Metallurgical Thermodynamics

D E P A R T M E N T O F

Metallurgical Engineering Institute of Technology Banaras Hindu University

Metallurgical Thermodynamics


MT – 2102 Credit:04

Instructors:

Dr. C. K. Behera and

Mr. J. K. Singh




Marks Distribution

Total Marks 100

Sessional Test - I 15

Sessional Test - II 15

Assignments / Attdn. 10

End Semester 60




Grading System

S 90 -100

A 80-89

B 70-79

C 60-69

D

E

F

50-59

40-49

< 40




Course Content

Basic Principles

Extensive and intensive properties, thermodynamic systems and processes. First Law of Thermodynamics, enthalpy, Hess’ Law, heat capacity, Kirchhoff’s law. Second Law of Thermodynamics, entropy, entropy change in gases, significance of sign change of entropy. Trouton’s and Richard’s rules. Driving force of a chemical reaction, combined statement of first and second laws of thermodynamics, Helmholtz and Gibbs free energies. Ellingham diagram, Equilibrium constants, van’t Hoff’s isotherm, Le Chatelier principle. Clausius-Clapeyron equation. Maxwell’s equations, Third Law of Thermodynamics.

Solution Thermodynamics

Solution, mixture and compound. Raoult’s law: activity, ideal solution, standard state. Partial molar quantities, Gibbs-Duhem equation, chemical potential, fugacity, activity and equilibrium constant. Free energy of mixing, excess and integral quantities. Regular solutions, -function. Dilute solutions: Henry’s and Sievert’s laws. Alternative standard states. Gibbs-Duhem integration

Statistical concept of entropy. Elements of Gibbs Phase Rule and its applications.

Experimental Techniques

Determination of thermodynamic quantities by different techniques, viz. calorimetry, chemical equilibria, vapour pressure and electrochemical: aqueous, fused and solid electrolytes; formation, concentration and displacement cells.




Suggested Reading

1. D.R. Gaskell: Introduction to Metallurgical Thermodynamics, McGraw-Hill.

2. L.S. Darken and R.W. Gurry: Physical Chemistry of Metals, McGraw-Hill.

3. G.S. Upadhyaya and R.K. Dube: Problems in Metallurgical Thermodynamics and

Kinetics, Pergamon.

4. J. Mekowiak: Physical Chemistry for Metallurgists, George Allen & Unwin.

5. J.J. Moore: Chemical Metallurgy, Butterworths.

6. R.H. Parker: An Introduction to Chemical Metallurgy, Pergamon.




Scope, Basic Concepts andDefinitions

Thermodynamics is that branch of science which deals with the study of the

transfer and conversion of energy from one form into other and its

conversion to work.

It deals with only conventional forms of energies like electrical, mechanical,

chemical. etc.

The non-conventional energy like nuclear energy related to atomic and sub-

atomic particle forms has to be dealt separately because in that case all

matter would have to be considered as per the famous Einstein’s equation :

E = mC2 .




Here the subject matter of discussion is chemical and/or metallurgical

thermodynamics alone.

The systems under discussions consisting of large no. of particles i. e

macro systems.

Classification

Thermodynamics may be broadly classified into three :

Classical: it treats a substance as continuum, ignoring behavior of

atoms and molecules. It consists of first, second and third laws of

thermodynamics





Statistical thermodynamics: The application of probability theory,

quantum theory and statistical mechanics allowed it to arrive at

macroscopic thermodynamic relations from atomistic point of view.

Irreversible Thermodynamics: Irreversible thermodynamics deals

with the application of thermodynamics to irreversible processes such

as chemical reactions.

The thermodynamics generally means classical thermodynamics





Chemical thermodynamics is based on the three laws of thermodynamics

systematically applied to various physico-chemical processes in physical

chemistry.

Broadly speaking, the application primarily chemical thermodynamics to

metals and materials lead to the development and growth of Metallurgical

thermodynamics or its later generalization as Thermodynamics of

materials.

Processing of ceramics and metals is carried out primarily at high

temperatures which led to the development of metallurgical

thermodynamics.





It is the ability to do work. This is too mechanical an answer.

The broader definition is : it is the capacity to bring about changes in the

existing materials as per the requirements.

Forms of energies

Mechanical: Kinetic, potential and configurational.

Thermal: Heat exchanged.

Electrical: Electrical energy = current x time x potential difference.

Chemical: Chemical energy = no. of chemical bonds x bond strength

Energy




System and Surrounding

Any portion of the universe selected for consideration is known as the system

or thermodynamic system.

A thermodynamic system must, of necessity be stable with respect to its

chemistry during its study. If the system is undergoing continuously some

chemical change cannot be considered as system

For example a live animate body like tree and human being are not system. All

inanimate aggregates are called systems as they have their fixed chemistry

The rest of the universe excluding system is called surrounding.





Classification of thermodynamic systems

Thermodynamic Systems

In terms of interaction with surrounding

Based on Material Distribution

Based on composition

Isolated

Neither matter nor energy is exchanged with surrounding

Open

Both matter and energy exchange occurs

Closed

Can exchange energy not the matter

Hom

ogen

eous

Het

erog

eneo

us

Single component

Multi component

Hom

ogen

eous

Het

erog

eneo

us

Hom

ogen

eous

Het

erog

eneo

us




Homogeneous and Heterogeneous system

Homogeneous system consists of single phase only.

Heterogeneous consists of more than one phase.




State of a system

As the position of a point in the space is described by its coordinates w.r.t

some prefixed axes, similarly the state of a system is described by some

experimentally determinable parameter which can lead to the complete

reproduction of the system.

These parameters are temperature, pressure and volume.

The minimum number of variable required to describe the state of the

system are called independent state variables.




Incase of a multicomponent system the independent state variables are

i) composition ii) two of the three variables T, P, V.

All other variables whose values get fixed with the specification of

independent state variables are referred to as dependent state

variables.

State variables are also known as state properties or state functions.

State of a system (cont.)




Extensive and Intensive State properties

If a state variable, whether dependent or independent, is a function of the

mass of the system it is known as extensive state property. For example:

Mass, Volume, weight, length, energy etc.

If a state variable is independent of the mass or size of the system it is

called intensive state property. For example: temperature, pressure,

conductivity, density, colour, odor, malleabilty, hardness, specific heat,

molar volume etc.




Product of an intensive and an extensive state variable is also

an extensive state variable.

Ratio of two extensive properties yields an intensive

properties. For example: Density = m / V

The general convention in chemical thermodynamics is to go

for molar properties, which are intensive and become

independent of quantity of matter and hence of more general

applicability.

Extensive and Intensive State properties




Equation of State

The state of the system can be described in the form of some

mathematical equation involving some state variables. The

analytical form as applicable to the system under consideration

is known as equation of state e.g. for an ideal gas PV = nRT.

The above relationship is an expression which correlates the P,

T and V. In fact this is found to be true in case of solids as well

as liquids though exact form of this relationship may not be

known.

The same can be described in generalized form:

F(P,V,T) = 0




Thermodynamic Processes

A material system may under go a change, under externally or

internally imposed constraints, in terms of their state variables

from the existing one to some different values. Such a change in

the state of the system is known as thermodynamic processes.

For example: Expansion of a gas from V1 to V2 may be called as

a thermodynamic process. Many a times such processes are

carried out under additionally imposed conditions and are

named accordingly. Such processes are:




Isothermal processes: These are the processes which proceed

without any change in temperature of the system e. g. melting

of ice or metal. dT = 0

Isobaric process: These are the processes which proceed

without any change in pressure of the system e.g. processes

carried out in open atmosphere. dP = 0

Isochoric Processes: These are the processes which proceed

without any change in volume of the system e.g. the processes

carried out in vessel of known volume . dV = 0





Adiabatic processes: These are the processes which

proceed without any exchange of heat by the system with

its surrounding e. g. the system is completely insulated

from the surrounding . dq = 0. for ideal gas PVγ = const. γ

= Cp/Cv

Polytropic Processes: Those processes which obey equation

PVn = const where n is any positive number between 1 and

γ.








Path and State Functions

• The property whose change depends on only the initial and

final states of the system not on the path adopted to bring about

the change is called state function. Mathematically therefore if

the property is a state function (X) then in a cyclic process,

when system under goes a change and returns to original state

then

Ф dX = 0

• If Y is not a state function

Ф δY ≠ 0

So Y is called a path function







Properties of State Function

• If a system has two independent variables say x and y and any other function or property can be expressed in its total differential form

dz = Mdx + Ndy where M and N may be function of x and y then the function z is

a state variable if and only if • For an ideal gas T = PV/R wherein P and V are independent

variables and T as dependent variable. It can be expressed in total differential as

YXx

N

y

M

dPP

TdV

V

TdT




Relationship among state variablesLet there be three state functions X, Y and Z and two of these as

independent state variables. Once Z and once Y as dependent

variables. Thus

Z = Z (X, Y)

Y = Y (X, Z)

Total differential can be written as

And

dYY

ZdX

X

ZdZ

XY

dZZ

YdX

X

YdY

XZ




Putting the value of dY in first expression leads to

equating the coefficients of dX and dZ on both

Sides, we get

dZZ

Y

Y

ZdX

X

Y

Y

Z

X

ZdZ

XXZXY

ZXYZXY

X

XXX

X

Y

Y

Z

X

Z0

X

Y

Y

Z

X

Zand

Z

Y

1

Y

Z1

Z

Y

Y

Z

Relationship among state variables




Similarly,

and

Finally after substitution we can write

The above two expressions correlate the partial differentials of three state

variables w.r.t one another and are popularly called reciprocity theorem.

Z

Z

Y

X

1

X

Y

Y

Y

Z

X

1

X

Z

1X

Z

Z

Y

Y

X

similarly

1Z

X

X

Y

Y

Z

YXZ

YZX

Relationship among state variables




Thermodynamic Equilibrium

Mechanical Equilibrium: if there is no pressure gradient in the

system.

Thermal Equilibrium: if there is no temperature gradient in the

system.

Chemical Equilibrium: if the rate of forward reaction is equal

to rate of backward reaction.

Complete thermodynamic equilibrium is thus that situation

where the system is in equilibrium with respect to all such

potentials like mechanical, thermal and chemical.




Reversible and Irreversible Processes

A process that can be reversed in its direction by an

infinetesimal change in one or more of the state variables is

said to be a reversible process.

A classical example of this is the gas cylinder and piston. If

the pressure of the gas is say P atm and (p+dp) is exerted

from outside on the piston, the gas inside the piston shall be

compressed. However, if the external pressure is (p-dp) then

the gas shall expand.




On the contrary a matchstick when it burns, the process

can not be reversed by changing one or other parameters.

Once burnt can not be reproduced by reversing the

process. This is typically a Irreversible process.

Other examples are mixing of two gases, mixing of two

liquids to form a solution or flow of electric current

through resistor.

All natural processes are irreversible.

Reversible and Irreversible Processes




Experimental Evidence Leading to First Law

• For number of thermodynamic cycle each consisting of number of processes

• If , however work and heat measured in same unit then for a thermodynamic cycle

W = q

184.4

i

i

q

W

It is impossible to produce energy of any kind or form without the disappearance of an equivalent amount of energy




System goes from state I to state II by

various path and returns to the initial

state by path r only. Then we can write

Wa +Wr = qa + qr

or Wa – qa = (qr - Wr)

Simillarly along other paths

Wb – qb = (qr - Wr)

Wc – qc = (qr - Wr)

Wd – qd = (qr - Wr)

Or qa – Wa = qb – Wb=

qc – Wc= qd – Wd

The difference between q and W shall remain constant as long as initial and final states are not changed.





From the previous discussion, the following conclusions can be drawn:

i) The amount of heat exchanged and work done for taking the system from state I to II is different for different paths thus these functions are path functions. These are denoted by symbol δq and δw for infinetesimal change.

ii) The difference between the heat input q and work done W can be equated to change in another variable, say U i.e

q - W = U

Since q-W is independent of path, U is a state function. It has

further been proved that for a thermodynamic cycle

ΔU = 0

It is denoted by dU for infinetesimal change.





Let us consider that the system goes from state I to II by absorption of

heat from the surrounding and doing work on the surrounding. As we

know that q and W cannot be equal we can consider two distinct cases:

i) q < W: system partly imparts energy for the work done.

ii) q > W: The heat is partly retained by the system itself and partly

returned to the surrounding in the form of work done.

In both the cases system acts as a reservoir of energy. This stored energy

in the system which can change during a thermodynamic process is called

internal energy and denoted by symbol U.





Internal Energy

It consists of

macroscopic kinetic energy due to motion of the system as a whole.

potential energy of the system due to its position in the force field.

kinetic energy of atoms and molecules in the form of translation, rotation and vibration.

energy of interaction amongst atoms and molecules

columbic energy of interaction amongst electrons and nucleii in atoms

energy contents of the electrons and nucleii of atoms




In conventional chemical thermodynamics which we shall be concerned

with only kinetic energies of atoms and molecules and interaction

amongst atoms and molecules i. e. items (3) and (4) are considered to be

important since changes occurring in them principally contribute to ΔU.

Absolute value of the energy is not known. All we can determine is

change in internal energy.

Internal energy will depend on temperature for a material of fixed mass,

composition and structure.

U is function of Temperature only.

Internal Energy




Statement of First law

For an infinetesimal process, the statement is:

dU = δq – δW

‘Sum of all forms of energy exchanged by a system

with its surrounding is equal to the change in internal

energy of the system which is a function of state’.

‘Energy can neither be created nor can be destroyed’




Significance of the First Law

It is based on the law of conservation of energy.

It brought in the concept of the internal energy.

It separates heat and work interactions between the system

and surroundings as two different terms.

It treats internal energy as a state property is an exact

differential.




Internal energy in terms of Partial Derivatives

For given homogeneous system consisting of given amount

of substance of fixed composition;

U = F(P,V,T)

If any two variables are independent third will be

automatically fixed. This also therfore can be stated as:

U = F(P,V); U = F(V,T) ; U = F(P,T)




From the theorem of partial derivatives

Internal energy in terms of Partial Derivatives




Enthalpy

If pressure is maintained constant during change of system

from state I to state II the work done

From the first law:

UII – UI = δq – P(VII – VI )

On rearranging these terms

(UII + PVII ) - (UI + PVI ) = δq

Or HII - Hi = ΔH = δq

Where H = U + PV is called entalpy. The heat content at

constant pressure is called enthalpy.

III

II

I

II

IVVPdVPPdVw




Enthalpy is a state property but heat is not.

As we know

dH = dU + PdV + VdP

As U is a function of state so one can write

Hence

= M dP + N dV (Say)

VPUH

dVV

UdP

P

UdU

PV

dVV

UPdP

P

UVdH

PV

Enthalpy




Differential of coefficient of dP w.r.t V at constant P is given by the

relation

Differential of coefficient of dV w.r.t P at constant V is given by the

relation

Hence the above two partial differentials are equal proving thereby that

the equation for dH is an exact differential equation and thus leading to

the conclusion that H is a function of state.

PV

U

V

M 2

1

VP

U

P

N 2

1

Enthalpy




As Internal energy of an ideal gas is a function of temperature only.

At constant temperature dT = 0 and therefore dU = 0 so

As we know:

For an ideal gas at Constant T, PV = constant, d (PV) = 0

0

TP

U

VPUH

TTT

P

VP

P

U

P

H

Enthalpy




In other words enthalpy of an ideal gas is independent of

pressure at constant temperature. Similarly, enthalpy is

independent of volume. Hence H is a function of T only for

the fixed mass of the substance.

0P

H

TTP

U

Enthalpy




Internal Energy Vs Enthalpy

Internal energy is all that energy stored in the system. But, What does

enthalpy mean? This can best illustrated by the example of calcination of

calcium carbonate.

CaCO3 = CaO + CO2

The enthalpy change of the process will be equal to invested

bond energy to break the bond between CaO and CO2 or internal energy

provided both CaO and CO2 are solids.

However if CO2 is allowed to form gas then breaking of one mole of CaCo3

nearly 22.4 ltrs. of CO2 will be formed. The expansion in volume will take

place.

In this process of expansion the system will do work equivalent to ∫ PdV as

mechanical work on the surrounding.




Hence in addition to the requirement of energy for breaking the bond

of Cao-CO2 additional energy equivalent to ∫ PdV will have to be

supplied to the system making a total of U + ∫ PdV and this is the

enthalpy change of the system on calcination.

For chemical processes where there is no significant change in volume

as in

2CaO3. SiO2 = 2CaO + SiO2

The ∫ PdV is practically absent and U and H are almost the same.

In chemical and metallurgical world, even if term ∫ PdV is absent, it

worthwhile to refer to H which is more appropriate.

Internal Energy Vs Enthalpy




Heat Capacity

In chemical and metallurgical processes, materials get heated or

cooled and therefore it is necessary to know the amount of heat

required to heat or amount of heat liberated on cooling a material

over a certain temperature range.

Different materials require different amounts of heat to get

heated through the same temperature rise. This is because the

materials have different heat capacities.

This is so because of the variation in the crystal structure of the

materials and their related parameters.




The heat capacity of a substance is the amount of heat

required to raise its temperature by one degree.

In thermodynamics the molar heat capacity (C) i.e. heat

capacity of 1 g-mole of a substance is most widely

employed. Thus

The molar heat capacity at constant volume is given by

Since at constant volume δq = dU.

T

qC

VV

V T

U

T

qC

Heat Capacity




Similarly molar heat capacity at constant pressure is given by

Since at constant pressure δq = dH. Further we can write

dH = CP dT

Or

CP > CV since CP includes heat required to do work against

pressure also besides raising temperature

PP

P T

H

T

qC

2

112

T

T PTT dTCHH

Heat Capacity




Interrelationship of CP and CV

And we know

differentiating w.r.t T at constant P we get

PVP

VVVP

VP

T

VP

T

U

T

U

T

U

T

PVU

T

U

T

HCC

dTT

UdV

V

UdU

VT

VPTPT

U

T

V

V

U

T

U




Putting the value of we get

For an ideal gas rate of change of U with V is zero at const T.

For ideal gas

So CP – CV = R

PT

U

PV

U

T

V

T

VP

T

V

V

UCC

TPPPT

VP

PT

VCC

P

VP

P

TRV

Interrelationship of CP and CV




Application of First law to Thermodynamic Processes

With the help of the first law, one is able to calculate the changes

taking place in internal energy and enthalpy of a system during a

thermodynamic process.

Processes which are frequently studied and to which this law will

be applied include: i) Isothermal process ii) Adiabatic process iii)

polytropic processes.

In all these cases working substance of the system shall be

considered to be an ideal gas




i) Isothermal Process: The process is carried out when dT = 0

thus dU = 0 . Hence

δq = δW = PdV

It is also true that in this case

If Vf > Vi i.e. for expansion of gas q is positive i.e.

the system will absorb heat from its surroundings

and produce an equivalent amount of work.

f

i

f

i

f

i

f

i

V

Vi

f

V

V

V

V

V

V

V

VTRVdTR

V

dVTRdV

V

TRPdVwq

lnln









If Vf < Vi i.e. for compression of gas q is negative i.e. the system will impart

heat to the surroundings at the same time absorbing mechanical energy from

them in the form of mechanical work done on it.

The change in enthalpy of the system will be:

Tf = Ti in the isothermal process.

Thus in an isothermal process with an ideal gas internal energy and enthalpy

remains unchanged and work done is equal to the heat exchanged.

0)(

)(

if

iiffif

TTR

VPVPUHHH





ii) Adiabatic Process: There is no exchange of heat between the system and

surrounding i. e. δq = 0. Hence first law takes the form

dU = - δW = - P dV

Or dU = CV dT = - P dV

This indicate that system will perform work at the cost

of its internal energy and therefore the lowering of

temperature of the system will result.

Adiabatic work done (w)

Change in Internal energy = CV (Tf – Ti)

ifV

T

TV TTCdTCf

i





Relationship between P and V and T and V in

adiabatic process.

Or

Integration of this equation under the limiting

condition

V = Vi at T = Ti

And

V = Vf at T = Tf

V

dVTRdTCV

V

dV

T

dT

R

CV

Adiabatic Process




And substituting the expression for R in above

Expression

Or where

Or Or

Also since by definition H = U + P V

Or dH = dU + P dV + V dP

As dU = - P dV

So dH = V dP

i

f

i

f

VP

V

V

V

T

T

CC

Clnln

11 iiff VTVT

V

P

C

C

ConstVT 1ConstVP

Adiabatic Process




Enthalpy change in a adiabatic process is

As for such process.

This on integration lead to the expression

f

i

f

i

V

V

P

PdVPdPVH

ConstVP

11

1

f

iii

V

VVPH

Adiabatic Process




Work done in reversible adiabatic expansion can

be deduced as follows

Or

Where

1

11

12

2

1

2

1

VVm

V

dVmPdVw

V

V

V

V

1

1122 VPVPw

.constamVP

Adiabatic Process




iii) Polytropic process: The general expression for this process is

The work done in this process is similar to adiabatic process but

.constVP n

11

11n

f

iii

ffii

V

V

n

VP

n

VPVPw

n

Polytropic Process




Change in Internal energy = CV (Tf – Ti)

Internal energy can also be expressed in terms of P and V as follows:

From the first law heat exchanged q is obtained as

For n = γ; q = 0 obtained from the above expression which is true for adiabatic

process.

1

1

1n

f

iiiV

ii

ffiiViiffV

ifV

T

TV

V

V

R

VPC

VP

VP

R

VPC

R

VP

R

VPC

TTCdTCUf

i

11

11n

f

iiiV

V

V

nR

VPRnCq

Polytrophic Process




Change in enthalpy of the system is given by

Substituting the expression for ΔU in above expression

Or

Above equation show that for n = 1 i.e. isothermal process with ideal gas ,

both

change in internal energy and enthalpy are equal to zero.

iiff VPVPUH

1

1n

f

iiiV

V

V

R

VPRCH

1

1n

f

iiiP

V

V

R

VPCH

Polytropic Process




Summary of ThermodynamicProcesses

Process characteristics P-V-T relationship

Work done Heat exchange

Isothermal dT = 0 PV = const RT ln (Vf / Vi) RT ln (Vf / Vi)

Isochoric dV = 0 P/T = const 0 Cv (Tf – Ti)

Isobaric dP = 0 V/T = const Pi(Vf – Vi) Cp (Tf – Ti) + Pi(Vf – Vi)

Adiabatic q = 0 PVγ = const (PfVf - Pi Vi) / (1- γ) 0

polytropic - PVn = Const (PfVf - Pi Vi) / (1- n) Cv (Tf – Ti).(γ - n)/(1-n)




Thermo chemistry

It may be defined as the branch of science which deals with the

study of heat exchanges associated either with chemical reaction

or physical changes in the state of matter such as melting,

sublimation or evaporation etc.

If heat is produced by a chemical reaction it is denoted by – ve

sign (Exothermic reaction)

If heat is absorbed during a chemical reaction it is denoted by

+ve sign (endothermic reaction).




Hess’s Law

It states that the total heat exchanged in a given chemical reaction, which may take place

under constant pressure, volume or temperature is the same irrespective of the fact

whether it is made to take place over a path involving formation of number of

intermediate products or over the one involving the formation of final product from the

reactant directly in one stage.

ΔH =ΔH1+ΔH2+ΔHx+ΔHy




Hess’s Law




Variation of heat Capacity with Temperature

Experimental data consists of CP as function of temperature.

For the liquid and solid, The P V term is very small. Hence H is

taken as equal to U and CP as equal to CV.

In other words no distinction is made between CP and CV so far as

applications are concerned.

It has been found that experimental CP Vs. T data for elements

and compounds fit best with an equation of type:

CP = a + b T + c T-2




Where a, b, c are empirically fitted constant and differ

from substance to substance.

The last term is the smallest and therefore often ignored.

In some cases, such as liquid metals, both bT and cT-2 are

usually ignored.

The above expression is also valid for diatomic and

polyatomic gases as well.

Variation of heat Capacity with Temperature




Variation of Enthalpy with Temperature

Change in enthalpy during the course of process (ΔH) is

equal to the heat supplied to the system (q) at constant

pressure.

The constant pressure restriction is mostly not important

for example H is not function of P for ideal gases.

Energies of solid and liquids are hardly affected by some

changes in pressure due to their very small molar volumes.

In other words, the VdP term is negligible.

In most metallurgical and materials processing, gases are

ideal and pressure is maximum a few atmosphere.

Therefore ΔH = q approximation is quite all right.




Classification of Enthalpy change

Absolute value of enthalpy change of a substance is not

known. Only we can measure are changes of enthalpy.

Enthalpy change occur due to various causes.

