic temperature/OC
~Mass% of anhydrous salt in
eutectic
NaCl
-21
23
KCl
-11
20
NH4Cl
-15
20
KI
-23
52
NaBr
-28
40
NaI
-32
39
Action of salt in melting ice formed on pedestrian paths and streets can be interpreted by the
phase diagram. If enough salt is added to ice, say at -2oC, to take the system to a point just above
a', where the stable state is solution, then we expect all the ice to melt, temperature of the system
remaining constant.
A good freezing mixture is one that has a low eutectic temperature, a high endothermic heat of
solution of the salt, the ability of the components to form an intimate mixture and cheap, non
toxic components. The most commonly used salt is common salt in a freezing mixture as it is
cheap and easily available. Both the melting of ice and the endothermic dissolution of salt
contribute to lowering in temperature of a freezing mixture. In the case of NaCl-ice mixture, the
lowering in temperature is almost due to the fusion of ice as the heat of solution of NaCl is very
low. CaCl2.6H2O – ice is a good freezing mixture as the eutectic temperature of this system is 55oC and CaCl2.6H2O has a high endothermic dissolution.
55
Dry ice (solid CO2) mixed with alcohol, acetone or ether forms a good freezing mixture with
temperatures reaching below -70oC.
The Ferric chloride - Water system
The ferric chloride – water is an example of a two component system in which the two
components are completely miscible in the liquid phase and form a number of congruent melting
compounds.
Ferric chloride forms four stable hydrates, Fe2Cl6.12H2O, Fe2Cl6.7H2O, Fe2Cl6.5H2O and
Fe2Cl6.4H2O.Fe2Cl6 is used instead of FeCl3 to avoid fractional number of molecules of water of
crystallization. When several congruent melting compounds are formed in a system, a maximum
is obtained for each as shown in the phase diagram for the system in Figure 10.3.8. The
description of the phase diagram is given in Table 10.3.8.1.
56
K
solution
I
J'
G
J''
J
C
H' H H''
E
b
c
d
e
f
F'
B'
B
j
k
D''
B''
C'
Water
i
F''
Fe2Cl6.5H2O
A
D
Fe2Cl6.7H2O
Fe2Cl6.12H2O
Temperature →
D'
g h
F
Fe2Cl6.4H2O
a
E'
G'
I'
Mole %Fe2Cl6 →
Fe2Cl6
Figure 10.3.8: The phase diagram for ferric chloride – water system
The freezing point of water is represented by A and the freezing point curve of ice by AB.
Systems at C, E, G and I are the congruent melting dodecahydrate, heptahydrate, pentahydrate
and tetrahydrate respectively. At the respective congruent meeting points, the solid hydrates are
in equilibrium with aqueous solutions of the same composition as the corresponding hydrates.
The eutectic temperatures are given by B, D, F, H and J and their values are -55oC, 27.4oC,
30oC, 55oC and 66oC respectively. The temperatures at the congruent melting points C, E, G and
I are 37oC, 32.5oC, 56oC and 78.5oC respectively. There are tie lines at B, D, F, H and J
connecting solidus lines.
57
Table 10.3.8.1 Description of the phase diagram for ferric chloride-water system
A
Freezing point of water (0oC)
C
Congruent melting point of dodecahydrate (37oC)
E
Congruent melting point of heptahydrate (32.5oC)
G
Congruent melting point of pentahydrate (56oC)
I
Congruent melting point of tetrahydrate (78.5oC)
B
Eutectic point: Ice D
Eutectic point: dodecahydrate F
Eutectic point: heptahydrate pentahydrate H
Eutectic point: pentahydrate tetrahydrate J
Eutectic point: tetrahydrate anhydrous salt CC'
Solid dodecahydrate
EE'
Solid heptahydrate
GG'
Solid pentahydrate
II'
Solid tetrahydrate
dodecahydrate solution B(-55oC)
heptahydrate solution D(27.4oC)
solution F(30oC)
solution H(55oC)
solution J(66oC)
58
AB
Freezing point curve of ice
BCD
Solubility curve of dodecahydrate
DEF
Solubility curve of hepta hydrate
FGH
Solubility curve of pentahydrate
HIJ
Solubility curve of tetrahydrate
JK
Solubility curve of anhydrous salt
Areas
ABB'A
Ice BCB'B
Dodecahydrate Solution (given by BC)
DCD'D
Dodecahydrate Solution (given by DC)
DED''D
Heptahydrate Solution (given by DE)
FEF'F
Heptahydrate Solution (given by FE)
FGF''F
Pentahydrate Solution (given by FG)
HGH'H
Pentahydrate Solution (given by HG)
HIH''H
Tetrahydrate Solution (given by HI)
Solution (composition given by AB)
59
JIJ'J
Tetrahydrate Right of J''JK
Anyhydrous salt Solution (given by JI)
Solution (given by JK)
Areas below
B'BB''
Ice+dodecahydrate
D'DD''
Dodecahydrate+ heptahydrate
F'FF''
Heptahydrate+ Pentahydrate
H'HH''
Pentahydrate+ tetrahydrate
J'JJ''
Tetrahydrate+ Anyhydrous salt
Area above
Liquid (solution of FeCl3 in water)
ABCDEFGHIJK
Isothermal evaporation of solution:
If a solution represented by point a (Figure 10.3.8) on being subjected to evaporation under
constant temperature conditions moves along abcd to the point k, a sequence of changes takes
place. These changes are given in Table 10.3.8.2.
Table 10.3.8.2 Isothermal evaporation of solution
a to b
Ferric chloride solution gets slightly more concentrated.
at b
Separation of Fe2Cl6.12H2O starts.
60
b to c
More of the dodecahydrate separates. Composition of the solution remains
unchanged at b. There is decrease in the volume of the solution.
at c
Solution disappears completely and the system has only Fe2Cl6.12H2O.
c to d
Dodecahydrate solution, composition given by d. As the system moves
towards d, more and more of the solution is formed, amount of solid left
decreases.
at d
Solid decahydrate disappears and there is only solution.
d to e
Unsaturated solution.
At e
Fe2Cl6.7H2O starts separating.
e to f
More heptahydrate separates. Solution composition remains unchanged at
e. Volume of the solution decreases.
At f
Complete solidification to Fe2Cl6.7H2O occurs
f to g
Again solution appears, solution having composition g,
solution increases, amount of Fe2Cl6.7H2O decreases.
At g
Fe2Cl6.7H2O disappears. There is only solution.
g to h
Unsaturated solution
At h
Fe2Cl6.5H2O starts forming.
h to i
More Fe2Cl6.5H2O separates. Volume of solution (composition h) decreases.
volume of the
61
At i
Complete solidification to Fe2Cl6.5H2O occurs.
i to j
Conversion of pentahydrate to tetrahydrate takes place.
At j
Conversion to tetrahydrate gets completed.
j to k
Conversion of tetrahydrate to anhydrous ferric chloride occurs.
If evaporation is continued, pure anhydrous ferric chloride is obtained.
Copper sulphate – water system
While discussing solid-liquid equilibria, we assumed the pressure on the system to be high
enough that no vapour was present. At lower pressures vapor, may be present if one or more of
the constituents is volatile. An important example of a solid-vapor equilibrium is the equilibrium
between salt hydrates and water-vapor. We will take up for discussion the copper sulphte-water
system
62
liquid
liquid+CuSO4.5H2O
sat soln+CuSO4.5H2O
c
CuSO4.H2O+CuSO4.
b
CuSO4.3H2O+CuSO4.H2O
Vap +
Liquid
CuSO4.5H2O+CuSO4.3H2O
Pressure/mmHg
a
Vap+CuSO4.3H2O
Vap+CuSO4.5H2O 7.85
d
4.32
e
f
g
Vap+CuSO4.H2O
Vap+CuSO4
H2O
0.017
h
Mass% CuSO4
i
CuSO4
Figure 10.3.9: Vapour pressure – composition graph of CuSO4-H2O system (25oC)
We will examine how the vapor pressure of the CuSO4- H2O system varies with the concentration
of CuSO4 at a fixed temperature. Figure 10.3.9 gives the vapor pressure – composition diagram of
this system at 25oC. Point a gives the vapor pressure of pure water at 25oC. The vapor pressure of
water gets lowered along the curve ab as anhydrous copper sulphate is added to liquid water. At b
the solution is saturated with respect to the pentahydrate, CuSO4. 5H2O. Along bc, there are 3
phases in equilibrium at constant temperature, CuSO4. 5H2O, saturated solution of CuSO4 in
63
water and water vapour. The variance of the system along bc is zero (invariant system), F=C+1P=2+1-3=0. As more anhydrous CuSO4 is added, the pressure does not change, but some of the
solution gets converted to the pentahydrate. At C all the water present has combined with the
added anhydrous CuSO4 to form the pentahydrate. Further addition of CuSO4 results in a sudden
drop of pressure to the value at de with the formation of some trihydrate, CuSO4. 3H2O. The
system is invariant along de, the three phases in equilibrium are CuSO4.5H2O, CuSO4.3H2O and
water vapor. Along de the pressure corresponding to the dissociation pressure of this system is
maintained constant till all the pentahydrate disappeared.
CuSO4.5H2O (S) CuSO4.3H2O(S) + 2H2O (g)
The standard equilibrium constant for this reaction is given by
K 0p = (peq,H2 O /p0 ) 2
Where peq,H O is the equilibrium vapour pressure (the dissociation pressure) of water over the
2
mixture of trihydrate and pentahydrate. The dependence of the vapor pressure on temperature can
be obtained by combining this equation with the van’t Hoff equation:
dlnK 0p
dT
=
∆rH 0
RT 2
Where ∆rH0 is the standard enthalpy change for the reaction (negative of the enthalpy of
hydration). The pressure to remain constant requires the presence of 3 phases, for example,
pressure is constant along de because 3 phases, CuSO4.5H2O, CuSO4.3H2O and water vapor
coexist in equilibrium. A single hydrate does not have a definite vapor pressure. For example,
CuSO4.3H2O and water vapour can coexist in equilibrium with any vapour pressure of water in
the range from e to f. At e the system is entirely made of CuSO4.3H2O.
With the addition of more CuSO4, some of the trihydrate is converted to the monohydrate
CuSO4.3H2O(s) CuSO4.H2O(s)+2H2O(g)
The pressure drops to a valve given by fg. Along fg, the system is invariant with the three phases,
CuSO4.3H2O, CuSO4.H2O and water vapor coexisting in equilibrium. At g it is all CuSO4.H2O,
the monohydrate. Addition of CuSO4 to the system results in the monohydrate getting converted
to the anhydrous CuSO4 with the pressure dropping to hi. The pressure remains constant at this
value till the monohydrate gets completely converted to the anhydrous CuSO4. Along hi, the
invariant system consists of CuSO4.H2O, CuSO4 and water vapor.
CuSO4.H2O(s) CuSO4(s) +H2O(g)
We can write expressions for K 0p and van’t Hoff equations for these equilibria as well.
64
Efflorescence and deliquescence
Each hydrated salt has a constant vapor pressure at a particular temperature. We have seen that
CuSO4.5H2O will be formed only if the pressure of the water vapor reaches a certain value. If the
vapor pressure falls below the dissociation pressure of the pentahydrate, this salt will undergo
dehydration. From this, it is clear that a crystalline salt hydrate will effloresce on exposure to air,
if the partial pressure of the water vapor in the air is lower than the dissociation pressure of the
hydrate. At room temperature, the pressure of water vapour in the air is ordinarily more than the
dissociation pressure of CuSO4.5H2O and hence it does not effloresce. Na2SO4.10H2O and
Na2CO3.10H2O are examples of salts which have their dissociation pressures greater than the
normal pressure of aqueous vapor in a room and hence these salts effloresce.
If the pressure of aqueous vapor in the air exceeds that of the saturated aqueous solution of the
salt, the salt on being exposed to air will get covered with a layer of saturated solution, that is, it
will deliquesce. NaOH.H2O and CaCl2.6H2O are examples of salts that have their saturated
solution vapour pressures lower than the pressure of the aqueous vapour in the air around room
temperature and hence are deliquescent.
Liquid-Liquid mixtures – ideal liquid mixtures
The ideal solution – Raoult’s law
The concept of the ideal gas has been very useful is discussing thermodynamics of gases. Also
many problems of practical interest are treated adequately by means of the ideal gas
approximation. It is fortunate that there is a similar concept, the ideal solution, to discuss the
theory of solutions. An ideal gas is considered to have no cohesive forces while an ideal solution,
uniformity of the cohesive forces. This means that intermolecular forces between A and A, B and
B and A and B are all the same in a solution of A and B. The vapor pressure of a constituent
above the solution, which is a good measure of the tendency of the component to escape from the
solution, gives an insight into the cohesive forces operating within a solution. Studying the
dependence of partial vapor pressures on temperature, concentration, etc. gives information
regarding the properties of the solution.
