Fizicka Hemija - Fazna Ravnoteza

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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)


This system has two phases, one solid, NH4Cl and one gaseous, a mixture of NH3 and HCl.

d) A solution of NaCl in water


e) A system consisting of monoclinic sulphur, rhombic sulphur and liquid sulphur


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.

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lH Oµ (P, T, Cl) =

2

νH Oµ (P, T, Cv)

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:

µ2

iH O (T, P, Ci) =

2

lH Oµ (T, P, Cl)

2

lH Oµ (T, P, Cl) =

2

vH Oµ (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, 2H OC , Csugar) = 0

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 1

1C and in phase 2 be 21C . Let the

concentration of component 2 in phase 1 be 12C and in phase 2 be 2

2C . The phase equations are:

for phase 1: f1 (T, P, 1 11 2C , C )=0 and

for phase 2: f2 (T, P, 1 22 2C , C )=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 1

1µ and 21µ , that of component 2 be 1

2µ and 22µ . The chemical potential equations are 1

1µ = 21µ

and 12µ = 2

2µ . Total number of variables are 6; P, T, 1 2 11 1 2C , C , C and 2

2C . 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

µ = µ = µ =1 2 31 1 1 ……………=µp1

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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µ = µ = µ =1 2 31 1 1 ……………=µp1

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. 1

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

XNH3 = XHCl

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+ + B-

The additional restricting equation is

AX + = 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:

m,Subl

m,v m,s

HdpdT T(V V )

∆⎛ ⎞ =⎜ ⎟ −⎝ ⎠

The variation of sublimation pressure with temperature is given by the Clausius-Clapeyron equation as:

lnp = – m,SublHI

RT∆

+

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:

m,Vap

m,v m,ll v

HdpdT T(V V )

∆⎛ ⎞ =⎜ ⎟ −⎝ ⎠

The Clausius – Clapeyron equation:

lnp = m,vapHI

RT−∆

+

(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:

m,fus

m,l m,ss l

HdpdT T(V V )

∆⎛ ⎞ =⎜ ⎟ −⎝ ⎠

As m,l m,sV -V 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 liquid-vapour curve, OA. This can be shown by using the Clapeyron equation as follows:

m,sub1

m,v m,ss v

HdpdT T(V V )

∆⎛ ⎞ =⎜ ⎟ −⎝ ⎠

m,vap

m,v m,ll v

HdpdT T(V V )

∆⎛ ⎞ =⎜ ⎟ −⎝ ⎠

Vm,v – Vm,s Vm,v – Vm,l

∆ = ∆ + ∆m,sub1 m,fus m,vapH H H

since m,subl m,vap∆H > ∆H

s v l v

dp dpdT dT

⎛ ⎞ ⎛ ⎞>⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

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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

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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:

m,fus

m,l m,ss l

HdpdT T(V V )

∆⎛ ⎞ =⎜ ⎟ −⎝ ⎠

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 Solid Liquid vapour P=3,F=0

8) Point A (31.1oC, 73 atm)

Critical temperature, critical pressure

22

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).

vapour

liquid solid

p

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.

α β

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.

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:

m,subl

m,v m,ss v

HdpdT T(V V )

∆⎛ ⎞ =⎜ ⎟ −⎝ ⎠

Where the variables refer to the α and β forms respectively. At the triple point, we can write

B

CE

D

A

T

p

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.

Figure 10.2.4.2: Vapour pressure temperature curve of a monotropic system

A

BC

D

EF

T

p

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

If a system on thesulphur will sublimthen the solid phasame system is suvapour pressure adisappear and the above and below th

As the heating ofconstant volume, tbecomes equal to the monoclinic forvapour. This is an remain unchangedheating at constanmonoclinic sulphu

A

B

C

D

E

F

liquid

Figure 10.2.

curve AB is allowe to keep the vapo

se will disappear abjected to an isotht a constant valuesystem will be all e curve AB compr

the system, rhomhe vapour pressurethat of monoclinicm and the systeminvariant system (P as long as the tht volume is continr. The system the

4.3: The phase

ed to expandur pressure atnd the systemermal compre. On the con

solid rhombicise of rhombi

bic sulphur of rhombic s sulphur. The at the point =3 and F=0) a

ree phases coued till the en has only 2

diagram for sulphu

keeping temperatu a constant value. If will be all vapourssion, then vapour tinuation of the pr

sulphur. Hence, wec sulphur and vapour

in equilibrium withulphur increases alon the rhombic formB has three phases, nd the temperature a

exist. These variablntire rhombic sulphuphases, monoclinic a

T

r

re constant, more of solid this process in continued, . If on the other hand the will condense to keep the ocess, vapour phase will conclude that the phases sulphur respectively.

vapour, is continued at ng AB and at the point B undergoes transition into rhombic, monoclinic and nd pressure of the system

es remain constant as the r gets converted into the nd 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.

m,rhom,subl

m,v m,rhomrhom v

∆HT(V -V )