Sensible heat: enthalpy change due to change of

temperature of a substance is known as sensible heat. It is

divided into:




i) change in enthalpy without any change in aggregation of the

substance:

As a universal convention, thermo chemical data books take

sensible heat at 298 K (250C) as zero for any substance. Hence

sensible heat at temperature T, per mole of a substance is given as:

298 K is known as reference temperature

dTCHHT

PT 298298





Putting the expression for heat capacity, we get

Where A, B, C and D are lumped parameters and functions

of empirical constants a, b, c

DTCTBTA

cTTa

dTTcTbaHH

Tb

T

T

12

2981122

2

298

2298

298298





ii) Enthalpy change due to changes in state of aggregation of

substance : These are isothermal processes. By convention

enthalpy change s for all isothermal processes are designated

by ΔH. For example,

ΔHm = enthalpy change of one mole of solid due to melting (i. e.

latent heat of fusion per mole )

ΔHv = enthalpy change of one mole of liquid due to vaporization

(i. e. latent heat of vaporization per mole )





Consider a pure substance A, which is undergoing

following changes during heating from 298 to T K

A (Solid) → A (Liquid) → A (Gas) → A (Gas) at 298 K at Tm at Tb at T

Tm and Tb are the melting and boiling points of A.Then

Cp (s), Cp (l) and Cp (g), are for solid liquid and

gaseous A. it is only applicable for pure substance.

dTgCH

dTlCHdTsCHH

m

b

b

m

m

T

T PV

T

T Pm

T

PT

)(

)()(298298





Heat of reaction (ΔH): This is the change of enthalpy that occurs

when a reaction takes place. By convention reaction is considered to

be isothermal. Consider the following reaction occurs at

temperature T:

A (pure) + BC (pure) = AB (pure) + C (pure)

Then

ΔH (at T) = ΔHAB (at T) + ΔHC (at T) - ΔHA (at T) - ΔHBC (at T)

Where ΔHAB , ΔHBC, ΔHA and ΔHc are molar enthalpies

of pure Ab, BC, A and C, respectively.





Heat of mixing (ΔHmix): This is the change of enthalpy that

occurs when a substance is dissolved in solvent.

This process is generalized as

A (pure) = A (in solution)

This process is also accompanied by a change of enthalpy

(ΔHmix), where

(ΔHmix) = HA (in solution) – HA (pure)

Again by convention, the process is assumed to be isothermal for

thermodynamic calculations





Some comments:

• For calculation of enthalpy change, reactions, dissolutions and

phase transformations has been assumed to occur

isothermally. Since enthalpy is a state property, it depends

only on initial and final states and not the path.

• For a reversible isothermal process, the temperature remains

constant all through. If the process is not reversible, then

temperature at the beginning and end of a process would be

same. In between, temperature can vary significantly.










Sign convention for ΔH:

The process accompanied by liberation of heat are called

exothermic. This happens if the enthalpy in the final state

(state 2) is lower than the initial state (State 1) i. e. H2 < H1 so

for the process

state 1 → state 2

we have ΔH = H2 – H1 < 0

Therefore ΔH is negative. The opposite is an

endothermic process which is characterized by

absorption of heat and positive value of ΔH.




Standard state of enthalpy

• The stable state of a substance changes with the temperature. The

stable state of H2O is ice which is below 0 oC, liquid water 0-100 oC and a stable gas at 1 atm pressure above 100 oC.

• Considering all these points a standard state has been defined as a

pure element or compound at its stable state at the temperature

under consideration and at 1 atm pressure. Thus the standard state

of H2O at 50 oC is pure water at 1 atm pressure.

• By convention enthalpy changes at standard state are denoted by

subscript ‘0’ e. g. 00TT HandH




As already mentioned, 298 K is the universal reference

temperature for compilation of sensible heats. By this

convention, sensible heat at standard state of a substance is

arbitrarily taken as zero at 298 K. This is solely for calculation

of sensible heats not ΔH0 for a process occurring at 298 K

Standard state of enthalpy




Kirchoff’s Law

Utility:

It allows us to calculate the enthalpy changes at various

temperatures, provided the enthalpy is known at some other

temperature, and the heat capacity of the reactants and

products are known in the range of temperatures under

consideration.

Derivation:

Consider a chemical reaction at temperature T1 whose enthalpy

change is ΔH1. Calculate the enthalpy change ΔH2 at

temperature T2.




The product of this reaction at T2 can be obtained in many

different ways. Let us, however, consider the reaction to be

carried out first by reacting the reactants (x+y) at T1 and then

raising the temperature of the products from T1 to T2 along ABC.

Kirchoff’s Law




The heat absorbed by this process will be

The second way of obtaining the same result is to raise the

temperature of the reactants from T1 T2 and then react them

together at T2 i. e. along ADC.The heat absorbed by this process is

2

11

T

T P dTCHz

2

1

yx

T

TPP2 dTCCH

Kirchoff’s Law




According to Hess’s law the heats absorbed during the two ways of

producing z must be the same since the initial conditions of the

reactants and final conditions of the products are in each case the

same.

=

Or

Or

Or

2

1

yx

T

TPP2 dTCCH 2

11

T

T P dTCHz

2

112

T

T PPP dTCCCHHyxz

2

112

T

T P dTCHH

2

112

T

T P dTCHH

Kirchoff’s Law




Since all the terms are known in the right hand of the equation, ΔH2

can be calculated. We know

Or

This is the statement of the Kirchoff’s law. It means the rate of

change of enthalpy of a process or a reaction with temperature is

given by the difference of the heat capacity at constant pressure of

products and reactants taking part in the reaction

2

112

T

T P dTCHH

P

P

CT

H

Kirchoff’s Law




Problem-1













Problem-2
















Enthalpy change:

H (a → c) = U (a→c) + (Pc Vc – Pa Va)

= - 9.13 + n R (Tc – Ta)

= - 9.13 + 4.09 x 8.3144 x (119 - 298)

= - 9.13 - 6.0870 kJ = - 15.2170 kJ






















Problem-3







Problem - 5

Calculate the heat of the following reaction at 1000K

Fe2O3(s) + 3C (s) = 2Fe(s) + 3CO(g) from the following data:

Cp(, Fe) = 4.18 + 5.92 x 10-3 T caldeg-1mol-1

Cp(CO) = 6.79 + 0.98 x 10-3T – 0.11 x 105 T-2 caldeg-1mol-1

Cp(Fe2O3) = 4.10 + 1.02 x 10-3 T – 2.10 x 105T-2 caldeg-1mol-1

The heats of formation of Fe2O3 and CO at 298 K are

-197000 and – 26400 cal/mol. respectively.




Solution

Cp = Cp (product) – Cp (reactant) = 2.Cp(-Fe)+ 3 Cp(CO) – Cp(Fe2O3) – 3Cp(C)

2 x Cp(-Fe) = 8.36 + 11.84 x 10-3T

3 x Cp(CO) =20.37 + 2.94 x 10-3 T -0.33 x 105 T-2

__________________________________________

Cp (product) = 28.73 + 14.78 x 10-3T – 0.33 x 105 T-2

Cp = 23.49 + 18.60 x 10-3T – 3.55 x 105 T-2

3Cp(C) = 12.30 + 3.06 x 10-3T – 6.30 x 105 T-2

__________________________________________

Cp (reactant) = 35.79 + 21.66 x 10-3T -4.85 x 105 T-2

Cp = -7.06 - 6.88 x 10-3T + 9.52 x 105 T-2




Solution

= 117800 – 7.06(1000-298) -6.88. (10002 – 2982) x 10-3 – 9.52 x 105 (1000-1 – 298-1)

= 117800 – 5850 = 111950 cals.

dTCHH p .1000

298

2981000

( )∫1000

298

253 dT)T10x52.9+T10x88.606.7+117800=




Problem-6

The standard heat of formation, of ammonia gas is -11.03

kcal/mol. at 250C utilizing the data given below derive a

general expression for heat of formation applicable in the

temperature range 273 – 1500K.

112633 deg10728.010787.7189.6, molcalTxTxNHC p

112632, deg100808.010414.1450.6 molcalTxTxNC p

112632, deg104808.0102.0947.6 molcalTxTxNC p




Solution

The reaction is

= -7.457 + 7.38 x 10-3T – 1.409 x 10-6T2

molcalHgNHgHgN /110302

3

2

1 0298322

223

, 2

3

2

1HCCCC pNpNHpp

∫298

273p273298 dT.CΔ+HΔ=HΔ

dTCΔ-HΔ=HΔ ∫298

273p298273

( )∫298

273

263 dTT10x409.1T10x38.7+457.7--11030




Solution

=-11030 + 7.457 (298 – 273) - x 7.38 x 10-3 (2982-2732) + x 10-6 (2983-2733)

= -11030 + 186.4 – 52.7 +2.9 = -10893 cal.

= -10893 – 7.457 T + 3.69 x 10-3T2 – 0.47 x 10-6T3 – (-7.457 x 273 + 3.69 x

10-3 x 2732 – 0.47 x 10-6 2733)

= -10893 +2036 – 275 + 10 – 7.457 T + 3.69 x 10-3T2 – 0.47 x 10-6T3

= - 9122 – 7.457T + 3.69 x 10-3T2 – 0.47 x 10-6 T3

= -9122 – 7.457T + 3.69 x 10-3T2 – 0.47 x 10-6 T3 cal/mol

For H1000K solution T = 1000 in the above equation

H1000K = -9122 -7457 + 3690 – 470 = -13359

T

T dTTxTxHH273

263273 10409.11038.7457.7




Solution

= - 11030 – 7.457 (1000-298) + 3.69 x 10-3 (10002-2982) – 0.47 x 10-6 (10003-2983)

= -11030 – 5234.8 + 3362.3 – 457.6 = -13360.1 cal/mol

∫1000

298p2981000 dTCΔ+HΔ=HΔ

( )dTT10x409.1T10x38.7+7457.7+11030-= ∫1000

298

363




Problem- 4

























Problem










CT-ISolution

Solution for Q-2: Fe + ½ O2 = FeO

Cal63200-=HΔ=HΔ FeO298

∫∫1123

10332ptr

1033

2981p2981123 dTCΔ+HΔ-dTCΔ+HΔ=HΔ

2532OpmagFepFeop1p T10×470-T10×424-93=C

2

1-C-C=CΔ ...,,,,

2532OpnonmagFepFeop2p T10×470-T10×51+920-=C

2

1-C-C=CΔ ...,,,,

kcal20863-=HΔ 1123 .




CT-ISolution

Solution for Q-3:

(a)

As U, T, ans S are dependent state functions

We can express G in the following manner

dG = dU + PdV + VdP – TdS – SdT

( ) ( ) dV+dP=dU PV∂U∂

VP∂U∂

( ) ( ) dV+dP=dS PV∂S∂

VP∂S∂

( ) ( ) dV+dP=dT PV∂T∂

VP∂T∂




CT-ISolution

( ) ( ) ( ) ( ) ( ) ( )NdV+MdP=

dPS-T-V++dVS-T-P+=dG VP∂T∂

VP∂S∂

VP∂u∂

pV∂T∂

pV∂S∂

pV∂u∂ ][][

( ) ( ) ( ) ( ) ( ) )( ∂P∂VT∂

pV∂T∂

vP∂S∂

∂P∂VS∂

pV∂S∂

vP∂T∂

∂P∂VU∂

vP∂M∂ 222

S--T--1+=

( ) ( ) ( ) ( ) ( ) )( ∂P∂VT∂

pV∂T∂

vP∂S∂

∂P∂VS∂

pV∂S∂

vP∂T∂

∂P∂VU∂

PV∂N∂ 222

S--T--1+=

So G is a function of state




CT-ISolution

(b)

dH = TdS + VdP

( )[ ] ( )NdV+MdP=

dVT+dPV+T=dH P∂VS∂

v∂PS∂

( ) ( ) ( ) 1+T+= ∂P∂VS∂

VP∂S∂

PV∂T∂

PV∂M∂ 2

)(

( ) ( ) ( ) )( ∂P∂VS∂

PV∂S∂

VP∂T∂

VP∂N∂ 2

T+=

So H is not a function of state




Second law of Thermodynamics

System which are away from equilibrium, upon initiation, shall

move towards equilibrium and such processes are referred to as

natural or spontaneous or irreversible processes.

Examples: heat flow from hotter to colder body or

depressurisation of inflated tube in a low pressure surroundings

or free fall under gravity and so on.

Introduction




The spontaneous change from an existing state to equilibrium

state is possible because in the existing state the system happens

to be at higher potential which is the driving force for the

change to occur.

Higher temperature is therefore the higher thermal potential

which makes heat flow from higher to a lower temperature.

Similarly higher pressure is a higher mechanical potential and

so on.

Second law of Thermodynamics




If the system is in equilibrium and if it is to be moved in the

reverse direction in the above examples it would be termed as

unnatural or non-spontaneous processes.

Therefore water can not be raised to an overhead tank unless

energy in the form of motor pump set is provided to the system

from the surroundings.




Similarly heat can not flow from a colder to a hotter body unless

aided by the surroundings in the form of compressor energy as in

the refrigerator.

But in chemical processes it is not readily possible to assess as to

what is a natural or unnatural process. This can only be evaluated

from the equilibrium state of the system.

The knowledge of equilibrium CO / CO2 ratio in contact with Fe

or FeO can only guide us as to how to prevent oxidation of iron or

effect reduction of iron oxide by providing a suitable CO / CO2

gas mixture as surroundings.




Heat Engines

•Heat engines operate in a cycle, converting heat to work then returning to original state at end of cycle.

•A gun (for example) converts heat to work but isn’t a heat engine because it doesn’t operate in a cycle.

•In each cycle the engine takes in heat Q1 from a “hot reservoir”, converts some of it into work W, then dumps the remaining heat (Q2) into a “cold reservoir”

HOT

COLD

Engine

Q1

Q2

W




This whole process for conversion of heat to work necessarily

produces a permanent change in the cold reservoir by way of

release of heat into it. This is called compensation.

It means the entire heat that is absorbed initially is not

converted to useful work but a part of it is rejected to a cold

reservoir as of necessity.

In other words ‘in a cyclic process it is just not possible to

convert all heat into mechanical work. In a non-cyclic process it

may be possible.




HOT

COLD

Engine

Q1

Q2

W

Efficiency of a heat engine

Definition:

cycleper input heat cycleper donework

efficiency

1QW

•Because engine returns to original state at the end of each cycle, U(cycle) = 0, so W = Q1 - Q2

•Thus:12

11

21QQ

QQQ




Efficiency of a heat engine

•According to the first law of thermodynamics (energy conservation) you can (in principle) make a 100% efficient heat engine.

BUT………….

•The second law of thermodynamics says you can’t:




Kelvin Statement of Second Law:

“No process is possible whose SOLE RESULT is the complete conversion of heat into work”

William Thomson, Lord Kelvin (1824-1907)




HOTCOLD

HOTTERCOLDER

WARMWARM

Q

HOTCOLDQ

Both processes opposite are perfectly OK according to First Law (energy conservation)

But we know only one of them would really happen – Second Law

Heat flow




Clausius Statement of Second Law:

“No process is possible whose SOLE RESULT is the net transfer of heat from an object at temperature T1 to another object at temperature T2, if T2 > T1”

Rudolf Clausius (1822-1888)




How to design a “perfect” heat engine

1)Don’t waste any workSo make sure engine operates reversibly (always equilibrium conditions, and no friction).

2)Don’t waste any heatSo make sure no heat is used up changing the temperature of the engine or working substance, ie ensure heat input/output takes place isothermally

Sadi Carnot (1796-1832)




HotSource

T1 Q1

Piston

Gas

a

b

Working substance (gas) expands isothermally at temperature T1, absorbing heat Q1 from hot source.

The Carnot Cycle (I): isothermal expansion

T1

P

V

T1T1

a

b

Q1Q1




The Carnot Cycle (II): adiabatic expansion

Gas isolated from hot source, expands adiabatically and temperature falls from T1 to T2.

Piston

Gas

b

c

Gas isolated from hot source, expands adiabatically, and temperature falls from T1 to T2

P

V

T1T1

a

b

T2T2c




The Carnot Cycle (III): isothermal compression

Piston

Gas

d

c

Q2

ColdSink

T2

Gas is compressed isothermally at temperature T2 expelling heat Q2 to cold sink.

T2

V

P

T1

a

b

T2cd

P

T1T1

a

b

T2T2cd

Q2Q2




The Carnot Cycle (IV):adiabatic compression

Gas is compressed adiabatically, temperature rises from T2

to T1 and the piston is returned to its original position. The work done per cycle is the shaded area. Piston

Gas

a

d

Gas is compressed adiabatically, temperature rises from T2 to T1 and the piston is returned to its original position. Work done is the shaded area.

V

P

T1

a

b

T2cd

W




Efficiency of ideal gas Carnot engine

12

11

21QQ

QQQ

•We can calculate the efficiency using our knowledge of the properties of ideal gases

V

P

T1

a

b

T2cd

W

Q2

Q1

V

P

T1

a

b

T2cd

W

V

P

T1T1

a

b

T2T2cd

WW

Q2Q2

Q1Q1




a

bab V

VnRTWQ ln11

Isothermal expansion (ideal gas)

P

V

T1T1

a

b

Q1Q1

Va Vb




Isothermal compression (ideal gas)

d

ccd V

VnRTWQ ln22

V

P

T1

a

b

T2cd

Q2

V

P

T1

a

b

T2cd

Q2

V

P

T1

a

b

T2cd

P

T1T1

a

b

T2T2cd

Q2Q2




a

b

d

c

VV

nRT

VV

nRT

QQ

ln

ln11

1

2

1

2

(2)

(1) 1

21

1

12

11

da

cb

VTVT

VTVT

Adiabatic processes

)2()1( d

c

a

b

VV

VV

2

2

1

1

1

21

TQ

TQ

TT

Efficiency of ideal gas Carnot engine

V

P

T1

a

b

T2cd

W

Q2

Q1

V

P

T1

a

b

T2cd

W

V

P

T1T1

a

b

T2T2cd

WW

Q2Q2

Q1Q1




So……..

h

c

TT

1

0h

h

c

c

h

h

c

c

TQ

TQ

TQ

TQ Conservatio

n of “Q/T”

For all Carnot Cycles, the following results hold:

What about more general cases?????




0h

h

c

c

h

h

c

c

TQ

TQ

TQ

TQ

Was derived from expressions for efficiency, where only the magnitude of the heat input/output matters. If we now adopt the convention that heat input is positive, and heat output is negative we have:

0h

h

c

c

TQ

TQ

The expression

In other words0

cycle

TdQ

For any reversible cycle:




The Carnot CycleThe Importance of the Carnot Engine

1.All Carnot cycles that operate between the same two temperatures have the same efficiency.

2. The Carnot engine is the most efficient engine possible that operates between any two given temperatures.




Entropy

To emphasise the fact that the relationship we have just derived is true for reversible processes only, we write:

0cycle

rev

TdQ

We now introduce a new quantity, called ENTROPY (S)

TdQ

dS rev

Entropy is conserved for a reversible cycle




Is entropy a function of state?

P

V

A

B

path1

path2

For whole cycle:

0 cycle

rev

TdQ

S

2 path1 path

2 path1 path

0

B

A

revB

A

rev

A

B

revB

A

rev

TdQ

TdQ

TdQ

TdQ

ABBABA SSSS 2) (path1) (path

Entropy change is path independent → entropy is a thermodynamic function of state

Reversible cycle




Irreversible processes

Carnot Engine

h

c

revh

cc T

TQ

Q

11

Irreversible Engine

h

c

irrevh

c

TT

Q

Q

11irrev

For irreversible case:

h

c

irrevh

c

h

c

irrevh

c

TT

Q

Q

TT

Q

Q

0

irrevc

c

irrevh

h

TQ

TQ

h

h

irrevc

c

TQ

T

Q





Following similar argument to that for arbitrary cycle:

0cycle

irrev

TdQ

P

V

A

BPath 1(irreversible)

Path 2(reversible)

For irreversible cycle

0)(

)(

)(

)(

rev

rev

irrev

irrev

A

B

rev

B

A

irrev

TdQ

TdQ

)(

)(

)(

)(

rev

rev

irrev

irrev

B

A

rev

B

A

irrev

TdQ

TdQ

Irreversible cycle





)(

)(

)(

)(

rev

rev

irrev

irrev

B

A

rev

B

A

irrev

TdQ

TdQ

)(

)(

)(

)(

rev

rev

irrev

irrev

B

A

B

A

irrev dST

dQ

TdQ

dS irrev

T

dQdS

Equality holds for reversible change, inequality holds for irreversible change

General Case




“Entropy statement” of Second Law We have shown that:

T

dQdS

For a thermally isolated (or completely isolated) system, dQ = 0

0dS

“The entropy of an isolated system cannot decrease”




What is an “isolated system”

The Universe itself is the ultimate “isolated system”, so you sometimes see the second law written:

“The entropy of the Universe cannot decrease”(but it can, in principle, stay the same (for a reversible process))

It’s usually a sufficiently good enough approximation to assume that a given system, together with its immediate surroundings constitute our “isolated system” (or universe)………




Entropy changes: a summary

For a reversible cycle:S (system) = S (surroundings) = 0S (universe) = S (system) + S (surroundings) = 0

For a reversible change of state (A→B):S (system) = -S (surroundings) = not necessarily 0S (universe) = S (system) + S (surroundings) = 0

For an irreversible cycle S (system) = 0; S (surroundings) > 0S (universe) = S (system) + S (surroundings) > 0

For a irreversible change of state (A→B):S (system) ≠ - S (surroundings) S (universe) = S (system) + S (surroundings) > 0




Other Types of EnginesSchematic diagram of a refrigerator:

Refrigerator removes heat from cold reservoir, puts it into

surroundings, keeping food in reservoir cold.

Heat transfer takes place from cold to hot body




Dependence of entropy on temperature

Dependence of entropy on temperature can becalculated at constant

volume as follow:

Dependence of entropy on temperature can becalculated at constant

pressure as follow:

TlndCSSS

TlndCdTT

C

T

PdV

T

dU

T

qdS

V

T

T12V

VVrev

2

1

TlndCSSS

TlndCT

dTCdH

T

1PVUd

T

1PVddU

T

1

T

qdS

p

T

T12P

PPrev

2

1




Volume dependency of entropy

We know;

and

So

As ‘S’ is function of state therefore the above

expression forms an exact differentials. Hence

PVddUT

1

T

qdS rev

dVV

UdT

T

UdU

TV

dVPV

U

T

1dT

T

U

T

1dS

TV

VTTVT

P

V

U

T

1

TT

U

T

1

V




PV

U

T

PT

TV

On simplification the above expression yields:

So we can write

And under isothermal condition dT = 0 lead to the following

expression:

dVT

PdT

T

U

T

1dS

VV

dVT

PdS

V

T




In case of gas one can directly integrate the above expression to calculate

the change in entropy during a process.

However, in case of solid and liquids, the equation of state is not known

and can also not be determined experimentally. Hence in such cases

this partial differential is required to be expressed in terms of some

experimentally determinable parameters. For this purpose the reciprocity

theorem is used to write the following:

VT

P

TPVV

P.

T

V

T

P




Isobaric coefficient of volumetric thermal expansion of a material ()

is given as

Isothermal compressibility of a material () is given as

So

PT

V

V

1

TP

V

V

1

dVdS

Which on integration will yield the change in entropy for a change in volume

from V1 to V2 under isothermal condition




Pressure dependency of entropy

We know:

From the definition of the enthalpy, one can get

Therefore,

The total differential dH can also be expressed as

dVPdUT

1dV

T

PdU

T

1

T

qdS rev

VdPdVPdUdH

dPT

VdH

T

1dS

dPP

HdT

T

HdH

TP




After substitution, we get

The equation is an exact differential because entropy being function of

state. Therefore,

On simplification this yields

dPT

V

P

H

T

1dT

T

H

T

1dS

TP

PTTPT

V

P

H

T

1

TT

H

T

1

P

PPTT

VV

P

H

T

1




• So after substitution we can write

• For an isothermal process the above expression yields,

• On integration, one can calculate the change in entropy for a

change in pressure from P1 to P2 under isothermal conditions.

dPT

VdT

T

H

T

1dS

PP

dPVdPT

VdS PT




Relationship between CP&CV

• We have derived the following relationship:

PTT

VTV

P

H

P

T

PT

V

U

VT

TPVV

P.

T

V

T

P

PT

V

V

1

T

P

V

V

1




• We know the following expression

• Elimination of from the above equation leads to the relation

• Or

TV

U

PV

VP T

V

T

PTCC

TV

CC2

VP

PV

U

T

V

T

VP

T

V

V

UCC

TPPPT

VP




Entropy Change in Gases

Ideal gas

dq/T = Cv .dT/T + R dV/V = dS

S = S2 –S1 =

On integration S = Cv ln + R ln

Entropy change can be calculated when temperature and volume change take place on heating an ideal gas

S1 p1

v1 T1

Change

S2 p2

v2 T2

∫ ∫2

1

2

1

T

T

v

vv v

dvR+

T

dTC

1

2

T

T

1

2

v

v




S = Cv ln (T2/T1) + R ln (T2p1/T1p2)

= Cv ln (T2/T1) + R ln (T2/T1 )+ R ln (p1/p2 )

Or S = Cp ln (T2/T1 )– R ln p2/p1

In an isothermal process i.e. at constant temp T = T1 = T2.