In 1886, Francois Marie Raoult first reported extensive data on vapour pressure of solutions.
This data fitted the relation:
pi=xi pi0
to a good extent, where pi is the partial pressure of the constituent i having a mole fraction xi
and pi0 is the corresponding vapor pressure of the pure constituent. Hence the statement that the
partial pressure of a constituent i, pi is proportional to its mole fraction, xi in the solution, is
known as the Raoult’s law.
A solution is said to be an ideal solution if its constituents follow Raoult’s law over the entire
range of composition, that is, the partial pressure of each and every constituent is given by
pi = xi pi0
(Eq. 10.4.1)
65
The ideal solution has two other important properties. The enthalpy of mixing the pure
constituents to form the solution, ∆ mix H is zero. The other property is that the volume of
mixing, ∆ mix V is also zero. In all real solutions, these properties are observed as the limiting
behaviour. If a solution is dilute with respect to all its solutes and if more solvent is added, then
the enthalpy of mixing approaches zero as the solution gets more and more dilute. Under similar
conditions, the volume of mixing also approaches zero for a real solution.
Thermodynamics of an ideal solution
Let us consider an ideal solution in equilibrium with its vapour at a constant temperature, T.
For each constituent, we have at equilibrium µi (sol) = µi (vap), where µi (sol) is the chemical
potential of constituent i in the solution and µi (vap) is the chemical potential of constituent i
in the vapor phase. The various constituents of an ideal solution, we know, follow the relation:
µi (sol) = µ 0i (l)+RTlnxi
(Eq.10.4.2)
Where µi (sol) is the chemical potential of ith constituent of the solution, µoi(l) is that of the pure
liquid constituent and xi is the mole fraction of the constituent in the solution. Let us calculate
now the changes in some thermodynamic functions when an ideal solution is formed by the
mixing of pure constituents.
(a)
The change in Gibbs Free energy on mixing, ∆Gmix
∆Gmix = Gfinal - Ginitial
= Σ niµi (sol) - Σ niµi(l)
i
i
0
)
= Σ ni(µi(sol) - µ i(l)
i
=
Σ niRTlnxi
=
ntotalRT Σ xilnxi
i
(Eq.10.4.3)
i
Where ntotal is the total amount of all the constituents in the solution.
(b)
The change in entropy on mixing, ∆Smix
⎛ ∂ (∆G mix ) ⎞
= −∆Smix
⎜
⎟
∂T
⎝
⎠ p,ni s
⎛ ∂ (n total RT Σxi ln x i
∆Smix = - ⎜
⎜
∂T
⎝
⎞
⎟
⎟
⎠ p,ni s
66
=
(c)
- ntotal R Σ xilnxi
(Eq.10.4.4)
i
The change in enthalpy on mixing, ∆Hmix
∆Gmix = ∆Hmix – T∆Smix
∆Hmix = ntotal RT Σ xilnxi-Tntotal R Σ xilnxi
i
i
∆Hmix = 0
Thus no heat is evolved or absorbed in the formation of an ideal solution.
(d)
The change in volume on mixing, ∆Vmix
∆Vmix =
⎛ ∂ (∆G mix ) ⎞
⎜
⎟
∂p
⎝
⎠T,n i s
∆Vmix =
⎛ ∂ ( Σ ntotal RTx i lnx i ) ⎞
⎜ i
⎟
⎜
⎟
∂p
⎝
⎠T,n i s
∆Vmix = 0
(Eq 10.4.6)
The change in volume on mixing is zero. The volume of an ideal solution is equal to the sum of
the individual volumes of its constituents.
Molecular interpretation of ∆mixH=0 and ∆mixV = 0
The enthalpy of mixing, ∆mixH of an ideal solution working out to be zero indicates that the
cohesive forces between the molecules of its constituents must be quite similar in nature. Thus
on mixing the environmental forces of each constituent remain almost unchanged. Two liquids A
and B mix to form an ideal solution if the intermolecular forces of attraction between A
molecules are similar to those between B molecules. Then the forces of attraction between A and
B molecules would also be similar. In an ideal liquid mixture of A and B, as the forces of
attraction are similar, a molecule
attracts B in the same way it attracts A without any
differentiation. Besides in an ideal solution, the volumes of different molecules must also be
similar because only then there would be no volume change on mixing the constituents.
The partial molar volume of a constituent in an ideal solution is equal to the molar volume of the
constituent in its pure form.
Raoult’s law
In a solution, partial pressure of a constituent i, pi is proportional to the mole fraction xi of the
constituent i.
pi = xip 0i
(Eq. 10.4.1)
67
Where p 0i is the corresponding vapour pressure of the pure constituent i.
If the constituents of a solution obey Raoult’s law over the entire range of composition, then the
solution is said to be an ideal solution. We seldom find solutions that follow Raoult’s law over a
wide range of composition. Ideality in solution implies, as we have discussed in the last section,
complete similarity in interaction between the constituents, which is rare to achieve in actual
practice. Solutions of isotopes, however, provide good examples of ideal solutions. There are a
few binary liquid solutions which are nearly ideal. A mixture of ethylene dibromide and
propylene dibromide was studied at 85oC. When the vapour pressure of the solution was plotted,
the experimental curve almost coincided with the theoretical curve obtained using Raoult’s law.
Other systems forming nearly ideal solutions include ethylbromide and ethyliodide, n-hexane
and n-heptane which were studied at 30oC.
Like the ideal gas law, Raoult’s law is also a limiting law. Real solutions follow Raoult’s law
more closely as the solution gets more and more dilute.
Henry’s law - Solubility of gases in liquids
Gases are soluble in liquids and the solubility depends on the nature of the gas, nature of the
liquid, temperature and pressure. Gases which interact chemically with the liquid have a high
solubility. For example, gases like NH3, CO2, HCl dissolve easily in water. Gases like hydrogen
and helium dissolve very little. For a given temperature and pressure, the more easily liquefiable
a gas is, the greater is its solubility. A gas in equilibrium with the dissolved gas in a liquid has 2
phases and 2 components. The number of degrees of freedom of this system is 2 (F=C+P2=2+2-2=2), that is, it is bivariant and the variables are temperature and pressure.
Solubility of gases can be expressed in terms of absorption coefficient or the coefficient of
solubility. Bunsen absorption coefficient (this was suggested by R. Bunsen) at a given
temperature is defined as the volume of gas, reduced to NTP, that has been dissolved by unit
volume of solvent under a partial pressure of 1 atm of the gas.
If vo is the volume of gas dissolved, reduced to NTP, by the volume V of the solvent under the
partial pressure p of the gas, then the absorption coefficient is given by:
α=
vo
vp
(Eq.10 .4.7)
The amount of gas dissolved by unit volume of solvent under a partial pressure of 1 atm is given
by
C=
vo/v m
α
=
vp
vm
where vm is the molar volume of the gas at NTP.
The coefficient of solubility (this was suggested by W. Ostwald) is the volume of gas measured
under given conditions of T and p which has been dissolved by unit volume of the solvent. If v is
68
the volume of gas dissolved by volume V of the solvent, then the coefficient of solubility, β is
given by:
β=
v
V
(Eq.10.4.8)
Relation between α and β
If we assume that the gas behaves ideally, we can write
(1atm)vo
pv
=
nRT nR(273K)
v=
β=
vo T
(273K)(p/atm)
v vo T/(273K)(p/atm)
=
V
V
β=
vo
T
vp (273K)
β=
αT
273K
(Eq.10.4.9)
Effect of temperature on the solubility of gases
A gas dissolves in a liquid generally with heat evolution. The solubility of a gas in a liquid
decreases with increase in temperature according to the Le-Chatelier principle. Experimentally,
this has been found to be true for most gases. The quantitative relation between solubility and
temperature at constant pressure, assuming the gas to be ideal, is given by the equation:
dlnC ∆H
=
dT RT 2
(Eq.10.4.10)
Where C is the molar concentration of the gas in the liquid and ∆H is the differential heat of
solution of 1 mol of gas in a saturated solution at temperature T.
Effect of pressure on the solubility of gases
The solubility of a gas in a liquid increases with increase in external pressure. Henry’s law gives
the quantitative relation between solubility and pressure. The law states that at a given
temperature, the mass of dissolved gas in a given volume of solvent is directly proportional to the
pressure of the gas with which it is in equilibrium. If a gas from a mixture of gases, is dissolving,
69
then the mass of gas dissolving in a given volume of solvent is directly proportional to the partial
pressure of the gas, other conditions remaining the same.
If m is the mass of the dissolved gas per unit volume of the solvent at equilibrium pressure p,
then, we write, according to Henry’s law
mαp
m = kp
(Eq: 10.4.11)
where k is the constant of proportionality. It follows from Henry’s law that the volume of an ideal
gas, v measured at the experimental pressure p, which is dissolved in a given volume of the
solvent at temperature T, has a constant value.
According to the ideal gas law, we have
pv=nRT=
m
RT where M is the molar mass of the gas.
M
⎛ m ⎞ RT
v= ⎜ ⎟
⎝p⎠ M
v=k
RT
M
(Eq. 10.4.12)
Thus, v is constant for a given gas at a given temperature. The volume of a gas dissolved in a
given volume of solvent at a given temperature is constant.
Henry’s law is applicable to only dilute solution of gases in liquids at high temperature and low
pressures, conditions when real gases are expected to show ideal behaviour. Deviations from
Henry’s law are expected to be large at low temperatures and high pressures.
Henry’s law and Raoult’s law
According to Henry’s law, mass m2 of a gas dissolved in a mass of solvent m1 at the gas pressure
p2 is given by
m2 = k p2
Dividing both sides by m1, we have
m2 k
=
×p 2 =k'p2
m1 m1
Dividing the two masses by their respective molar masses, we get
(Eq. 10.4.13)
70
m 2 /M 2 ⎛ M1
= ⎜ k'
m1/M1 ⎝ M 2
⎞
⎟ ×p 2 =k"p2
⎠
n2
=k''p 2
n1
(Eq.10.4.14)
The mole fraction of the gas in the solution is x2, which is given by x2 =
n2
n1 +n 2
For a dilute solution of the gas in the liquid,
n1 〉〉n 2
x2 ≅
n2
n1
Writing the equation 10.4.14 in terms of mole fraction, we have
x2 = k''p2
(Eq. 10.4.15)
In the cases of gases which dissolve very little, the solution is saturated with respect to the gas
and yet is a dilute solution. Hence from equation (10.4.15), we can say that the solubility of a gas
in terms of mole fraction is directly proportional to the pressure of the gas, x2 being regarded as
the solubility under a pressure p2. p2 may be considered as the vapor pressure of a volatile solute
with a mole fraction x2 in the solution.
p2 =
1
x2
k''
= Khx2
Where Kh =
(Eq. 10.4.16)
1
and is known as Henry’s law constant.
k''
Kh has units of pressure. In the special case of a system obeying Henry’s law over the entire range
of concentrations, infinitely dilute solution (pure solvent, x2 = 0) to pure liquid solute (x2=1), the
constant k'' is given as
k'' =
x2
p2
=
1
p02
where po2 is the vapor pressure of the pure liquid solute (x2=1). Thus, the equation 10.4.15 can be
written as
71
p
x2 = 2 or p2 = x 2p02
p0
2
10.4.17)
(Eq
This equation (10.4.17) is the Raoult’s law statement as applied to a volatile solute. We may,
hence, conclude that Raoult’s law is a special case of Henry’s law. All systems that follow
Raoult’s law must obey Henry’s law but the reverse is not true unless Henry’s law is applicable
over the entire range of concentration.
Raoult’s law for solvent and Henry’s law for solute
We can show that if Henry’s law is applicable to the volatile solute, then Raoult’s law is
applicable to the volatile solvent. We take the Duhem-Margules equation:
d log p1 d log p 2
=
d log x1 d log x 2
(Eq. 10.4.18)
Applying Henry’s law to the volatile solute, we write
p2 =Khx2
(Eq.10.4.16)
log p2 = logKh + log x2
d log p2 = d log x2
d log p 2
=1
d log x 2
(Eq. 10.4.19)
Substituting from Eq.10.4.19 into Eq.10.4.18, we get
d log p
1 =1
d log x
1
(Eq. 10.4.20)
On integrating the above equation, we obtain
log p1 = log x1 + constant
or p1 = Kx1
(Eq. 10.4.21)
when x1=1, p1= p10 , where p10 is the vapor pressure of the pure solvent. Thus K= p10
Equation 10.4.21 becomes
p1=x1 p10
(Eq.10.4.22)
72
By applying Henry’s law to the solute, we could derive equation 10.4.20 which in turn has given
us equation 10.4.22. Hence, one can conclude that Raoult’s law is applicable to the solvent as
long as Henry’s law is obeyed by the solute. If the solute obeys Henry’s law only in dilute
solutions (i.e. low concentration range), then Raoult’s law will be applicable to the solvent also in
the same concentration range. When Henry’s law is not obeyed by the solute, Raoult’s law is also
not obeyed by the solvent.
