chemical equilibrium and thermodynamics

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Chemistry (Chemical Thermodynamics and Phase Equilibria)
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1. CHEMICAL THERMODYNAMICS
The word thermodynamics implies ‘Flow of Heat’
Thermodynamic
Thermos dynamics
Heat Flow
Greek word
It deal with energy change accompanying all types of physical and chemical process.
The laws of thermodynamics apply only to matter in bulk and not to individual atoms or molecules.
Thermodynamics can only predict whether a given process including a chemical reaction, is feasible
under a given set of conditions.
For example, thermodynamics predicts that hydrogen and oxygen gases would react at ordinary
temperature to yield liquid water. But it does not tell whether the reaction will be fast or slow. We know
only from experiment that in the absence of a catalyst, the reaction is extremely slow.
The study of thermodynamics is based on three generalization derived from experimental results.
These generalization are known as first, second and third law of thermodynamics based on human
experience and there is no formal proof for them.
THERMODYNAMICS TERMINOLOGY
Some terms are mainly used in thermodynamics are given below :
(A) System : A system is defined as a specified portion of matter which is separated from the
rest of the universe with a bounding surface.
(B) Surrounding : The rest of universe which can interact with the system, is surroundings.
(C) Isolated System : A system which can exchange neither energy nor matter with its
surrounding is called isolated system.
(D) Closed System : A system which can exchange only energy but not matter with its
surroundings is called a closed system.
(E) Open System : A system which can exchange energy and matter with its surrounding is
called as open system.
(F) Homogenous and Heterogenous System :
1. Homogenous System : A system is said to be homogenous when it is completely
uniform throughout. A Homogenous system is made of one-phase only.
For example : a pure single solid, liquid or gas, mixture of gases and a true solution.
2. Heterogenous System : A system is said to be Heterogenous when it is not
uniform throughout. A heterogenous system consists of two or more phase.
For example : Ice in contact with water, two or more immiscible liquids, insoluble solid in
contact with a liquid, a liquid in contact with vapour etc.
(G) Thermodynamic Properties : These are of two types :
1. Extensive Properties : The properties whose magnitude depends upon the quantity
of matter present in the system are called Extensive Properties.
For example : mass, volume and energy.
Chemical Thermodynamics
and Phase Equilibria
UNIT-4

104
2. Intensive Properties : The properties which do not depends upon the quantity of
matter present in the system are called Intensive Properties.
For example : temperature, pressure, density, concentration, specific heat etc.
(H) State Functions or State Variable : Fundamental properties which determine the state
of a system are referred to as state variables or state functions or thermodynamic
parameters.
The change in the state properties depends only upon the manner in which the change
has been brought about Pressure, Temperature, Volume, Mass, Enthalpy, Free Energy,
Internal Energy, Entropy are the most important state variables.
(I) Thermodynamic Process : The operation by which the thermodynamic system change
from one state to another, is called a process.
The various types of the processes are :
(a) Isothermal Process : A process is termed as isothermal if temperature remains constant
during each stage of the process.
For isothermal process; dT = 0.
(b) Adiabatic Process : A process is termed as adiabatic if no exchange of heat takes place
between the system and surroundings.
For adiabatic process; dQ = 0
(c) Isobaric Process : A process is termed as isobaric if pressure of the system remains
constant during each step of the process.
For isobaric process; dp = 0
(d) Isochoric Process : The process termed as isochoric if volume of the system remains
constant i.e, dv = 0
(e) Cyclic Process : The process is termed as cyclic, when a system undergoes a number
of different processes and finally returns to its initial state, for cyclic process dE = 0, and
dH = 0
REVERSIBLE AND IRREVERSIBLE PROCESS
(i) Reversible Process : A process which occurs infinitesimally slowly so that driving force
is only infinitesimally greater than the opposing force is called a reversible process.
A reversible process cannot be realised in practice because it would require infinite time for
its completion. Hence, almost all processes occurring in nature or laboratory are irreversible.
(ii) Irreversible Process : A process which does not take place in the above manner is said
to be irreversible process.
An irreversible process is spontaneous in nature. It is real and can be performed in
practice. All natural precesses are irreversible in nature.
Irreversible Process
1 It is ideal process It is spontaneous process
2 It takes infinite time It takes finite time
3 The driving force is infinite greater than the
opposing force
The driving force is much greater than the
opposing force
4 It is in equilibrium at all stages It is in equilibrium in the initial and final stages
only
5 Work obtained is maximum Work obtained is not maximum
6 It is difficult to realise in practice It can be performed in practice
Reversible Process