ST = - R ln p2/p1 = R ln p1/p2 = R ln (v2/v1)

For an isobaric process p1 = p2 = p

Sp = Cp ln (T2/T1)

For an isochoric process v1 = v2 = v

Sv = Cv ln (T2/T1)





As the system absorbs heat, its entropy will increase e.g. with

melting and boiling.

S= S2 – S1 =

S is the area beneath the

curve between temperature,

T1 and T2(graphical method).

2

1

lnT

T

TCpd





Significance of sign of entropy change of a process in closed system

• Two identical copper vessel touching each other, full of water and each at different temperature.

• Vessels are completely insulated so that heat can neither enter nor leave this system.

• T1 – T2 is the measure of irreversibility of this process. This also denote the how much heat will flow

I II

T1 T2

q

Insulation

T1 > T2




Entropy change of vessel I = - q1 / T1

Entropy change of vessel I = q / T2

Total change of entropy will be the sum of entropy changes of the two

vessels. Thus

When q is +ve and T1 – T2 is +ve then the entropy

change for a real irreversible process in a closed

system must also be positive.

21

21ocessPr

1221ocessPr

TT

TTqS

T

q

T

qSSS




T1 > T2 ; Sirr > 0 (+ve) : The entropy change of a real process is greater

than zero(+ve).

T1 = T2 ; Srev = 0 (zero): Dynamic equilibrium exists between two vessels

and there is no heat transfer.

T1 < T2 ; Sirr < 0 (-ve). : Entropy change is negative and the process

proceeds in the reverse direction.

Sign of entropy change shows the direction of flow of heat energy.




Calculation of entropy change

• Entropy is a state property. Hence the basis and procedure for

calculation of entropy changes are similar to those for the

enthalpy changes.

• Hess’s law and Kirchoff’s law are applicable here too.

• A pure substance at its stablest state also constitutes standard state

for entropy at that temperature which is designated as .0TS




Following significant difference are to be noted

between entropy and enthalpy.

– Entropy changes are to be calculated only through the

reversible path. This restriction is not there for any other state

property, including enthalpy.

– Absolute value of entropy can be determined. Thermodynamic

data sources provide the values of entropy of a substance

for pure substance. This is in contrast with the energy where

only changes are available in the data sources.




Entropy changes associated with phase transformation

For a pure substance, reversible phase changes

(melting, boiling etc.) at a constant pressure occurs at

a constant temperature. Therefore,

for melting

for boiling

in general for phase trans.

m

0m0

m T

HS

V

0V0

V T

HS

tr

0tr0

tr T

HS




A (Solid) → A (Liquid) → A (Gas) → A (Gas)

at 298 K at Tm at Tb at T

Tm and Tb are the melting and boiling points of A. Then

dT

T

gC

T

H

dTT

lC

T

HdT

T

sCSS

T

T

0P

V

0V

T

T

0P

m

0m

T

298

0P0

2980T

b

b

m

m




Entropy changes in chemical processes

By nature, reactions and mixing are irreversible. Like enthalpy

changes, entropy changes are considered only for the isothermal

process. However, they cannot be calculated from enthalpy as done for

the phase transformation in view of irreversibility. Consider the

reaction A + BC = AB + C

The entropy change of the reaction at temperature T

)Tat(S)Tat(S)Tat(S)Tat(SS 0BC

0A

0C

0AB

0T




The entropy change of the reaction with temperature

can be calculated as

dTT

CS

dTT

CdT

T

CS

SSSSSS

2

11

2

1

2

11

121212

T

T

P0T

T

T

ttanreacPT

T

productP0T

ttanacRe

0T

0T

product

0T

0T

0T

0T




Various interpretations of entropy

for an infinitesimal, isothermal reversible process. Entropy is times arrow i. e. a fundamental indicator of

time. Entropy has the relationship with heat not available for

work. Entropy is a measure of disorder of a system.

T

qdS rev




Combined statement of first and second law

If a system is capable of doing only mechanical work the first law equation can be put

as:

dU = δq – P dV. It is not exact differential. Holds true for the reversible processes.

The second law for the reversible process gives

dS = δq rev / T or δq = T dS

The two laws therefore can lead to

dU = TdS – P dV or TdS = dU + PdV

This is the combined statement of first and second law.

It includes only those terms which are state functions only and

hence is exact differential equation.




For irreversible process second law gives

dS > 0

Or therefore, TdS > dU + PdV

For a unnatural or non spontaneous process

dS < 0

Or therefore TdS < dU + P dV




Thermodynamic Potentials

A system by itself, in isolated state or in contact with

surrounding shall stay in equilibrium unless acted upon by some

constraints.

If a system tends to move as a result of being not in equilibrium

or as a result of external constraint, there must be a driving

force making the system move from within itself or under the

applied constraints.

This driving force is referred to as thermodynamic potential

driving the system to change to a new state.




Is this driving force or potential the same under all conditions of the system or

it varies from condition to condition of the system ?

Heat flow s from higher to lower temperature. The flow is possible due to

higher thermal potential.

Similarly the higher pressure is the driving mechanical potential.

Also higher electrical voltage is driving the electrons under the influence of

electrical potential as voltage.




However the question remains as to what is that potential, forcing a

thermodynamic process to take place?

It is easy to imagine that a system with higher associated energy will be relatively

unstable as compared to the one having lower energy.

In other words, when a natural changes take occurs the system moves from higher

to lower energy levels or moves from higher to lower thermodynamic potentials.

It also means that the energy is a potential driving the system for change to occur.

It also means that for equilibrium to exist the potential of all the systems or sub-

systems within it must be at the same potential.




So far the internal energy, enthalpy and entropy have been evolved as

energy parameters but it is not yet known as to whether these act as

potential in driving a particular process?

It must be noted that for natural process the system moves from lower

to a higher entropy level hence entropy can not be considered as a

driving force.




Can internal energy and enthalpy qualify as potentials capable of driving the

process towards equilibrium if already at higher levels ?

If internal energy and enthalpy are potential terms then changes in them will have

to be zero at equilibrium and these will have to decrease for natural changes and

increase for unnatural changes.

Let us therefore see if criteria can be evolved mathematically to evaluate what

constitute as thermodynamic potential and if so under what conditions?




Potentials under constant volume and constant entropy conditions

Mathematically these conditions are expressed as

dV = 0 and dS = 0

i) TdS > dU + P dV for natural process

So under above condition we can write

dU < 0

ii) TdS = dU + P dV for the equilibrium process

So under above conditions we can write

dU = 0

iii) TdS < dU + P dV for unnatural process


dU > 0




As internal energy decreases for natural process, remains same at

equilibrium and increases for unnatural process thus fully qualifies to be

called as thermodynamic potential under the conditions of constant

volume and entropy.

As a corollary, the internal energy must be the function of entropy and

volume can be expressed:

U = F (S, V)

Total differential of internal energy is given by

dVV

UdS

S

UdU

SV




We know: dU = TdS – P dV

On comparison this leads to

and

By specifying U as function of S and V it is possible to evaluate T and

P thereby describing the system fully.

By describing U as a function of any other two variables, it is not

possible to describe the system fully.

TS

U

V

PV

U

S




Potentials under constant pressure and constant entropy conditions

Mathematically these conditions are expressed as

dP = 0 and dS = 0



dU + d (PV ) < 0

Or d (U + P V) < 0

Or dH < 0



dU + d (PV ) = 0

Or dH = 0




iii) TdS < dU + P dV for unnatural process


dU + d (PV ) > 0

Or dH > 0

So on the whole dH > = < 0 for unnatural, equilibrium

and natural processes as the case may be.

Therefore, the term enthalpy qualifies as being called

a potential term under constant pressure and entropy

conditions.




Enthalpy can therefore be described mathematically as: H = F (P, S)

Total differential of enthalpy can be written as

Also by definition: dH = d (U + P V)

= dU + P dV + V dP

= T dS + V dP

On comparison we get

dPP

HdS

S

HdH

SP

VP

HandT

S

H

SP




Mathematically these conditions are

expressed as

dV = 0 and dT = 0



dU - TdS < 0

Or d ( U – T S) < 0

Let us define new mathematical term

A = U – T S

then dA < 0

Potentials under constant volume and constant temperature conditions






d ( U – T S) = 0

Or dA = 0

iii) TdS < dU + P dV for the unnatural process


d ( U – T S) > 0

Or dA > 0

The newly defined function qualifies for being referred as a

potential term under constant volume and temperature

conditions




‘A’ being the function of all state variables ‘A’ also

must be state function. Expressing as before

A = F (V, T)

Total differential of ‘A’

Now A = U – T S

dA = dU – TdS – S dT

And since, TdS = dU + P dV

Therefore dA = - S dT – P dV

On comparison we get

dVV

AdT

T

AdA

TV

PV

AandS

T

A

TV




This function ‘A’ was first defined by Helmholtz and

therefore is known as Helmholtz free energy.

We know dA = dU – T dS

And for reversible process qrev = T dS

An therefore qrev = dU – dA

From the First law dU = qrev – W

And hence qrev = dU + W

Comparing the two equations for qrev we get

- dA = W

Or dA = - W

That the change in Helmholtz free energy in a process is equal

to the amount of work done by the system on the surrounding

or is equal to the amount of work the system is capable of doing.




These are by far the most commonly adopted

conditions in chemical and metallurgical engineering

practices. It means

dP = 0 and dT = 0



dU + d (P V) - TdS < 0

Or d ( U + P V– T S) < 0

Let us define new mathematical term

G = U + P V– T S

then dG < 0

Potentials under constant pressure and constant temperature conditions






d ( U + P V– T S) = 0

Or dG = 0

iii) TdS < dU + P dV for the unnatural process


d ( U + P V– T S) > 0

Or dG > 0

The function defined above thus qualifies for being

called as a potential term under constant pressure

and temperature conditions




‘G’ is a function of all state variable. so it is a state function.

The function ‘G’ is referred to as Gibbs free energy.

Now G = U + P V – T S

And H = U + P V

Then G = H – T S

Or G = H – T S for a finite change

Or dG = dH – T dS – SdT in differential form

At constant pressure dP = 0 and dH = q = TdS

And hence dG = - SdT

Or

STd

Gd

P




Again dG = dH – T dS – SdT

On putting dH = dU + P dV + V dP

dG = dU + P dV + VdP – T dS – SdT

Putting in this dU + P dV = T dS

dG = T dS + V dP – T dS – S dT

= V dP – S dT

At constant temperature when dT = 0 ; dG = V dP

Or

Since G is a state function any change in G can be represented as G = U – T S + (P V)

= - Wrev + (P V)The change of Gibbs free energy during a process is equal and opposite in sign to the net reversible work obtainable from the process occurring under isothermal condition when corrected for change in volume under isobaric condition.

VPd

Gd

T




Since G is a state function any change in G can be

represented as

G = U – T S + (P V)

= - Wrev + (P V)

The change of Gibbs free energy during a process is

equal and opposite in sign to the net reversible work

obtainable from the process occurring under

isothermal condition when corrected for change in

volume under isobaric condition.




Important thermodynamic relations

For closed and isolated systems have fixed mass and

composition and the reversible work is done against

pressure.

dU = T dS - PdV

dH = T dS + V dP

dA = - S dT – P dV

dG = - S dT + V dP




Criteria for thermodynamic equilibria

Differential form Finite difference form

(dU)S, V = 0

(dH)S, P = 0

(dA)T, V = 0

(dG)T, P = 0

(U)S, V = 0

(H)S, P = 0

(A)T, V = 0

(G)T, P = 0

Since it is easy to maintain temperature and pressure

constant the Gibbs free energy criteria is employed in

chemical and metallurgical processes. However, in other

areas, other criteria are also employed.




Maxwell’s equations

YXx

N

y

M

From the properties of the exact differential

equation: dZ = M dX + N dY,

PT

VT

PS

VS

T

V

P

S

T

P

V

S

S

V

P

T

S

P

V

T

Maxwell’s relations are used

frequently in thermodynamics

for the calculation of changes in

thermodynamic variables for

different processes.




The Driving Force of a Chemical Reaction

Why does a chemical reaction take place?

It was believed that the energy change accompanying a

reaction can be measured directly by the enthalpy change

at constant pressure, or change in intrinsic energy (U) at

constant volume.

The reasoning behind this would be apparent – if a system

losses energy as a result of a chemical reaction, that

reaction will take place spontaneously – and the greater

the quantity of heat lost, the greater the driving force

behind the reaction.




C (s)+ O2 (g) = CO2(g) with evolution of heat energy, i.e. cal/mol which happens to be a driving force in this

case. 2 Fe (s) + O2 (g) = 2FeO(s) and

3FeO(s) + 2 Al (l) = 2Fe(l) + Al2O3(s) Both have negative heats of reaction at 16000C and both take place

spontaneously On the other hand, if we consider the reaction; ZnO(s) + C(s) = Zn(g) + CO (g) and This reaction will not take place at 250C but if the system is heated

to 11000C, carbon will reduce zinc oxide to produce zinc metal.

940500298 H

8320001373 H 568000

298 H

The Driving Force of a Chemical Reaction (cont.)




we cannot use heats of formation as a criterion of their

tendency to take place. We must therefore search for a

more consistent rule for the driving force of a reaction.

Consider the reaction: ZnO(s) + C(s) = Zn(g) + CO (g)

At 11000C, i.e above the boiling point of Zn (9070C), the

reaction will take place between phases indicated above.

At 250C, zinc is a solid (melting point 419.60C) so that the

reaction is

ZnO(s) + C(s) = Zn (s) + CO (g)





The most obvious difference in these reactions apart from

their temperature, is the difference in the physical state,

means a difference in the state of order of a system and

consequently in the entropy of the system.

S0 at 250C should be much less than that 11000C, because

at 11000C two molecules of gas are being produced from

two solid molecules whereas at 250C two solid moles only

produce one gaseous mole.





A solid to gas transformation means a comparatively large entropy increase and the following figures for the reduction of ZnO by C:

Why not use entropy change of a reaction as a measure of the driving force behind a chemical reaction?

The idea is immediately contradicted if we consider the formation of the oxide of any metal:

We know that this reaction is spontaneous – an oxide film forms readily on iron at room temperature. Thus a positive entropy change is not the criterion of a chemical reaction.

degmol/cal68=SΔanddegmol/cal46=SΔ 1373298

molcal17-=SΔsFeO=gO2

1+sFe 0

2982 deg/,)()()(





The Second Law of thermodynamics states that a spontaneous process is always accompanied by an increase in entropy of the system and its surroundings.

The surroundings receive e a quantity of heat - H at constant temperature and pressure

entropy increase of the surroundings = If S is the entropy change of the system, total entropy

change of the system and surroundings = S –

For the 2nd Law to be obeyed (S - ) must be +ve

T

H

T

H

T

H

ve+=T

HΔ-SΔT





we can say that (H – T S) must always be –ve for a reaction to proceed spontaneously in order that the total entropy change of the system and surroundings can be +ve, when the reaction proceeds.

We should examine the factor (H – T S) for the above three reactions in question. For the reduction of ZnO by C at 250C

The reaction cannot proceed because this factor is +ve, and the total entropy change of the system and its surroundings is –ve

This is negative so that the reaction will take place spontaneously at 11000C. This explains why zinc oxide smelting must be carried out at temperature of the order of 11000C in order that reduction of the oxide by carbon can proceed.

( ) molcal43100+=46298-56800=SΔT-HΔ 298298 deg/

( ) cal10200-=681373-83300=SΔT-HΔC1100at 137313730





For the oxidation of iron at 250C

Again it is negative, and the reaction takes place spontaneously at 250C.

Consistent Rule The driving force of a reaction can be calculated as (H -

TS); the more negative this factor, the greater the driving force and if the factor is +ve, the reaction will not proceed spontaneously.

G = H - TS called Gibbs Free energy

.571401729863200298298 calSTH





It is the maximum work available from a system at

constant pressure other than that due to a volume change.

Most metallurgical processes work at constant pressure.

We had already seen the fundamental importance of the

factor (H-TS), so that G is a measure of the ‘driving

force’ behind a chemical reaction. For a spontaneous

change in the system, G must be negative, the more

negative, the greater the driving force.





Thermodynamic Mnemonic Square

In order to memorize important thermodynamic relationship

involving the different thermodynamic potentials Max Born

suggested a diagram known as Thermodynamic Mnemonic

Square.

Four sides of the square are leveled

in alphabetical order starting from the

top and moving in clock-wise direction

with four thermodynamic potentials

namely A, G, H, U.

The other four primary functions namely V, T, P, S are placed at the four corners in such a way

that each thermodynamic potential is surrounded by the condition under which it acts

U

V T

S PH

G

A




We can write the total differentials of each of

potentials with help of this diagram as follows:

Differential of any potential is equal to the sum of the differentials of its

adjoining variables with their coefficients equal to diagonally opposite

variables.

This coefficients are taken to be positive if arrow points away from the

variable and negative if it points towards the variable.

dG = - S dT + V dP




Maxwell’s relations can also be read from the

diagram as follows:

It is required to consider the corners only

The square is rotated in anti clock-wise direction in such a way that the

function whose partial differential is to be arrived it appears on the top

left hand corner of the square.

The partial derivative of this variable w.r. t variable on the bottom left

hand corner keeping the lower right hand corner variable as constant is

equal to the partial derivative of the top right hand corner variable w.r.t

bottom right hand corner variable with bottom left hand corner variable

as constant.




Negative sign on the right hand side of the equation is put if the arrows

are placed unsymmetrically with respect to vertical axis drawn from

centre of the square on rotation.

VSS

P

V

T




Standard state of free energy

Standard state is pure element or compound at 1 atm pressure and at its stablest state at the temperature under consideration.

Free energy change of the reaction are generallycalculated when the reactants and products arepresent in their standard states is called the standard free energy change.

The standard free energy is designated by G0T.

Like enthalpy, we cannot measure the absolute value of free energy but change in free energy is quite possible to measure.




Therefore, there must be some reference point with respect to

which the actual values of various substance can be calculated.

The free energies of stable form of the elements at 298 K and 1

atm pressure are arbitrarily assigned as zero value.

The free energy of formation of compound are calculated on

the basis of the above assumption and the value is described as

standard free energy of formation.




This quantity is generally reported at 298 K and for compound say

MO, it would be written as G0298, MO.

The Hess’s law is applicable for calculation of free energy change as it

is a state property.

The standard free energy of formation of compound and the standard

free energy of compound are same.

M + ½ O2 = MO

G0298 = G0

298,MO - G0298,M - ½ G0

298, O2

G0298,MO = G0

298,MO

00




knownbemustproductsandttanreacofenergiesfreedardtansthe

reactionaofchangeenergyfreedardtansthecalculatetoinorder

GbGaGdGc

GGG

bygivenisK298atGchangeenergyfreedardtansthe

DdCcBbAa

reactiontheFor

0B,298

0A,298

0D,298

0C,298

0ttanacRe,298

0oductPr,298

0298

0




Some thermodynamic relationships

By definition G = H – TS = U + PV –TS

differentiating dG =dU + PdV + VdP – TdS – S.dT (1)

Assuming a reversible process involving work due only to

expansion at constant pressure

According to First Law: dU = dq – PdV and from Second

Law: dq = TdS

dU = TdS – PdV (2)

From (1) and (2) we get dG = (TdS – PdV) + PdV + VdP –TdS –

SdT

or dG = VdP – SdT (3)





At constant pressure, dP = 0, so that = - S

and at constant temperature, dT = 0 and dG = VdP

from PV = RT for ideal gas, dG =

Integrating between the limits PA and PB at constant

Temperature

if GA is the free energy of the system in its initial state and GB

the free energy in its final state when the system undergoes a

change at constant pressure

PT

G

dPP

RT

)4(ln BP

AP A

BAB P

PRT

P

dPRTGGG




but G = GB – GA and S = SB – SA and so that d (G) = -S.dT

on substitution


dTSdGanddTSdG BBAA .( ) ( )dTS-S=G-Gd ABAB

ST

G

P

TPT

GHSTHG

,

known on Gibbs – Helmholtz equation




Calculation of G0 at high temperature

It is possible to calculate G0 of a reaction at high

temperature from the H0 and S0 values at 298 K

(available in the literature ) in this way

It must be remembered that if any transformation

takes place between 298 and T K in reactants and

products must be introduced in the above equation.

T

298

P0298

T

298

P0298

0T

0T

0T

0T

dTT

CSTdTCHGor

STHG




Fugacity

The variation of Molar Gibbs free energy of a

closed system of fixed composition, with pressure

at constant temperature is given by

dG = V dP

For one mole of ideal gas

dG = (RT / P) dP = RT d ln P

For isothermal change of pressure from P1 to P2

at T

G (P2, T) – G (P1, T) = RT ln (P2 / P1)




Fugacity contd……

AS Gibbs free energy do not have absolute values,

it is convenient to choose an arbitrary reference

state from which changes in Gibbs free energy can

be measured. This reference state is called

standard state and chosen as being the state of

one mole of pure gas at one atm pressure and the

temperature of interest.

The Gibbs free energy of the ideal gas in the

standard state G (P = 1, T) is designated as G0 (T).




Fugacity contd..

Thus the Gibbs free energy of 1 mole of gas at any

other pressure P is given as

G (P, T) = G0 (T) + RT ln P

Simply

G = G0 + RT ln P

So the molar Gibbs free energy of the ideal gas is

a linear function of the logarithm of the pressure of

the gas




Fugacity contd…

If the gas is not ideal then the relation of molar free energy and

logarithm of pressure is not linear.

A function is invented which when used in place of pressure gives a

linear relationship.

This function is called ‘Fugacity ( f )’ is partially defined as

dG = RT dln f




In addition, as the real gas and the ideal gas

behave the same at very low pressure, it is

obvious that (f / P) → 1 as P → 0.

after integration from standard state to any arbitrary state we can write

G - G0 = RT ln (f / f0)

Where G0 is the Gibbs free energy of the real gas

in the standard state. f is the fugacity at a

specified state and f0 is standard state fugacity.

Fugacity is the measure of escaping tendency of a gas.

Fugacity is the effective pressure corrected for non-ideality

Fugacity contd…




Activity

Activity (a) is defined as

So we can write

0i

ii0 f

faor

f

fa

i0ii

0 alnRTGGOralnRTGG

Partial molar free energy of component i in the state of interest. free energy of component I

in the standard state.




The standard state of a substance are generally chosen as the pure solid

or liquid form of the substance at 1 atm pressure and temperature under

consideration or as the gases at 1 atm pressure at the temperature under

consideration.

The activity of a substance in its standard state is seen to be unity.




Equilibrium Constant

Let us consider a general chemical reaction at

constant temperature and pressure

Capital letters for chemical element and small

letters for the number of gram mole. The general

free energy change of the reaction may be written as

RrQqMmLl

MLRQ GmGlGrGqG




The standard free energy change can be written as

Subtracting these two equations,

Or

Or

0M

0L

0R

0Q

0 GmGlGrGqG

M0ML

0L

R0RQ

0Q

0

GGmGGl

GGrGGqGG

ML

RQ0

alnRTmalnRTl

alnRTralnRTqGG

JlnRTaa

aalnRTGG

mM

lL

rR

qQ0

The parameter J is called activity quotient




Or

Which is known as van’t Hoff isotherm.

Let us consider the equilibrium in which all the reactant and

products are in equilibrium. In this case the activity product

defined as K, the thermodynamic equilibrium constant.

At equilibrium G = 0,

Or

eq

mM

lL

rR

qQ

aa

aaK

KlnRTJlnRTG .eqat0

K

JlnRTG

JlnRTGG 0




Since the standard state of a substance is at 1atm

pressure the G0 of a reaction is function of

temperature only. Therefore, the equilibrium

constant K also function of temperature only.

(J / K) < 1 reaction is in forward direction.(J / K) > 1 reaction is in backward direction.




For this process, mass, composition and pressure are

assumed to be constant only variation of temperature

is considered.

Derivation of Gibbs – Helmholtz Equation

P22

2

P

P

T

G

T

1

T

H

T

G

obtainweTbybothsidesDeviding

T

GTHSTHGr0

ST

G

Gibbs – Helmholtz Equation




This is one of the form of Gibbs – Helmholtz Equation.

The alternative form can be derived as follows.

2

P

22P

T

H

T

TGor

T

H

T

G

T

G

T

1or

2P

2

P

2

P

PP

P

T

G

T

G

T

1TG

T

G

T

1T

T

1

TGor

GT1

T

T

G

T

1G

T1

G

T

1

T

1

TG




This is the alternative form of Gibbs – Helmholtz Equation .