Kh(B)
pB0
xB=0
xA=1
xB →
←xA
H2O
xB=1
xA=0
Pressure
p0A
xB=0
xA=1
xB →
←xA
Mass%CuSO4
(a)
xB=1
xA=0
CuSO4
(b)
Figure 10.4.1: Graphs showing applicability of Henry’s law to the solute and Raoult’s law to the
solvent (a) for liquid B (b) for liquid A.
If we take an example, say a solution of liquid A in liquid B and plot partial pressure versus mole
fraction curves, we can understand the above discussion better. The partial pressure of B is
plotted against its mole fraction in solution to give the curve shown in Figure 10.4.1. The
Raoult’s law line is obtained by joining the points p=0 for xB=0 and p= poB for xB=1. This line
meets the partial vapour pressure curve tangentially in the region where xB→1, implying thereby
that Raoult’s law is applicable only in this region, xB → 1, amount of B much more than A, that B
is the solvent. A line drawn tangent to the curve at xB=0 (low values of xB) gives the Henry’s law
line. Extrapolation of this line to meet the xB=1 axis gives the hypothetical vapour pressure which
would be observed if Henry’s law were applicable over the entire range of composition, upto
xB=1. It follows from equation 10.4.16 at xB=1.
pB =Kh
73
Thus Henry’s law constant, Kh has a value pB, the vapour pressure pure B would have if it obeyed
Henry’s law instead of Raoult’s law. According to Raoult’s law, the constant Kh would have been
equal to p0B . Hence we can say that the Henry’s law constant for a non ideal system is different
from Raoult’s law constant.
Figure 10.4.1. (b) gives a plot of partial vapour pressure of liquid A versus the mole fraction of A
in the solution. The dotted lines give the Raoult’s law and Henry’s law graphs. Raoult’s law line
meets the partial vapor pressure curve around xA → 1 indicating that Raoult’s law is applicable to
the solvent. As the Henry’s law line meets the curve in the region xA→0, we can say that Henry’s
law is applicable to the solute. We can also see from the figure that Raoult’s law is applicable to
the solvent over the same range where Henry’s law is applicable to the solute.
Ideal and ideally dilute solutions
An ideal solution is one whose constituents follow Raoult’s law in the entire range of
composition. Each constituent also follows Henry’s law and the Henry’s law constant is equal to
the vapor pressure of the pure constituent. An ideally dilute solution is a real solution in the limit
x1 → 1 and x2 → 0. Such a solution has few solute molecules and these are surrounded by solvent
molecules. As solute molecules are quite separated, they do not interact with one another. The
laws governing such ideally dilute solutions are Raoult’s law for the solvent and Henry’s law for
the solute. Thus, the vapor pressures of solvent and solute of ideally dilute solutions are given by:
Raoult’s law for solvent: p1 = x1 p10
(Eq. 10.4.23)
Henry’s law for solute: p2 = x2Kh
(Eq. 10.4.24)
These equations can be used to derive chemical potential expressions for the solvent and solute of
an ideally dilute solution. When the solution and its vapor are in equilibrium, then according to
thermodynamics we have,
µ1(l) = µ1(v) and µ2(l) = µ2(v)
(Eq 10.4.25)
We have for the vapor phase, µ = µo + RT lnp.
Writing for the vapor phase using this equation in equation 10.4.25, we get
0
+RTlnp1
µ1(l) = µ1(v)
µ2(l) = µ 02(v) +RTlnp2
Substituting for p1 and p2 from equations 10.4.23 and 10.4.24 into the above equations, we have
0
µ1(l) = (µ1(v)
+ RTlnp10 ) + RTlnx1
µ2(l) = (µ 02(v) + RTlnK h ) + RTlnx 2
74
0
µ1(l) = µ1(1)
+ RTlnx1
(Eq. 10.4.26)
µ2(l) = µ 2(hs) +RTlnx 2
(Eq. 10.4.27)
Where “hs” stands for the hypothetical state. It is seen from equation 10.4.26 that the standard
state of solvent in ideally dilute solution is the pure liquid solvent. The standard state for solute is
the hypothetical state of solute where its vapor pressure is Kh which is obtained by extrapolating
the experimental vapor pressure of solute in the region x2→o to a value of x2 → 1. This
hypothetical state has the properties which the pure solute would have if it retained its limiting
low concentration properties in solution in its pure state. The actual properties of the pure solute
will be different from that of the hypothetical state as the solute molecule is surrounded by other
solute molecule in the pure state whereas in the hypothetical state solute molecule is surrounded
by solvent molecules only.
Ideal binary liquid system – calculation of total pressure and composition of the vapor
phase in equilibrium with the liquid.
(i)
Total vapor pressure of an ideal binary liquid system: As the system is ideal, both the
constituents follow Raoult’s law over the entire range of composition. The partial
pressure exerted by the vapor of these constituents over the solution is given by
pA = xA p0A
(Eq.
10.4.28)
pB = xB p0B
(Eq.
10.4.29)
where xA and xB are the mole fractions of A and B, the two constituents in the solution, p0A and
p0B are their vapor pressures in the pure state. The total pressure over the solution is the sum of
the partial pressures.
p = p A + pB
= xA p0A + xB p0B
In a binary solution xA+xB=1 and hence, we write xB=1-xA
p=xA p0A + (1-xA) p0B
= xA p0A + p0B
—
xA p0B
= p0B +xA ( p0A — p0B )
(Eq. 10.4.30)
The total vapor pressure over the solution is thus a linear function of xA with intercept equal to
p0B and slope equal to p0A — p0B . The slope can be positive or negative depending upon the relative
magnitudes of p0A and p0B .
75
Case 1: If p0A > p0B , that is liquid A is more volatile than liquid B, p0A — p0B is positive, then the
total vapor pressure, p increases with the increase in xA, the mole fraction of the more volatile
constituent.
Case 2: If p0A < p0B , that is liquid B is more volatile than liquid A, p0A - p0B is negative, then the
total vapor pressure, p decreases with increase in xA, the mole fraction of the less volatile
constituent.
Plots of pA as a function of xA, pB as a function of xB and p as a function of xA, for the case
p0A > p0B are shown in the figure 10.4.2.
p0A
+ pB
p=p A
Pressure
xA=0
xB=1
pA
p0B
pB
xA→
←xB
xA=1
XB=0
Figure 10.4.2: Plots of pA, pB and p versus mole fraction in the liquid state, temperature remaining
constant.
76
(ii)
Composition of the vapor phase in equilibrium with the liquid phase: Composition of the
vapor phase is obtained using the Dalton’s law of partial pressure. If yA and yB are the mole
fractions of the liquids A and B in the vapor phase, then
pA
p
and yB = B
p
p
yA =
Substituting for pA, pB and p from equations 10.4.28, 10.4.29 and 10.4.30 into the equations for
yA and yB, we write
0
x A pA
yA=
0
0
0
p B + x A (p A - p B )
(Eq. 10.4.31)
0
x BpB
yB =
0
0
0
p B + x A (p A - p B )
writing xA=1-xB and simplifying, we get
yB =
x B p0B
p0A + x B (p0B - p0A )
(Eq. 10.4.32)
We will now get an expression for p, the total pressure in terms of yA and yB, the mole fractions
of A and B, respectively in the vapor phase.
From eq. 10.4.31, we write
xA=yA (p 0B +x A p0A -x A p0B )/p0A
On rearranging, this yields
xA =
y A p0B
p +y A (p 0B -p 0A )
0
A
(Eq. 10.4.33)
Substituting for xA from equation 10.4.33 into eq. 10.4.30, we get
p= p0B +(p0A -p 0B )
p=
y A p0B
p +y A (p 0B -p 0A )
0
A
p0B p 0A
p0A +(p 0B -p 0A )y A
Figure 10.4.3 shows a plot of p vs yA for a solution having poA poB .
(Eq.10.4.34)
77
poA
p
p oB
yA→
yB=1
←yB
yA=0
yA=1
yB=0
Figure 10.4.3: A plot of total vapour pressure, p against yA, the mole fraction of A in the vapor
phase. Temperature is kept constant.
The reciprocal of equation 10.4.34 gives a linear equation relating
1 1
1 1
= +(
- )y A
p p oB p oA p oB
1
and y A
p
(Eq. 10.4.35)
78
The reciprocal of total pressure varies linearly with yA, the mole fraction of constituent A in the
vapour phase, as shown in figure 10.4.4.
Equation 10.4.35 can be rearranged to the more convenient, symmetrical form.
1 yA yB
=
+
p p0A p 0B
(Eq.10.4.36)
1/p
1/p 0B
1/p 0A
yB=1
←yB
yB=0
yA=0
yA→
yA=1
Figure 10.4.4: A plot of 1/p verus yA, mole fraction of the constituent A in the vapour phase.
Temperature is kept constant.
79
Application of phase rule to a binary liquid system
Let us first understand the diagram in the figure 10.4.5 which has the plots of figures
10.4.2 and 10.4.3 drawn together. The total vapor pressure is plotted against the mole fraction of
one constituent in the liquid phase xA in one plot and that in the vapor phase yA in the other plot.
p 0A
l v
pressure
p0B
Figure 10.4.5: Vapour pressure – composition diagram, temperature is kept constant
The total pressure is an equilibrium vapor pressure and hence the system would exist only as
liquid if the pressure on it is more than the equilibrium pressure. It would exist only as gas, if the
pressure is less than the equilibrium pressure. This means any point above the pressure - xA plot
represents a liquid and any point below the pressure - yA plot represents vapor. The upper curve
is known as the liquidus curve as above this curve only the liquid phase exists. The lower curve is
called the vaporous curve as below this curve only the vapor phase exists. Along the two curves,
two phases, liquid and vapor are in equilibrium. At any point between the two curves, both liquid
and vapor coexist in equilibrium. Hence, the enclosed region is the liquid-vapor region.
To describe a two component system, the phase rule shows, since C=2 that variance F=4-P. The
variance of the system will depend on the number of phases present in the system. Above the
liquidus curve and below the vaporous curve, the system exists as only liquid or only gas
respectively and as P=1, the system is trivariant. Three variables must be specified to describe the
system. These are temperature, pressure and composition. As temperature is held constant,
pressure and composition values only need to be stated.
If two phases, liquid and vapor, are present in equilibrium, the variance of the system is 2 (F=42=2). Since the temperature is fixed, one other variable, any one of p, xA, yA suffices to describe
the system. As xA+xB=1 and yA+yB=1, the variables could as well be xB and yB instead of xA and
yA. If the pressure is chosen to describe the two phase system, the intersections of the horizontal
80
line (the tie line) at that pressure (Figure 10.4.6) with the liquid and vapor curves yield the values
of xA and yA directly. If the variable chosen is xA, then the intersection of the vertical line at xA
with the liquid curve yield the values of the pressure p. The value of yA can then be obtained from
the pressure value (Figure 10.4.6). For a given composition, the points on the liquidus and
vaporous curves represent, respectively, the maximum and the minimum pressure within which
the two phases can coexist in equilibrium (Figure 10.4.5).
p 0A
liquid
p
l
a
xA
XA
v
pressure
p 0B
vapour
Pure B
Composition
yA
Pure A
81
Figure 10.4.6: Vapor pressure – composition diagram – the lever rule. Temperature is kept
constant
A point within the liquid and vapor curves, we know, represents two phases, liquid and vapor in
equilibrium with each other. For a state point given by a (Figure 10.4.6), the mole fraction of the
constituent A in the liquid phase is given by the point l, that is, xA and that in the vapor phase by
the point v, that is yA. Any state point on the tie line lv represents the same compositions of liquid
and vapor phases, namely, xA and yA, respectively. If the state point a lies near the liquid line, the
system consists of relatively more liquid than vapor. If on the other hand, the point a is near the
vapor line, then the amount of liquid is relatively small compared to that of the vapor.