105
2. THE FIRST LAW OF THERMODYNAMICS
It is also known as “the law of conservation of energy.” It states that :
(i) Energy can neither be created nor destroyed, but it can be converted from one form to
another.
(ii) The total energy of the universe is constant.
(iii) Whenever energy in one form disappears, an equal amount of energy in some other form
must appear.
(iv) It is impossible to construct a perpetual motion machine, i.e., a machine which can
produce energy without consuming energy.
Consider a system, if in a given process, the quantity of heat transferred from the surroundings
to the system is q and the work done in the process is W, then the change in internal energy.
U is given by
U Q W  
This is the mathematical statements of the first law of thermodynamics. In this statement, Q is
the heat absorbed and W is the work done on the system.
In case Q is the heat absorbed and w the work by the system, then the relationship becomes
U Q W  
If work is done on the system, W is taken as positive so that U = Q + W if, however work is
done the system, W is taken as negative so that U = Q – W
Enthalpy [H]
1. Heat change at Constant Pressure
(p = 0)
according to first law
Q = E + W ...(1)
at constant pressure the work of expansion (W) may be replaced by PV, where V is the
change in volume.
So equation (1) taken the form
QP
= Ep + PV ....(2)
QP
= (E2
– E1
) + P(V2
– V1
)
QP
= (E2
+ PV2
) – (E1
+PV1
) ....(3)
The term (E2
+ PV2
) is a measure of the heat constant or total energy of the system in the final
state and (E1
+ PV1
) is the total energy of the initial state.
Since the quantities E, P, V are state function. It is covenant to replace E + PV term by another
state function H. Which is called as enthalpy.
So H = E + PV ....(4)
for infinitesimal change we shall have
dH = dE + PdV + VdV
at constant pressure dP = 0
dH = dE + PdV
from equation (2)
dH [dQ]P .... (5)

106
So the enthalpy is the heat absorbed at Constant pressure
Qp
= H2
– H1
or p pQ H  ...(6)
at Constant pressure heat absorbed is equal to change in enthalpy.
2. Heat Change at Constant Volume (V = 0)
According to equation (1)
Q = E + W
Q = E + PV  W = PV
at Constant volume
V = 0 ...(7)
So Qv
= Ev
...(8)
That is heat absorbed at constant volume is equal to change in internal energy of system.
p
p
p
dQ dH
C or dH Cp.dT
dT dT
 
   
 
...(9)
v
v v
v
dQ dE
C or dE C .dT
dT dT
 
   
 
...(10)
Where :
CP
= Heat capacity at Constant P
CV
= Heat Capacity at Constant V
The difference between CP
and CV
is equal to work done by 1 mole of gas in expansion when
heated through 1°C
Work done by gas at constant pressure = PV
for 1 mole of gas PV = RT
When temperature is raised by 1°C, the volume become V + V ;
So,
P (V + V) = R (T + 1)
or PV = R
Hence ; Cp
– CV
= PV = R
Ratio of heat capacity
Cp
Cv
 
  
 
depends on atomicity of gas.
Atomicity Cv Cp
Cp
Cv
 
1. Monoatomic 3
/2
R 5
/2
R = 1.66
2. Diatomic 5
/2
R 7
/2
R = 1.40
3. Polyatomic 3 R 4 R = 1.33
Let n1
and n2
moles of two non-reacting gases A and B are mixed than the heat capacity of the
mixture may be calculated as -
1 v 1 2 v 2
v mixture
1 2
n (C ) n (C )
(C )
n n