⎛ ⎞ =⎜ ⎟⎝ ⎠

dpdT

m,mono,subl

m,v m,monomono v

HdpdT T(V V )

∆⎛ ⎞ =⎜ ⎟ −⎝ ⎠

m,r hom,subl m,rhom mono m,mono,sublH H H→∆ = ∆ + ∆

m,rhom,subl m,mono,subl∆H >∆H

r hom v mono v

dp dpdT dT

⎛ ⎞ ⎛ ⎞>⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

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 co-exist, 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 r v P=2 F=1

BC

sublimation curve of monoclinic sulphur m v P=2 F=1

CD vaporization curve of liquid sulphur l v P=2 F=1

BE Transition line of rhombic to monoclinic r m P=2 F=1

CE Fusion line of monoclinic sulphur m l P=2 F=1

29

EF Fusion line of rhombic sulphur r l P=2 F=1

BG Metastable sublimation of rhombic S

r v P=2 F=1

CG Metastable vaporization curve of liquid S

l v P=2 F=1

GE Metastable fusion line of rhombic to liquid

r l P=2 F=1

B Triple point (95.5oC, 0.01mmHg) r m v P=3 F=0

C Triple point (119.2oC, 0.025mmHg) m l v P=3 F=0

E Triple point (151oC, 1290atm) r m l P=3 F=0

G Metastable triple point (114.5oC, 0.03mmHg)

r l v P=3 F=0

Area to the left of ABF

Rhombic sulphur P=1 F=2

Area above CD and right of CEF

Liquid sulphur 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

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.

a

c d

e

liquid cooling

liquid

solid cooling

Tem

pera

ture

→

34

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).

liquid cooling

solid starts separating

solid mixture cooling

second solid starts separating

Time →

Tem

pera

ture→

b

c d

e

a

35

Fig 10.3.3.3: Cooling curve of a liquid mixture of two components in which supercooling has occurred

a

liquid cooling

solid starts separating

solid mixture cooling

second solid starts separating

Time →

Tem

pera

ture

b b'

f

c d

e

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

37

Figure 10.3.4: The phase diagram for bismuth-cadmium system

Tem

pera

ture

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


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


Area CEBC Solid cadmium in equilibrium with liquid having composition given by the curve BC


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

System

A m.pt./oC B m.pt./oC

Eutectic tempera-ture/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

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

liquid

liquid + solid Ag

solid silver + eutectic

Composition → Pure Pb Pure Ag

Tem

pera

ture

→

A

B

C

D E

G H I 2

1

a

a'

a''

43

B (303oC, 2.5 mass % Ag)

Eutectic point C=2, P=3, F=0 Fixed T and composition

AB Crystallization of lead begins


BC Crystallization of silver begins


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


Area CEBC Solid silver in equilibrium with liquid having composition given by the curve BC


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

Figure 10.3.6: T phase diagram for magnesium – zinc system

The melting points of pure zinrespectively. AB is the freezing pwith liquid containing zinc and mSimilarly, ED is the freezing poiliquid containing magnesium and

If a liquid having the compositioAt the point C, a solid compounThis temperature, the congruent liquid phase freezes. On further cpattern of the liquid of compositcompound formed is Mg (Zn)2 w

CB and CD give the freezing popoint of Mg(Zn)2 and the tempera(separate) from various liquids,

Zn+Melt Mg(Zn)2+Melt

Mg+Mg(Zn)2 Zn+Mg(Zn)2

Tem

pera

ture

→ Mg

+ Melt

Pure Zn Pure Mg

A

B

C

D

E

F G

H I

J K Composition →

'

liquid

he C

c and pure magnesium are represented by points A and E oint curve of zinc. In the area ABFA, solid zinc is in equilibrium agnesium, the composition of which is given by the curve AB.

nt curve of magnesium. Solid magnesium is in equilibrium with zinc, composition of the liquid phase lying on the curve DE.

n c' is cooled, the liquid merely cools till it reaches the point C. d, having the same composition as the liquid starts separating. melting point of the compound, remains constant till the entire ooling, the temperature of the solid decreases. Thus, the cooling

ion c' is similar to that of a pure component. The formula of the hich corresponds to the composition c'.

int curves of Mg (Zn)2. Addition of zinc depresses the freezing tures at which the solid compound Mg (Zn)2 will begin to freeze

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)


D Eutectic point (Mg, Mg(Zn)2, liq of composition D)


AB Freezing point curve of Zn, crystallization of Zn begins


BC and CD Crystallization of Mg(Zn)2 begins


47

ED Crystallization of Mg begins


Area ABFA Zn liquid (composition given by AB)


Area CBGC Mg(Zn)2 liquid (composition given by CB)


Area CHDC Mg(Zn)2 liquid (composition given by CD)


Area EDIE Mg liquid (composition given by ED)


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


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

A m.pt./oC B m.pt./oC

Compound 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 NaCl-H2O system is given in Figure 10.3.7.