The relative amounts of liquid and vapor present are calculated by the lever rule. The mole
fraction XA corresponding to the point a represents the mole fraction of the component A in the
entire system consisting of liquid and vapor phases.
xA
=
n A(l) +n A(v)
Eq.
n A(l) +n A(v) +n B(l) +n B(v)
10.4.37
al
=
xA-xA=xA -
av
=
yA(v)-xA =
n A(l)
(Eq.10.4.38)
n A(l) +n B(l)
n A(v)
n A(v) +n B(v)
- xA
(Eq.10.4.39)
Multiplying equation 10.4.38 by nA(l)+nB(l), we get
(nA(l)+nB(l)) al = xA (nA(l)+nB(l)) – nA(l)
(Eq.10.4.40)
Multiplying equation 10.4.39 by nA(v)+nB(v), we get
(nA(v)+nB(v)) aν = nA(v) – xA (nA(v)+nB(v))
Subtracting equation 10.4.41 from equation 10.4.40, we write
(nA(l)+nB(l)) al – (nA(v)+ nB(v))av =
xA
(nA(l)+nB(l) + nA(v)+nB(v))-(nA(l)+nA(v))
On substituting from equation 10.4.37, this equation becomes:
(nA(l)+nB(l))al – (nA(v)+nB(v)) av = 0
(Eq.10.4.41)
82
al n A(ν) +n B(ν)
=
av n A(l) +n B(l)
al Amount in the vapor phase
=
av Amount in the liquid phase
Or nliq × al = nvap × av
Eq.10.4.42
This equation is the lever rule; the number of moles of liquid times the length from “a” to
the liquid curve is equal to the number of moles of vapor times the length from a to the vapor
curve. Thus if “a” lies very close to v, then av is small and nvap>>nliq and the system consists
mainly of vapor. Similarly, when “a” lies close to l, the system consists mainly of liquid. If “a”
coincides with l, then the vapor phase has just started forming with mole fraction yA and if it
coincides with v, then the last drop of liquid phase with mole fraction xA is just going to be
converted into the vapor phase.
Isothermal fractional distillation of ideal binary solution
When an ideal binary solution is in equilibrium with its vapor phase, the latter is richer in the
more volatile component as compared to the liquid phase. It is on this fact that isothermal
fractional distillation of an ideal mixture of liquids is based. The mole fraction of component A in
the vapor phase, yA is given by
yA =
x A p0A
p +x A (p 0A -p0B )
yA =
x A p0A
p0B +x A p 0A -x A p0B
yA =
x A p0A
x A p0A +p 0B (1-x A )
yA =
x A p 0A
(as xB = 1-xA)
x A p0A +x B p 0B
0
B
yA
1
=
x A x A +x B (poB /poA )
(Eq.10.4.31)
(Eq.10.4.43)
If constituent A is more volatile than B, poA poB , then p0B / p0A <1 and so also the denominator of
equation 10.4.43 because xA+xB can only be equal to one. The result of this is that the ratio yA/xA
is more than 1 or in other words yA > xA, that is, the mole fraction of A, the more volatile
component, is more in the vapor phase as compared to its mole fraction in the liquid phase. Let
us start with the liquid phase represented by a in the diagram (figure 10.4.7) and subject it to a
reduction in pressure keeping temperature constant. The state of the system moves along the
83
vertical line a a' a''. The system exists in one phase (liquid) as the pressure reduced till the
pressure corresponding to l is reached. At this point, vapor of composition given by v starts
forming. The vapor phase obtained is richer in A, the more volatile component. With the removal
of A in the vapor phase, the solution gets richer in B resulting in the composition of the liquid
phase moving along ll'. The pressure of the system must be lowered to continue evaporation and
the overall state of the system still moves along the same vertical line, say, by the point a'. The
composition of the liquid phase is given by l' and the vapor phase by v' at the point a'. The
relative amounts of liquid and vapor is given by the lever rule as:
a'ν'
Amount in the liquid phase
=
a'l'
Amount in the vapor phase
Figure 10.4.7: Effect of reducing pressure at constant temperature on the ideal binary mixture
Further reduction in pressure brings the system to the state point v'' where only a trace of liquid of
composition l'' is present and the vapor has composition given by X. As the pressure is further
decreased, the system moves on the vertical line towards a''. As the state point of the system falls
below v'', it has only one phase, the vapor phase and the vapor expands as the pressure is
reduced. With this discussion we have noted the effects produced when the pressure of the
system is reduced isothermally.
Isothermal fractional distillation of a binary liquid mixture involves collecting the vapor over the
solution and condensing it to get a new liquid. This liquid is richer in the more volatile
component. Again the vapor formed over this liquid is collected and condensed to get yet another
new liquid. This process of collecting the vapor and condensing it is repeated several times till the
84
vapor has only the more volatile constituent and the liquid, the less volatile constituent. Thus with
this process, a separation of constituents is achieved. This method is especially useful in the case
of liquid mixtures which decompose on subjecting to normal distillation. As the method is
cumbersome, it is used only when other methods of separation are not available.
Konowaloff’s rule
If component A is more volatile than component B, that is, poA poB , the total vapor pressure of the
system increases with increase in the mole fraction of A, xA (Eq. 10.4.30). It is also seen that the
vapor phase is richer in the component A than the liquid phase (Eq. 10.4.43).
These two conclusions are combined in a statement known as the Konowaloff’s rule which is as
follows: The vapor phase is richer in the component whose addition to the liquid mixture results
in an increase of total vapor pressure.
Temperature composition diagram
In a binary liquid system, if two phases are in equilibrium, then the variance of the system is 2.
Hence the system gets defined if any two out of the three variables, temperature, pressure and
composition, are fixed. In the previous sections we have seen how pressure varied with
composition, temperature remaining constant. If on the other hand, the pressure is kept constant,
then the temperature of the system is a function of composition.
In the diagrams given in the earlier sections, the temperature was constant and the equilibrium
pressure of the system was a function of either xA or yA, according to equations 10.4.30 or
10.4.34. In these equations, the values of pOA and pOB depend only on temperature. Now we want
to keep the equilibrium pressure p constant. Hence for a system, this given value of p can be
obtained by changing the temperature of the system. We also know that the temperature at which
the vapor pressure of a liquid (or the total vapor pressure of a solution) becomes equal to the
external pressure is its boiling point. If the external pressure is 1 atm, then the boiling point is the
normal one. A liquid which has a higher vapor pressure has a lower boiling point and vice versa.
The equations relating temperature, T and mole fractions xA and yA are not such simple ones as
between pressure and composition but they can be derived using the Clausius-Clapeyron
equation. Applying this equation separately to both the constituents, we have
⎪⎧ ∆H A,m,Vap ⎛ 1 1 ⎞ ⎪⎫
⎜ 0 - ⎟⎬
R
⎝ TA T ⎠ ⎪⎭
⎩⎪
(Eq. 10.4.44)
⎧⎪ ∆H B,m,Vap
R
⎪⎩
(Eq. 10.4.45)
p0A /atm =
exp ⎨
p0B /atm =
exp ⎨
⎛ 1 1 ⎞ ⎫⎪
⎜ 0 - ⎟⎬
⎝ TB T ⎠ ⎪⎭
Where TA0 and TB0 are the normal boiling points of constituents A and B, respectively. T is the
normal boiling point of a solution of known composition and p0A /atm and p0B /atm be the vapor
pressures of pure constituents, A and B at the temperature, T.
is
The total vapor pressure of the solution having a known composition given by xA and xB,
85
p=xA p0A +xB p0B
By definition, the total pressure p is equal to 1 atm (the external pressure).
⎪⎧ ∆H A,m,Vap ⎛ 1 1 ⎞ ⎪⎫
⎜ 0 - ⎟ ⎬ + xB exp
R
⎪⎩
⎝ TA T ⎠ ⎪⎭
1 = xA p0A +xB p0B = xA exp ⎨
⎧⎪ ∆SA,m,vap
⎩⎪ R
1=xA exp ⎨
⎛ TA0
⎜1⎝ T
⎞ ⎫⎪
⎟ ⎬ + xB exp
⎠ ⎭⎪
⎧⎪ ∆SB,m,vap
⎨
⎪⎩ R
⎛ TB0
⎜ 1⎝ T
⎞ ⎫⎪
⎟⎬
⎠ ⎭⎪
⎪⎧ ∆H B,m,Vap ⎛ 1 1 ⎞ ⎪⎫
⎨
⎜ 0 - ⎟⎬
R
⎪⎩
⎝ TB T ⎠ ⎭⎪
(Eq. 10.4.46)
Where ∆SA,m,vap and ∆SB,m,vap are the molar entropy changes for the liquid to vapor conversion for
the constituents A and B, respectively and have a value 88JK-1 if the liquids are assumed to obey
Trouton’s rule. Since TA0 , TB0 and xA are known, this equation can be solved to get T, the boiling
point of the solution.
Composition of the vapor phase at temperature T can be obtained from the relations.
yA =
⎧⎪ ∆Sm,vap
x A p 0A
=x A exp ⎨
0
0
x A p A +x B p B
⎩⎪ R
and y B =
⎛ TA0
⎜1⎝ T
⎞ ⎫⎪
⎟⎬
⎠ ⎭⎪
⎧⎪ ∆Sm,vap ⎛ TB0 ⎞ ⎪⎫
x B p0B
=x
exp
⎨
⎜1- ⎟ ⎬
B
x A p 0A +x B p0B
⎪⎩ R ⎝ T ⎠ ⎪⎭
(Eq.10.4.47)
The plot of boiling points versus composition for the ideal solution at constant pressure
corresponding to the figure 10.4.5 is shown in figure 10.4.8.
X
86
Figure 10.4.8: Boiling point versus composition curve, constant pressure conditions,
corresponding to the figure 10.4.5.
This figure resembles figure 10.4.5 except that neither curve is a straight line and the liquid-vapor
region is tilted in the opposite direction. It is clear from the nature of the equations 10.4.46 and
10.4.47 that the graphs cannot be straight lines. As constituent A has the higher vapor pressure in
fig 10.4.5, therefore it has the lower boiling point which explains the liquid-vapor region tilt in
the opposite direction in the figure 10.4.8. As the liquid phase under a constant pressure is stable
at lower temperatures, the region beneath the lower curve is liquid and the lower curve is the
liquidus curve. Similarly, it can be concluded that the region above the upper curve is vapor and
the upper curve is the vaporous curve.
On heating, under constant pressure, a liquid mixture represented by the point a, the temperature
of the liquid increases until point l is reached. At this temperature T, the first trace of vapor
appears, composition of which corresponds to the point v. The vapor, thus is richer in component
A as compared to the liquid, A being the lower boiling component. This is the basis for the
separation of volatile mixtures by distillation. As heating is continued, the system moves towards
point a', the liquid composition changes continuously along ll' and the vapor composition varies
continuously along vv'. At the point a', the composition of the liquid and vapor phases are given
by the points l' and v' respectively and the relative amounts of the two phases by the lever rule as
Amount in liquid phase a'v'
=
Amount in vapor phase a'l'
Further increase in temperature takes the system to the point v'' where the last trace of liquid of
composition l'' is left. The vapor phase in equilibrium has the composition X. With a slight
increase in temperature, the liquid phase disappears and the system, which is only vapor expands
on heating as it reaches the state point a''.
87
Fractional distillation of an ideal binary solution
The process of separation of constituents of a binary liquid system by isobaric distillation is more
convenient to carry out than isothermal distillation and hence the isobaric distillation method is
often employed. The principle of separation by this method can be understood by following the
temperature-composition diagram shown in figure 10.4.9. When the solution to be distilled,
represented by the point a, is heated to boiling, then the vapor pressure of the solution becomes
equal to the external pressure, which is usually 1 atm. The composition of the vapor at the boiling
point (temperature corresponding to l') is given by v', which is richer in the constituent A, the
more volatile constituent. The residual liquid becomes richer in B and hence boils at a slightly
higher temperature. The vapor formed is removed and condensed separately to yield a distillate b.
If this distillate is again heated to boiling, the vapor emerging has composition v'' and is still
richer in the more volatile constituent A. By repeating several times this process of collecting the
vapor, condensing it to get a new distillate and heating the new distillate to its boiling point,
finally vapor containing only A is obtained. The residual liquid obtained at any stage is mixed
with that of the previous stage and treated in the same way when the residual liquid gets richer in
B and after several repetitions, a residual liquid containing only B is obtained. Thus, we see that
by the process of distillation, it is possible to separate the two constituents of a binary liquid
mixture; vapor containing the more volatile constituent and the liquid, the less volatile
constituent. Each horizontal line in the zig-zag steps (fig.10.4.9) is known as the theoretical plate.