We can write Gibbs – Helmholtz Equation for a process as

H

T

HT

T

1

TG2

2

P

H

T

1

TG

T

H

T

TG

P

2P

Utility Evaluation of enthalpy of reaction from

free energy change. Evaluation of free energy change from

calorimetric data.




.etemperaturoneatH

andGorestemperaturtwoatGeitherofknowledgethefrom

edminerdetareThese.ttanconsegrationintareIandHWhere

TIT

1

2

cT

2

bTlnTaH

TdT

c

2

b

T

a

T

HT

TdTdTcTbaT

1

T

HT

TdTdCT

1

T

HTGor

TdT

H

T

G

0

00

0

20

320

222

0

P2200

T

2

0T

0T




Trouton’s and Rechard’s RulesRechard’s rule: it states that the ratio of latent heat of melting

to the temperature of the normal melting point of the F.C.C

metal is approximately 9.61 J / K and that of BCC metal is

approximately 8.25 J / K. i.e

Trouton’s Rule: it states that the ratio of latent heat of boiling to the temperature of

the normal boiling point of the F.C.C metal is approximately 87 J / K.

metalC.C.BforK/J25.8T

H

metalC.C.FforK/J61.9T

H

m

m

m

m

K/J87T

H

V

V




Problem

The initial state of one mole of an ideal gas is P = 10 atm and

T = 300 K. Calculate the entropy change in the gas for

a) Reversible isothermal decrease of pressure to 1 atm.

b) Reversible Adiabatic decrease of pressure to 1 atm.

c) Reversible constant volume decrease of pressure to 1 atm.

Solution

.ltrs62.241

462.210

P

VPVVPVP

processisothermalperasK300TTandatm1P)a

.ltrs462.210

30008207.0

P

TRVTRVP

K300Tandatm10P

2

1122211

122

1

11111

11




K/J14.19462.2

62.24ln314.8

V

VlnR

dVV

RdV

T

P

T

dVPUd

T

qS

1

2

V

V

V

V

2

1

2

1

rev2

1

2

1

processadiabaticfor0qas0T

qdS)b rev

rev

K/J71.28300

30ln314.8

2

3

T

TlnC

dTT

C

T

Ud

T

dVPUd

T

qS

K3008207.0

462.21

R

VPT,R

2

3C.,atm1P)c

1

2V

T

T

V2

1

2

1

2

1

rev

222V2

2

1




Problem

Calculate the entropy change of the universe in isothermal

freezing of 1 gm-mole of super cooled liquid gold at 1250 K

from the following data for gold.

Tm = 1336 K,H0m = 12.36 kJ/mol,

CP(s) = 23.68 + 5.19 x 10-3 T J / mol / K, CP(l) = 29.29 J / mol / K

gSurroundinSystemUniverse SSS

Solution

.propertystateaisentropy

cesinchoosewepathreversiblewhichtoasdifferencenomakesIt

.pathreversibleaalongonlycalculatedbetoisS,But

.etemperaturfreezingmequilibriutheatoccuringnotisitcesin

,leirreversibbutisothermalisprocessfreezingcasethisIn

S)i

System

System




)2State()1State(

K1250atK1336atK1336atK1250at

)s(Au)s(Au)l(Au)l(Au

ispathreversiblesimplestTheCoolingFreezingHeating

1

1250

1336

31336

1250

1250

1336

P

m

0m

1336

1250

P12System

KJ327.9

TdT

T1019.568.23

1336

12360Td

T

69.29

TdT

)s(C

T

HTd

T

)l(CSSS




Hence.K1250atreversiblyfreezingduringreleasedheatabsorbsngSurrroundi

.capacityheatiniteinfhasitandK1250atisgsurroundintheifonlypossibleisThis

.K1250atbetoassumedbeenhavesystemtheofstatefinalandinitialtheBoth

S)ii gSurroundin

T

H

T

qS System

gSurroundin

J12460

TdT1019.568.2312360Td69.29

Td)s(CHTd)l(CHHH

1250

1336

31336

1250

1250

1336

P0m

1336

1250

P12System

1SystemgSurroundin KJ967.9

1250

12460

T

HS

1Universe KJ64.0967.9327.9S

Suniverse is positive since the process is irreversible.




Problem

one gram of liquid ThO2 at 2900 0C is mixed with 5 gm of ThO2

at 3400 0C adiabatically. a) what is the final temperature

b) What is the entropy change of the system and surrounding

Combined c) is the process spontaneous? Assume CP to be

Independent of temperature.

f

H

f

C

T

T

PH

T

T

PC

f

fPfP

coldPHotP

T

TdC5)S(bodyhotofchangeentropy

T

TdC)S(bodycoldofchangeentropy

K3589T

T3673C13173TC5

TdCmTdCm

bodycolderbyabsorbedheatbodyhotterbyreleasedHeat)a

Solution




P

C

fP

H

fP

T

T

P

T

T

P

CHSystem

C00752.0

T

TlnC

T

TlnC5

T

TdC

T

TdC5

SS)S(systemtheofchangeentropyTotal)b

f

C

f

H

eoustansponisprocesstheSo

0SSS

.llyadiabaticaoutcarriedisprocessas

zeroisgsurroundintheofchangeEntropy)c

gSurroundinSystemUniverse




Problem

Calculate the standard entropy change of the following reaction

at 1000 K.

Pb (l) + 0.5 O2(g) = PbO (s)

and also calculate the entropy change of the Universe.

Given: Tm, Pb = 600K, H0m, Pb = 4812 J mol-1, H0

PbO,298 = -219 kJ

S298, PbO = 67.78 J K-1, S298, Pb = 64.85 J K-1, S298, O2 = 205.09 J K-1

CP, PbO(s) = 44.35 + 16.74x10-3 T J K-1mol-1

CP,Pb(s) = 23.55 + 9.75X10-3 T J K-1mol-1

CP,Pb(l) = 32.43 – 3.09X10-3 T J K-1mol-1

CP,O2 = 29.96 + 4.184X10-3 T - 1.67x 105 T-2 J K-1mol-1




1

0O,298

0Pb,298

0PbO,298

0298

2

P0298

2

01000

KJ615.9909.2052

185.6478.67

S2

1SSS

)2.......(..........)s(PbO)g(O2

1)s(Pb

reactiontheforvaluesCandSfrom

)1......(..........)s(PbO)g(O2

1)l(Pb

reactiontheforS

evaluationofconsistsphysicallyoblemPrThe

Solution

2




11253P

11253P

molKJT10836.0T1089.481.5)2(C

molKJT10836.0T1072.1705.3)1(C

1

22

53

22

5

30298

1000

600

P

m

0m

600

298

P0298

01000

KJ079.99

07.0956.196.202.835.035.513.2615.99

600

1

1000

1

2

10836.060010001089.4

600

1000ln81.5

600

4812

298

1

600

1

2

10836.0

2986001072.17298

600ln05.3S

TdT

)1(C

T

HTd

T

)2(CSS

writecanwe,K600atmeltsPbthatNoting




kJ244.218

73.5580.15642324481220.14179.240210.21910219

600

1

1000

110836.06001000

2

1089.4

600100081.54812298

1

600

110836.0

2986002

1072.1729860005.3H

HHTd)2(CHTd)1(CHH

isK1000atreactionofheatorreactionthebyreleasedHeat

3

5223

5

223

0298

0PbO,298

0298

1000

600

P0m

600

298

P0298

01000

1

gSurroundinSystemUniverse

101000

gsurroundin

KJ119218079.99

SSS

KJ2181000

244.218

T

HS

.capacityheatearglhavingK1000atisoundingsurrthethatgminAssu




Problem

From the data given for different phases involved, calculate the

change in entropy of one mole of manganese when it is heated

from 298 K to 1873 K under one atmospheric pressure.

(CP)-Mn = 5.7 + 3.4 x 10-3 T – 0.4 x 105 T – 2 cal / deg/ mol

(CP)-Mn = 8.3 + 0.73 x 10-3 T cal / deg/ mol

(CP )γ-Mn = 10.70 cal / deg/ mol

(CP )δ-Mn = 11.3 cal / deg/ mol

(CP )Mn, l = 11.0 cal / deg/ molTransformation Temperature / K Latent heat in Cal / mol.

→→γ

γ →δ

δ→l

990

1360

1410

1517

535

525

430

3500




The total change in entropy will be equal to the sum of the

changes in entropy due to the rise of temperature and those

due to the phase transformation occurring on heating. Thus

dT

T

C

1517

3500

dTT

C

1410

430dT

T

C

1360

525dT

T

C

990

535dT

T

CS

1873

1517

MnlP

1517

1410

MnP1410

1360

MnP

1360

990

MnP990

298

MnP

MnlMnMnMnMn 151714101360990




dTT

113071.2dT

T

3.113049.0

dTT

7.103860.0dT

T

T107.03.8

5404.0dTT

T104.0T104.37.5S

1873

1517

1517

1410

1410

1360

1360

990

3

990

298

253

.mol/K/cal3655.193188.23071.28265.0

3049.03863.03860.08945.25404.04010.9S




Problem

Using the following values, calculate the standard free energy change per

mole of the metal at 1000 K for the reduction of molybdic oxide and chromic

oxide by hydrogen:

.mol/kcal5.45G

.mol/kcal0.120G

.mol/kcal5.205G

0)g(OH,1000

0)s(MoO,1000

0)s(OCr,1000

2

3

32

mol/kcal5.45G;OHO2

1H

mol/kcal0.120G;MoOO2

3Mo

.mol/kcal5.205G;OCrO2

3Cr2

Solution

01000222

0100032

01000322




.veischangeenergyfreetheasetemperatur

thisatreducedbecanmolybdenum,But.etemperatur

thisathydrogenbyreducedbenotcanit,veisoxide

chromiumofreductiontheforchangeenergyfreethecesin

.mol/kcal5.161205.453G

OH3MoH3MoO

.mol/kcal5.342

5.2055.453G

OH3Cr2H3OCr

aswrittenbecanreactionreductionThe

01000

223

01000

2232




Problem

The standard heat of formation of solid HgO at 298 K is – 21.56

kcal / mol. The standard entropies of solid HgO, liquid Hg and

O2 at 298 K are 17.5, 18.5 and 49.0 cal /deg/mol, respectively.

Assuming that H0 and S0 are independent of temperature,

calculate the temperature at which solid HgO will dissociate

into liquid Hg and O2.

/moldeg//cal49S

/moldeg//cal5.18S

/moldeg//cal5.17S

mol/kcal56.21H

Solution

0)g(O,298

0)l(Hg,298

0)s(HgO,298

0)s(HgO,298

2




C572K8455.25

21560TT5.25215600

mol/calT5.2521560STHG

0GwhendissociatetostartwillHgOsolidThe

STHGwritecanWe

ToftindependenareSandHcesin

KJ5.25492

15.185.17

S2

1SSS

.mol/kcal56.21HH

)s(HgO)g(O2

1)l(Hg

aswrittenbecanreactionThe

0

0298

0298

0T

0

0298

0298

0T

00

1

o)g(O,298

o)l(Hg,298

o)s(HgO,298

0298

0)s(HgO,298

0298

2

2




Problem

For the reaction WO3 + 3H2(g) =W(s) + 3H2O(g)

a)Calculate G0 and K at 400, 700 and 1000K

b)What is the maximum moisture content of H2 needed for the

reaction given:

WO3(s) = W(s) +

3H2 +

Solution:

WO3(s) = W(s) +

H2+

WO3(s) + 3H2(g)W(s) + 3H2O(g),

calsTTTGgO 7.91log2.10201500),(2

3 02

.1.1358900,)()(2

1 022 calsTGgOHgO

calsTTTG 4.52log2.10248000

calsTTTGgO 7.91log2.10201500),(2

3 02

..,)()( calsT339++176700-=GΔgOH3=gO2

3 022




Problem-2

a) i)

ii)

iii)

K400= 1.26x10-8

K700= 2.5x10-2

K1000= 0.22

b)

KatG 00 400 .144604004.52400log400.10248000400 calsxxG

KatG 00 700 .cals5122=GΔ 0700

KatG 00 1000 .cals3000=GΔ 01000

RTGeKorRT

GKeiKRTG

00

0 ln..ln

3

2

2

H

OH

p

pK




solution

+ = 1

(0.00247+1) = 1

% moisture = 0.24 %

At 700K, K700 = 2.317x10-3

( ) 3-318-31400

H

OH10x4722=10x2531=K=

p

p

2

2 ..

OHp2 2Hp

2Hp

...

atm99750=0024721

1=p

2H

atm0024660=p∴and OH2.

( ) 2960=10x52=p

p 312-

H

OH

2

2 ..




solution

(0.296 +1) = 1

at 1000K K1000 = 0.22

pH2O= 0.3776

% H2O at 1000K = 37.76 %

2Hp

22830=pand77160=1321

1=p OHH 22

...

%.% 8322=K700atOH∴ 2

( ) 60670=220=p

p31

H

OH

2

2 ..

62230=60671

1p =H2

..




Problem-3

What is the maximum partial pressure of moisture which can

be tolerated in H2 – H2O mixture at 1atmospheric total

pressure without oxidation of nickel at 7500C?

SolutionEquation under question Ni + H2O = NiO(s) +H2

= 450 + 10.45 T cals.

..)()()( calsT5523+58450-=GΔsNiO=gO2

1+sNi 0

12

..)()()( calsT113+58900-=GΔgOH=gO2

1+gH 0

2222

02

01

0r GΔ-GΔ=GΔ




solution

G 01023 = 450 + 10.45 x 1023

= 11140

= - RT lnK

ln k = -

=-5.48006

K = + 0.0041669 =

from 1 & 2 :

1023987.1

11140

OH

H

p

p

2

2

(1) 0041669.022 OHH pp (2) 1

22 OHH pp

1=p+p 00416690 OHOH 22. 99585.0

0041669.1

1

2OHpor

00415.02 Hp Hence maximum tolerable partial pressure of

moisture is 0.99585 atm




problem

The standard free energy change for reaction at 1125K NiO(s) +

CO(g) = Ni(s) + CO2(g) is -5147 cals. Would an atmosphere of 15% CO2,

5% CO and 80% N2 oxidise nickel at 1125K?

Solution:

G0 = -RTlnK = -4.575 x 1125 log K = -5147

K = 10

K for the above reaction is

In the given atmosphere i.e J/K is <1 so no oxidation of nickel

0001=1125x5754

5147=K .

.log

NiO

Ni

CO

CO

a

a

p

p=K 2 . NiONi aa 1

10=p

p=

CO

CO2

J=3=p

p

CO

CO2




Problem

For the reaction NiO(s) + H2(g) = Ni(s) + H2O(g)

calculate the equilibrium constant at 7500C from the following

data:

Could pure nickel be annealed at 7500C in an atmosphere

containing 95% H2O and 5% H2 by volume without oxidation?

..),()()( Tcals5523+58450-=GΔsNiO=gO2

1+sNi 0

2

..),()()( Tcals113+58900-=GΔgOH=gO2

1+gH 0

222




solution

G0 for the required reaction would be G0 =

G0 = -RT lnK = -4.575T log K

(aNi = 1 = aNiO)

J/K <1 rxn in forward direction

So no oxidation will take place

01

02 GG

2

2

2

2

H

OH

NiO

Ni

H

OH0

p

p=

a

a

p

p=

T5754

GΔ=K .

.log

382=1023x5754

11140= .

.

222

2 240..240 HOHH

OH ppeip

pK

J=19=5

95=

p

p given

2

2

H

OH:




Problem

Calculate the standard enthalpy and entropy changes at 298 K

for the reaction

2 Cu (s) + ½ O2 = Cu2O (s)

G0 = - 40500 -3.92 T log T + 29.5 T cal / mol

T

702.1

T

40500

T

TG

,atingDifferenti

5.29Tln702.1T

40500

T

G

T5.29TlnT702.140500

T5.29TlnT4343.092.340500G

Solution

2

0

0

0




K/cal10.18

298ln702.1798.27S,K298at

Tln702.1798.27Sor

Tln702.1798.27

5.29Tln702.1702.1ST

G

.mol/kcal99.39

29870.140500HK298at

mol/calT70.140500Hor

T

702.1

T

40500

T

H

writecanwe,equationHGBy

0298

0

00

0298

0

22

0




Problem

Enthalpy of the solid platinum at any temperature is given by

the expression:

Derive an expression for the change in free energy when one

mole of platinum is heated from 298 K to any temperature T K.

.mol/cal1785T1064.0T8.5HH 23298T

CTdT

H

T

G

.Tetemperatur

atsystemanyofequationHGtoAccording

Solution

2T




mole/cal69.9903C298T1064.0TlnT8.5TCGor

C2982981064.0298ln29885.51785HG

GGG

bygivenbethereforewillK298from

TetemperaturtoheatingonGchangeenergyfreeThe

TCT1064.0TlnT8.51785HGor

CT1064.0Tln8.5T

1785H

T

Gor

CdTT

17851064.0

T

8.5

T

H

T

G

Thus

23

23298298

298T

23298T

3298T

23

2298T




Problem-1

Calculate (1) the elevation of the boiling point of zinc when the external pressure is 2 atm. and (ii) the depression of the freezong point when the external pressure is 50 atm. The latent heat of vaporization of zinc is 27.3 kcal/g atom and the normal temperature of boiling is 9070C. The corresponding heat of fusion is 1.74 kcal/gatom and the normal temperature of fusion is 419.50C. The density of solid zinc is 7.0 g/cc and that of liquid zinc is 6.48 g/ce at 1 atm. pressure.

Solution

Using the Clausius-Clapeyron equation for this problem we have

To solve this problem we have also to know the values of the volume of 1

gatom of zinc in the vapour and liquid states. The volume occupied by 1 gatom of a perfect gas at STD in 22400 cc.

V

S

dT

dP e

zincliquid

ofatomg1 ofvolumetheis vandvapourzincofgatom1ofvolumetheisvwhere) g l

lge

e

vvT

L

dT

dP




The volume occupied by 1 g.atom of zinc vapours at 9070C

(1180K) is

Volume occupied by 1 gatom of liquid zinc =

Substituting these values in the equation dP = 1 atm.

1 cc atm = 0.024212 cals.

( ) .cc1180273

22400=vg

.09.1048.6

38.65cc

( )e

lge

L

v-vTdP=dT

.

glglg vvvvv




solution

cal 273×27300

Katmcc 1180×22400×1180×1=

..

K3101= cal 273×27300

Kcal 0242120×22400×1180×1180×1= .

.

7

38.65

,48.6

38.65 (ii)

s

lsl

f

f

v

vvvT

L

dT

dP

slf

f

vvL

TdPdT

.

.024212.0 1.1740

0.7

38.65

48.6

38.655.69249

calatmcccal

Katmcc




Problem-2

( ) 0242120×349-089101740

5692×49= ...

.

0242120×750×1740

5692×49= ..

.

= 0.354 K Ans




Problem-2

The latent heat of vaporization of zinc is 27.3 k cal/mole at the boiling point of

9070C. Find the vapour pressure over pure zinc at 8500C.

Solution:

for l g transformation we have

P

RTvvv

vT

L

vvT

L

dT

dPlg

ge

e

lge

e

; .

KTpKTatmpRT

LP

dT

dP1123,?,1180.,1 22112

21

12

1

22

ln1

TT

TT

R

L

p

por

TR

L

P

dP




Solution

ln p2 – 0 = = -0.5909869

p2= 0.55378 atm

11231180

11801123

987.1

27300




Lechatelier’s Principle

Statement

It states that if an equilibrium in a system is upset,

the system will tend to react in a direction that will

reestablish the equilibrium.

Factors

i) Concentration of reactant and product

ii) Pressure

iii) Temperature




Effect of concentration

Consider the reaction

A + B = C + D

At equilibrium for which

If the concentration of any one of the reactant or product is altered the

concentration of others must alter to keep K constant.

If for example [D] is increased by adding more D to the system then [C] will

decrease (by reaction with D to form A and B) preserving the value of K.

In general, If the product is added, the system will shift towards left to reestablish

the equilibrium and converse is true for the reactant.

BA

DCK




If the reactant is removed, the equilibrium will shift towards reactant and

same is true for the product.

This shift of equilibrium may be used to advantage in some commercial

processes since continuous removal of product will drive the reaction to

completion. This is more easily achieved if one of the product is a gas e.g.

CaCO3 = CaO + CO2

In the lime kiln removal of CO2 in an air current drives the reaction

towards right.

Other example is the production of magnesium:

2 MgO(s) + Si(s) = 2Mg (g) + SiO2(s)

At the temperature employed the magnesium vapour can be evacuated.




Effect of Pressure

Changes in pressure have no effect on the position of equilibrium when

only solids or liquids are involved or in gaseous reactions involving no

volume change e.g.

FeO(s) + CO(g) = Fe(s) + CO2((g)

Since equal volumes of gas appear in both sides.

However, the equilibrium position for a reaction in which changes in

gaseous volume occur may be displaced by pressure




We can consider following examples

N2(g) + 3 H2(g) = 2 NH3(g)

1 vol + 3 vol = 2 vol

Here in this case there is an overall volume decrease( 4 → 2).

If the system is subjected to a pressure increase the system will move

in such a direction as to lessen this increase in pressure .

By moving to the right the volume diminution results in reduction in

pressure i. e increasing pressure drives the reaction to the right.

Ammonia is commercially produced at 350 atm.




Effect of Temperature

‘K’ is unaffected by concentration and pressure change but dependent

on temperature.

Exothermic reaction:

N2 +3 H2 = 2 NH3 H = - 92.37 kJ

If the temperature is increased the system reacts in such a way as to

oppose this constraint by removing heat. Which can be done by

shifting to the left.

Exothermic reactions are favoured by low temperature




Endothermic reaction:

ZnO + C = Zn + CO H = +349 kJ Increase in temperature will again shift the equilibrium in the

direction which absorbs heat

i. e. to the right in this case. Endothermic reactions are therefore, favoured by high temperatures.




Clausius-Clapeyron Equation

It is extremely important for calculating the effect of change of pressure on the equilibrium transformation temperature.Derivation Let us consider a single solid substance in equilibrium with its liquid at its

temperature of melting and under one atm pressure.

There is a natural tendency for the molecule to pass from solid into liquid or vice versa.

The number of atoms passing at any time from one state to other will depend on the temperature and pressure.




Let us assume that the molar free

Energy of the solid at constant T, P

Is GA and of the liquid is GB; then

If GA > GB the solid metal can

decrease its free energy by

dissolving, i. e. if

GA > GB Solid → Liquid; solid melts since G = -ve

GA = GB Solid = Liquid; Equlibrium since G = 0

GA < GB Solid ← Liquid; liquid solidifies since G = +ve

Me (s)

GA

Me (l)

GB




The condition for a dynamic equilibrium between the solid and the liquid metal is

GA = GB

and dGA = dGB

The change in the free energy of either phase may caused by the change in temperature and pressure of the phases. dG = f ( T, P)

dP.P

GdT.

T

GdG

T

A

P

AA

dP.P

GdT.

T

GdG

T

B

P

BB




At equilibrium dGA = dGB

Or

Or

Or

Or

dP.P

GdT.

T

G

T

A

P

A dP.P

GdT.

T

G

T

B

P

B

dPvdTSdPvdTS BBAA

dTSSdPvv ABAB

dT.SdP.v

tr

tr

T

HSas

V

S

Td

Pd

ABtr

tr

vvT

H

Td

Pd




Or

If v is +ve as is normally in the case of melting and Htr will be +ve quantity i. e as

the pressure increases the transition temperature should also increase. In other

words the equilibrium melting point shall increase with pressure and vice versa.

However, for ice – water system v is – ve and hence (dP / dT) is –ve and which is

fully exploited in the game of skating on ice wherein the skate pressure increase

shall decrease the melting point of ice and hence help skating by providing fluid for

lubrication.

vT

H

Td

Pd

tr

tr

Clausius – Clapeyron Equation




Applications of Clausius – Clapeyron Equation

.lyrespectiveliquidand

vaporofvolumesmolarthearevandvand

ionvapourizatofheatlatenttheisHwhere

vvT

H

Td

Pd

EquilibiaVapourLiquid

liqvap

V

liqvap

V




Applications of Clausius – Clapeyron Equation

Contd….