50

Figure 10.3.7: The phase diagram for sodium chloride – water system

A C

B

C'

D

E F

G

H I

J

Unsaturated solution

a

a'

ice + solution

a''

a''' b'

b

c'''

c''' d'''

d''

d'

d

G'

e''

e'

e

NaCl + solution

NaCl +

NaCl.2H2O

NaCl.2H2O +

solution

ice + NaCl.2H2O

Mass % NaCl

Tem

pera

ture

O 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 solution (composition given by AB)

Area to the right of DCGJ NaCl solution (composition given by CD)

Area of BCGFB NaCl.2H2O solution (composition given by BC)

Area below GJ NaCl and NaCl.2H2O

Area above ABD Unsaturated solution

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

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.

Fe2C

l 6.12

H2O

Fe2C

l 6.7H

2O

Fe2C

l 6.5H

2O

Fe2C

l 6.4H

2O

Tem

pera

ture

→

solution

a b c d e f g h i j k

D' D D''

F' F F''

H' H H''

B' B B''

A

C

E

G

I

K

J'' J J'

C' E' G' I'Water Mole %Fe2Cl6 → Fe2Cl6

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 dodecahydrate solution B(-55oC)

D Eutectic point: dodecahydrate heptahydrate solution D(27.4oC)

F Eutectic point: heptahydrate pentahydrate solution F(30oC)

H Eutectic point: pentahydrate tetrahydrate solution H(55oC)

J Eutectic point: tetrahydrate anhydrous salt solution J(66oC)

CC' Solid dodecahydrate

EE' Solid heptahydrate

GG' Solid pentahydrate

II' Solid tetrahydrate

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 Solution (composition given by AB)

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)

59

JIJ'J Tetrahydrate Solution (given by JI)

Right of J''JK Anyhydrous salt 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

ABCDEFGHIJK

Liquid (solution of FeCl3 in water)

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, volume of the 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.

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

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

i

liquid

liquid+CuSO4.5H2O

Vap + Liquid

b sat soln+CuSO4.5H2O

c

d e

f

g

h

Vap+CuSO4.5H2O

Vap+CuSO4.3H2O

Vap+CuSO4.H2O

Vap+CuSO4

7.85

4.32

0.017

CuS

O4.5

H2O

+CuS

O4.3

H2O

CuS

O4.3

H2O

+CuS

O4.H

2O

CuS

O4.H

2O+C

uSO

4.

Pres

sure

/mm

Hg

a

H2O Mass% CuSO4 CuSO4

63

water and water vapour. The variance of the system along bc is zero (invariant system), F=C+1-P=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

0pK =

2

0 2eq,H O(p /p )

Where 2eq,H Op is the equilibrium vapour pressure (the dissociation pressure) of water over the

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:

0 0p

2

dlnK ∆rH=dT RT

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 0pK 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=xi0ip

to a good extent, where pi is the partial pressure of the constituent i having a mole fraction xi and 0

ip 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 = xi0ip (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

= iΣ niµi (sol) -

iΣ niµi(l)

= iΣ ni(µi(sol) - 0

i(l)µ )

= iΣ niRTlnxi

= ntotalRT iΣ xilnxi (Eq.10.4.3)

Where ntotal is the total amount of all the constituents in the solution.

(b) The change in entropy on mixing, ∆Smix

mixmix

(∆G )∆S

Tip,n s

∂⎛ ⎞ = −⎜ ⎟∂⎝ ⎠

mix∆S = -i

total

p,n s

(n RT ln x

T

i ix∂ Σ⎛ ⎞⎜ ⎟⎜ ⎟∂⎝ ⎠

66

= - ntotal R iΣ xilnxi (Eq.10.4.4)

(c) The change in enthalpy on mixing, ∆Hmix

∆Gmix = ∆Hmix – T∆Smix

∆Hmix = ntotal RTiΣ xilnxi-Tntotal R

iΣ xilnxi

∆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 = i

mix

T,n s

(∆G )p

⎛ ⎞∂⎜ ⎟∂⎝ ⎠

∆Vmix = i

i ii

T,n s

( RTx lnx )

p

totalnΣ∂⎛ ⎞⎜ ⎟⎜ ⎟∂⎝ ⎠

∆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+P-2=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 = ovvp

(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= o mv /vvp

= α

mv

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 = vV

(Eq.10.4.8)

Relation between α and β

If we assume that the gas behaves ideally, we can write

o(1atm)vpv =nRT nR(273K)

ov Tv=

(273K)(p/atm)

β = ov T/(273K)(p/atm)v =V V

vo Tβ=vp (273K)

αTβ=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:

2dlnC ∆H=dT RT

(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

m α p

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

mpv=nRT= RTM

where M is the molar mass of the gas.

m RTv=p M

⎛ ⎞⎜ ⎟⎝ ⎠

RTv=kM

(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 (Eq. 10.4.13)

Dividing both sides by m1, we have

22

1 1

m k= ×pm m

=k'p2

Dividing the two masses by their respective molar masses, we get

70

2 2 12

1 1 2

m /M M= k' ×p

m /M M⎛ ⎞⎜ ⎟⎝ ⎠

=k"p2

22

1

n=k''p

n (Eq.10.4.14)

The mole fraction of the gas in the solution is x2, which is given by x2 = 2

1 2

nn +n

For a dilute solution of the gas in the liquid,

n1 2n⟩⟩

x2 ≅ 2

1

nn

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 = 21 xk''

= Khx2 (Eq. 10.4.16)

Where Kh = 1k''

and is known as Henry’s law constant.