For example, there are two theoretical plates in going from liquid of composition a to c.
88
vapour
TBO
v'
Temperature
l'
v''
l''
a
b
c
TAO
liquid
Pure B
Composition
Figure 10.4.9: Distillation of an ideal binary solution at constant pressure
Pure A
89
This process of separating the constituents by distillation as carried out in batches is extremely
laborious and time consuming. These difficulties are overcome by using a fractionating column
which carries out the distillation in a continuous manner. The type of column illustrated in the
figure 10.4.10 is a bubble-cap column. The column is heated at the bottom and there is a
temperature gradient along the length of the column. The top of the column is cooler than the
bottom. The various plates of the column are thus at different temperatures. The principle of a
bubble-cap column can be explained using the temperature-composition diagram shown in figure
10.4.11.
Figure 10.4.10: Bubble-cap distilling column
90
Let the temperature of the boiling liquid at the bottom be To. The composition of the vapor
emerging is vo. When this vapor passes through the first plate, it is cooled to temperature T1. Its
state hence moves to the point a. Some of the vapor condenses to form liquid of composition l1
and the remaining vapor has composition v1. The vapor has more of the more volatile component
A and the liquid has more of the less volatile component B. Next, the vapor of composition v1
passes through the second plate which is at temperature T2 (T2<T1). The vapor then cools to T2
and the state of the system moves from v1 to b. Again a part of the vapor condenses to give liquid
of composition l2 and the remaining vapor has composition v2. The vapor has become richer in
the more volatile constituent. This happens at every plate of the column. As the vapor moves up
the column, it is being cooled and this cooling results in the condensation of the less volatile
component preferentially. Thus the vapor gets more and more rich in the more volatile
component as it moves upward from one plate to the other in the column.
91
TB0
vapour
T0
ν0
l0
ν1
a'
l1
T1
a
l2
b'
Temperature
T2
b
T3
ν2
l3
liquid
TA0
Pure B
Composition
Pure A
Figure 10.4.11: Scheme of redistribution of constituents in the distilling column at constant
pressure
Similarly, as the liquid flows down the column, its temperature increases and a redistribution of
constituents takes place. Liquid of composition l3, let us say, flows from plate 3 to plate 2. The
temperature of the liquid has gone up from T3 to T2 and the state of the system has changed from
l3 to b'. A part of the liquid evaporates to yield vapor of composition v2 which is richer in the
92
more volatile constituent, A. The remaining liquid has a composition l2 and thus has more of the
less volatile constituent B. This again happens at every plate of the column. As the liquid flows
down the column, it is being heated, the heating preferentially vaporizing the more volatile
component leaving the liquid enriched in B.
If sufficient number of plates are used, it is possible to separate the two constituents of a binary
liquid mixture, the more volatile component in the vapor form at the top of the column and the
less volatile component in the liquid form at the bottom.
Non ideal solutions of liquids in liquids
We have seen that two liquids having more or less identical intermolecular forces of attraction
when mixed form ideal solutions and there are very few such liquid pairs. Most liquid pairs form
non ideal solutions. They show deviations from Raoult’s law and the nature of deviation depends
on the types of liquids mixed. Some liquid mixtures show positive deviations while others show
negative deviations from Raoult’s law.
In a mixture of two liquids A and B, if A-B attractions are weaker than A-A and B-B attractions,
such mixtures show positive deviations from Raoult’s law. The vapor pressure-composition
curves of the constituents and that of the mixture lie above those of the ideal curves. The
tendency of the molecules A and B to escape into the vapor phase is greater in such a non ideal
solution than in an ideal solution of A and B. The extent of positive deviation depends upon many
factors, which include differences in polarity, in intermolecular forces of attraction, in length of
the hydrocarbon chain, of molecules A and B. Greater these differences, larger is the positive
deviation and the total vapor pressure versus composition plots show maxima. Most liquid pairs
show positive deviations. Some examples are : carbon tetrachloride and heptane at 323K, ethyl
ether and acetone at 293K and 303K, heptane and ethyl alcohol at 323 K and acetone and carbon
disulfide at 308 K.
Carbon tetra chloride – heptane mixture shows only slight deviation from ideal behaviour as both
the constituents are non polar and have weak intermolecular interactions. The other liquid pairs
show greater deviation. The additional factor that alcohol exists as associated molecules in the
liquid state contributes to the heptane ethyl alcohol pair showing very large deviations.
Binary liquid mixtures, in which A-B attractions are stronger than those of A-A or B-B, show
negative deviations from Raoult’s law. The vapor pressure curves of the constituents and the
mixture of these liquid systems lie below those of the ideal curves. The tendency of the molecules
A and B to leave the solution phase and move into the vapor phase is less in these solutions as
compared to that in an ideal solution of A and B. If there is any kind of association or chemical
interaction between molecules of A and B, then the deviations are larger and total vapor pressure
versus composition curves show minima. Liquid pair mixtures showing negative deviations are
not very common. A few examples are : pyridine-formic acid, chloroform-acetone and aqueous
solutions of hydrochloric, nitric or perchloric acids. In the chloroform - acetone system, there is
partial association between the two constituents due to hydrogen bonding. In the case of acid
solutions, the volatile acids form ionic solutions.
93
The Duhem-Margules equation
The Duhem-Margules equation relates the partial vapor pressure of the two constituents of a
binary liquid mixture with their corresponding mole fractions. This relation can be derived
thermodynamically starting with the Gibbs-Duhem equation.
nAdµA+ nBdµB = 0
nAdµA = -nBdµB
Dividing both sides of nA+nB, we have
nA
-n B
dµ A =
dµ B
n A +n B
n A +n B
xAdµA = -xBdµB
where xA and xB are mole fractions of A and B, respectively.
Dividing by dxA, we get
xA
dµ A
dµ
=-x B B
dx A
dx A
xA+xB = 1
dxA+dxB = 0
dxA=-dxB
xA
dµ A
dµ
=x B B
dx A
dx B
(Eq.
10.4.48)
The chemical potential of any constituent of a liquid mixture, assuming that vapor behaves
ideally, is given by
µi(sol)= µ0i +RT lnp
(Eq 10.4.49)
Where p is the partial pressure of the constituent. Writing this equation for A and differentiating
with respect to xA at constant temperature and total pressure, we have
dµ A
dlnp A
=RT
dx A
dx A
Similarly, for the component B, differentiating with respect to xB gives
(Eq.10.4.50)
94
dµ A
dlnp B
=RT
dx B
dx B
(Eq. 10.4.51)
Multiplying equation 10.4.50 by xA and equation 10.4.51 by xB and applying the equation
10.4.48, we get
xA dlnp A = xB dlnp B
dx A
dx B
dlnp A dlnp B
=
dlnx A dlnx B
(Eq.10.4.52)
Equation 10.4.52 is known as the Duhem-Margules equation. This equation relates the partial
pressures of the two constituents with their corresponding mole fractions and is applicable to both
ideal and non ideal liquid mixtures. The only assumption made was that the vapor behaved
ideally.
The Duhem-Margules equation can be used to show that (a) if one component of a binary liquid
mixture behaves ideally, the other one must also do so and (b) if one shows positive (negative)
deviation from ideality, the other must also show the same deviation. The following derivation
shows the above.
(a) If component A is taken to behave ideally, then according to Raoult’s law, we have
pA = p0A xA
lnpA= ln p0A +ln xA
dlnpA = d lnxA
dlnp A
=1
dlnx A
Using Duhem – Margules equation, we write
dlnp B
=1
dlnx B
dlnpB=dlnxB
Integrating this equation, we get
lnpB=lnxB+I, where I is the constant of integration. When xB=1, pB= p0B and
95
I=ln p0B
lnpB=lnxB+ln p0B
lnpB = ln (xB p0B )
pB = xB p0B
Component B also obeys Raoult’s law which means it behaves ideally.
(b)
Let the component showing positive deviation be A.
pA(real)>pA(ideal) = poA xA
lnpA(real)>lnpA(ideal) = ln poA + lnxA
dlnp A(real)
dlnp A(ideal)
dx A
dx A
xA
dlnp A(real)
dx A
dlnp A(real)
dlnx A
xA
=
1
xA
dlnp A(ideal)
dx A
=1
1
According to Duhem-Margules equation, we have
dlnp B(real)
dlnx B
1
On integration, we get
pB(real)>pB(Ideal) = poB x B
Hence, we can say that the second component also exhibits positive deviation from ideality. It can
be shown along similar lines that if one component exhibits negative deviation, the other will also
do so.
Composition of the vapor phase: Konowaloff’s Rule
We start with the Duhem-Margules equation:
96
dlnp A dlnp B
=
dlnx A dlnx B
(Eq.10.4.52)
x A dp A x B dp B
=
p A dx A p B dx B
(Eq.10.4.53)
xA+xB = 1
dxA+dxB = 0
dxA=-dxB
Substituting for dxB in terms of dxA in equation 10.4.52, we write
x A dp A x B dp B
+
=0
p A dx A p B dx A
(Eq. 10.4.53)
p = pA+pB
dp dp A dp B
=
+
dx A dx A dx A
Substituting for dpA/dxA from equation 10.4.53 into the above equation, we write
x dp B p A dp B
dp
=- B
+
dx A p B dx A x A dx A
dp dp B
=
dx A dx A
⎡ x B pA ⎤
⎢1⎥
⎣ pB x A ⎦
(Eq.10.4.54)
If we assume that addition of a component increases its partial vapor pressure, then dpA/dxA and
dpB/dxB are positive and dpB/dxA is negative. Hence the change in the total vapor pressure with
the addition of component A, dp/dxA will be positive or negative depending on the sign of the
term inside the brackets in equation 10.4.54.
If xBpA>xApB, then dp/dxA is positive, that is, pA/pB>xA/xB. Using Dalton’s law of partial
pressures, we can write pA/pB = yA/yB. Hence yA/yB>xA/xB.
The ratio of the mole fractions in the vapor phase is more than the corresponding ratio in the
liquid phase. We can infer from this that the vapor phase is richer in A than is the liquid phase in
equilibrium with it. Thus, the vapor is richer in the component whose addition to the liquid
mixture results in an increase in the total vapor pressure (dp/dxA is +ve).
If xBpA<xApB, then dp/dxA is negative or dp/dxB is positive. pA/pB<xA/xB and yA/yB < xA/xB. We
can also write yB/yA>xB/xA.
97
The ratio of mole fractions of B and A in the vapor phase is greater than the corresponding ratio
in the liquid phase. The vapor is richer in B than the liquid in equilibrium with it. Thus, the vapor
is richer in the component addition of which to the liquid mixture increases the total vapor
pressure (dp/dxB is +ve).
General conclusion drawn from the above discussion is:
The vapor phase is richer in the component whose addition to the liquid mixture results in an
increase of total vapor pressure.
Since this rule was stated empirically by D.P.Konowaloff, it is known as the Konowaloff’s rule.
Temperature-composition diagrams of non ideal solutions
Non ideal solutions that depart only slightly from ideality have boiling point versus composition
curves very similar to those of the ideal solution (Figure 10.4.8). The boiling point of the solution
lies in between those of the pure constituents. From the diagram, we can see that the boiling point
of the solution increases or decreases in a regular manner with the composition.
PA0
PB0
Figure 10.4.12: Pressure – composition curves of a non ideal solution exhibiting positive
deviation from ideality (constant temperature)
Non ideal solutions that show very large deviation from ideality exhibit either maxima or minima
in their vapour pressure – composition curves (Figures 10.4.12 and 10.4.13). As boiling point is
98
Pressure
inversely related to vapour pressure, the boiling point – composition curves of these solutions
exhibit either minima or maxima. The general appearance of a temperature versus composition
curve is expected to have a one to one inverse correspondence with that of the vapour pressure –
composition curve. The temperature versus composition curve for a given solution can be drawn
in accordance with Konowaloff’s rule.
p
Raoult’s law
limiting line
Henry’s law
limiting line
xB=1
xA=0
xA
xB
xB=0
xA=1
Figure 10.4.13: Pressure-composition curves of a non ideal solution exhibiting negative deviation
from ideality (constant temperature).
At the maximum or minimum point of the vapour pressure – composition curve (Figures 10.4.12
and 10.4.13), dp/dxA is zero. From equation 10.4.54, we see that either dpB/dxA is equal to zero or
xBpA = xApB.
99
dpB/dxA cannot be zero as this would mean that the partial pressure would remain
constant inspite of a change in composition of the solution. Hence
xBpA = xBpB
pA x A
=
pB x B
The ratio of mole fractions of A and B in the vapor phase is the same as that in the liquid phase.