2V

2V

vap

vap

V

liqvap

TR

H

Td

Plnd

TR

HP

Td

Pd

getweonsubstitutiafterP

TRv

gasideal

anasbehavesvaporthethatgminAssu

vT

H

Td

Pd

hencevv




ttanconsegrationintisCwhere

CTR

HPln

.eqaboveofegrationint

thenttanconsisHgminAssu

V

V

concernedervalinttemptheovervalue

meanthebewillmethodthisbycalculatedmetal

liquidofionvapourizatofheatThe.Ciserceptint

andR

His

T

1.vsPlnplottheofslopeThe V




.etemperaturoftindependen

isHthatgminAssu.lyrespectiveTandT

tempsthetoingcorrespondPandPitslimthe

withindonebecanequationCCofnIntegratio

V21

21

.knownareionvapourizatofheatand

etemperaturanotheratpressurevapourtheifetemperaturany

atpressurevapourthecalculatetousedbecanequationThis

T

1

T

1

R

H

P

Plnor

T

Td

R

HPlnd

12

V

1

2

T

T2

V

P

P

2

1

2

1




Solid – Vapour (sublimation) Equilibria

On the basis of assumptions similar to those made

in liquid – vapour equilibria, one can obtain the

similar expression for solid – vapour equilibria.

Where Hs is the heat of sublimation

2S

TR

H

Td

Plnd




Solid – liquid (Fusion) equilibria

Applying C – C equation to so solid liquid equilibria

Hf is the molar heat of fusion, vliq and vsolid are the

molar volumes of liquid and solid respectively.

This equation may be applied to calculate the change

in melting point of a metal with change of pressure.

f

solidliqf

solidliqf

f

H

vvT

Pd

Td

vvT

H

Td

Pd




Solid – Solid equilibria

The rate of change of transition temperature at

which two crystalline forms of a solid are in

equilibrium with pressure is given by following

expressions:

Where Htr is the molar heat of transition, v and

v are the molar volume of the indicated forms of

solid measured at Ttr.

trT

tr

tr

tr

tr

H

vvT

Pd

Td

vvT

H

Td

Pd




Ellingham Diagram

Ellingham diagrams are basically graphical representation of G0 vs. T

relations for the chemical reactions of chemical and metallurgical

engineering interest.

H. J. T. Ellingham in 1944, was first to plot the standard free energy of

formation of oxides against temperature and these later became known as

Ellingham diagram.

Later on the same plotting was applied for sulphides, chlorides, fluorides

etc.

Oxide diagrams are mostly used in metallurgy.







Features of Ellingham diagram

1) Formation reaction for oxides may be generalized as

322

2

yx

yx2

OAl3

2)g(OAl

3

4

NiO2)g(OiN2

areexamplesSpecific

.lyrespective,oxidesmetaland

metalforsymbolsgeneralareOMandM.compound

specifictheondependwillyandxofvaluestheWhere

OMy

2gOM

y

x2




activity

unitat.g.estatespureorstatedardtans

theirinareproductand)metal(ttanreac

themeansenergyfreedardtanS.oxygenof

moleperareGofvaluetheThus.reaction

formationoxidetheinbasisthesconstitute

Oofmole1diagramEllinghamtheIn

.involvedmolesof.notheondepend

wouldreactionaofGvaluehence

propertyextensiveanisenergyFree

0

2

0




2) Since G0 = H0 - T S0

The variation of H0 not being large, it can be treated as constant over a wide

range of temperatures. S0 however, changes with temperature particularly

when gaseous phases are involved with the condensed phases.

The plot of G0 against temperature is a straight line. So the slope of the line is

- S0 . Let us consider the reaction

M(s) + O2 (g) = MO(s)

S0 = S0 (oxide) - S0 (metal) - S0 (oxygen gas)

since S0 (oxide) and S0 (metal) are practically the same, the

entropy change (S0) arises predominantly due to disappearance of one mole of

oxygen gas and hence it is -ve. So the slope of the line is positive.




The entropy changes of various such metal oxidation process are expected to

be substantially of the same value. Therefore, the most of the oxide lines

slope upwards and parallel to each other.

3) The plot of G0 against temperature is a straight line as long as there is no

phase change (melting, boiling, phase transformation) in either the reactant

and products. The reason is that when such phase change takes place, there

is a change in entropy and since the change entropy is the slope of the line,

therefore when such changes takes place, the straight line will change its

slope. Let us consider the reaction




let us consider the melting of reactant. The reaction is

M(l) + O2(g) = MO(s)

S0 = S0 (oxide) - S0 (metal (l)) - S0 (oxygen gas)

Since S0 (metal (l)) > S0 (metal (s))

S0 = more negative than when both were solid

Hence the slope of the line will be further upwards from

the temperature of melting.

if the reactant boils. We can write

S0 (metal (g)) > S0 (metal (l))

S0 = more negative than when metal oxide was solid

and metal was liquid

Hence the slope of the line will be further upwards from

the boiling point..




let us consider the melting of product. The reaction is

M(s) + O2(g) = MO(l)

S0 = S0 (oxide(l)) - S0 (metal (s)) - S0 (oxygen gas)

Since S0 (oxide (l)) > S0 (oxide (s))

S0 = les negative than when both were solid

Hence the line will be bend downward with +ve slope from

the temperature of melting.

if the product boils. We can write

S0 (oxide (g)) > S0 (oxide (l))

S0 = less negative than when metal oxide was liquid

and metal was solid

Hence the line will be bend further downward with +ve slope from the temperature

of melting.







4) The intercept of the straight line with the ordinate at absolute zero gives

approximately the value of H0, since from the equation G0 = H0 - T S0

when T = 0, G0 = H0

5) since G0 must have a negative value for the reaction to take place, it can be seen

from this plot that all metals shown below the negative areas of G0 are oxidized

spontaneously by oxygen while those above are not e. g. gold

6) When a line touches the positive regions of G0 this means beyond that

temperature no further oxidation can take place or the oxide formed in that

region is unstable or oxides will start decomposing when G0 = 0. i. e Ag2O will

decompose at 15000C.




7) The stability of an oxide is directly related to its - G0 values; less stable

oxides have a small - G0 and more stable oxides have a high - G0 value.

8) An oxide can be reduced by only those metals below it in Ellingham

diagram; the reverse cannot take place. For example at 8000C, Cr2O3 can be

reduced by aluminium but Al2O3 cannot reduced by chromium:

4/3 Al + O2 = 2/3 Al2O3 G0 = - 800 kJ

4/3 Cr + O2 = 2/3 Cr2O3 G0 = - 500 kJ

2/3 Cr2O3 + 4/3 Al = 2/3 Al2O3 + 4/3 Cr; G0 = - 300 kJ







9) G0 of an oxide may be greater than another at low temperature but

becomes less than the other at higher temperature. Thus while the MnO is

reduced by Na below 2200C, the reverse is true above that temperature.

10) The line of the reaction C + O2 = CO2 runs nearly horizontally on the chart

i.e its slope is zero or practically there is no entropy change for this

reaction. This can be seen from the fact that the initial and final volumes

are practically the same which is the one volume of oxygen gas and one

volume of CO2, respectively; entropy of the solid being negligible.




11) The line for the reaction 2 C + O2 = 2CO runs downwards i.e. it has a negative

slope, this is due to the large increase in entropy: two volume of gaseous CO are

formed from one volume of oxygen gas. In this case G0 become more negative

as the temperature increases.

12) That the line for CO formation runs downward is of great importance in

pyrometallurgy. It enables almost all the metal – metal oxide lines meet C – CO

lines at high temperature. This makes most the oxides unstable beyond point of

intersection. This is called reversion of stability. This means that reduction of

most metal oxides by carbon is possible at high temperature.




13) Carbon monoxide can reduce all oxides above the CO2 lines. For example at 7000C NiO can be reduced by CO

2 CO + O2 = 2 CO2 G0 = - 95 kcal.

2 Ni + O2 = 2 NiO G0 = - 75 kcal.

2CO + 2NiO = 2Ni + 2CO2 G0 = - 20 kcal.

14) All oxides above H2O lines can be reduced by H2 e.g at 7000C

2 H2 + O2 = 2 H2O G0 = - 92 kcal.

2 Co + O2 = 2 CoO G0 = - 75 kcal.

2CoO + 2 H2 = 2Co+ 2 H2O G0 = - 17 kcal.




Nomographic scale on Ellingham diagram

The diagram already described can be made more useful

by superimposing grids or nomographic scales around

them.

1. PO2 grid : Consider the reaction involving the oxidation of pure solid metal to

pure solid oxide

2 M (s) + O2 (g) = 2 MO (s)

G0 = - R T ln (1 / PO2) = R T ln PO2

Hence knowing G0 and T the corresponding equilibrium PO2 can be calculated

for any such reaction.

When G0 = 0, PO2 = 1 atm and the equilibrium PO2 values radiate from the point

‘O’ on the G0 axis. For the fixed values of PO2 the G0 is plotted as a function

of T







2) CO / CO2 grid: For the reaction

2 CO + O2 = 2 CO2

G0 = - 135000 + 41.57 T cal / mole

G0 = - R T ln K = - R T ln (P2CO2 / P2

CO . PO2)

RT ln PO2 = G0 - 2 R T ln (PCO / PCO2)

= - 135000 + 41.57 T - 2 R T ln (PCO / PCO2)

RT ln PO2 can be plotted as a function of temperature for fixed values of CO /

CO2. When T = 0 , RT ln PO2 = - 135000 cal / mole. The equilibrium (PCO /

PCO2) values radiate from the point ‘C’ on the G0 axis.







3. H2 / H2O grid: For the reaction

2 H2 + O2 = 2 H2O

G0 = - 118000 + 26.57 T cal / mole

RT ln PO2 = G0 - 2 R T ln (PH2 / PH2O)

= - - 118000 + 26.57 - 2 R T ln (PH2 / PH2O)

RT ln PO2 can be plotted as a function of temperature for fixed values of H2 / H2O.

When T = 0 , RT ln PO2 = - 118000 cal / mole. The equilibrium (PH2 / PH2O) values

radiate from the point ‘H’ on the G0 axis.







Use of Nomographic Scale




Disadvantages of Ellingham diagram

1) The diagram is applicable only for the substances present in their standard states. But practically activity may not be unity.

2) Compound whose formation lines are represented in the diagram are assumed to be stoichiometric only which often is not true.

3) The information regarding rate of the reaction can not be obtained.

4) The diagram do not show the condition under which the reaction tend to occur.

5) Where oxide formation lines in the diagram are close together accurate measurement and subsequent calculation is difficult.

6) The possibility of formation of intermetallic compound between reactants and products is ignored




Non-standard state condition

If any one of the constituents of the reaction is not present in its standard state is called non-standard state condition.

The activity of that substance will be less than one. It can be best illustrated by the reduction MgO by silicon.

2 MgO(s) + Si(s) = 2 Mg(g) + SiO2(s)

G01473 = + 272 kJ

So there appears to be very little chance of using silicon as a reducing agent to produce magnesium from magnesia.




From the van’t Hoff isotherm, the actual free energy change of the this

reaction is given by

If pMg and aSiO2 can be lowered sufficiently, G can be made negative even

though G0 is positive.

PMg is lowered by working at a pressure of 10 – 4 atm.

aSiO2 is lowered by adding sufficient CaO to form orthosilicate (2 CaO. SiO2). A

basic slag would give aSiO2 < 0.001.

Si2MgO

SiO2Mg0

a.a

a.plnTRGG 2




Third law of Thermodynamics

Enunciation of third lawThe change in entropy with constant pressure and constant

volume is given by following equations

Integration of these two leads to the following

dTT

CdSand

dTT

CdS

V

P

V,0

T

0

V

P,0

T

0

P

SdTT

CS

SdTT

CS

Where S0, P, and S0,V are the integration constant.




These equations also state that for calculation of absolute value of entropy of

any substance, these constants should be known.

In order to evaluate the constant, Nernst analyzed the low temperature data

on free energy and enthalpy of chemical reactions.

The analysis lead to the formulation of first, the Nernst heat theorem and later

the third law of thermodynamics.




Nernst heat theorem

• Nernst collected enthalpy of

formation of various reactions

by calorimetric method and

determined free energies as a

function of temperature by

emf method.

• These free energies and

enthalpy when plotted against

temperature yielded curves as

shown.




It is seen from the figure that at low temperature that both

enthalpy and free energy of a reaction approaches each

other asymptotically with the asymptote laying parallel to

temperature axis. This lead Nernst to arrive at the following

conclusions:

As

Therefore, one can write

0

T

HLimii

0T

GLimi

0T

0T

S

T

G

P

0SLim0T




Statement of Nernst heat theorem

It states that both entropy change and temperature

coefficients of enthalpy and free energy changes for the

reaction tends to zero as temperature approaches 0 0K.

Entropy of a pure substance

A pure substance is a pure element or pure compound.

Consider the formation of compound AB from elements A

and B i.e

A(s) + B(s) = AB(s)

For which

At absolute zero temperature S0 = 0 it is possible provided

either i)

Or ii)

0B

0A

0AB

0 S-S-S=SΔ

)( 0B

0A

0AB S+S=S

zerolyindividualallareSandSS 0B

0A

0AB ,




From a probability point of view, the first alternative is very unlikely as a general feature. It may be true, by chance, in a few cases.

Hence alternative (ii) is accepted as of general validity and it may be concluded that “entropies of pure solids at T = 0 are zero”

Therefore, the absolute value of entropy of a pure substance at any temperature can be determined by taking T = 0, S = 0 as the lower limit.




Various statements of third law

“All reactions involving substances in the condensed state

S is zero at absolute zero”

“The entropy of any homogeneous crystalline substance

which is in complete internal equilibrium may be taken as

zero at 0 0K”

“The entropy of any phase whose quantum states and

atomic arrangement corresponds to a unique lowest energy

state at the absolute zero is zero”




Experimental verification of the third law

Third law is not so obvious as first and second law; hence it is required to

describe some of the evidence by which it was established.

The so called direct verification of the third law involves the application of the

second law, namely the principle that the entropy of the system is a function of

state and the net entropy change experienced by a system undergoing a cycle is

therefore zero.




• Let us consider the cycle shown in the figure.

• The starting point is a system composed of reactants at the absolute zero.

• The first step of the cycle consists in heating the reactants to temperature T.

• The reaction is allowed to completion at T in the second step.

• The products are cooled to absolute zero in the third step.

• The fourth step which imaginary, the reverse reaction is allowed to proceed

00

% reacted

I

II

III

IV

T

100

0K




We have from the second law

SI + SII + SIII + SIV = 0

Or SIV = - (SI + SII + SIII )

Since the entropy change in the first three steps measured experimentally, the

entropy change at absolute zero can be determined.

Let us consider the following transformation reaction

)white(Sn)gray(Sn C190

EU04.11Tlnd)white(CS

EU85.1292

541

T

QS

EU11.9Tlnd)gray(CS

0

292

PIII

r

rII

292

0

PI




The entropy change at absolute zero ‘S0’ is

S0 = SIV = - (SI + SII + SIII )

= - (9.11 + 1.85 – 11.04) = 0.08 EU

This is probably smaller than the experimental error and thus it has been

shown that in this case S0 is zero within the experimental error.




Consequences of the third law

Third law leads to following important consequences

1. CP, CV and will be zero at absolute zero.

0C0Tat,finiteisT

Sas

T

STCwritecanwe

CT

HandS

T

GknowweAs

ST

ST

T

H

T

G

STHG

P

P

P

P

P

PP

PPP

Prove that CV is zero at absolute zero




00T

V

0P

Sso0S,0Tas

P

S

T

V

equations'wellmaxtoaccording

T

V

V

1

P

0T

TP

P




2. Unattainability of absolute zero temperature

let us consider a carnot cycle in which heat is absorbed at temperature T (T > 00K)

delivered to cold reservoir maintained at 00K. Entropy change during such a cycle

will be equal to zero i. e.

SI + SIII = 0

where SI is the entropy change during isothermal expansion and SIII is the entropy

change during isothermal compression.




According to third law the entropy change during isothermal

volume change (SIII ) at 00K is zero. Hence SI will be zero.

But this not true as both heat exchange and temperature for

this is finite non-zero and positive.

Hence one cannot design a cycle in which the temperature of

the working substance descend to the absolute zero.

Thus it can be concluded that the temperature of 00K cannot

be attained.




3. Calculation of entropy and free energy from calorimetric data.Entropy of a substance at constant volume is given by

From the knowledge of heat capacity which can be

determined by calorimetric techniques, one can

calculate the absolute value of entropy of a component

at any temperature T.

T

0

VV

V,0

T

0

V,0VV

TlndCS,therefore

0SlawthirdperAs

STlndCS




Third law also helps in the calculation of free energy during the

chemical reaction from the calorimetric data obtained in the

form of enthalpy of a reaction as a function of temperature. We

know G = H – T S

Where H0 is the enthalpy of reaction at 00K.

Or

All the quantities present in the above expression can be

determined by calorimetric techniques. Thus, one can calculate

the change in free energy of any reaction at desired temperature

T

0

P0

T

0

P TdCHHandTdT

CS

T

0

PT

0

P0 TdT

CTTdCHG




ProblemCalculate the standard free energy change G0 and free energy change G for the following reaction at 1000 K. Ca (a = 0.9 ) = Ca (a = 0.5 )

SolutionG0 is calculated by measuring the free energy change of both reactants and products in their standard states. Since calcium is solid at 1000 K the standard state for both reactant and products is pure solid Calcium as indicated and thus G0 = 0




we know G = G0 + RT ln K

= G0 + RT ln (aproduct / aproduct)

= G0 + RT ln (aCa0.5 / aCa0.9) = 0 + 8.314 x 1273 ln (0.5 / 0.9) G = - 4.88 kJ / molProblemConsider the equilibrium reaction of pure solid CaSi at 1000 K with Ca and Si dissolved in a solvent. Suppose that activity of calcium is 0.5 in the solution and CaSi is pure. Find

i) aSi in equilibrium with Ca (a = 0.5) and pure CaSi

ii) PCa in equilibrium with dissolved Ca.

iii G1000K of the reaction: Ca (a = 0.5) + Si(a= 0.4) = CaSi (a = 0.8)

Given : Ca(s) + Si(s) = CaSi (s); G01000k = - 172 kJ / mol

P0Ca = 14.6 N / m2




Solution

i)For the reaction: Ca(s) + Si(s) = CaSi (s)

G0 = - RT ln K

K = exp (- G0 / RT)

= exp( 172000 / 8.314 x 1273 ) = 8.92 x 108

Or K = (aCaSi / aCa. aSi) = 8.92 x 108

since CaSi exists in pure states, aCaSi = 1 and

putting aCa = 0.5 we get aSi = 2.24 x 10-9

ii) It is given that aCa = 0.5, since aCa = PCa / P0Ca .

We can obtain PCa if the value of P0Ca is known.

Since our standard states for Ca is pure solid at

1000K, P0Ca is simply the vapour pressure of pure Ca at

1000 K. It is given that P0Ca = 14.6 N / m2, So PCa =

7.3 N / m2




iii) For the reaction

Ca (a = 0.5) + Si(a= 0.4) = CaSi (a = 0.8)

G = G0 + R T ln K

= - 172000 + 8.314 x 1273 ln (0.8 / 0.5 x 0.4)

= - 160 kJ / mol

Problem

Solid TiO2 is converted into gaseous TiCl4 by treatment with

chlorine in presence of carbon. Calculate the thermodynamic

utilization of chlorine gas at 10000C. Assume the oxidation of

carbon into CO and total pressure as 1 atm.

Given: TiO2(s) + 2 C (s) + 2 Cl2 (g) = TiCl4(g) + 2 CO (g)

G0 = - 318 kJ / mol at 10000C.




Solution

G0 = - R T ln K

Or – 318 000 = - 8.316 x 1273 ln K

Or K = 1.11 x 1013 = (P2CO . PTiCl4) / (aTiO2 . a2

C . P2Cl2)

Assuming TiO2 and carbon as pure solid aTiO2 = 1= aC

From the stoichiometry of the reaction production of one mole

of TiCl4 leads to the production of 2 moles of CO i.e.

PCO = 2 PTiCl4

Making all the substitution

P3TiCl4 / P2

Cl2 = 0.275 x 1013 i.e PTiCl4 >> PCl2

So as a first approximation let us take PTiCl4= PT = 1atm. Then

we get PCl2 = 6 x 10 -7 atm. This is truly negligible. Therefore,

the thermodynamic utilization of Cl2 is 100%.




Problem

A gas consisting of 60.2% H2 , and 39.8 % H2O at 1

atmospheric pressure is in equilibrium with pure γ – iron at

9100C. At the same temperature the gas composition in

equilibrium with an iron – nickel alloy (0.71 atom fraction of

iron) is 51.9% H2 and 48.1% H2O. Determine the activity of

iron in the alloy.

SolutionIn equilibrium with pure iron the reaction is:

Fe (s) + H2O(g) = FeO(g) + H2(g)

For which: K = (aFeO . PH2) / (aFe . PH2O)




Fe and FeO are in their standard state (aFe = 1 = aFeO)

K = (PH2 / PH2O)Fe, Pure

The corresponding reaction with the alloy is:

Fe (Fe - Ni) + H2O(g) = FeO(g) + H2(g)

The activity of pure FeO is again unity. But the activity of iron is

less than unity in the alloy. Thus :

K = (PH2 / aFe . PH2O)Fe – Ni

However the value of the equilibrium constant is unchanged on

the solution of the iron in the alloy, then:

(PH2 / PH2O)Fe, Pure = (PH2 / aFe . PH2O)Fe – Ni

Or aFe = (PH2 / aFe . PH2O)Fe – Ni / (PH2 / PH2O)Fe, Pure

Substituting the gas composition

aFe = (0.519 / 0.481) / (0.602 / 0.398) = 0.713




ProblemChromium plates are bright annealed at 7270C in wet

hydrogen atmosphere. The pressure of wet hydrogen is 1 atm.

i) Calculate the permissible water content in the hydrogen if there is to be no oxidation

at 7270C.

ii) Will annealed chromium plates be oxidized when cooled to 2270C in the furnace

atmosphere as calculated in (i).

Given: 2 Cr (s) + 3 H2O(g) = Cr2O3(s) + 3 H2

G0 = - 91050 + 22.80 T cals.

Solution

i)For the reaction: 2 Cr (s) + 3 H2O(g) = Cr2O3(s) + 3 H2

G01000K = - 68250 cal = - R T ln K




K = exp( - G01000K / 1.987 x 1000) = 8.26 x 10 14

K = (PH2 / PH2O)3 = 8.26 x 10 14

(PH2 / PH2O) = 9.384 x 10 4

As PH2 + PH2O = 1

After solving above equations we get:

PH2O = 1 / (1 + 9.384 x 10 4 ) = 1.065 x 10 – 5 atm.

Since total pressure is 1 atm, the volume percent of H2O

is1.065 x 10 – 3. This the maximum water content of H2 to

avoid any oxidation of chromium.




ii) G0500K = - 79650 cal = - R T ln K

K = 6.57 x 10 34

The activity quotient is :

J = (PH2 / PH2O)3 = (1 / 1.065 x 10 – 5 )3 = 8.25 x 1014

J / K = 8.25 x 1014 / 6.57 x 10 34 < 1 so forward reaction

will oxidise the Cr




ProblemThe lowest temperature at which copper oxide (Cu2O)can

dissociate in a vacuum of 10 – 5 mm Hg.

Given : Cu2O(s)= 2 Cu(s) + ½ O2(g)

G0 = 40500 + 3.92 T log T – 29.5 T cal

Solution:

For the reaction: Cu2O(s)= 2 Cu(s) + ½ O2(g)

G = G0 + (RT /2) ln PO2 = G0 -2.288 T log PO2

PO2 = (0.2 x 10 -5 ) / 760 = 2.635 x 10 – 9 atm.

At equilibrium G = 0

G0 = -2.288 T log (2.635 x 10 – 9 ) = 19.6 T

Or 40500 + 3.92 TD log TD – 29.5 TD =19.6 TD

Or 3.92 TD log TD = 49.1 TD – 40500

This is best solved graphically and TD = 1085 K.




Problemi. At what temps will carbon reduce a) SnO2(s) b) Cr2O3 (s) and c) SiO2(s)?

ii. Steam blown through hot coke gives rise to the fuel gas mixture called water gas (CO+H2): C + H2O = CO + H2. calculate the temperature the coke must be maintained for the reaction to be feasible?

iii. At what temperature the Ag2O just begin to decompose at one atmospheric pressure?

iv. In what temperature range can hydrogen be used to reduce SnO2 to Sn?

v. Deduce the standard free energy change for the reduction of Al2O3 by Mg at 1000 0C.

vi. Explain the reasons for the change in slope of the following lines:




2Mg + O2 = 2 MgO at 1100 0C.

2Pb + O2 = 2 PbO at 1470 0C

4 Li + O2 = 2 Li2O at 1300 0C

VII. Estimate the standard free energy change for the following reaction at 1200 K. a) reduction of copper (I) oxide with hydrogen and b) reduction copper oxide (I) with carbon.

VIII. At what temperature is the reaction: 4/3 Cr + O2 = 2/3 Cr2O3

at equilibrium when PO2 is 10-14 atm.