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'' = x2p2

= 01

p2

where o2p is the vapor pressure of the pure liquid solute (x2=1). Thus, the equation 10.4.15 can be

written as

71

x2 = 0p2 or p = x p2 2 2p02

(Eq

10.4.17)

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 p d log p1 2=d log x d log x1 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 p2d log x2

= 1 (Eq. 10.4.19)

Substituting from Eq.10.4.19 into Eq.10.4.18, we get

d log p1d log x1

= 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= 01p , where 0

1p is the vapor pressure of the pure solvent. Thus K= 01p

Equation 10.4.21 becomes

p1=x101p (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.

F

If fraplRamthis linwoxB

Pres

sure

Kh(B)

0Bp

0Ap

xB=0 xB → xB=1 xA=1 ←xA xA=0

igure 10.4.1: Graphs showing applicability of Henry’s law to thsolvent (a) for liquid B (b) for liquid

we take an example, say a solution of liquid A in liquid B and ction curves, we can understand the above discussion bette

otted against its mole fraction in solution to give the curveoult’s law line is obtained by joining the points p=0 for xB=

eets the partial vapour pressure curve tangentially in the regionat Raoult’s law is applicable only in this region, xB → 1, amounthe solvent. A line drawn tangent to the curve at xB=0 (low vale. Extrapolation of this line to meet the xB=1 axis gives the hypuld be observed if Henry’s law were applicable over the en

=1. It follows from equation 10.4.16 at xB=1.

pB =Kh

H2O Mass%CuSO4

(a)

xB=0 xB → xB=1 xA=1 ←xA xA=0

e solute and Raoult’s law to the A.

plot partial pressure versus mole r. The partial pressure of B is shown in Figure 10.4.1. The 0 and p= o

Bp for xB=1. This line where xB→1, implying thereby t of B much more than A, that B ues of xB) gives the Henry’s law othetical vapour pressure which tire range of composition, upto

CuSO4

(b)

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 0

Bp . 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 = x101p (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

µ1(l) = 01(v)µ +RTlnp1

µ2(l) = 02(v)µ +RTlnp2

Substituting for p1 and p2 from equations 10.4.23 and 10.4.24 into the above equations, we have

µ1(l) = 0 01(v) 1 1(µ + RTlnp ) + RTlnx

µ2(l) = 02(v) h 2(µ + RTlnK ) + RTlnx

74

µ1(l) = 01(1) 1µ + RTlnx (Eq. 10.4.26)

µ2(l) = 2(hs) 2µ +RTlnx (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 0Ap (Eq. 10.4.28)

pB = xB 0Bp (Eq. 10.4.29)

where xA and xB are the mole fractions of A and B, the two constituents in the solution, 0Ap and

0Bp are their vapor pressures in the pure state. The total pressure over the solution is the sum of

the partial pressures.

p = pA + pB

= xA 0Ap + xB 0

Bp

In a binary solution xA+xB=1 and hence, we write xB=1-xA

p=xA 0Ap + (1-xA) 0

Bp

= xA 0Ap + 0

Bp — xA 0Bp

= 0Bp +xA ( 0

Ap — 0Bp ) (Eq. 10.4.30)

The total vapor pressure over the solution is thus a linear function of xA with intercept equal to 0Bp and slope equal to 0

Ap — 0Bp . The slope can be positive or negative depending upon the relative

magnitudes of 0Ap and 0

Bp .

75

Case 1: If 0Ap > 0

Bp , that is liquid A is more volatile than liquid B, 0Ap — 0

Bp 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 0Ap < 0

Bp , that is liquid B is more volatile than liquid A, 0Ap - 0

Bp 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 0Ap > 0

Bp are shown in the figure 10.4.2.

p=p +pA

B

pB

pA

Figure 10.4.2: Plots of pA, pB and p versus mole fraction in the liquid state, temperature remaining constant.

xA=0 xB=1

xA→ ←xB

xA=1 XB=0

0Bp

0Ap

Pres

sure

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

yA = App

and yB = Bpp

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

yA= 0

0 0 0

x pA Ap + x (p - p )B BA A

(Eq. 10.4.31)

yB =0

0 0 0

x pB Bp + x (p - p )B BA A

writing xA=1-xB and simplifying, we get

yB =0

0 0 0

x pB Bp + x (p - p )B BA A

(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 0 0 0 0B A A A B A(p +x p -x p )/p

On rearranging, this yields

xA = 0

A B0 0 0A A B A

y pp +y (p -p )

(Eq. 10.4.33)

Substituting for xA from equation 10.4.33 into eq. 10.4.30, we get

p=0

0 0 0 A BB A B 0 0 0

A A B A

y pp +(p -p )

p +y (p -p )

p = 0 0B A

0 0 0A B A A

p pp +(p -p )y

(Eq.10.4.34)

Figure 10.4.3 shows a plot of p vs yA for a solution having o oA Bp p .