The composition of the vapor phase is the same as that of the liquid phase with which it is in
equilibrium. The vaporous curve and the liquidus curve meet at the point of maximum or
minimum in a temperature – composition diagram so that both the phases have the same
composition at this point.
The temperature – composition diagram of a nonideal solution exhibiting positive deviation from
ideality (corresponding to the total vapor pressure versus composition of figure 10.4.12) is given
in figure 10.4.14. We can see the system exhibiting a minimum in the curve.
vapour
Boiling point
TB0
B
T
x1
y1
TA0
y2
liquid
xA
A
x2
M
xB
Figure 10.4.14: Temperature – composition diagram for a system exhibiting a minimum in the
curve (pressure constant)
100
On the other hand, a non ideal solution exhibiting negative deviation from ideality shows a
maximum in the temperature – composition plots (figure 10.4.15). This corresponds to the total
vapor pressure – composition plot of figure 10.4.13.
M
B
y1
vapour
x1
TB0
y2
Boiling point
x2
TA0
A
liquid
Pure B
xA
xB
Pure A
Figure 10.4.15: Temperature – composition diagram for a system exhibiting a maximum in the
curve (pressure constant)
Fractional distillation of non ideal solution
As binary liquid solutions that deviate only slightly from ideal behaviour exhibit temperature –
composition diagrams very similar to that of the ideal solution, the constituents of these solutions
can be separated by fractional distillation. Nonideal solutions exhibiting minima or maxima in
their temperature composition diagrams yield only one of the constituents in the pure form on
fractional distillation.
a)
Solutions exhibiting a minimum in the boiling point:
If we take a liquid mixture of composition x1 (figure 10.4.14) and distill it, it will start boiling at
temperature T and the vapor formed will have a composition y1. The vapor thus has more of
constituent A while the liquid left, more of B. As the distillation continues, the composition of the
liquid phase moves along x1B, i.e., towards pure B, and that of the vapor phase along y1M,
towards the point M. At the point M, the vapor phase and the liquid phase have the same
composition. Thus, condensing the vapor and distilling the resultant liquid will not yield any
further separation of the constituents. The liquid at this point M would boil at a constant
101
temperature as if it is a single constituent and is called an azeotrope (from the Greek: to boil
unchanged).
The fractional distillation of a solution of composition x1 will yield vapor of composition M and
pure liquid B, collecting the vapor from the top and the pure liquid B from the bottom of the
fractionating column.
Similarly if a liquid mixture of composition x2 is taken and distilled, vapor of composition M is
obtained from the top and pure liquid A from the bottom of the fractionating column.
b)
Solution exhibiting a maximum in the boiling point:
Let us consider a liquid mixture of composition x1 (Figure 10.4.15). When this mixture starts
boiling, vapor of composition y1 is obtained. The vapor is richer in B and the liquid left, in A.
During distillation, vapor phase composition moves along y1B and the liquid phase composition
along x1M. At the end of the distillation, collected vapor on condensing will yield pure B and the
residual liquid will yield an azeotrope of composition M.
Similarly a liquid mixture of composition x2 on subjecting to fractional distillation will yield
vapor of pure A and an azeotrope of composition M.
For a given pair of liquids and at a given pressure, the composition of the azeotrope and its
boiling point have definite values. Their values change with variation in external pressure
showing thereby that the azeotrope is a mixture of liquids, not a compound formed by the
combination of the two liquids.
The hydrochloric acid-water mixture exhibits a maximum in its boiling point – composition
diagram and hence forms an azeotrope. The normal boiling points of water and hydrochloric acid
are 100oC and -85oC respectively. The azeotrope has 20.22% by mass of water and a boiling point
of 108.6oC when the external pressure is 1 atm. As the external pressure increases from 500 torr
to 800 torr, the mass % of HCl in the azeotrope decreases from 20.92 to 20.16 and boiling point
increases from 97.58oC to 110.01oC. Examples of other maximum boiling mixtures are given in
the table 10.4.1.
Table 10.4.1 A few examples of azeotropic mixtures
Type
Minimum
Components
Azeotrope*
A
B
Mass% A
b.pt/0C
Water (100oC)
Ethanol(78.3oC)
4.0
78.2
Water (100oC)
Ethylacetate(79.6oC)
11.3
73.4
102
Maximum
Chloroform(61.2oC)
Methanol(64.7oC)
79.5
55.7
Carbondisulphide(46.3oC)
Acetone (56.1oC)
67.0
39.3
Water (100oC)
Hydrochloric acid
(85oC
20.2
108.6
Chloroform(61.2oC)
Acetone(56.1oC)
78.5
64.4
Water (100oC)
Nitric acid (86oC)
32.0
120.5
* Values are at 1 atm pressure
The ethanol-water system exhibits a minimum in its boiling point – composition diagram and
hence forms an azeotrope. The normal boiling points of the components are : ethanol 78.3oC and
water 100oC. The azeotrope boils at 78.2oC and has 4% by mass of water. Examples of other
minimum boiling mixtures are included in the table 10.4.1.
Partially miscible liquids
Partially miscible liquids are those that do not mix in all proportions at all temperatures. These
are miscible only in a limited range of concentrations. Phenol-water, aniline-hexane, methanolcarbondisulphide, triethylamine-water are few examples of liquid pairs that show partial
miscibility.
Phenol-water system
A small quantity of phenol when added to water at room temperature dissolves to give a solution
but with the additions of more phenol, at some point phenol does not dissolve any more and two
solutions are formed. These, a saturated solution of phenol in water and a saturated solution of
water in phenol, known as conjugate solutions, are in equilibrium with each other. At a given
temperature, the compositions of these two solutions are fixed and are independent of the relative
amounts of the solutions. If the addition of phenol is continued, the added phenol gets dispersed
into the conjugate solution in such a way that the compositions of the two phases remain
unaltered, only the relative amounts of the phases change. The amount of the phase phenol in
water decreases and that of the phase water in phenol increases. When an extremely small amount
of phenol in water is left, addition of phenol results in the system again becoming a single
solution, a saturated solution of water in phenol. Addition of more phenol to this single phase
system only makes the solution more unsaturated in water. Liquid pairs, which show the same
behaviour as the phenol-water system, have been observed to exhibit large positive deviations
from Raoult’s law. In these liquid systems there occurs a limited solubility of one liquid into
another and hence two saturated solutions are formed over a certain range of composition. The
103
liquids are completely miscible beyond this range. Thus, we may say that there exists a
miscibility gap in the system.
b''
Temperature
c'
b'
b
c''
a
A
Pure A
d''
c
d
d'
B
Composition
Pure B
Figure 10.5: Temperature-composition diagram for phenol-water system (upper CST) A=water,
B=phenol, constant external pressure (1 atm).
We have seen that at a given temperature the compositions of the conjugate solutions remain
constant and are independent of the relative amounts of the two solutions. According to the phase
rule, F=C-P+2=2-2+2=2, the two degrees of freedom are any two out of pressure, temperature
and composition of either of the two conjugated solutions. One of these two degrees of freedom is
the external pressure which has a fixed value and which remains constant. Hence stating the value
of one other parameter, temperature or composition of either of the conjugate solutions, describes
the system completely. If, for example, temperature is stated, then compositions of the two
solutions are fixed. If a graph is plotted between temperature and composition of the two
conjugate solutions at that temperature, a curve is obtained as shown in figure 10.5. We see from
the diagram that the compositions of the two solutions get closer as the temperature of the system
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is increased. This is due to the increase in mutual solubility of the two components, phenol and
water, with increase in temperature. At a certain temperature, the composition of the two
solutions become identical and both the solutions merge into each other forming a single solution.
This temperature is known as the consolute temperature or the critical solution temperature
(CST). If the temperature of the system is above this CST, then the two liquids are miscible in all
proportions. A system represented by a point outside the cuve ACB is a single solution and one
represented by a point within the curve ACB is made up two conjugate solutions in equilibrium
with each other. The compositions of the two solutions are given by the two points on the
solubility curve at the given temperature.
Let us consider a system consisting of phenol and water represented by the point a at temperature,
toC. As this point lies within the curve, this system consists of two conjugate solutions in
equilibrium whose compositions are given by b and d on the solubility curve ACB. For a system
at any point on the line bd, the compositions of the two solutions are given by the points b and d.
The relative amounts of the two solutions present in the system would vary as we move from a
point near a to a point near d. The relative amounts of the two solutions present in any system can
be calculated using the lever rule, for example, in the system represented by a, we have:
ba Total amount of substances in solution of composition d
=
da Total amount of substances in solution of composition b
A system represented by a point outside the curve has only one phase, P=1. The number of
degrees of freedom of this system, F=C-P+2=2-1+2=3. As the pressure is maintained constant
(usually 1 atm), the values of two variables, temperature and composition of the solution need to
be mentioned to describe the system completely.
If we next take a system represented by a point inside the curve, this system exists as two
conjugate solutions in equilibrium with each other. The number of phases in this system is two
and hence F=C-P+2=2-2+2=2. As pressure is constant, we need to give the value of one variable,
temperature or composition of one solution (usually temperature) to define the system
completely. If the temperature of the system is stated, then the compositions of the two solutions
are fixed. These are given by the two points on the solubility curve corresponding to the
temperature stated.
A state point on the solubility curve corresponding to the CST, c has one restricted condition of
identical composition of the two solutions. Hence, the degrees of freedom F=(C-r)-P+2=(2-1)2+2=1. As pressure is kept constant, this system becomes invariant. The critical solution
temperature and the corresponding composition on the curve (point c) have fixed values for a
given value of pressure (usually the pressure chosen is 1 atm). For the phenol-water system, the
CST is 65.9oC and the composition at the CST is 66% by mass of water.
Effects produced on changing the composition of the system keeping temperature and
pressure constant.
We start with a system represented by b' in the diagram (Figure 10.5). This system consists of an
unsaturated solution of B in A. As addition of B is continued, it dissolves in A till the point b is
reached. At b, a phase separation takes place, two solutions appear, a saturated solution of B in A,
present in larger amount having composition given by b and a saturated solution of A in B,
present in traces (as it has just started forming) having a composition given by the point d. As
105
more B is added, the system continues to have the above two solutions, composition remaining
the same, relative amounts changing till the point d is reached. The amount of saturated solution
of B in A decreases while that of A in B increases, the ratio of the amounts at any point being
given by the lever rule. At the point b, the solution of B in A is present in traces (as it is going to
disappear) while solution of A in B is present in larger amounts. Further addition of B to the
system reduces it to a single solution, an unsaturated solution of A in B. With the addition of
more B, the solution gets more unsaturated in A and the state point moves towards d'.
The system can move from b' to d' without the formation of two solutions if it is heated to a
temperature above the CST, say b'', the required amount of B added so that the system moved to
d'' and then cooled to d'. All along these state transformations, the system would remain a single
solution.
Effect of changing the temperature of the system keeping the composition and pressure
constant.
Let us start with a system represented by a in the diagram (Figure 10.5). This system has two
solutions, B in A given by the composition b and A in B given by the composition d. As the
system is heated, it moves along the vertical line ac', the composition of the solution in B in A
moves along bc' while that of the solution A in B moves along dc''. The relative amounts of the
two solutions also vary, the amount of B in A increases while that of A in B decreases. The
system continues to exist as two solutions till the point c' is reached where there is only a trace of
solution of A in B. The system becomes a single saturated solution of B in A. The temperature at
which this happens is known as the mutual solubility temperature (MST). A further rise in
temperature results in the system becoming an unsaturated solution of B in A. The CST, hence, is
the maximum of all MSTs.
A few examples of partially miscible liquid pairs which show an increase in mutual solubilities
with increase in temperature are given in table 10.5.1.
Table 10.5.1
A few examples of systems showing upper CST
System
CST/oC
Composition
Water-phenol
65.9
66 mass % of water
Aniline-hexane
59.6
48 mass% of aniline
Cyclohexane-methanol
49.0
71 mass % cyclohexane
Isopentane-phenol
63.5
51 mass% isopentane
106
Methanol-carbondisulphide
49.5
20 mass% methanol
Triethylamine-water system
Temperature
There are some partially miscible liquid pairs, such as triethylamine-water, which exhibit increase
in mutual solubilities with increase in temperature. Figure 10.5.1 gives the temperaturecomposition diagram for the triethylamine-water system. The two liquids are completely miscible
at or below 18.5oC and only partially miscible above this temperature. As the curve confining the
area of partial miscibility exhibits a minimum, the temperature at or below which the liquids are
completely miscible is known as the lower critical solution temperature or lower consolute
temperature. It is difficult to determine the composition of the solution corresponding to the CST
as the curve is quite flat; it appears to be about 30% by mass of triethylamine.