107
Difference between H and E is significant when gases are involved in a chemical reaction.
H = E + P V
H = E + nRT
Here,
n = Number of gaseous moles of products – Number of gaseous moles of reactants.
Using the above relation we can interrelate heats of reaction at constant pressure and at constant
volume.
INTERNAL ENERGY
Every system having some quantity of matter is associated with a definite amount of energy. This
energy is known as internal energy. The exact value of this energy is not known as it includes all types
of energies of molecules constituting the given mass of matter such as translational, vibrational, rotational,
the kinetic and potential energy of the nuclei and electrons within the individual molecules and the manner
in which the molecules are linked together, etc. The internal energy is denoted by E.
E = EEtranslational
+ Erotational
+ Evibrational
+ Ebonding
+ Eelectronic
+ ....
Accurate measurements of some forms of energy which contribute to the absolute value of
internal energy for a given substance in a given state is impossible. But one thing is certain that the
internal energy of a particular system is a definite quantity at the given moment, irrespective of the
manner by which it has been obtained. Internal energy like temperature, pressure, volume, etc., is a state
function, i.e., total of all possible kinds of energy of a system is called its internal energy*.
It is neither possible nor necessary to calculate the absolute value of internal energy of a system.
In thermodynamics, one is concerned only with energy change which occurs when the system moves
from one state to another. Let E be the difference of energy of the initial state (Ei
) and the final state
(Ef
), then
E = Ef
– Ei
E is positive if Ef
> Ei
and negative if Ef
< Ei
.
A system may transfer energy to or from the surroundings as heat or work or both.
Characteristics of Internal Energy
(i) Internal energy of a system is an extensive property.
(ii) Internal energy is a state property.
(iii) The change in the internal energy does not depend on the path by which the final state
is reached.
(iv) There is no change in internal energy in a cyclic process.
Expansion of an Ideal Gas and Changes in Thermodynamic Properties
With the help of the first law of thermodynamics, it is possible to calculate changes in
Thermodynamic Properties such as Q, W,V and H, when an ideal gas undergoes expansion. The
expansion may be isothermal or Adiabatic.
1. Isothermal Expansion
A. Calculation of H 
We know that
H = U + PV
H = (U+PV)
or H = U +(PV) PV nRT
H = U = nRT
Since for an isothermal process,

108
T = 0
U = 0
hence, H O 
Calculation of Q and W
B. According to first law of thermodynamics.
U = Q + W
Since U = 0,
hence,
Q W
i.e, the work is done at the expense of the heat absorbed.
Work done in Reversible Isothermal Expansion
During expansion, Pressure decreases and volume increases, these two Parameters. Are assigned
opposite signs.
The work done by the gas in an infinitesimal expansion is thus given by -
dW = – PdV
The total work w done by the gas in expansion of ideal gas from volume V1
to volume V2
is
therefore,
2
1
v
v
W PdV 
for an ideal gas,
P =
nRT
V
So,
2
1
v
v
dv
W nRT
V
  
Integrating,
2
e
1
V
W nRTlog
V

or
2
1
V
W 2 303nRTlog
V
 
Since in an ideal gas,
P1
V1
= P2
V2
or
2 1
1 2
V P
V P

So,
1
2
P
W 2 303nRT log
P
  
Since during expansion V2
is more than V1
and P2
is less than P1
, hence the work W comes out
to be negative.
Isothermal Compression work of ideal gas may be derived similarly and it has exactly the same
value with positive sign.

109
W’compression
= 2.303 nRT log
1
2
V
V
= 2.303 nRT log
2
1
P
P
Since during Compression, the initial volume V2
is more than the final volume V1
and also since
the initial pressure P2
is less than the final Pressure P1
, hence, the work done W comes out to be
positive
Work done in irreversible isothermal expansion :
(i) expansion against zero external pressure or free expansion :- Since in free expansion the
external pressure in zero.
Pext
= 0
than, irrW 0
(ii) expansion against Constant external pressure :- suppose the volume of the gas increases
from V1
to V2
Wirr
=
2
1
V
ext
V
P dV 
Wirr
= – Pext
(V2
– V1
)
Since, V =
nRT
P
So Wirr
= - Pext
2 1
nRT nRT
P P
 
 
 
irr ext
2 1
1 1
W P nRT
P P
 
   
 
Since Pext
is less than p, the work done during intermediate isothermal expansion is
numerically less than the work done during reversible isothermal expansion in which Pext
is almost equal to P.
2. Adiabatic Expansion
In adiabatic expansion no heat is allowed to enter or leave the system.
Hence,
Q = 0
thus, according to first law of thermodynamics
E = Q + W
Q = 0
So, E = W
The molar specific heat capacity at Constant volume of an ideal gas in given by-
Cv
=
v
dE
dT
 