77

p

oBp

oAp

Figure 10.4.3: A plot of total vapour pressure, p against yAphase. Temperature is kept co

The reciprocal of equation 10.4.34 gives a linear eq

Ao o oB A B

1 1 1 1= +( - )yp p p p

yA=0 yA→ ←yB

yB=1

, the mole fraction of A in the vapor nstant.

uation relating A1 and yp

(Eq. 10.4.35)

yA=1 yB=0

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.

A B0 0A B

y y1 = +p p p

(Eq.10.4.36)

Fig

0B1/p

1/p

ure 10.4.4: A plot of 1/p verus yA, mole fraction of the constituent A iTemperature is kept constant.

yA=0

yB=1

yA→

←yB

0A/p
1
n the vapour phase.

yA=1

yB=0

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.

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=4-2=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

l v

0Ap

0Bp

pres

sure

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).

liquid

vapour

p l a v

pres

sure

0Bp

0Ap

Pure B xA XA yA Pure A Composition

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 = A(l) A(v)

A(l) A(v) B(l) B(v)

n +nn +n +n +n

Eq.

10.4.37

al = xA-xA=xA - A(l)

A(l) B(l)

nn +n

(Eq.10.4.38)

av = yA(v)-xA = A(v)

A(v) B(v)

nn +n

- 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)) (Eq.10.4.41)

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

82

A(ν) B(ν)

A(l) B(l)

n +nal =av n +n

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 = 0

A A0 0 0B A A B

x pp +x (p -p )

(Eq.10.4.31)

yA =0

A A0 0 0B A A A B

x pp +x p -x p

yA =0

A A0 0

A A B A

x px p +p (1-x )

yA = 0

A A0 0

A A B B

x px p +x p

(as xB = 1-xA)

Ao o

A A B B A

y 1=x x +x (p /p )

(Eq.10.4.43)

If constituent A is more volatile than B, o oA Bp p , then 0

Bp / 0Ap <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:

Amount in the liquid phaseAmount in the vapor phase

= a'ν'a'l'

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, o oA Bp p , 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 O

Ap and OBp 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

0Ap /atm = exp A,m,Vap

0A

∆H 1 1-R TT

⎧ ⎫⎛ ⎞⎪ ⎪⎨ ⎬⎜ ⎟⎪ ⎪⎝ ⎠⎩ ⎭

(Eq. 10.4.44)

0Bp /atm = exp B,m,Vap

0B

∆H 1 1-R TT

⎧ ⎫⎛ ⎞⎪ ⎪⎨ ⎬⎜ ⎟⎪ ⎪⎝ ⎠⎩ ⎭

(Eq. 10.4.45)

Where 0AT and 0

BT are the normal boiling points of constituents A and B, respectively. T is the normal boiling point of a solution of known composition and 0

Ap /atm and 0Bp /atm be the vapor

pressures of pure constituents, A and B at the temperature, T.

The total vapor pressure of the solution having a known composition given by xA and xB, is

85

p=xA 0Ap +xB

0Bp

By definition, the total pressure p is equal to 1 atm (the external pressure).

1 = xA0Ap +xB

0Bp = xA exp A,m,Vap

0A

∆H 1 1-R TT

⎧ ⎫⎛ ⎞⎪ ⎪⎨ ⎬⎜ ⎟⎪ ⎪⎝ ⎠⎩ ⎭

+ xB exp B,m,Vap0B

∆H 1 1-R TT

⎧ ⎫⎛ ⎞⎪ ⎪⎨ ⎬⎜ ⎟⎪ ⎪⎝ ⎠⎩ ⎭

1=xA exp 0

A,m,vap A∆S T

1-R T

⎧ ⎫⎛ ⎞⎪ ⎪⎨ ⎬⎜ ⎟⎪ ⎪⎝ ⎠⎩ ⎭

+ xB exp 0

B,m,vap B∆S T

1-R 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 0

AT , 0BT 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.

0 0m,vapA A A

A A0 0A A B B

∆Sx p Ty = =x exp 1-

R Tx p +x p⎧ ⎫⎛ ⎞⎪ ⎪⎨ ⎬⎜ ⎟⎪ ⎪⎝ ⎠⎩ ⎭

and 0 0

m,vapB B BB B0 0

A A B B

∆Sx p Ty = =x exp 1-

R Tx p +x p⎧ ⎫⎛ ⎞⎪ ⎪⎨ ⎬⎜ ⎟⎪ ⎪⎝ ⎠⎩ ⎭

(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.