Pure A Composition
Pure B
Figure 10.5.1: Temperature – composition diagram for triethylamine – water system. (lower CST)
A=water, B=triethylamine, constant external pressure (1 atm)
Liquid pairs that show a lower critical solution temperature invariably tend to form loosely bound
compounds with each other. This increases solubility at low temperatures. These liquid systems
have been found to be a hydroxy compound and an amine or a ketone or an ether. Hence the
possibility of the liquids associating through hydrogen bonding is high. The extent of such
association decreases with increase in temperature of the system and this is probably the reason
for the decrease in mutual solubility with rise in temperature.
The detailed treatment given to the phenol-water system in the last section applies to this system
as well except for the fact that phenol-water shows an upper CST whereas triethylamine-water a
lower CST. Few examples of liquid pairs that show lower CST are given in the Table 10.5.2.
107
Table 10.5.2 A few examples of systems showing lower CST
System
Triethylamine-water
γ-Collidine-water
β-picoline-water
CST/oC
18.5
6
49
Nicotine-water system
Some partially miscible liquid pairs show both upper and lower critical solution temperatures.
Nicotine-water is one such system and its temperature-composition diagram is given in figure
10.5.2. The upper CST is at 208oC and the lower is at 60.8oC. Within the enclosed area the
liquids are only partially miscible while outside the area they are completely miscible. Otherwise
this system could be discussed along lines similar to the phenol-water system. The composition
of the system, 34% nicotine, is the same at both upper and lower CST. At point A, about 95oC,
nicotine is least soluble in water while at point B, about 130oC, water is least soluble in nicotine.
108
C Upper CST
Temperature
B
A
C' Lower CST
Pure A
Composition
Pure B
Figure 10.5.2.: Temperature – composition diagram for nicotine-water system. A=water,
B=nicotine, constant external pressure >> 1 atm
109
For many liquid pairs, the upper CST values are very high and hence these can be realized only
under high pressure. In many cases the upper CST values are above the critical temperature of the
liquids and the system vaporizes much before the CST is reached. In the case of liquid mixtures
such as ethyl acetate-water and ether-water, it is not possible to obtain both the upper and lower
CSTs. As the solubilities of these liquids in water increases with decreasing temperature, a lower
CST is expected but is not obtained experimentally as water freezes before CST is reached.
Similarly for the CHCl3-H2O mixture, the upper CST cannot be realized as it is above the critical
temperature of CHCl3.
Effect of impurities on CST
Critical solution temperature is markedly affected by the presence of impurities in the system. If
the impurity present or added is soluble in both the liquids, then the CST is lowered as shown in
the diagram of figure 10.5.3. The impurity increases the mutual solubility of the liquids by
distributing itself between the two conjugate solutions in a definite manner. The impurity keeps
exchanging between the conjugate solutions, thus acting as a cementing force and increasing the
mutual solubility of the liquids. For example, when succinic acid is added to the phenol-water
system, its CST gets lowered.
Temperature
110
Pure A
Composition
Pure B
Figure 10.5.3: Effect of impurity on CST. Impurity is soluble in both the liquids. Dotted curve
with impurity. A=water, B=phenol, constant external pressure (1 atm)
On the other hand, if the impurity present or added is soluble in only one liquid, then the CST is
raised as shown in the diagram of the figure 10.5.4. Such an impurity affects the solubility of the
other liquid in the liquid in which the impurity is soluble. This affects the mutual solubility of the
liquids. For example, addition of 0.1 mol of KCl per dm3 of water raises the CST of the phenol-
111
Temperature
water system by about 8oC. If 0.1 mol of naphthalene per dm3 of phenol is added to the same
system, the CST is elevated by about 20 degrees Celsius.
Pure A
Composition
Pure B
Figure 10.5.4: Effect of impurity on CST. Impurity is soluble in one of the liquids. Dotted curve
with impurity. A=water, B=phenol, constant external pressure (1 atm).
112
Effect of pressure on CST
Temperature
Application of phase rule to partially miscible liquid pairs at the CST showed that for a given
external pressure, the CST has a definite value. As the external pressure changes, the CST also
changes. Increase in external pressure increases the mutual solubility of the two components.
Hence, it is expected that the upper CST will decrease and lower CST will increase with increase
in external pressure. The loop confining the area of partial miscibility gets smaller (figure 10.5.5.)
with increase in pressure and finally gets reduced to a point. At this stage, the two liquids become
completely miscible in all proportions and form a single solution.
Pure A
Composition
Pure B
Figure 10.5.5: Effect of pressure on CST. Dotted curve at low pressure, solid curve at high
pressure.
113
Immiscible Liquids
As immiscible liquids are not soluble in each other, the properties of either liquid is not affected
by the addition of one liquid to the other. Hence each liquid will behave independently of the
other. As a result in an immiscible liquid mixture of two liquids, each liquid exerts the vapor
pressure corresponding to the pure liquid at the given temperature. The total vapor pressure of the
mixture is thus the sum of the vapor pressures of the two pure liquids.
p=p0A +p0B
Where p is the total vapor pressure and p0A and p0B are the vapor pressures of pure A and pure B
respectively. As long as both the liquids are present, the total pressure remains constant and is
independent of the relative amounts of A and B.
The temperature at which the total vapor pressure of a liquid system becomes equal to the
external pressure is its boiling point. As a binary mixture of immiscible liquids can attain any
given total pressure at a lower temperature than either liquid alone, it follows that any mixture of
two immiscible liquids must boil at a temperature less than the boiling point of either of the two
liquids. As there is no change in the total vapor pressure with change in overall composition of
the liquid mixture, the boiling point will remain constant as long as both the liquids are present.
However, when one of the liquids has boiled away, the boiling temperature will increase abruptly
from that of the mixture to the boiling point of the left over liquid.
The composition of the vapor formed when such a liquid mixture boils can be obtained by using
the Dalton’s law of partial pressures.
p0A
p0B
=
y A n A w A /M A
=
=
y B n B w B /M B
w A M A p0A
=
wB
M B p0B
(Eq.10.4.55)
(Eq.10.4.56)
Where yA and yB are the mole fractions of A and B, respectively, in the vapor phase; nA and nB
are the number of moles of A and B in any given volume of vapor. As the ratio of partial
pressures at a given temperature is constant, nA/nB must also be constant, that is, as long as both
liquids are present, the vapor phase composition at all times is constant. The mass of any
constituent distilling over depends on both its vapor pressure and molecular mass.
Distillation of immiscible liquids is used in industry and in the laboratory for the purification of
those organic liquids which have a high boiling point or which tend to decompose below their
normal boiling points.
Steam distillation
If water is used as one of the immiscible liquids, then the distillation is known as steam
distillation. The immiscible mixture of the liquid and water is either heated directly or by passing
steam through the liquid. The vapors are collected, condensed and then separated. From the
114
expression for wA/wB (Eq. 10.4.56), we can see that the condensed vapors will contain a high
proportion of the liquid provided the molar mass of the liquid is high and the liquid has
appreciable vapor pressure at the boiling temperature of the mixture. In this way, it is possible to
distil liquids having high boiling point at temperatures below 100oC, the boiling point of water.
Distillation of immiscible liquids can also be used to determine the approximate molar mass of a
constituent if the ratio of vapor pressures and masses of the two liquids are obtained and the
molar mass of the other constituent is known (Eq.10.4.56).
Nernst distribution law – thermodynamic derivation, applications
Walther Hermann Nernst, a German physical chemist carried out a large number of experiments
to study the distribution of numerous solutes between suitable immiscible solvents. He
generalized his observations into a law, the Nernst distribution law, which states:
A solid or liquid distributes itself between two immiscible solvents in such a way that at constant
temperature the ratio of its concentrations in both the solvents is constant. The constant, KD, is
known as the distribution coefficient or the partition coefficient. The value of the partition
coefficient, KD depends only on the temperature of the system. It is independent of the amount of
solute taken and also of the relative amounts of the two immiscible solvents.
C1
=K D
C2
(Eq. 10.4.57)
Where C1 and C2 are the equilibrium molar concentrations of the solute in the solvents 1 and 2
respectively. Such a distribution of solute between two immiscible or only slightly miscible
solvents can be achieved with any solute for which a pair of immiscible solvents can be found.
For example iodine is soluble in both carbon tetrachloride and water. If a solution of iodine in
carbon tetrachloride is shaken with water, which is immiscible with carbon tetrachloride, iodine
distributes itself between the carbon tetrachloride and water layers in such a way that at
equilibrium,
2O
CH
I
2
4
Cccl
I2
= constant. This constant, KD has a value of 0.117 at 25oC. Bromine distributing between
water and CS2, phenol between water and amyl alcohol are few other examples.
According to the phase rule, for a system of two immiscible liquids with a solute distributed
between them and at equilibrium, the number of degrees of freedom
F =
=
C-P+2
3-2+2=3
Temperature and pressure, held constant, account for two out of the three degrees of freedom.
The third variable, concentration of solute in one of the layers, if stated, defines the system
completely. That is, the concentration of the solute in the other layer will have a definite value
and the ratio of the two concentrations thus will be constant.
115
The distribution law equation as given by equation 10.4.57 strictly holds good only if the
solutions in the two immiscible solvents are ideal and dilute, the mutual miscibility of the two
solvents is not affected by the presence of the solute, and the solute has the same molar mass in
both the layers.
Thermodynamic derivation of the distribution law
If a solute soluble in two immiscible solvents A and B distributes itself between the solvents at a
constant temperature, then at equilibrium, µA = µB, where µA and µB are the chemical potentials
of the solute in the solvents A and B respectively. If both the solutions are ideal dilute solutions,
then we have
µA = µ0A + RTlnxA and
µB = µ0B + RTlnxB
(Eq.
10.4.58)
Substituting from equation 10.4.58 into µA = µB, we get
µ0A + RTlnxA=µ0B+RTlnxB
ln
xA
1
=
(µ0B-µ0A)
xB
RT
(Eq.10.4.59)
At a given temperature and pressure, the right hand side of equation 10.4.59 is a constant.
Hence, we have
ln
xA
= constant
xB
xA
= Constant
xB
As the solutions are assumed to be dilute, the rato of mole fractions can be replaced by the ratio
of their molar concentrations. Thus, we can write
CA
= Constant = KD
CB
(Eq. 10.4.57)
Modifications of the Nernst distribution law expression
Nernst in the year 1891 reported that the distribution law statement is applicable only when the
solute has the same molar mass in both the layers. If the solute undergoes dissociation or
association, then the law does not apply to the total concentration in the two phases but applies
only to the concentrations of the species common to both the layers. For example, in the
distribution of benzoic acid between water and chloroform, the ratio of the total concentration of
benzoic acid in the two liquids does not have a constant value. This is because benzoic acid is
116
associated into dimers in the chloroform layer and is slightly dissociated into benzoate and
hydrogen ions in the aqueous layer. In the aqueous layer, there is the ionization reaction:
C6H5COOH C6H5CO O + H+
There is dimerisation taking place in the chloroform layer
2C6H5COOH (C6H5COOH)2
As the distribution law is applicable only for the concentration of the benzoic acid, the species
present in both the phases, the ratio taking total concentrations is not constant. If the ratio is
calculated taking the concentration of undissociated benzoic acid in the aqueous layer and
monomer molecules in the organic layer, then it has a constant value.
concentration of undissociated acid in water
= 0.442 at 40oC
concentration of monomer in chloroform
Some examples of systems in which the solute ionizes in the aqueous layer and dimerises in the
organic layer are: benzoic acid between water and benzene, salicylic acid between water and
benzene or chloroform and acetic acid between water and carbon tetrachloride
a)
Solute exists as normal molecules in liquid A and associated molecules in liquid B.
Let the solute be Q and at constant temperature when the system is in equilibrium, its
concentration be CA in liquid A. in liquid B, it associates according to the equation:
Qn
nQ (Eq. 10.4.58)
Let the total concentration in liquid B be CB and ξ a be its degree of association. Then the
concentrations of monomer and associated solute Q are given by:
[Q] = CB – n ξ
[Qn] = ξ
a
a
The equilibrium constant, Keq for the association reaction is given by :
Keq =
Keq =
[Q n ]
(Eq. 10.4.59)
[Q]n
ξa
B
(C -nξ a ) n
(CB-n ξ a)n =
ξa
K eq
(Eq.10.4.60)
117
⎛ ξ
CB-n ξ a = ⎜ a
⎜ K eq
⎝
1
⎞n
⎟
⎟
⎠
If the solute in liquid B is present almost in the associated form, then (Qn) = ξ a=CB/n
1
⎛ CB ⎞ n
CB-n ξ a = ⎜⎜
⎟⎟
⎝ nKeq ⎠
KD =
CA
CA
= B
[Q] C -nξ a
(Eq.10.4.61)
(Eq.10.4.62)
Substituting from equation 10.4.61 into the expression for KD, we have
KD =
1
CA
1
(C B ) n
CA
1
(CB ) n
=
(n Keq) n
(Eq.10.4.63)
KD
(Eq.10.4.64)
(nK eq
1
)n
= K'D
At a given temperature, Keq has a constant value. For a given association reaction, n is constant.