IX. Calculate the equilibrium oxygen pressure between Al2O3 and Al at 1000 K. Could a vacuum of 10-10 mm Hg prevent the oxidation of aluminum?

X. Suppose CaO is placed in a vacuume in which the partial pressure of oxygen is 10-5 mm Hg. Will CaO be reduced.

XI. Which is the suitable material for the steam pipe Ni or Al?

XII. What is the equilibrium CO / CO2 ratio for the following reaction: MnO + CO = Mn + CO2

XIII. At what temperature is the reaction PbO + H2 = Pb + H2O at equilibrium when the H2 / H2O ratio is 1/104?




ProblemCalculate the vapour pressure of liquid silver at its melting

point, 960 0C, making use of Trouton’s rule. Boiling point of

silver is 2210 0C.

Solution:

atm10459.246.10)273960(314.8

216021plnor

46.10TR

HplnSo

.46.10betooutcomesC

K2483Tat1PconditionthebyevaluatedbecanC

CTR

Hpln

equilibriavapourliquidtheforknowWe

.mol/Jk021.216248387HHence

K/J87T

H

rules'TroutontoAccording

5

V

V

V

b

V




ProblemThe equilibrium vapour pressure in atm. of a solid is given by the

expression: ln p = - 4.085 + 1.57x10 -4 ln T – 2.83x10- 3 T

Calculate the heat of sublimation of solid at 300 0C.

Solution:

getweionaboveequattheinthisputting

1083.2T

1057.1

Td

plndSo

T1083.2

Tln1057.1085.4pln

solidtheforthatgivenisit

Td

plndTRH

TR

H

Td

plnd

equilibriavapoursolidtheforknowWe

34

3

4

2S2

S




ProblemCalculate the melting point of ice under a pressure of 2 atm.

From the following data.

Density of water at 00C = 1.0 gm /cc

Density of ice at 00C = 0.9174 gm /cc

Latent heat of fusion of ice = 80 cal / gm

Assume that the above data to be independent of temperature

and pressure and also that the ice melts at 0 0C under 1 atm.

Pressure.

K573at.cal085.1846isationlimsubofHeat

.cal085.1846

1083.2107399.2573987.1

1083.2)273300(

1057.1RTH

372

34

2S




Solution

CPH

VVTln

equilibrialiquidsolidtheforknowWe

c.c62.199174.0

18

density

eightmolecularwC0iceofvolumeMolar

c.c181

18

density

eightmolecularwC0atwaterofvolumeMolar

M

sl

0

0

6095.5CorC1107.2273lnSo

atm1P,K273TconditiongiventheFrom

CP107.2

CP293.411880

62.1918Tln

atmc.c293.41cal1

atmc.ctocalfromconversionthefor

5

5




ProblemCalculate the rate of change of melting point of iron with

pressure with the help of the following:

M. P of iron = 1535 0C.

Density of solid iron at melting point = 7.86 gm/cc

Density of liquid iron at melting point = 7.55 gm/cc

Latent heat of fusion of iron = 3.3 kcal / mol

Atomic weight of iron = 56

K98.272Tor

6094.56095.5)2107.2(Tln

getweandequationabovethein.atm2Pand

6095.5CofvaluetheputcanWe..atm2

aticeofintpomeltingofncalculatiotheFor

5




ProblemRhombic sulphure transform to monoclinic sulphure at 368.5

K. with an enthalpy changes of 96 cal / mol. test the validity of

the third law of thermodynamics for this transition from the

following data:

i) CP of the rhombic sulphure at 15 K = 0.3 cal/deg/mol

ii) Area under the curve drawn between (Cp / T ) and T for rhombic sulphure between T = 15 K to 368.5 K is 8.71 cal/deg/mol.

.atm/reedeg1088.31808

293.413300

1246.74172.7T

H

VV

Pd

Td

equilibrialiquidsolistheforknowwe

atmc.c293.41cal1

mol/cc4172.755.7

56

density

.wtmolecularironliquidofvolumeMolar

mol/cc124.786.7

56

density

.wtmolecularironsolidofvolumeMolar

3m

m

sl




iii) For monoclinic sulphure: S368.5 – S0 = 9.07 cal /deg / mol

Assume that the entropy contribution of rhombic sulphure

below 15 K follows Debye’s equation.

Solution:

0

0

% reacted

I

II

III

IV

T

100

0K

SR → SM




5.368

15

P15

0P

5.368

15

P15

03

5.368

15

P15

0

2

5.368

15

P15

0

3

I

3P

5.368

15

P15

0

P

dTT

CC

3

1

dTT

CTa

3

1

dTT

CdTaT

dTT

CdT

T

TaS

TaCK15belowobyedislaws'Debye

dTT

CdT

T

C




5.368

15

P15

0

3

I

3P

5.368

15

P15

0

P

5.368

0

P05.368I

dTT

CdT

T

TaS

TaCK15belowobyedislaws'Debye

dTT

CdT

T

C

dTT

CSSS

stepfirstduringchangeEntropy




2605.05.368

96

T

HS

mole/deg/cal81.871.83.03

1

dTT

CC

3

1

dTT

CC

3

1

tr

trII

5.368

15

P15P

5.368

15

P15

0P




.verifiedislawthirdtheThus.zerois

zeroabsoluteatreactiontheofchangeentropythethatmeansit

moldeg//cal0005.0SSSS

lawondsectoAccording

07.9)SS(SSS

IIIIIIIV

05.3685.03680III




Statistical Interpretation of Entropy

Statistical Thermodynamics: It deals with the interpretation and derivation of thermodynamic

properties based on the properties of the particles constituting

the system and their distribution in the system.

Elementary Statistical ConceptsMicrostates: In a gaseous system, each particle shall have

three position coordinates namely x, y, and z and three velocity

coordinates vx, vy, and vz. If at any instant one is able to specify

value of all these six coordinates for each of the particle

constituting the system, such a complete specification is said to

define the microstate of the system.




• A set which defines all the possible microstates of a system is called an ENSEMBLE.

• A set of microstates having same characteristics is called a MACROSTATE.

let us consider an example a system consisting of three

distinguishable balls A, B and C contained in a box consisting

of two compartments I and II. Let us construct an ensemble.

Position of

the Particles

Microstates No.

1 2 3 4 5 6 7 8

Compartment I A, B,C A,B A,C B,C A B C -

Compartment II - C B A B,C A,C A,B A,B,C




These eight microstates can be grouped into the following four

macrostates:

I. All the three particles lie in compartment I – only one No.1 microstates

corresponds to this macrostates.

II. Two particles present in compartment I and only one in compartment II:

three microstates namely 2, 3, 4 correspond to this macrostates.

III. one particle present in compartment I and two in compartment II: three

microstates namely 5, 6, 7 correspond to this macrostates.

IV. All the three particles present in compartment II corresponds to Microstate

no.8




In actual practice a system consists of a large no. of particles and hence for every

macrostates there are large no. of microstates and no. of such macrostates are also

large. For the calculation microstates for a macrostates, one uses the technique of

permutation and combination.

Next, how to calculate the no. of microstates or maximum no. of possible

arrangement of particles in a system.

This can be illustrated by an example: let us consider first where we have, say,

eight object numbered 1 through 8 and four boxes.




Let us attempt to answer the question: in how many different ways can these

objects are placed in these boxes assuming that each box is large enough to contain

the all of the objects.

We may put object labeled 1 in any of the four boxes. Similarly, since each event is

independent we may do likewise for the object 2 etc.

The total no of ways of putting object 1 is four . The total no of ways of putting

object 2 is four, etc up through object 8.

The total no of ways of arranging all the balls is therefore 48.




In the situation we will be dealing with (atoms in crystals) we can differentiate

atoms. We can in principle distinguish a gold atom from a copper atom. However

we can not distinguish within the given type one from other.

The situation discussed above therefore is not completely applicable to our case: we

shall not be interested in which object is in given box but, how many of the same

type of object are in each box.

To illustrate, let us calculate the number of ways for which there are two balls in

each box. Such a total is called the total no. of microstates in the given macrostate.




Thus we need to calculate how many ways we have of arranging two objects in box 1 two of the remaining six in box 2 and two of the remaining four in box 3 etc.

The first step is to calculate the no of ways of picking two objects out of eight for box 1. This is done by combination of N objects taken n at a time i.e 8C2 = 8! / (2! 6!).

We have six objects left. So no. of ways of putting two object out of remaining six in box 2 is 6C2 = 6! / (2! 4!).

Similarly for box 3 we have four objects left and 4C2 = 4! / (2! 4!) and so on..

Now we are ready to answer: the total no of ways putting two balls in each box is (W)

W = 8C2 x 6C2 x 4C2 x 2C2 = N! = 8! = 2520

n1!n2!n3!n4! (2!)4




What is the probability of achieving the macrostate in which we have two objects in each box. The answer is:

2520 / 48.

The probability of a system to be present in a particular macrostate is directly proportional to the no. of microstates which it has.

In other wards we can say a system will spend highest time fraction in the macrostate which has highest no of microstates (W).

It can be further simplified as it is this macrostate of highest probability which determines all the macroscopic properties of the system.




Entropy and Most Probable Macrostate

Properties of a system depends on the most probable macrostate i. e the

one having highest no. of microstates (W).

It has been proved that for the isolated system of constant volume have

maximum entropy at equilibrium.

So one can conclude that both W and entropy must be related to each

other.

Let such a relation be expressed as:

S = f(W)




In order to arrive at the analytical form of this function, let us

consider a system divided into two sub-system A and B. The

no of microstates W in any macrostate will be equal to the

product of the microstates WA and WB of the sub-system.

Hence

W = Wa. WB

Total entropy of the system is

S = SA+ SB

As we can write

SA = f (WA)

SB = f (WB)

So f(WA. WB) = f(WA) + f(WB)




B

B

BA

BAA

B

B

B

A

B

BA

BA

BA

B

W

Wf

WW

WWfW

yieldstionsimplificaonwhich

W

Wf

W

Wf

W

WW

WW

WWf

givesWtorespectwithantionDifferenti

)(

).(

).(.

)()().(

).(

).(

0

.

)(

).(

).(

).(

).(.

.

)(

).(

).().(

).(

).(.

...

2

2

2

2

2

2

A

B

BBA

B

BA

BAB

BA

BAA

BA

B

BA

BA

A

BA

BA

BAA

A

W

Wf

WWW

Wf

WW

WWfW

WW

WWfW

or

WW

Wf

WW

WWf

W

WW

WW

WWfW

givesWtrwationDifferenti




for a perfect crystal S = 0 for W = 1, we get C2 = 0

and C1 can be determined from the throttling of one mole of ideal gas. The gas expands by throttling process to double its volume. The throttling process is isenthalpic process.

21

21

2

2

2

2

ln

ln)(

..sec

0)(

)(

)(

)(.

0).(

).(

).(

).(.

CWCSThus

CWCWf

iseqdifforderondthetosolutiongeneralA

W

Wf

W

WfW

or

WW

WWf

WW

WWfWW

BA

BA

BA

BABA




Let us calculate the W in the initial and final state

2ln

2/

ln

RSSTherefore

PPthrottlingofprocessthisIn

P

PRdP

P

RSSS

isprocessthisduringchangeentropyThe

if

fi

i

fP

P

if

f

i

WKSgetweCofvaluetheputting

KN

RC

NNN

NNNNNN

Cor

WWCSS

NNNNeiionapproximatsterlingtoAccordingN

NW

NN

NW

ifif

i

f

ln

22ln

2

22ln

2ln

2ln

;lnln

ln!ln.

1!

!

!)2/(!)2/(

!

1

11

1




Entropy of Mixing for a Binary Solution

Component A(N0 –n) atoms

Component Bn atoms

SolutionN0 atoms

(N0-n) A + n B = solution [(N0- n) A, n B]

Sm

Sm = SA,B – SA – SB

Or Sm = K (ln WA,B – ln WA – ln WB)

In 1 mole of solution there are N0 lattice sites

(Avogadro's no.)




WA,B is the no.of ways of arranging (N0 – n) A atoms and n B atoms on N0 sites so

Since, in pure metal the atoms are indistinguishable so

WA = WB= 1. Thus

entropy of mixing

of ideal solution

)!(!

!

0

0, nNn

NW BA

BBAAm

BA

m

m

xxxxRSSo

N

nxand

N

nNxas

nN

N

N

nN

N

n

N

nKNS

ionapproximatsterlingApplying

nNn

NKS

lnln

lnln

)!(!

!ln

00

0

0

0

0

0

000

0

0




THERMODYNAMICS OF SOLUTIONS

A solution may be defined as a homogeneous phase composed of

different chemical substances, whose concentration may be varied

without the precipitation of a new phase.

It differs from a mixture by its homogeneity and from a compound

by being able to possess variable composition.

Solution may be gaseous, liquid or solid. It may be classified as

binary or ternary solution depending on whether it contains two or

three components.

A binary solution has two chemical substances (elements or

compounds), e.g. molten cadmium and zinc miscible in all

proportions.




Composition of solution

Composition of solution may be expressed in a number of ways. In metallurgy usually by weight, mol or atom percent, for example, if w1 and w2 are weights of the solvent and solute in the solution:

where WA and WB are weights of components A and B having atomic/molecular weights MA and MB, respectively.

100%21

2

ww

wsolutewt

100%/%

B

B

A

A

A

A

Mw

Mw

Mw

Amolatom




Composition of a solution

If nA number of moles of A and nB number of moles of B form a solution, A-B. atom fractions of A and B are given as

and xA + xB = 1

BA

AA nn

nx

BA

BB nn

nx




Raoult’s law

The law states that the relative lowering of the vapour pressure a solvent due to the addition of a solute is equal to the mole fraction of the solute in the solution.

Imagine A and B forming a solution, each one exerts its own vapour pressure, pA and pB in the solution, respectively.

Suppose p is the total pressure of the solution and xA and xB are

atom/mole fractions. If p0A and p0

B are the partial pressure of

pure A and pure B, respectively at the same temperature at which solution exists.

0

0

BA

AA xp

pp

0

0

AB

BB xp

pp




Ideal Solution

B

A

A xp

p

01

BB

BAB

A

A xp

pandxx

p

p

001

)(. and

)(...0

B

0

BBBB

AAAAA

xporxpp

xporxppei

A solution which obeys Raoult’s law is called an ideal solution.

BBAA BBAA 2

1




Ideal Solutions




Non-ideal or real solutions

Deviations from Raoult’s law occur when the attractive forces between the molecules of components A and B of the solution are stronger or weaker than those existing between A and A or B and B in their pure states.

For example, if there were a stronger attractive force between components A and B in solution than the mutual attraction between molecules of A-A and of B-B, there would be less tendency for these components to leave the solution is observed.

In this case the vapour pressure whould be less than that predicted by Raoult’s law . This is called negative deviation from Raoult’s law.









Using the same argument if the attractive force between A and

B was less than that between A and A or B and B in their pure

states there would be a greater tendency for these components

to leave the solution as a gas, thereby increasing the vapour

pressure above the liquid. This is represented in Fig and is

called positive deviation from Raoult’s law.

Immiscible liquids exhibit positive deviation since the

attractive force between the components in the liquid is low.




Activity

We have already defined activity as

For ideal solutions we have:

For an ideal solution it will be seen that aA = xA, xA the mole fraction of A in the solution.

if the solution deviates from Raoult’s law we write

where γA is a fraction greater or less than unity for a positive or negative deviation, respectively is called Raoultian activity coefficient.

For a pure substance: xA = 1 and γA = 1, so that we have unit activity of the substance A which is said to be in its ‘standard state’

0A

AA p

pa

AAA xpp .0

AAA xa .




Activity

If we return to our original argument that the vapour pressure

of a substance is a measure of its attraction to the solution in

which it exists and hence a measure of its availability for

reaction, perhaps with another phase, we can state that this

fundamental definition of activity, as that fraction of molar

concentration “available” for reaction, is universally

applicable.




Activity

Let us now consider the physical significance of both Raoult’s

ideal and non-ideal behaviour of binary systems.

1. Raoult’s ideal behaviour: If in a solution composed of A and B (atoms or

molecules) the attractive forces between A and B are the same as between

A and A or between B and B, then the activities of A and B in the

solution at all concentrations will be equal to their mole fractions and the

solution is said to be ideal. The system Bi-Sn can serve as an example of

such a solution at a particular temperature. In this case the net attractive

force between Bi and Sn in the solution can be represented by the

equation: BBAABBAABA

2

1)(




Activity

2. Positive deviation: When the net attractive force between the substances

A and B is less than between A-A and B-B, then the solution of A and B

exhibits positive deviation from Raoult’s law. In this case the Raoultian

activity coefficient is always greater than unity except when approaching

the concentration of xA 1 Pb and Zn liquid solutions show such

behaviour at temperature above 1071K. The heat of solution in systems

showing positive deviation is endothermic. In general a solution, A-B

exhibiting positive deviation has

))}({(2

1BBAABA




Activity

3. Negative deviations occur when the attractive force between the two

components A and B is higher than between A-A and B-B. For example

Mg-Bi system shows such a behaviour. Negative deviations generally

indicate a tendency for compound formation (Mg3Bi2). The heat of

solution for systems exhibiting negative deviatias is usually exothermic.

Occasionally both negative and positive deviations from Raoult’s law

occur in the same binary system. An outstanding example of this

behaviour is found in the Zn-Sb, Cd-Bi and Cd-Sb systems. In general in

case of negative deviation in the system, A-B we have

BBAABA 2

1




Activity

Fig.: Relationship between activity and mole fraction of substance

A in three solutions-one ideal, the second showing a positive

deviation and the third a negative deviation from ideal behavior.




PARTIAL MOLAR QUANTITIES

In dealing with solutions, one of the first questions which naturally arises is

how to express molar quantities of a substance in solution when two liquids

are mixed, the total volume of the solution is, in general, not equal to the sum

of the individual volumes before mixing, this reflects the difference of the

interatomic forces in the pure substance and in the solution.

The problem presented is not solved directly but is avoided by the

introduction of partial molar quantities. Since the same general treatment is

applicable to any extensive thermodynamic quantity such as volume, energy,

entropy and free energy, we shall use the symbol Q to represent any one of

these.





Prime is used to indicate any arbitrary amount of solution rather than one mole, molar quantities are represented as unprimed.

Thus Q is the total quantity of solution, Q the molar. If n1, n2 , n3, ………are number of moles of components 1,2

3,……. respectively in the solution, we have

)1(..........321

nnn

QQ





Let us imagine that to an arbitrary quantity of a solution an infinitesimal number of moles, dn1, of component 1 is added at constant temperature and pressure without changing the amount of other constituents. The corresponding increment in the property Q is dQ, the ratio

Is known as the as the partial molar quantity of components 1 and designated as

......3,2,,1 nnTPn

Q

)2(,.......3,2,,1

1 nnTPn

QQ





may be represented equally well as the increment of Q when 1 mol of

the first component is added to a very large quantity of the solution for

example, if the volume increase accompanying the addition of 1 gm

mol/atom of Cu to a large quantity of a liquid alloy is observed to be 8.5

cc, the partial molar volume of copper in the alloy at the particular

composition, temperature and pressure is 8.5 cc. This is written as

We know Q’ =Q’(T,P, n1, n2, ……)

From the fundamentals of partial differentiation we have at constant

pressure and temperature

1Q

ccVCu 5.8

)3(..........2

3,121

3,21

dn

n

Qdn

n

QQd

nnnn

)4...(..............................' 332211 dnQdnQdnQdQ





If we add to a large quantity of solution n1 moles of component 1, n2

moles of component 2 etc the increment in Q after mixing is given as

If we now mechanically remove a portion containing n1 + n2 + n3 ….moles,

the extensive quantity Q for the main body of solution is now decreased

by ( n1 + n2 + …) Q.

Since at the end of these processes the main body of the solution is the

same in composition and amount as it was initially, Q has the same value

finally as initially and the increment in Q accompanying the individual

addition is equal to the decrement accompanying their mass withdrawal

...........2211 QnQn





Dividing by (n1 + n2+………) and noting that

on multiplying by (n1+ n2 + n3 ……..) we can write as

In a binary solution we can write equations 5 and 6 as

.......................)( 221121 QnQnQnn

ii x

nnn

n

321

)5(..........2211 QxQxQ

)6..(....................2211 QnQnQ

)6(

)5(

2211

2211

QnQnQ

QxQxQ





On differentiating equation 6, we get

Subtracting (4) from (7) we get

Equation (9) is known as one of the forms of Gibbs – Duhem equation.

)7(22112211 dnQdnQQdnQdnQd

)8(02211 QdnQdn

)9(02211 QdxQdx




Method of obtaining partial molar quantities from molar quantities

Differentiate equation (5)

Combining equation (9) and (10)

Multiplying (11) by (x1/dx2)and putting dx1 = -dx2,(x1 + x2 = 1)

Adding this to (5) we get

)10(22112211 dxQdxQQdxQdxdQ

)11(2211 dxQdxQdQ

2111212

111

21 . QxQxQx

dx

dxQx

dx

dQx

222212

1 QQxQxdx

dQxQ




Method of obtaining partial molar quantities from molar quantities

Similarly,

)12(12

22

12 dx

dQxQ

dx

dQxQQ

)12(11

11 adx

dQxQQ




Tangent intercept method

1Q

2Q

BD F

G

H

0

EA C

↑Q

1 x2 2

CHCFCHCF

x

CFCHxCF

dx

dQxQQ

xx

yyslope

x

CFCH

dx

dQ

22

222

12

12

22

1).1(

)1(

)(1

AGADAGAD

x

ADAGxAD

dx

dQxQQ

xx

yyslope

x

ADAG

dx

dQ

11

111

12

12

11

1).1(

)1(

)(1




Method of obtaining one partial molar quantities from another partialmolar quantities

Integration of this equation from x1 =1 to x1 = x1 will result




Method of obtaining one partial molar quantities from another partialmolar quantities

If the input in the calculation is the functional relationship

between and ⎯ x2, then

If the input in the calculation is the functional relationship

between and ⎯ x1, then




Problem







Obtaining Molar propertyfrom Partial molar properties




Problem

The enthalpies of mixing of Cd-Sn alloys at 5000C are given below:

xCd 0 0.1 0.3 0.5 0.7 0.9 1.0

HMCal/mole 0 298.2 652.4 800.0 620.5 251.5 0

Calculate the values of the partial molar enthalpies of mixing of cadmium and tin in a Cd-Sn alloy containing 60 at % cadmium.

Solution:

To find the value of partial molar enthalpies of mixing of cadmium,

Draw a tangent at xCd = 0.6. The values of and are obtained by intersection of the tangent with the axes at xCd = 1 and xSn = 1, respectively.

CdMM

SnMCd xvsHplotHeitinofthatandH ,..

MCdHM

SnH

./1360/320 molcalHandmolcalH SnCd




Problem

In the formation of liquid brass: (1-x) Cu(l) + x Zn(l) = Cu-Zn (l), the molar heat of formation is given by H = - 7100 x (1-x) cals, where x is the atom fraction of zinc. Derive expressions for partial molar heat of mixing of Cu and Zn in the liquid brass as a function of composition.




Solution

H = - 7100 x (1-x)

=

=

Zn

ZnZn x

HxHH

.1

xxdx

d

x

H

x

H

Zn

17100

xxxdx

d2171007100 2

xxxxH Zn 217100117100

xxx 21710071001

2171001171002117100 xxxxxx




solution

= -7100 x (1-x) – x [ -7100 (1-2x)]

= -7100 x [1-x – (1-2x)]

= -7100 x [1-x -1 + 2x] = -7100 x (x) = -7100 x2 Ans.

x

HxH

x

HxH

x

HxH

x

HxHH

ZnZn

CuZn

CuCuCu

.1




A 30% mole by methanol –water solution is to be prepared.

How many m3 of pure methanol (molar volume

=40.7x10-3m3/mol) and pure water (molar volume = 18.068x10-

6m3/mol) are to be mixed to prepare 2m3 of desired solution.

The partial molar volume of methanol and water in 30%

solution are 38.36x10-6 m3/mol and 17.765x10-6 m3/mol

respectively.

Problem




Methanol =0.3 mole fraction

Water=0.7 mole fraction

V=0.3 x38.36x10-6+0.7x17.765x10-6

=24.025x10-6 m3/mol

For 2 m3 solution

mol10246.8310025.24

2 36

Solution




Number of moles of methanol in 2m3solution=83.246x103x0.3= 24.97x103mol

Number of moles of water in 2m3solution

=83.246x103x07= 58.272x103mol

Volume of pure methanol to be taken

= 24.97x103 x 40.7x10-3 =1.0717 m3

Volume of pure water to be taken

= 58.272x103 x 18.068x10-6 =1.0529 m3

Solution




Find weather the equation given below is thermodynamically consistent

)xx10(xxx150x100G 212121

111 x

G)x1(GG

112 x

GxGG

101x16x35x18G 12

13

11

150x8x18G 21

312

Problem




16x70x54dx

Gd1

21

1

1

12

11

2 x16x54dx

Gd

0)1(1

21

1

11

dx

Gdx

dx

Gdx

0)x16x54)(x1()16x70x54(x 12

1112

11

It satisfies the GD equation, the above equation is consistent.