Figure 10

This figure resemregion is tilted in10.4.47 that the fig 10.4.5, therefthe opposite direat lower temperliquidus curve. Sthe upper curve i

On heating, undeof the liquid incappears, composA as compared separation of volpoint a', the liqucontinuously aloby the points l' a

Amount in liquid Amount in vapor

Further increase composition l'' iincrease in tempon heating as it r

X

86

.4.8: Boiling point versus composition curve, constant pressure conditions, corresponding to the figure 10.4.5.

bles figure 10.4.5 except that neither curve is a straight line and the liquid-vapor the opposite direction. It is clear from the nature of the equations 10.4.46 and

graphs cannot be straight lines. As constituent A has the higher vapor pressure in ore it has the lower boiling point which explains the liquid-vapor region tilt in ction in the figure 10.4.8. As the liquid phase under a constant pressure is stable atures, the region beneath the lower curve is liquid and the lower curve is the imilarly, it can be concluded that the region above the upper curve is vapor and s the vaporous curve.

r constant pressure, a liquid mixture represented by the point a, the temperature reases until point l is reached. At this temperature T, the first trace of vapor ition of which corresponds to the point v. The vapor, thus is richer in component to the liquid, A being the lower boiling component. This is the basis for the atile mixtures by distillation. As heating is continued, the system moves towards id composition changes continuously along ll' and the vapor composition varies ng vv'. At the point a', the composition of the liquid and vapor phases are given nd v' respectively and the relative amounts of the two phases by the lever rule as

phase a'v'=phase a'l'

in temperature takes the system to the point v'' where the last trace of liquid of s left. The vapor phase in equilibrium has the composition X. With a slight erature, the liquid phase disappears and the system, which is only vapor expands eaches 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

Figu

vapour OBT

Pure B

Tem

pera

ture

re 10.4.9: Distillation of an ideal binary solution at constant pressure

liquid a b c

l'

l''

v'

v''

OAT

Composition 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

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

vapour

0BT

0AT

T1

T3

T2

T0 l0

l1

l3

l2

a' a

b'

b

ν0

ν1

ν2

liquid

Tem

pera

ture

Pure B Composition Pure A

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

A BA B

A B A B

n -ndµ = dµ

n +n n +n

xAdµA = -xBdµB

where xA and xB are mole fractions of A and B, respectively.

Dividing by dxA, we get

A BA B

A A

dµ dµx =-x

dx dx

xA+xB = 1

dxA+dxB = 0

dxA=-dxB

A BA B

A B

dµ dµx =x

dx dx (Eq.

10.4.48)

The chemical potential of any constituent of a liquid mixture, assuming that vapor behaves ideally, is given by

µi(sol)= 0iµ +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

A A

A A

dµ dlnp=RT

dx dx (Eq.10.4.50)

Similarly, for the component B, differentiating with respect to xB gives

94

A B

B B

dµ dlnp=RT

dx dx (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 A

A

dlnpdx

= xB B

B

dlnpdx

A

A

dlnpdlnx

= B

B

dlnpdlnx

(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 = 0Ap xA

lnpA= ln 0Ap +ln xA

dlnpA = d lnxA

A

A

dlnpdlnx

= 1

Using Duhem – Margules equation, we write

B

B

dlnpdlnx

= 1

dlnpB=dlnxB

Integrating this equation, we get

lnpB=lnxB+I, where I is the constant of integration. When xB=1, pB= 0Bp and

95

I=ln 0Bp

lnpB=lnxB+ln 0Bp

lnpB = ln (xB 0Bp )

pB = xB 0Bp

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) = oAp xA

lnpA(real)>lnpA(ideal) = ln oAp + lnxA

A(real) A(ideal)

A A A

dlnp dlnp 1=dx dx x

A(real) A(ideal)A A

A A

dlnp dlnpx x =1

dx dx

A(real)

A

dlnp1

dlnx

According to Duhem-Margules equation, we have

B(real)

B

dlnp1

dlnx

On integration, we get

pB(real)>pB(Ideal) = oB Bp x

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

A B

A B

dlnp dlnp=

dlnx dlnx

(Eq.10.4.52)

A A B B

A A B B

x dp x dpp dx p dx

= (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

A A B B

A A B A

x dp x dp+ = 0

p dx p dx (Eq. 10.4.53)

p = pA+pB

A B

A A A

dp dpdp = +dx dx dx

Substituting for dpA/dxA from equation 10.4.53 into the above equation, we write

B B A B

A B A A A

x dp p dpdp =- +dx p dx x dx

B B A

A A B A

dp x pdp = 1-dx dx p x

⎡ ⎤⎢ ⎥⎣ ⎦

(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.

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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.

Nin

0BP

0AP

Figure 10.4.12: Pressure –deviation

on ideal solutions that show very their vapour pressure – compos

composition curves of a non ideal solution exhibiting positive from ideality (constant temperature)

large deviation from ideality exhibit either maxima or minima ition curves (Figures 10.4.12 and 10.4.13). As boiling point is

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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.