 
 
or dE = Cv
.dT

110
vW = E = C .dT
Reversible Adiabatic Expansion
Let P be the external pressure and V be the increase in volume.
Then, the external work done by the system is given by
W = – PV
According to first law of thermodynamics
U = – PV
Since U = Cv
dT
hence, Cv
dT = – PdV
for very small change in reversible process.
Cv
dT = – PdV = – RT
dV
V
(for 1 mole of gas)
or Cv
dT/T = – RdV/V
or Cv
d(In T) = – Rd(In V)
Integrating,
Cv
In(T2
/T1
) = – R ln (V2
/V1
) = R In(V1
/V2
)
or In (T2
/T1
) = (R/Cv
) In(V1
/V2
)
we know that :
Cp
– Cv
= R
and Cp
/Cv
= ,
We get
In(T2
/T1
) = (–1) In(V1
/V2
) = In(V1
/V2
)(–1)
 T2
/T1
= (V1
/V2
)
or T1
/T2
= (V2
/V1
)
or
1
1 1 2
2 2 1
P V V
P V V

 
  
 
P1
V1
 = P2
V2

PV
= Constant
Thus knowing  , V1
, V2
(or P1
, P2
) and the initial Temperature TT1
and final temperature T2
can
be readily evaluated.
Irreversible Adiabatic Expansion
(i) Free Expansion
Since Pext
= 0
So W = 0
According E = W = 0
Since the internal energy is a function of temperature,
therefore T = 0
Thus in a free adiabatic expansion :
T = 0
E = 0
W = 0
and H = 0

111
(ii) Intermediate expansion
Since
W = –Pext
(V2
–V1
)
V =
RT
P
(for 1 mol of gas)
V2
=
2
2
RT
P
1
1
1
RT
V
P

putting the value
W = –Pext
2 1
2 1
RT RT
P P
 
 
 
= –Pext
2 1 1 2
1 2
T P TP
R
PP
 
 
 
or W = Cv
(T2
– T1
)
  2 1 1 2
v 2 1 ext
1 2
T P TP
W C T T RP
PP
 
    
 
Thus knowing Cv
, T1
, P1
, P2
and Pext
, we can readily calculate T2
, the final temperature of the gas.
Reversible Isothermal Expansion of Real Gas
We shall not derive expression for W, U, H and Q for the reversible isothermal expansion of
real gas.
(A) Work of expansion :
the work done in the expansion is given by.
–dW = PdV

2
1
v
v
W PdV  
for the van der Walls gas.
2
2
an
P (V nb) nRT
V
 
   
 
So that
2
2
nRT an
P
V nb V
 

Hence,
2
1
v 2
2
v
nRT an
W dv
V nb V
 
   
 


112
2 2
1 1
v v 2
2
v v
nRT an
dv dv
V nb V
 
 
22
1 2 1
V nb 1 1
nRTln an
V nb V V
   
     
   
(B) Internal Energy Change
For a vander walls gas, the term an2
/V2
is the internal pressure of the gas, further, the quantity
T
U
V
 
 
 
is also called the internal pressure of the gas.
Thus,
2 2
T
U
an / V
V
 
  
 U = an2
(dV/V2
) (at constant pressure)
Integrating,
2
1
V
2 2
2 1
2 2 1V
dU 1 1
U U U U an an
V V V
 
         
 
 
(C) Heat Change
from the first law of thermodynamics
U = Q + W
or Q = U – W
Substituting the value for W and U
2
1
V nb
Q nRT ln
V nb
 
  
 
(D) Enthalpy Change (H)
We know that
H = U + PV
H1
= U1
+ P1
V1
(Initial state) ...(1)
H2
= U2
+ P2
V2
(Final state) ...(2)
The enthalpy change
H = H2
– H1
= (U2
+ P2
V2
) – (U1
+ P1
V1
)
= (U2
– U1
) + (P2
V2
– P1
V1
)
= U + (P2
V2
– P1
V1
) ...(3)
For the Vander walls gas
2
2
nRT an
P
V nb V
 

...(4)
Multiplying both side of equation by v, we get
2
nRTV an
PV
V nb V
 

...(5)

113
Substituting the expression for P2
V2
and P1
V1
as obtained from Eq.5 into Eq.3, we obtain
2
2 1 1 2
1 1 1 1
H U nRT an
V nb V nb V V
   
         
    
...(6)
Now,
   
 
2 1 1 22 1
2 1 2 1
V V nb V V nbV V
V nb V nb V nb (V nb)
  
 
   
 
  
1 2
2 1 2 1
nb V V nb nb
V nb V nb V nb V nb

  
    ...(7)
Substituting Eq. 7 into Eq. 6, we have
2
2 1 1 2
nb nb 1 1
H U nRT an
V nb V nb V V
   
        
    