G D equation




The Free Energy Change due to the Formation of a solution

The pure component i, occurring in a condensed state at the

temperature T exerts an equilibrium partial pressure, pi0 . When

occurring in a condensed solution at temperature T, it exerts a

low equilibrium pressure pi. Consider the following steps:

1. evaporation of 1 mole of pure condensed i to pure i at the

pressure pi0 at T.

2. change in the pressure of 1 mole of vapour i from pi0 to pi at the

temperature T.

3. Condensation of 1 mole of vapour i from the pressure pi into the

condensed solution at T.





The difference in molar free energy between pure i and i in

the solution = Ga + Gb +Gc.

However as steps a and c are equilibrium processes, Ga and

Gc are both zero.

The overall free energy change for the three step process thus

equals Gb which can be written as

Gb = Gi(in sol)-Gi (pure) = RT ln pi – RTln pi0

= RT ln (pi /pi0) = RT ln ai

but Gi (in solution) is simply the partial molar free energy of i

in the solution, (pure) is the molar free energy of

pure i, Gi0 .

iG and iG





The difference between the two is the free energy change accompanying the dissolution of 1 mole of i in the solution.

The quantity is designated as the partial molar free energy

of the solution of i. Hence , If at constant T and P, nA number of mole of A and nB of moles

of B are mixed to form a binary solution, free energy before mixing = nAGA

0 + nBGB0

free energy after mixing = The free energy change due to mixing, GM, referred to the

integral free energy of mixing, is the difference between the two quantities, i.e.

MiGΔ

iiiM

i aRTGGG ln0

BBAA GnGn





= RT (nA ln aA + nB ln aB)

In terms of one mole of solution,

i. e. integral molar free energy

for an ideal solution ai = xi, GM,id = RT (xAln xA + xB ln xB)

0000BBBAAABBAABBAA

M GGnGGnGnGnGnGnG

MBB

MAA GΔn+GΔn=

BBAAM axaxRTG lnln




Properties of RaoultianIdeal solution

Change in volume accompanying the formation of ideal solution.

GM,id = RT (xAln xA + xB ln xB)

We can write,

A

idMA xRTG ln

,

B

idMB xRTG ln

,

i

CompT

iV

P

G

,

0

,

0

i

CompT

i VP

G

0

,

0)(ii

CompT

iiVV

P

GG

M

i

CompT

Mi

VΔ=P∂

GΔ∂

,





As xi is not a function of pressure, then

Integral volume of the solution (mixing)

i

idMi xRTGsolutionidealanfor ln,

,

0,

idM

iV

0,,, idM

BB

idMAA

idM VxVxV 0=VΔor idM ,

00BBAABBAA

M VnVnVnVnV

( ) )( 0BBB

0AAA V-Vn+V-Vn=

0=VΔn+VΔn= MBB

MAA





The heat of formation of an ideal solution

For a component in the solution, G - H equation

For the pure component

2, -

/

T

H

T

TG i

compPi

2

0

,

0

T

HH

T

T

GG

ii

CompP

ii

( )2

oi

compP

0i

T

H-=

T∂

TG∂,

/





is the partial Molar heat of solution (mixing) of i In an ideal solution,

Integral molar heat of mixing,

2

,

T

H

T

TG

Mi

CompP

Mi

MiH

i

idMi xRTG ln

,

2

,

,

ln

T

H

T

xRidM

i

CompP

i

0,,, idM

BB

idMAA

idM HxHxH

0,

idMiHor

0, idMH





The entropy of formation of an ideal solution

For the formation of a solution

For an ideal solution

GM,id = RT (xAln xA + xB ln xB)

ST

G

ompP

,

M

ompP

M

ST

G

,

BBAA

ompP

idMidM xxxxR

T

GS lnln

,

,,

iidM

i xRS ln,




Properties of Real Solution

Real solutions do not obey Raoult’s law (ai = γi. xi)

Integral molar free energy of mixing :

Partial molar free energy of mixing:

BBAAM axaxRTG lnln

idMXS

BBAABBAAM

GG

xxxxRTxxRTG,

lnlnlnln

idM

iXS

i

ii

iM

i

GG

xRTRT

aRTG

,

lnln

ln




Properties of Real SolutionChange in volume accompanying the formation of Real solution.

Partial molar volume of mixing

Integral Molar Volume of mixing

P

aRT

P

GV

i

CompT

M

iM

i

ln

,

P

ax

P

axRTV B

BA

AM lnln

MBB

MAA

M VxVxV





Change in Entropy accompanying the formation of real solution.

Partial molar entropy of mixing

Integral molar entropy of mixing

i

i

CompP

M

iM

i

aRP

aRT

T

GS

lnln

,

BBAAB

BA

AM axaxR

T

ax

T

axRTS lnln

lnln

MBB

MAA

M SxSxS





Change in enthalpy accompanying the formation of real solution.

Partial molar enthalpy of mixing

2

,

T

H

T

TG

Mi

CompP

Mi

CompP

iMi

CompP

i

Mi

T

aRTH

T

aR

T

H

,

2

,

2

ln

ln





Integral molar enthalpy of mixing

MBB

MAA

M HxHxH

compP

BB

AA

M

T

ax

T

axRTH ,

2 lnln




Regular Solution

In the case of non ideal solutions it is still possible to assume random

mixing in certain cases but the enthalpy of mixing will no longer be zero

because there will be heat changes due to changes in binding energy.

This assumption of random mixing can only be made where there is a

small deviation from ideal behaviour, so that the enthalpy of mixing is

quite small. Solutions of this type are called regular solution. For regular

solutions the entropy of mixing is the same as for ideal solution, so that

where and are the actively coefficient of A and B, respectively.

MMM STGH BBAA axaxRT lnln BBAA xxxxRT lnln

BBAA xxRT lnln




Regular Solution

In 1895 Margules suggested that activity coefficients, A and B of a binary solution at any given temperature can be represented by power series equations as follows:

By the application of the Gibbs – Duhem equation xA d lnγA = - xB d ln γB, it can be shown, for the system to hold over the entire range of composition, .

By similar comparison of co-efficients of the power series, Margules further demonstrated that the variation of the γ’s can be represented by the quadratic terms only, when

..........3

1

2

1ln

..........3

1

2

1ln

33

221

33

221

AAAB

BBBA

xxx

xxx

011

22




Regular Solution

For regular solutions Hildebrand in 1929 established that

α- function is defined as:

From comparison of the above expressions we get

αA = αB = α and

α is independent of compositionwhich indicates that Tl – Sn system follows regular solution model.

2BA

2AB xα′=γRTandxα′=γRT lnln

22 lnln ABBBAA xandx

RT




Properties of Regular Solution

Thermodynamic properties of solutions may be divided into two parts: ideal and excess.

Properties of regular solutions can be discussed by the excess parts.

QReal = Qid + QXS The change in property during mixing:

Hence we can write for integral molar free energy of mixing as

XSidMRM Q+QΔ=QΔ ,,

XSidMRM G+GΔ=GΔ ,,

idMRMXS GΔ-GΔ=GΔ ,,





it may be shown as

ΔGxs = ΔHxs - TΔSxs = (ΔHM,R – ΔHM,id ) – T (ΔSM,R – ΔSM,id )

= ΔHM,R (as ΔHM,id =0 and ΔSM,R = ΔSM,id )

ΔGxs = ΔHM,R

XS

p

XS

ST

G

idMRMXS GΔ-GΔ=GΔ ,,

BBAA axaxRT lnln BBAA xxxxRT lnln

RMBBAA HxxRT ,lnln

XSBB

XSAA GxGx

MBB

XSB

MAA

XSA HRTGandHRTG lnln





Hence for regular solution

This means GXS for a regular solution is independent of

temperature

BBAAXS xxRTG lnln

22ABBA xxxxRT

BABA xxxxRT

BAxxαRT=

BAMXS xxRTHG

RTwherexx BA





This can also be shown as

As SXS for a regular solution is zero, then GXS and ΔHM are independent of temperature. Similarly, is also independent of temperature.

This equation is of considerable importance and of practical use in connecting activity data at one temperature to activity data at another temperature. Hence for a regular solution we have.

XS

p

XS

ST

G

XSG

22211 lnln BAA

XSA xTRTTRTG

2

1

1

2

ln

ln

T

T

eTtemperaturat

eTtemperaturat

A

A





2211

2

1

1

2

2

1

12

22

2

2

1

1

2

)(

)(

Tat )1(

ln

Tat )1(

ln

)1(;ln

ln

TTorT

T

T

T

T

T

x

x

getwexbydividingT

T

Tetemperaturat

Tetemperaturat

A

A

A

A

AA

A

For strict adherence to this model αT should be independent of temperature. Thus Tl – Sn is not strictly regular in behaviour





Change in volume accompanying the formation of regular solution.

Partial molar volume of mixing

Integral molar volume of mixing

ompT

i

CompT

M

iM

i

P

aRT

P

GV

,

,

ln

MBB

MAA

M VxVxV

ompT

BB

AA

M

P

ax

P

axRTV

,

lnln





Change in Entropy accompanying the formation of real solution.

Partial molar entropy of mixing

Integral molar entropy of mixing

i

M

i xRS ln

BBAAM xxxxRS lnln

MBB

MAA

M SxSxS





Change in enthalpy accompanying the formation of real solution.

Partial molar enthalpy of mixing

Integral molar enthalpy of mixing

XSii

Mi GRTH ln

MBB

MAA

M HxHxH

BBAAM xxRTH lnln




Regular solution behaviour:

1. linear variation of ln γA vs xB2 (slope = ) at a given

temperature indicates regular solution behaviour of the system: A-B.

2. However for strict adherence to the model, T should be independent of T but not so in many cases. In general we find ln varying linearly with x but iT decreases with temperature.




Thermodynamic properties of Solutions




Thermodynamic properties of Solutions

Kk QxQ .

idMMXS QQQ ,




GIBBS – DUHEM INTEGRATION

Thermodynamic equations for calculation of excess free energy and

integral molar free energy of a solution need activity coefficient and

activity of all the components of the solution. However,

experimental techniques viz. chemical equilibria, vapour pressure

and electrochemical can measure activity of only one component.

In order to get activity of the second component in a binary solution

we must couple activity and atom/mole fractions of both the

components with the aid of Gibbs-Duhem equation as follows:

Q is any extensive property. ,0iQdXi




GIBBS – DUHEM INTEGRATION –METHOD -I

Since activity of a component is related to the partial molar free energy, we can write Gibbs-Duhem equation as under:

)1(0 MB

MAA GxdGdx

)2(0lnln BBAA adxadx

)(lnln 3adx

x=ador B

A

BA

AA

A

AA

xx

x

BA

BxxA ad

x

xa

1

ln|ln




GIBBS – DUHEM INTEGRATION METHOD -I




GIBBS – DUHEM INTEGRATION – METHOD -II

(4)

Eq.(4) Eq.(2)

(5)




GIBBS – DUHEM INTEGRATION

A

BBAB x

xbutaxas 0ln,1,1

0ln,0,0 A

BBBBB x

xforfinitealsoisandfiniteisaxas




GIBBS – DUHEM INTEGRATION – METHOD -III

As a further aid to the integration of the Gibbs Duhem equation, the is introduced as

-function is always finite because For the components of a binary solution

B is known as a function of composition:

On differentiation we get

21

ln

i

ii

x

11 ii xas

,ln

2B

AA x

2

ln

A

BB x

and

2ln ABB x

function

,..2ln 2BAAABB dxdxxd





substituting this into

On integration

BA

BA d

x

xd lnln

BABABBBAA

BAAB

A

BA dxxdxxdx

x

xdxx

x

xd .2..2ln 2

xAx

x

xxat

xat

ABABABBA

A

A

AAB

AB

dxxdxx1 1

)(2ln





(A)





Thus ln at xA=xA is obtained as – xBxAB minus the area under the plot of

B vs xA from xA=xA to xA = 1. Since B is everywhere finite, this

integration does not involve a tail to infinity.




Problem

The activity coefficient of zinc in liquid Cd-Zn alloys at 4350C

have been expressed as ln Zn = 0.87 x2 Cd – 0.30 x3 Cd

(a) Calculate the activity of cadmium in a 30 at % Cd at 4350C.

(b) Develop a corresponding equation for the activity coefficient of cadmium in the alloy system at this temperature.




Solution

From Gibbs-Duhem equation we can write as

xZn d ln Zn +xCd d ln Cd = 0

on integration we get

Zn

xx

x Cd

ZnCd d

x

xCdCd

Cd

lnln1

CdCdCd

xx

x Cd

Cd dxxxx

xCdCd

Cd

2

1

90.0-74.1-1

CdCdCd

xx

x

Cd dxxxxCdCd

Cd

2

1

90.0-74.1)-1(




Solution

= -1.74(0.3-0.1) + 1.32 (0.09-1)-0.3 (0.027-1)

Cd

x

x

CdCd dxxxCd

Cd

3.0

1

290.064.2-74.1

301

3Cd

2CdCd x30+x321-x741-= .]...[

[ ] 30

13Cd

2CdCd x30-x321+x741-=

....

3090=2920+2011-2181=γCd ....ln

362.1 Cd..... Ans40860=30×3621=xγ=a CdCdCd




Zn

xx

x Cd

ZnCd d

x

xb

CdCd

Cd

lnln)(

Solution

CdCdCd

xx

x Cd

Cd dxxxx

xCdCd

Cd

2

1

90.0-74.1-1

CdCdCd

xx

x

Cd dxxxxCdCd

Cd

2

1

90.0-74.1)-1(

CdCdCdCdCdCd dxxxdxxx )9.064.2-74.1()9.0-74.1(-1 2

CdCd

xx

x

Cd dxxxCdCd

Cd

)90.0-64.274.1-( 2

1




Solution

Cdx=Cd

1=Cd

x

x3Cd

2CdCD x30-x321+x741- ]...[

formdesiredthegettoxxput

xxx

xxx

ZnCd

CdCdCd

CdCdCd

1

3.032.174.172.0

)3.032.174.1(3.032.174.132

32

3Zn

2ZnZn x-130-x-1321+x-1741-720= )(.)(.)(..

)331(3.032.164.232.174.174.172.0 322ZnZnZnZnZnZn xxxxxx

720+741-321+30-741+642-90x+321+90-x+x30= Zn2Zn

3Zn ....)...()..(.

... Ansx420+x30= 2Zn

3Zn




Solution

Alternatively, we can also solve as

)-)(9.0-74.1()9.0-74.1(ln 2 CdCdZnCdCdCdCd

ZnCd dxxxdxxx

x

x

ZnZn

Zn

CdCd

Cd

xx

x

ZnZnZnZnZn

xx

x

ZnCdZn dxxxdxxdxxx01

)-1(9.0-74.1).9.0-74.1(

ZnZnZnZnZnZn dxxdxxdxx 29.09.074.1

320

3

022 3.042.0]

3.90.0[]42.0[9.084.0 ZnZn

xZnxZnZnZnZnZn xx

xxdxxdxx ZnZn




Problem

At 746K the activity coefficient of lead in liquid Pb-Bi alloy is expressed as Making use of the Gibbs-Duhem equation develop the corresponding equation for the activity coefficient of bismuth in the alloy at 746K. (a) Calculate the activity of lead at 746K and 1000K in the Pb-Bi alloy containing 50at % lead (b) Calculate the integral molar heat of mixing /excess free energy of the alloy containing 40at% Pb at 746 K (c) What is the integral molar free energy of mixing of the above alloy in (b) at 1000K (d) Calculate the difference in change in free energy when 1 gatom of lead dissolves in a very large amount of the above alloy at 746 and 1000K

.)(.ln 2PbPb x-1740-=γ




Solution

(a) For Pb-Bi system the Gibb-Duhem equation may be written as:

on integration we get

0=γdx+γdx PbPbBiBi lnln

Pb

xxat

xat Bi

PbBi d

x

xBiBiPb

BiPb

ln-lnln

1ln

PbPbPb

PbPb

dxxd

x

)1(48.1ln

)1(74.0ln 2

PbPbPb

Pb dxxx

x-148.1

-1




Solution

= - 0.74(1-xBi)2

So the solution is a regular solution

)1(

0

2

1 248.1-48.1

BiPb

Pb

Bi

Bi

xx

x

Pb

x

x

PbPb

xdxx

.)746(415.083.05.0,83.0)746( AnsKaK PbPb




Solution

for regular solution

22 Pb

Bi

Bi

PbBiPb x

γ=

x

γ=α=αb

lnln)(




Solution

at 746 K

= RT xAxB

= - 1.987 746 0.74 0.4 0.6

= -263.3 cal/mole. Ans.

2Bi

2BiPb x740-=xα=γ .ln

2Pb

2PbBi x740-=xα=γ .ln

BibIPbPbMXS xxRTHG lnln




(C) for regular solution 1T1 = 2T2

1 = -0.74 (746 = T1)

At 1000K

550-=1000

746×740-=

T

Tα=K1000α

2

112 .

.)(

4350=50x87150=a

87150=γ==>x550 -=xα=γ

Pb

Pb2Bi

2BiPb

...

..ln

4350=50x87150=a

87150=γ==>x550-=xα=γ

Bi

Bi2Pb

2PbBi

...

..ln




Solution

=1.987 1000 (0.5 ln 0.435 + 0.5 ln 0.435)

= 1.987 1000 ln 0.435 = 1987 (-0.8324)

= -1654 cal/m. Ans.

(d) difference in free energy =

= RT ln aPb (1000K) –RT ln aPb(746K)

= 1.987 (1000 ln 0.435 – 746 ln 0.415)

= 1.987 [1000(-0.8325) – 746 (-0.8795)]

= 1.987 (-176.4) = -350.5 cal/mole. Ans.

BiBiPbPbM axaxRTG lnln

)()( K746GΔ-K1000GΔMPb

MPb




Problem

Al-Zn alloys exhibit the following relationship at 4770C:

RT ln γZn = 1750 (1-xZn)2 where R and T an expressed in cal/deg.gatom

and K, respectively.

i) Develop the corresponding expression ln Al

(ii) Calculate the heat of mixing of the alloy containing 40 at % Zn at 4770C. What would be excess molar free energy of the alloy at this temperature?

(iii) Calculate the integral molar free energy of the above alloy at 5070C.




Solution

appropriate form of the Gibbs-Duhem equation.

dx )x-(1RT

3500- ln,)1(

1750ln ZnZnZn

2 dxRT ZnZn

0lnln ZnZnAlAl dxdx

ZnAl

ZnAl d

x

x ln.ln

ZnZn

xx

x Zn

Zn dxxRTx

xAlAl

Al

)1(3500

11

AlAl

Al

AlxZnx

Znx

xx

x

ZnZnAl

x

RTdxx

RT 1

2 )1(

02.

35003500

2)1(1750

AlxRT




Solution

(ii) The above form of equation show that is independent of composition. Thus Al-Zn system follows regular solution model, hence

xZn = 0.4, xAl = 0.6

= 1750 0.24 = 420 cal/gatom Ans.

AlZn

Al

Zn

ZnZn RTx

1750

1

ln2

ZnAlXSM xxRTGH

4.06.01750

. RT

RT




Solution

(iii) ).ln( ZnZnAlAlM axaxRTG

.780273507,750273477

solutions,regular for

21

2211

KTKT

TT

174.1987.13

7

750

17501750

)1(

ln2750

RRTxZn

Zn

129.1987.178

175

780

750

750

1750

780

750750780

xx

R

27801

ln

Zn

Zn

x

xZn=0.4




Solution

γAl= 1.19798

aAl = 1.19798 0.6 = 0.71879

40644.036.0129.11ln 2780)780( ZnZn x

50146.1)780( Zn

6005.04.050146.1. ZnZnZn xa

2780 )1(

ln

Al

Al

x

18664.016.0129.1)6.01(ln 2780 Al

)6008.0ln4.0718979.0ln6.0( RTG M




Solution

=1.987 780 (0.6 (-0.3301856) + 0.4 (0.5075)

= 1.987 780 [-0.1981 – 0.2038] = 1.987 780 (-0.4019)

= -622.8 cal.mole. Ans.




Problem

In liquid Fe-Ni solution at 1873 K the activity of nickel as a function of

composition is listed below:

xNi 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

aNi 0.067 0.137 0.208 0.287 0.376 0.492 0.620 0.776 0.89

(i) Calculate the activity of iron in an alloy containing 60 at% iron by Gibbs - Duhem integration.

(ii) Calculate in the above alloy at 1873 K.XS

iMi

MFe

MNi GGGG ,,,




Solution

Solution:

By third method of integration calculate Ni

xNi 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

aNi 0.067 0.137 0.208 0.287 0.376 0.492 0.620 0.776 0.89

γNi 0.67 0.685 0.69 0.72 0.75 0.82 0.986 0.96 0.99

(1-xNi)2 0.81 0.64 0.49 0.36 0.25 0.16 0.09 0.04 0.01

Ni -0.49 -0.59 -0.75 -0.92 -1.14 -1.24 -1.35 -1.07 -1.01

Making use of function we have

= - (-0.92 0.6 0.4 – 0.08) = 0.301 i.e. γNi = 1.35, aFe = 0.6 1.35 = 0.81

FeFe

Fe

xx

x

FeNiNiFeNiFe dxxx1

..ln

][ FeNiNiFeNi dxxx




Solution

./19438)287.0ln(1873314.8ln moleJaRTG Ni

MNi

./3281)81.0ln(1873314.8ln moleJaRTG Fei

MFei

./9744)3281(6.0)19438(4.0)( moleJGxGxG FeFeNiNiM

)35.1ln6.072.0ln4.0(1873314.8)lnln(, FeFeNiNiMid

MXS xxRTGGG

= 758 J/mole




Problem

At 1200K the activity coefficient of zinc in liquid brass is

expressed as

ln γZn = - 1.929 (1-xZn)2

(a) Calculate the integral molar heat of mixing and the excess free energy of brass containing 40 at % copper at 12000 K.

(b) What is the integral molar free energy of mixing of the above alloy at 1300 K. Check your answer.

(c) Calculate the difference in change of free energy when one g. atom of liquid Zinc dissolves in a large amount of liquid brass at 1200 and 1300 K.




Solution

The variation of activity coefficient with composition shows that Zn =-1.929, is independent of composition. This indicates that brass behaves regularly at 1200K.

Zn=Cu

we know that for regular solution we have

Integral molar heat of mixing (or excess molar free energy) of brass,

T = 1200 K , xCu = 0.4, xZn = 0.6

929.1)1(

ln

)1(

ln22

CuCu

Cu

Zn

ZnZn xx

)lnln( BBAAXSM xxRTGH

)lnln( CuCuZnXSM xxZnRTGH

30864.04.0929.1929.1ln 22 CuZn x




Solution

= - 2384.4 (0.6 0.30864 + 0.4 0.6944)

= -2384.4 0.4629 = -1103.8 cal/mol.

or GXS= (RT) xCu xZn = (-4600) 0.24 = -1104 Ans.

4406.06.07344.07344.0 ZnZn aand

6944.06.0929.1929.1ln 22 xxZnCu

1997.04.04993.04993.0 CuCu aand

)ln4.0ln6.0(1200987.1 CuZnXSM GH




Solution

(b) for brass is needed at 1300 K, hence aCuand aZn should be first calculated at 1300K by

making use of properties of regular solution.

MBBAA

M GceaxaxRTG sin),lnln(

2211)1200(929.1 TTandKatCuZn

1300.1200. 13001200

781.113

12929.1

13

12.12001300

2849.0)4.0(781.1.ln 221300 CuZn x

45125.06.07520.07520.0 ZnZn aand

6412.06.0781.1.ln 221300 ZnCu x




Solution

= 1.987 1300 (-1.10034) = -2842.3 cal/mol.

21071.04.05267.0,5267.0 CuCu a

)21071.0ln4.045125.0ln6.0(1300987.1 MG

5.110324.0)781.1(1300987.1.)1300( ZnCuXS xxRTG

)(Re./1104)1300()1200( solutiongularmolcalGG XSXS

./5.1738

)4.0ln4.06.0ln6.0(1300987.1)lnln(,

molcal

xxxxRTG CuCuZnZnidM

./8.1103)5.1738(3.2842 molcalGGHG Mid

MMXS




Solution

Difference in change in free energy =

=RT ln aZn(1300) – RT ln aZn (1200)

=1.987 (1300 ln 0.4513 – 1200 ln 0.4406)

=1.987 (-1034.31 + 984.54) = -98.9 cal/mol. Ans.