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.

Henry’s law limiting line

Raoult’s law limiting line Pr

essu

re p

xB=1 xA=0

xB=0 xA=1

xA

xB

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

A A

B B

p x=

p x

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.

Figure 10.4.14: Temperature – composition diagram for a system ecurve (pressure constant)

0BT

0AT

vapour

liquid

T x1

y1

y2

M

B

Boi

ling

poin

t

x2

A

xA xB

xhibiting a minimum in the

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.

Figure 10.4.15: Temperature – composition diagram for a systemcurve (pressure constant)

Fractional distillation of non ideal solution

As binary liquid solutions that deviate only slightly from ideal becomposition diagrams very similar to that of the ideal solution, thecan be separated by fractional distillation. Nonideal solutions extheir temperature composition diagrams yield only one of the cofractional distillation.

a) Solutions exhibiting a minimum in the boiling point:

If we take a liquid mixture of composition x1 (figure 10.4.14) andtemperature T and the vapor formed will have a composition y1constituent A while the liquid left, more of B. As the distillation coliquid phase moves along x1B, i.e., towards pure B, and that otowards the point M. At the point M, the vapor phase and thecomposition. Thus, condensing the vapor and distilling the resufurther separation of the constituents. The liquid at this point

vapour

liquid

y1 x1

x2

A

y2

B

M

0BT

Boi

ling

poin

t

Pure B xA xB

0AT

exhibiting a maximum in the

haviour exhibit temperature – constituents of these solutions hibiting minima or maxima in nstituents in the pure form on

distill it, it will start boiling at . The vapor thus has more of ntinues, the composition of the f the vapor phase along y1M, liquid phase have the same

ltant liquid will not yield any M would boil at a constant

Pure A

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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 Components Azeotrope*

A B Mass% A b.pt/0C

Water (100oC) Ethanol(78.3oC) 4.0 78.2 Minimum

Water (100oC) Ethylacetate(79.6oC) 11.3 73.4

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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

Maximum

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, methanol-carbondisulphide, 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

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liquids are completely miscible beyond this range. Thus, we may say that there exists a miscibility gap in the system.

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

Tem

pera

ture

A B

b' b d d'

c' c''

c

a

b'' d''

Pure A Pure BComposition

104

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

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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

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Methanol-carbondisulphide 49.5 20 mass% methanol

Triethylamine-water system

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 temperature-composition 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.

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.

Tem

pera

ture

Pure A Pure BComposition

107

Table 10.5.2 A few examples of systems showing lower CST System CST/oC Triethylamine-water 18.5 γ-Collidine-water 6 β-picoline-water 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

Figure 10.5.2.: Temperature – composition diagram for nicotine-water system. A=water, B=nicotine, constant external pressure >> 1 atm

Pure A Composition Pure B

C Upper CST

C' Lower CST

A

B

Tem

pera

ture

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.

110

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-


Tem

pera

ture

111

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.

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).


Tem

pera

ture

112

Effect of pressure on CST

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.

Figure 10.5.5: Effect of pressure on CST. Dotted curve at low pressure, solid curve at high pressure.


Tem

pera

ture

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.

0 0A Bp=p +p

Where p is the total vapor pressure and 0Ap and 0

Bp 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.

0A A A A A0

B B B BB

p y n w /M= = =

y n w /Mp (Eq.10.4.55)

0A A A

0B B B

w M p=

w M p (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.

1D

2

C=K

C (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,

2

2

4

2

H OIcclI

C=

C 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.

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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

lnA

B

xx

= 1RT

(µ0B-µ0A) (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

lnA

B

xx

= constant

A

B

xx

= Constant

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

A

B

CC

= Constant = KD (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 waterconcentration of monomer in chloroform

= 0.442 at 40oC

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:

nQ Qn (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 a

[Qn] = ξ a

The equilibrium constant, Keq for the association reaction is given by :

Keq = n

n[Q ][Q]

(Eq. 10.4.59)

Keq = a

B na

ξ(C -nξ )

(Eq.10.4.60)

(CB-nξ a)n = a

eq

ξK

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CB-nξ a =

1n

a

eq

ξK

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

If the solute in liquid B is present almost in the associated form, then (Qn) = ξ a=CB/n

CB-n ξ a =

1B nC

nKeq⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

(Eq.10.4.61)

KD = A A

Ba

C C=[Q] C -nξ

(Eq.10.4.62)

Substituting from equation 10.4.61 into the expression for KD, we have

KD = A

1B n

C

(C )

(n Keq)1n (Eq.10.4.63)

A

1B n

C

(C )

= D1n

eq

K

(nK )

= K'D (Eq.10.4.64)

At a given temperature, Keq has a constant value. For a given association reaction, n is constant.