1
Hence KD /(nKeq) n is equated to K'D, a constant known as the apparent distribution coefficient.
The value of n can be obtained by plotting a graph between logCA and log CB, and knowing n, K'D
can be determined.
Alternatively if only two sets of data are available, then n can be calculated using equation
10.4.65.
n=
logCB2 -logC1B
A
logCA
2 -logC1
(Eq.10.4.65)
where C1A and C1B are the concentrations of the solute in liquid A and liquid B respectively
obtained from the first set, and CA2 and CB2 are the values from the second set.
The distribution coefficient, KD and the equilibrium constant, Keq can be obtained by substituting
for [Q] from equation 10.4.62 and for [Qn] from equation 10.4.59 into the expression:
CB = [Q] + n[Qn]
CB =
CA
+nK eq [Q]n
KD
118
⎛ CA ⎞
CA
+nK eq ⎜⎜
⎟⎟
KD
⎝ KD ⎠
CB =
CB
C
=
A
n-1
CA nK eq A n
+
(C )
K D KnD
1 nK eq A n-1
+
(C )
K D K nD
(Eq.10.4.66)
A plot CB/CA, the ratio of total concentration of Q in liquid B to the concentration of Q in liquid
A, versus (CA)n-1 has (KD)-1 as the intercept and nKeq/K nD as the slope. From these, the values of
KD and Keq can be calculated.
b)
Solute exists as normal molecules in liquid A and dissociates in liquid B.
Let the solute dissociate in liquid B according to the equation:
Q
E+F
If ξ is the extent of reaction in the system at equilibrium and CB is the initial
concentration of Q, then the equilibrium concentrations are:
[Q] = CB - ξ
[E] = [F] = ξ
The equilibrium constant, Kdiss for the reaction is given by:
Kdiss =
[E][F] ξ 2
=
[Q] CB -ξ
[Q] = CB—ξ =
ξ2
K diss
The distribution law expression for the system is :
KD =
KD =
CA
CB -ξ
where CA is the concentration of Q in the liquid A
CA
ξ 2 /K diss
(c) solute exists as dimers in liquid B and dissociates in liquid A.
(Eq. 10.4.67)
119
KD =
c A -ξ d
(Eq.10.4.68)
c B -2ξ a
Where CB is the total concentration of the solute in liquid B and CA is the initial concentration of
the solute in liquid A, ξd and ξa are the extent of dissociation and extent of association of the
solute in the solvents A and B respectively.
(d)
Solute exists as normal molecules in liquid A but reacts chemically with liquid B. The
reaction between the solute Q and the liquid B is represented as:
Q+nB Q.nB
The equilibrium constant of this reaction is given by:
Keq =
[Q.nB]
(Eq. 10.4.69)
[Q][B]n
Let the concentration of Q in liquid A be CA and the total concentration in liquid B be CB.
CB = [Q] + [Q.nB]
Substituting for the concentration of Q.nB from equation 10.4.69, we get
CB = [Q] +Keq [Q][B]n
CB = [Q] {1+Keq[B]n}
CB
[Q] =
1+K eq [B]n
(Eq. 10.4.70)
The distribution coefficient, KD is given by:
KD =
CA
C
B
C A CA (1+K eq [B]n )
=
CB
[Q]
=
KD
1+K eq [B]n
(Eq.10.4.71)
Keq is a constant at a given temperature and as liquid B is present in large amount, [B]~ constant.
Hence,
CA
= Constant
CB
(Eq. 10.4.72)
120
Applications of the distribution law
The distribution law forms the basis of the solvent extraction process. In this process a solute in
one solvent is extracted with another solvent in which it is more soluble. The solute is then
recovered by various methods from the extracting solvent. This process is important in the
laboratory and in industry. In the laboratory aqueous solutions of various organic compounds are
extracted with solvents such as ether, chloroform, carbon tetrachloride, etc. For example, aniline
is prepared by reduction of nitrobenzene in aqueous solution, extracted by using ether as the
extracting solvent and recovered by evaporating ether.
In industry, this process is also used for removing unwanted constituents of a product. For
example, the harmful ingredients in petroleum oils are removed by shaking these up with
immiscible solvents in which the impurities are more soluble.
The law finds extensive applications in analytical chemistry. For example, in qualitative analysis,
presence of iodide ion is confirmed by oxidizing iodide to iodine using chlorine water, shaking
this solution with carbondisulphide and observing the violet color of the carbondisulphide layer
due to iodine. Iodine is about 400 times more soluble in carbon disulphide than in water.
Similarly bromide can also be confirmed by observing the orange color of bromine in the
carbondisulphide layer.
Distribution measurements have also been used to determine the equilibrium constants of various
reactions such as complex formation, hydrolysis of ions etc.
a)
Determination of equilibrium constant by carrying out a distribution experiment.
Let us discuss the method by taking potassium iodide- iodine reaction as an example.
KI+I2 KI3
I+I2 I3
As this is an aqueous phase reaction, we choose a solvent immiscible with water and in which one
of the reactants or products of the reaction under study is soluble. Such a solvent for this reaction
is say, carbondisulphide or carbon tetrachloride. If say, carbon tetrachloride is selected, then the
distribution coefficient of iodine, KD between water and carbon tetrachloride is determined first.
4
Cccl
I ,1
2
2O
CH
I ,1
=KD
(Eq
2
10.4.73)
H2O
4
Where Cccl
I 2 ,1 is the concentration of iodine in CCl4 that is in equilibrium with CI2 ,1 , the
concentration of iodine in water.
To determine the equilibrium constant of the complex formation reaction, iodine is distributed
between a solution of KI of concentration CKI and carbon tetrachloride. The equilibrium
4
in the aqueous
concentrations of iodine in the two layers are obtained. Let these be CI2 and Cccl
I2
and CCl4 layers respectively. The equilibrium constant, Keq for the complex formation reaction is
given by:
121
Keq
[KI3 ]
[I 2 ][KI]
=
(Eq. 10.4.74)
CI2, gives the total iodine in the aqueous layer (KI solution), the total of free iodine and the
complexed iodine (as KI3). Hence, we write
CI2, = CI2 ,free +CKI3
(Eq. 10.4.75)
The concentration of free iodine in the aqueous layer can be obtained by using the distribution
coefficient, KD and the concentration of iodine in the CCl4 layer.
4
CCCl
/K D
I
CI2 ,free =
CKI3
(Eq. 10.4.76)
2
=
CI2 - CI2,free
4
CKI3 = CI2 -CCCl
/K D
I
(Eq. 10.4.77)
2
CKI
=
CKI, free+ CKI3
CKI, free =
CKI- CKI3
Substituting from equation 10.4.77, we write
4
/K D ⎤
CKI, free = CKI - ⎡CI2 -CCCl
I2
⎣
(Eq. 10.4.78)
⎦
Substituting from equations 10.4.76, 10.4.77 and 10.4.78 into 10.4.74, we get
Keq
=
4
CI2 -(CCCl
/K D )
I
2
4
4
(CCCl
/K D )[CKI -(CI2 -CCCl
/K D )]
I
I
2
(Eq.10.4.79)
2
Thus experimentally measuring the various concentrations and KD, the value of Keq can be
calculated.
b)
Solvent extraction
When a solute dissolved in a solvent is extracted by using another solvent immiscible with the
original and in which the solute is more soluble, it is important to know the volume of extracting
solvent required and the number of times the extraction is to be performed to achieve a certain
extent of recovery. If the solute has the same molar mass in the extracting solvent as in the
original solvent, it is possible to calculate the mass of solute extracted after every extraction. It
can be shown that it is better to extract with small volumes of solvent several times than once
with the entire volume.
122
We have, say, a solution of volume V1 cm3 containing w gram of a substance and an extracting
solvent of volume nv2cm3. Let the partition coefficient of the substance in favour of the original
solvent by KD. First let us calculate the mass left unextracted in the original solvent after n
extractions taking v2 cm3 of the extracting solvent each time. If the mass of the substance left in
v1cm3 of the original solvent after the first extraction is w1 gram, then the mass of the substance
extracted by v2cm3 of the extracting solvent is (w-w1) gram.
w1/v1
=K D
w-w1/v 2
(Eq. 10.4.80)
⎛ w-w1 ⎞
w1
=K D ⎜
⎟
v1
⎝ v2 ⎠
w1 K D w K D w1
=
−
v1
v2
v2
w1 K D w1 K D w
+
=
v1
v2
v2
⎛ v +K v ⎞
w
w1 ⎜ 2 D 1 ⎟ =K D
v2
⎝ v1v 2 ⎠
⎛ v +K v ⎞
w1 ⎜ 2 D 1 ⎟ =K D w
v1
⎝
⎠
⎛ K D v1 ⎞
w1 =w ⎜
⎟
⎝ K D v1 +v 2 ⎠
(Eq. 10.4.81)
Equation 10.4.81 gives the mass of solute left in the original solvent, w1 gram, after the first
extraction. If w2 gram of substance gets left unextracted in the original solvent after the second
extraction with another v2cm3 of the extracting solvent, then (w1-w2) gram gets extracted.
w 2 /v1
= KD
w1 -w 2 /v 2
Rearranging the above equation, we get
⎛
K D v1 ⎞
⎟
K
⎝ D v1 +v 2 ⎠
w2 = w1 ⎜
Substituting for w1 from equation 10.4.81, we write
⎛ K D v1 ⎞
w 2 =w ⎜
⎟
⎝ K D v1 +v 2 ⎠
2
(Eq.10.4.82)
123
If after n extractions, each time with v2cm3 of the extracting solvent, wn is the mass of the
substance left in the volume v1 cm3 of the original solvent, we have
⎛ K D v1 ⎞
w n =w ⎜
⎟
⎝ K D v1 +v 2 ⎠
n
(Eq. 10.4.83)
If instead of extracting n times with volume v2 cm3 each time, we take nv2cm3 in one lot and
carry out the extraction, then let w' gram be the mass of substance left in the original solvent and
w-w' be the mass that has got extracted.
⎛ w'/v1 ⎞
⎜
⎟ =K D
⎝ w-w'/nv 2 ⎠
Rearranging the above equation, we write
⎛ K D v1 ⎞
w'=w ⎜
⎟
⎝ K D v1 +nv 2 ⎠
(Eq. 10.4.84)
Let us compare wn and w'.
⎛
⎜
⎛ K D v1 ⎞
1
w n =w ⎜
⎟ =w ⎜
⎜ 1+ v 2
⎝ K D v1 +v 2 ⎠
⎜ K v
⎝
D 1
n
⎞
⎟
⎟
⎟
⎟
⎠
n
⎛
⎞
⎜
⎟
w
⎟
wn = ⎜
⎜ 1+ nv 2 +...... ⎟
⎜ K v
⎟
⎝
⎠
D 1
(Eq.10.4.85)
⎛
⎜
1
w'=w ⎜
⎜ 1+ nv 2
⎜ K v
D 1
⎝
(Eq.10.4.86)
⎞
⎟
⎟
⎟
⎟
⎠
From equations 10.4.85 and 10.4.86, we can say that w'>wn or wn<w', the mass remaining
unextracted is less if the extraction is done many times taking a small volume of the extracting
solvent each time. In other words, the mass extracted is more in the case of multiple extraction
than in a single extraction.
A complete extraction of the solute can never be achieved however large the number of
extractions be as some solute will always be left behind in the original solvent in accordance with
the distribution law.
124
Suggested readings
1.
2.
3.
4.
5.
6.
Elements of physical chemistry (Oxford University
Press) 4th Edition
Physical chemistry
A textbook of physical chemistry (Vol.3)
The phase rule
Physical chemistry
Physical chemistry
7.
Physical chemistry
Peter Atkins and Julio de
Paula
G.W.Castellan
K.L.Kapoor
Alexander Findlay
G.K. Vemulapalli
Robet J. Silbey
Robert A. Alberty
Ira N. Levine