)()( 1200G-1300GMZn

MZn




Dilute Solution –Henry’s Law




Henry’s Law




Henry’s Law

The constant is equal to the slope of the curve at zero concentration of A, designated by coefficient of the solute A at infinite dilution).

Like Raoult’s law, Henry’s law is valid within a concentration range where the extent varies from one system to another, but it is valid only at low concentration.

AAAAAAA

AA

AA

AA

xaorxconstaeixp

k

p

phavewepbydividing

kxpei

xp

...

)1(..

000

activity(0A

AAA xa .0




Henry’s Law

aA

xA

γ0A → Henry’s law constant




Henry’s Law

In concentratrated solution the standard has been defined as unit atmospheric pressure and unit activity i.e. pure substance at any temperature.

In dilute solutions relative standard states other than pure substance being used. Henry’s law offers two such standard states, called alternative standard states.

(1) Infinitely dilute, atom/mole fraction standard state.

(2) Infinitely dilute, wt% (w/o or %) standard state.




Solubility of Gases

It is important to note that the validity of Henry’s law depend upon the proper choice of solute species. For example, consider

(a) solution of nitrogen in water

(b) solution of nitrogen in liquid iron

In the first case nitrogen dissolves molecularly as N2

As solubility of N2 in water is low according to Henry’s law we have

)waterin dissolved(22 NgN 2

2

N

N

p

aK

22 NN xka 2

2

N

N

p

kxKor

222)lub( NNN pkp

k

Kilitysox




Solubility of Gases

Thus the solubility of nitrogen in water is proportional to the partial pressure of nitrogen gas in equilibrium with water.

Solubility can be expressed as mole fraction, cc per 100 g of water or any other unit.

(2) In the second case under consideration nitrogen dissolves atomically in solid or liquid metals:

)(2)(2 FeinNgN

22

2)(

2 )(

N

N

N

inFeN

p

kx

p

aK

22')lub()( NNinFeN pkp

k

Kilitysoxor




Solubility of Gases –Sievert’s law

Since all the common diatomic gases N2, O2, H2 etc. dissolve atomically in metals, the general expression for solubility is given as:

This is known as Sievert’s law and can be stated as – solubility of diatomic gases in metals is directly proportional to the square root of partial pressure of the gas in equilibrium with the metal.

2NpkS




Problem

At 8000C, 100g of silver dissolves 3.3 cm3 (STP) of oxygen at one

atmosphere pressure. How much oxygen does silver dissolve from air at

8000C?

Solution:

According to Sieverts law:

In air pO2= 0.21 atm,

2OpkS 3.313.3.. kkei

351.121.03.32

cmpkS O




Problem

The following solubility of oxygen in 100 g of silver at 10750C have been

measured.

a) Show whether these observations agree with Sievert’s law for the

solubility of oxygen in metals.

b) How much oxygen does 100g of silver absorb at 10750C from air?

c) What pressure of air corresponds to one atm of O2 with respect to the solubility of oxygen in silver at 10750C.

1203760488128)(2

mmHgpO

Oxygen dissolved cm3/100gAg 81.5 156.9 193.6 247.8




Solution

2

solubilitytan'

O

sp

ktconssSievert

2037.7128

5.81Sk

1025.7488

9.156sk

0226.7760

6.193sk

1444.71203

8.247Sk




Solution

Almost constant values of ks demonstrate that solubility of oxygen in

silver is proportional to the square root of partial pressure of oxygen in

equilibrium. Hence the observations agree with Sievert’s law. Average

value of ks = 7.1183.

(b) In air, pO2 = 0.21 atm. = 0.21 760

AggmpercckilitySo S 10093.896.1591183.776021.0lub

7629.4

)21.0,1(21.0

1)(

2

Oairair patmppc




Problem

At 15400C liquid iron dissolves 0.04% nitrogen in equilibrium with

nitrogen gas at one atmospheric gas pressure and 0.23% oxygen in

equilibrium with oxygen gas at one atmosphere gas pressure. At that

temperature nitrogen pentoxide gas was passed over liquid iron such

that equilibrium was attained with fully dissociated gas at a net pressure

of one atmosphere. What is the nitrogen and oxygen contents of the melt?




Solution

N2 gas : O2 gas

1 mol : (5/2) mols

2 mol : 5 mols.

2 vol : 5 vol.

ONgOgNgON 52)(2

5)()( 2252

atm7

5and atm

7

222 ON pp

04.0104.02

NNNNN KKpKS

pKOS NN )N ed(dissociat 52'

.%02138.07

204.0 Ans




Solution

23.0

123.0

O

O

OO

K

K

pKS

%19439.07523.0)N issociated( 52 OdSO




Alternative Standard States

1. Infinitely dilute, atom fraction standard stateHenerian standard state is obtained from Henry’s law which, strictly being limiting law obeyed by the solute in the dilute solution is expressed by

If the solute obeys Henry’s law over a finite concentration range, then

tconslawsHenerytheisandstatedards

RoultiantrwAofactivitytheisawhere

xasx

a

A

A

AAA

A

tan'tan

..

0

0

0

AAA xa .0





Henerian standard state is obtained by the extending the Henry’s law line to xA = 1.

This state represents pure solute in the hypothetical, nonphysical state in which it would exist as a pure component if it obeyed Henry’s law over the entire composition range (i.e., as it does for a dilute solution)

Having defined the Henrian standard state, the activity of A in solution with respect to the Henrian standard state is given by:

activityheneriantheish

andstatedardsheneriantheisThis

1=xatγ=1=h

A

A0AA

tan

.

tcoefficienactivityheneriantheisfwhere

xfh

A

AAA .





In the range of composition in which the solute obeys Henry’s Law, fA =1 and solute exhibit the Henerian ideality

hA = xA

In the range of composition in which the solute obeys Henry’s Law, fA =1 and γA = γA

0

tconsxA

A

AA

AA

A

A

Affx

x

h

a

tan.

.

tconsx

AA

A

Ah

a

tan

0





The free energy change accompanying the transfer of one mole of solute A from pure substance standard state (Raoultian standard state) to Henerian standard state, that is:

A (in the Raoultian standard state) → A( in the Henerian standard state)

is given by

The partial molar free energy of the solute at constant concentration is independent of standard state. The value of ΔGA

0 remains unchanged if is added and is subtracted from the right hand side of the above equation

000

)()()(

RAHAGGHRGA

)(RAG )(HAG

0)(

0)(

)()(000

)()(

)()()(

HARA

RAHA

GGGG

GGGGHRG

HARA

HARAA





AAA aTRGGBut ln0

0

tan

0

ln

ln)(,

A

tconsxA

AA

RT

h

aRTHRGHence

A





2. Infinitely dilute, wt% standard state.

The use of this standard state eliminates the necessity of converting weight

percentages, obtained via chemical analysis, to mole fractions for the

purpose of thermodynamic calculations. This standard state is particularly

convenient to use in metallurgical systems containing dilute solutes. This

standard state can formally be defined as:

0.%.%

0.%1.%

AwtA

A

Awtaor

AwtasAwt

a





If the concentration up to 1 weight-percent of solute A, then aA

= 1 at wt%A =1 and this 1 weight-percent solution is then the

standard state

W.r. t 1 weight-percent standard state, the activity of solute A

is given by

Where fA(1wt%) is the 1 wt.% activity coefficient and in the

range of composition in which A obeys the Henry’s law fA(1wt%)

Awtfh wtAwtA .%.%)1(%)1(

Awth wtA .%%)1(





We can also write

In the range of composition in which the solute obeys Henry’s Law, fA(1wt%) =1 and γA = γA

0 , therefore,

ncompositiotconswtA

AA

wtA

A

Awtf

x

h

a

tan%)1(%)1( %.

.

ncompositiotcons

AA

wtA

A

Awt

x

h

a

tan

0

%)1( %

.





We know that

Where MA and MB are the molecular weight of A and B. the first term in

the denominator is small compared to the second and the relation may be simplified as

BA

AA

M

Awt

M

AwtM

Awt

x.%100.%

.%

A

BA

A

B

B

AA

M

M

Awt

xTherefore

M

MAwt

M

M

Awt

x

.100.%,

.100

..%100

.%





The free energy change accompanying the transfer of one mole of solute A from pure substance standard state (Raoultian standard state) to 1 wt.% standard state, that is:

A (in the Raoultian standard state) → A( in the 1 wt.% standard state)

is given by

A

BA

constxWtA

AA

M

MRTRT

h

aRTwtRG

A

.100lnln

ln%).1(

0

.%)1(

0





The free energy change accompanying the transfer of one mole of solute A from Henerian standard state to 1 wt.% standard state, that is:

A (in the Henerian standard state) → A( in the 1 wt.% standard state)

is given by

A

B

AA

BA

AAA

M

MRT

RTM

MRTRT

HRGwtRGwtHG

.100ln

ln.100

lnln

)(%).1(%).1(

00

000




Problem

Calculate the free energy change when the standard state of manganese is

transferred from pure liquid state to infinitely dilute wt% solution of

manganese in iron at 16270C melting point of Mn =12450C, at wt of Mn =

54.94, Fe = 55.85. Assume ideal behaviour of the solution.

Solution:

T = 1627 + 273 = 1900, MMn = 54.94, MFe = 55.85

Mn (pure substance standard state) Mn (dilute wt% standard state)

)(10 idealMn

Mn

FeMn

wtMn

MnMn M

MRT

h

aRTHRG

100lnln)( 0

%)1(

0

molJ /72455

94.54100

85.551ln1900314.8




Problem

From experimental measurements of the equilibria between H2-H2O gas

mixtures pure silica and silicon dissolved in liquid iron, the free energy

accompanying the transfer of standard state from pure silicon to the

infinitely dilute, wt.% solution of silicon in iron i.e.:

Si (pure, 1) Si (wt% dil. in Fe) has been expressed as G0 = -28500 –

5.8T cal./mol. At 16000C, the activity coefficient of silicon in iron ,

relative to pure silicon as the standard state is 0.0014 at 1 atomic% Si.

Calculate the activity coefficient of silicon, relative to the wt% standard

state at this concentration.




Solution

G0 = -28500 – 5.8T = -28500 -5.8 1873 = -39360 cal.

wt% Si = 0.50

Si (pure, 1) Si (1wt% in Fe)

This is the activity coefficient of Si at infinite dilution, relative to pure Si as the standard state.

85.55%100

09.28%

09.28%

01.0%SiwtSiwt

Siwtsiatom

3936007.28100

85.55.ln

100ln RT %)1(

000

Si

Si

FeSiSi RT

M

MwtRG

00128.00 Si




Solution

Hence the activity coefficient of Si at 1 at% Si, relative to pure Si as the

standard state is calculated as follows:

000014.001.00014.0.)( SiSiSi xpurea

Si

FeSi

wtSi

Si

M

M

h

a

100

0

%)1(

FeSi

SiSiSi M

Mawth

0

100.%)1(

55.0

85.5500128.0

09.28100000014.0

FeSi

SiSi

MSiwt

MSiwt

MSiwt

x)%100(%

%




Solution

50.0%85.55

)%100(09.28

%09.28

%01.00

Siwtor

SiwtSiwt

Siwtr

hSi = fSi . wt% Si

.1.150.0

55.0

%Ans

Siwt

hf Si

Si




Problem

The Raoultian activity coefficient of Al at infinite dilution, in

liquid Fe-Al alloys is reported to be 0.063 at 16000C. Calculate the

standard free energy of formation of Al2O3(s) at 16000C for each of the

three standard states for solution of Al in Fe.

MAl = 26.98 and MFe = 55.85

given 1. 2 Al(l) + 3/2O2(g) = Al2O3(s) G0 = -1682927 – 323.24TJ

2. 2 Al(l, H) + 3/2O2(g) = Al2O3(s)

3. 2 Al(l, wt %) + 3/2O2(g) = Al2O3(s)

0Al




Solution

(1)

G0 = -1082927 +323.24 1873 = -1077498.5 J

Total free energy change for reaction (2) =

)1()()(2

3)(2 322 sOAlgOlAl

)3(),()(

)2()()(2

3),(2 322

HlAllAl

sOAlgOHlAl

00 ln)(

)3(

AlAl

AlAl RT

h

aRTHRG

reactionFor

0)(

0)(

0)( 2 HRRH GGG




Solution

= -1077498 – 2 8.314 1873 ln (0.063)

= -1077498 – (-86102) = -991396 J.

Total fre energy change for the reaction (4) =

)5(%)1,()(

)4()()(2

3%)1,(2 322

wtlAllAl

sOAlgOwtlAl

00)( ln2 AlR RTG

Al

FeAl

wtAl

AlAl M

MRT

h

aRTwtRG

reactionFor

100ln%)1(

)5(0

%)1(

0

0%)1(

0)(

0%)1( 2 wtRRwt GGG

Al

FeAlR M

MRTG

100ln2

00

)(




Solution

= -1077498 – 2 8.314 1873 ln

= -1077498 – 2 8.314 1873(-6.6422)

= -1077498+206866.8

= -870631J Ans.

98.26100

85.55063.0




Problem

Vanadiam melts at 1720oC. The raoultion activity coefficient of vanadian

at infinite dilution in liquid iron at 1620oC is 0.069. Calculate the free

energy change accompanying the transfer of standard state from pure solid

vanadium to the infinitely dilute wt% solution of V in pure iron at 1620oC.

Given: heat of fusion of V = 4800 cal/g atom

MFe = 55.85 and MV = 50.95.

Solution:

.1893069.01993 0 KatKT Vf

V

..deg/4084.21993

4800molcal

T

HS

f

ff




Solution

Free energy of fusion of vanadium (s l) at the operating temperature 1893K.

We calculate Gf for fusion because the melting point of V is more than the operating temperature (at which V is in solid state).

V (pure l) V (wt%)

= -27033 cal/mol

For energy change V (pure solid) V infinitely dilute wt% solution

= ΔGf + ΔGV0 (R →H) = 240.9 – 27033

= -26792.1 cal/mol.

./9.2404084.218934800 molecalSTHG fff

95.50100

85.55069.0ln1983987.1

100ln%)1(

00

V

FeVV M

MRTwtRG




Chemical Potential

The general equation for the free energy change of a system with temperature and pressure dG = VdP – SdT, does not take into account any variation in free energy due to concentration changes.

We know

From the fundamentals of partial differentiation we have

The coefficient called the ‘chemical potential’ and is

denoted by hence

....),,,,( 21 nnPTGG

......2....,,2

1....,,1,,

2

1

1

2

dnn

Gdn

n

GdP

P

GdT

T

GGd

nexceptnPT

nexceptnPTniTniP

i

nexceptnPTi

dnn∂

G∂+dPV+dTS=

i

1

∑....,,

'''.

inexceptnTPin

G

.......,, 1




Chemical Potential

is an intensive variable

This gives a new sets of fundamental equations for the open systems.

i

nExceptnPTi

i

n

G

.....,, 1

ii dndPVdTSGd

ii dnVPddTSAd

ii dndPVSTdHd

ii dnVPdSTdUd




Physical Meaning of Chemical potential Consider the change in free energy (dG/) of a system produced by the

addition of dnA mole of component A at constant pressure and temperature. The change in free energy of a system is given by is the partial molar free energy of component A in solution

Chemical potential of either 1 g mol or 1 g atom of a substance dissolved

in a solution of definite concentration is the partial molar free energy. Thus

AAAA dnGdnGd

solutionofquantitylargeafor,,

A

BnTPAA n

GG

solutionof mole onefor,,

A

BnTPAA n

GG




Equality of chemical potentialamongst phases at equilibrium

We know:

At constant T and P:

Consider two phases(I and II) in the system. Then,

Consider moving an infinitesimal of quantity dn1 from phase

I to phase II. Then,

ii dndPVdTSGd

∑ ii dnμ=G′d





Therefore, total free energy change of the system is

For equilibrium at constant temp and pressure

Hence,

It can be generalized for all components at constant T and

P when phase I and II are at equilibrium as





Where P is the total no. of phases in the system




Phase Rule

Phase(P)

A phase is defined as any homogeneous and physically distinct part of a

system which is separated from other part of the system by a bounding

surface. For example, at 273.15K, three phases ice, water and water vapour

can exist in equilibrium. When ice exists in more than one crystalline

form, each form will represent a separate phase because it is clearly

distinguishable from each other.

Components(C)

The number of components in a system at equilibrium is the smallest

number of independently variable constituents by means of which the

composition of each phase present can be expressed directly or in the form

of a chemical equation.




Phase Rule

As an example let us consider decomposition of calcium carbonate :

CaCO3 (s) = CaO(s) + CO2(g)

According to the above definition, at equilibrium this system will consist of two components since the third one is fixed by the equilibrium conditions.

Thus we have three phases – two solids (CaCO3 and CaO) and a gas (CO2) and the system has only two components.

If CaO and CO2 are taken, the composition of calcium carbonate

phase can be expressed as xCaO + xCO2 giving xCaCO3 (by the

chemical reaction).

The composition of the three phases could be expressed equally by taking CaCO3 and CaO or CaCO3 and CO2 as the components.




Phase Rule

The dissociation of any carbonate, oxide or similar compounds involves

two components; the same is true in the case of salt hydrate equilibria, for

example : CuSO4.5H2O(s) = CuSO4.3H2O(s) + 2H2O(g) when the simplest

components are evidently CuSO4 and H2O.

In the slightly more complicated equilibrium : Fe(s) + H2O(g) = FeO(s) +

H2(g) it is necessary to choose three components in order that the

composition of each of the three phases can be expressed.

The composition of the two solid phases could be given in terms of Fe and

O, but these alone are insufficient to define the gaseous phase which is a

mixture of hydrogen and water vapour, a third component, viz., H2O is

necessary.




Phase Rule

The water system for example consists of one component, viz., H2O each

of the phases in equilibrium i.e. solid, liquid and vapour may be regarded

as being made of this component only.

Degrees of freedom(F)

The number of degrees of freedom is the number of variable factors, such

as temperature, pressure and concentration that need to be fixed in order

that the condition of a system at equilibrium may be completely defined

when referring to its equilibrium phase diagrams.




Derivation of the PhaseRule Equation












Application of phase ruleto single component system




Application of phase ruleto two component system

For two component system: F = 3 – P as the pressure is kept

constant

P=1 F= 2 can vary T and composition - bivariant

P=2 F= 1 can vary either T or P – monovariant

P=3 F= 0 no free variables it is a fixed point. - invarient




Two componentsEutectic system




Phase rule in Reactive Components

Consider a system consisting of N chemical species and

there are P number of phases.

In this case the number of components differ from number

of species.

Let us consider there are three out of N species are

chemically active and participate in the following reaction:

AB(s) = A(g) + B(g).

The number of total variables = P(N-1) + 2

Total number of constraints due to phase equilibrium= N(P-

1).

There is another additional constraints: AB(s) = A(g) + B(g).




Phase rule in Reactive Components Additional, in the absence of A(g) and B(g) in the starting

reactant mixtures, stoichiometric consideration requires that PA= PB.

Some times, special constraints are placed on the system. For example, the system under consideration, the partial pressure of A has been fixed at 2 atm.

So this way total no. of constraints are = N(P-1)+1+1+1. F = [P(N-1)+2] – [N(P-1)+1+1+1] = (N-2) – P +1 = C – P

+1 Generalizing, for a system in which there are ‘r’

independent chemical equilibria, ‘s’ stoichiometric relation and ‘t’ special constraints we have

F = (N – r – s - t) – P +2 = C- P + 2 –t

where C = N - r - s




Application of Phase rulein Reactive ComponentsProblem: A system is composed of a solid phase CaCO3, a solid phase CaO,

and a gas phase CO2 . The following equilibrium occurs:

CaCO3(s) = CaO(s) + CO2(g)

How many components are there and what are the degrees of freedom?Solution:

Species: CaCO3(s) , CaO(s), CO2(g) : N =3, Phases : two solid and a gas phase P = 3. No. of independent reaction equilibria r = 1. There is no stoichiometric or special constraints.

So s = 0 and t = 0 C = N-r-s = 3-1-0= 2F = C-P+2-t = 2-3+2-0 = 1Either temperature or pressure must be specified.




Application of Phase rulein Reactive ComponentsProblem:A pure solid NH4Cl is introduced into an evacuated chamber. It is then allowed to decompose and equilibrium has been established by following reaction:

NH4Cl(s) = NH3(g) + HCl(g)

Calculate the number of components and degrees of freedom.Solution:

N = 3 (NH4Cl(s) , NH3(g) , HCl(g))

P = 1 solid (NH4Cl(s) ) + 1 gases (NH3(g) + HCl(g ) = 2r = 1

s = 1 as P NH3(g) = P HCl(g)

t = 0C = N – r – s = 3 – 1 -1 = 1F = C – P + 2 – t = 1 -2 + 2 – 0 = 1




Application of Phase rulein Reactive Components

Problem:Show that the system in which the reaction

Mn(s) + 2/3 AlCl3(g, 1atm) = MnCl2(l) + 2/3 Al (l)

is at equilibrium is invariant.

Solution: N = 4 (Mn(s), AlCl3(g), MnCl2(l), Al (l))

P = 1 solid (Mn) + 2 liquids(MnCl2(l)and Al (l)) + 1 gas (AlCl3(g)) = 4

r = 1s = 0

t = 1 (1 atm of AlCl3(g)) C = N –r – s = 4 – 1 – 0 = 3F = C – P + 2 – t = 3 – 4 + 2 – 1 = 0This is an invariant system




Application of Phase rulein Reactive ComponentsProblem: Consider reduction of FeO with CO under standard

conditions i.e. P = 1 atm. FeO(s) + CO(g) = Fe(s) + CO2(g).

Calculate the number of components and degrees of freedom.

Solution:In this system we have P = 3 (i.e. two solids FeO and Fe and a gaseous phase CO+CO2) and N = 4, r = 1, s = 0 and

t = 1 (PCO + PCO2 =1 atm)C = N –r –s = 4 -1 -0 = 3F = C - P + 2-t = 3 – 3 + 2 -1 = 1Thus the above system has only one degree of freedom,

either temperature or pressure.




Phase Diagram

Graphical representation of what phases are present

in materials systems at various temperatures, pressures and compositions are called phase

diagrams




Solid Solution




Solid Solution




Types of Solid Solubility




Phase Diagram

Solidus

The phase boundary between solid and two phase

region.

Liquidus

The phase boundary between liquid and two phase

region.

Solvus

The solid state phase boundary between terminal solid

solution and two phase region.




Single – Component System – Variation of Free energy with Temperature

S(l) > S(s), So slope of the line

for liquid H2O is greater than

solid H2O

)()(

)()(

l

P

l

s

P

s

ST

G

ST

G




Single – Component System – Variation of Free energy with Temperature

Where ΔS is the change in

molar entropy which occurs

as a result of the change of

state. The slope od the line is –ve

which shows that at all temp

SH2O(l) > SH2O(s)

ST

G

P




Single – Component System – Variation of Free energy with Pressure

V(l) < S(s) for H2O So slope

of the line for solid H2O

is greater than liquid H2O

at all pressure

)()(

)()(

s

T

s

l

T

l

VP

G

VP

G




Variation of Free energies of solid liquid and vapour H2O with Temperature and Pressure




Binary Phase Diagrams

Isomorphous System: The system which exhibits

complete solid solubility and liquid solubility is

called an isomorphous sytem. The crystal

structure of both the components as well as solid

solution are same.

Eutectic system: The system which exhibit limited

solid solubility or terminal solid solution is called

an eutectic system.




Isomorphous system




Eutectic system




Experimental Determination of Liquidus and Solidus– Cooling Curve Method

T = f(t) cooling curves measurements for several compositions

Pure metal : melts /solidifies /allotropic transformation takes place at one temperature.

Binary solutions: melts /solidifies over a range of temperature.

The temperature at which the start of solidification takes place is called liquidus.

The temperature at which the end of solidification takes place is called solidus.




Experimental Determination of Liqidus and Solidus– Cooling Curve Method




Experimental Determination of Solvus– Cooling Curve Method

The common methods for the determining the solvus are microscopic examination and X-ray diffraction methods.

A series of small ingots of alloys of different compositions are prepared and homogenized.

They are annealed at various temperatures for prolonged time (a few days) and then quenched.

High temperature phases may be retained on quenching.

Subsequent, metallographic studies and X-ray diffraction reveals the various phases present at that temperature for given alloys




Experimental Determination of Solvus– Cooling Curve Method

A phase boundary is first bracketed between two compositions.

The exact location of the boundary is determined by studying a more alloys of closely varying compositions in the boundary region.




Free energy – Composition Diagram

The intercepts of the two axes by the tangent of the Gibbs free energy curve of the α phase at the composition represent

similarly, for the β phase

2X

21 and




Free – Energy Composition Diagram for Binary Systems

Isomorphous System




Free – Energy Composition Diagram for Binary Systems

Eutectic System




Documents

Metallurgical Thermodynamics