Hence KD /(nKeq)1n 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=B B2 1A A2 1

logC -logClogC -logC

(Eq.10.4.65)

where A1C and B

1C are the concentrations of the solute in liquid A and liquid B respectively obtained from the first set, and A

2C and B2C 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 = A

neq

D

C +nK [Q]K

118

n-1A A

eqD D

C C+nKK K

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

CB = A

eq A n

D

nKC + (C )K Kn

D

Beq A n-1

A nD D

nKC 1= + (C )KC K

(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/Kn

D 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 = 2

B[E][F] ξ=

[Q] C -ξ

[Q] = CB—ξ = 2

diss

ξK

The distribution law expression for the system is :

KD = A

BC

C -ξwhere CA is the concentration of Q in the liquid A

KD = A

2diss

Cξ /K

(Eq. 10.4.67)

(c) solute exists as dimers in liquid B and dissociates in liquid A.

119

KD = A

dB

a

c -ξc -2ξ

(Eq.10.4.68)

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 = n[Q.nB][Q][B]

(Eq. 10.4.69)

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}

[Q] = B

neq

C1+K [B]

(Eq. 10.4.70)

The distribution coefficient, KD is given by:

KD = AC

[Q] =

A neq

B

C (1+K [B] )C

A

BCC

= Dn

eq

K1+K [B]

(Eq.10.4.71)

Keq is a constant at a given temperature and as liquid B is present in large amount, [B]~ constant. Hence,

A

BCC

= Constant (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

2

2

2

cclI ,1H OI ,1

C

C=KD (Eq

10.4.73)

Where 4

2

cclI ,1C is the concentration of iodine in CCl4 that is in equilibrium with 2

2

H OI ,1C , 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 concentrations of iodine in the two layers are obtained. Let these be

2IC and 4

2

cclIC in the aqueous

and CCl4 layers respectively. The equilibrium constant, Keq for the complex formation reaction is given by:

121

Keq = 3

2

[KI ][I ][KI]

(Eq. 10.4.74)

2,IC gives the total iodine in the aqueous layer (KI solution), the total of free iodine and the

complexed iodine (as KI3). Hence, we write

2,IC = 2 3I ,free KIC +C (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.

2I ,freeC = 4

2

CClDIC /K (Eq. 10.4.76)

3KIC =

2IC -2,freeIC

43 2 2

CClKI I DIC = C -C /K (Eq. 10.4.77)

CKI = CKI, free+ 3KIC

CKI, free = CKI- 3KIC

Substituting from equation 10.4.77, we write

CKI, free = CKI - ⎡ ⎤⎣ ⎦

42 2

CClI DIC -C /K (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

2 2

4 422 2

CClI DI

CCl CClD KI I DI I

C -(C /K )

(C /K )[C -(C -C /K )] (Eq.10.4.79)

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.

1 1D

1 2

w /v=K

w-w /v (Eq. 10.4.80)

1 1D

1 2

w w-w=K

v v⎛ ⎞⎜ ⎟⎝ ⎠

1 D D 1

1 2 2

w K w K w=

v v v−

1 D 1 D

1 2 2

w K w K w+

v v v=

2 D 11 D

1 2 2

v +K v ww =Kv v v

⎛ ⎞⎜ ⎟⎝ ⎠

2 D 11 D

1

v +K vw =K w

v⎛ ⎞⎜ ⎟⎝ ⎠

D 11

D 1 2

K vw =w

K v +v⎛ ⎞⎜ ⎟⎝ ⎠

(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.

2 1

1 2 2

w /vw -w /v

= KD

Rearranging the above equation, we get

w2 = w1 D 1

D 1 2

K vK v +v

⎛ ⎞⎜ ⎟⎝ ⎠

Substituting for w1 from equation 10.4.81, we write

2D 1

2D 1 2

K vw =wK v +v

⎛ ⎞⎜ ⎟⎝ ⎠

(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

nD 1

nD 1 2

K vw =w

K v +v⎛ ⎞⎜ ⎟⎝ ⎠

(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.

1D

2

w'/v=K

w-w'/nv⎛ ⎞⎜ ⎟⎝ ⎠

Rearranging the above equation, we write

D 1

D 1 2

K vw'=w

K v +nv⎛ ⎞⎜ ⎟⎝ ⎠

(Eq. 10.4.84)

Let us compare wn and w'.

n

nD 1

n2D 1 2

D 1

K v 1w =w =wvK v +v 1+

K v

⎛ ⎞⎜ ⎟⎛ ⎞ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎜ ⎟⎝ ⎠

nw =2

D 1

wnv1+ +......

K v

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

(Eq.10.4.85)

2

D 1

1w'=wnv1+

K v

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

(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. Elements of physical chemistry (Oxford University

Press) 4th Edition Peter Atkins and Julio de Paula

2. Physical chemistry G.W.Castellan 3. A textbook of physical chemistry (Vol.3) K.L.Kapoor 4. The phase rule Alexander Findlay 5. Physical chemistry G.K. Vemulapalli 6. Physical chemistry Robet J. Silbey Robert A. Alberty 7. Physical chemistry Ira N. Levine

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