...(8)
Substituting for U from internal energy change, we have
2 2
2 1 2 1
1 1 1 1
H n bRT 2an
V nb V nb V V
   
       
    
Comparison of Work of Expansion of an Ideal Gas and a vander Waals Gas. We know that for
an ideal gas, numerically
Wideal
= nRT ln (V2
/V1
) ...(9)
and for a vander Waals gas, from work of expansion numerically
22
vdw
1 2 1
V nb 1 1
w nRT ln an
V nb V V
   
     
   
...(10)
If V  nb, then Eq.10 reduces to
22
vdw
1 2 1
V 1 1
w nRT ln an
V V V
   
     
   
...(11)
Hence, numerically

2
2 2 1
ideal vdw
2 1 1 2
an (V V )1 1
w w an
V V V V
  
     
 
...(12)
Since for the expansion of a gas, V2
> V1
, it is evident from Eq. 12 that numerically, the work in
the reversible isothermal expansion of an ideal gas is greater than that for a vander Waals gas.
The Joule Thomson effect :
“When the adiabatic expansion of the gas from higher pressure to lower pressure by passing
through a porous plug, it gets cooled appreciably. This effect is known as Joule Thomson effect.
P1 P2
v1
A
v2
BG
Porous PlugJoule-thomson effect

114
The cooling effect is due to decrease in the kinetic energy of gaseous molecules since a part of
this energy is used up is overcoming the vander walls force of attraction existing between the molecules
during expansion. So gas become Cooler.
Joule-Thomson Coefficient (J.T
). The experimental technique used by Joule and Thomson for
deriving the mathematical relation between the fall of pressure of a gas on expansion and the resulting
lowering of temperature is illustrated schematically in Fig.
A tube made of a non-conducting material is fitted with a porous plug G in the middle and two
pistons A and B on the sides, as shown. The tube is thoroughly insulated to ensure adiabatic conditions.
A volume V1
of the gas enclosed between the piston A and the porous plug G at a pressure P1
is forced
slowly through the porous plug by moving the piston A inwards and is allowed to expand to a volume V2
at a lower pressure P2
by moving the pisten B outward, as shown.
 Work done on the system at the piston A = + P1
V1
Work done by the system at the piston B = – P2
V2
 Net work done by the system = –P2
V2
+ P1
V1
Since the expansion of the gas has taken place adiabatically, the system is not in a position to
absorb heat from the surroundings. The system, therefore, performs work at the expense of its internal
energy. Consequently, the internal energy of the system changes, say, from U1
to U2
.
 –P2
V2
+ P1
V1
= U2
– U1
or U2
+ P2
V2
= U1
+ P1
V1
or H2
= H1
or H = 0.
Thus, the Joule-Thomson expansion of a real gas occurs not with constant internal energy but
with constant enthalpy. This is, therefore, called an isenthalpic process.
Joule-Thomson Coefficient :
J.T
H
T
P
 
    
...(1)
Case - I If J.T
has positive value, the gas cools on expansion.
Case - II If J.T
has negative value, the gas warms on expansion
for e.g.- H2
, He etc.
Case - III If J.T
has zero value. No Joule Thomson effect is observed.
for example - Ideal gas
Joule-Thomson Coefficient for an ideal gas :
Since
H = U + PV
The enthalpy is the function of temperature and pressure.
H = f (T, P)
p T
H H
dH dT dP
T P
    
    
    
...(2)
we know that p
P
H
C
T
 
  
p
T
H
dH C dT dP
P
 
    
...(3)
but Joule - thomson effect is an isoenthalpic process

115
So,
dH = 0
0 = Cp
dT +
T
H
dP
P
 
  
p
T
H
C dT dP
P
 
    
p
H T
T H
C
P P
    
        
J.T
pH T
T 1 H
P C p
   
         
...(4)
Since
H = U + PV
The Eq. - 4 may be written as -
 
J.T
pH T
U PvT 1
P C P
   
           
p T T
1 U (PV)
C P P
     
           
p T T
1 U V (PV)
C V P P
      
             
...(5)
Since for an ideal gas, (U/V)T
is zero

T T
U U
0
V P
    
       
...(6)
Also, Since for an ideal gas, PV is constant at constant temperature,
T
(PV)
P
 
  
= 0, Hence
J.T 0  ...(7)
Thus the Joule - Thomson Coefficient for ideal gas is thus zero
Joule-Thomson Coefficient in a Real Gas
The Joule - Thomson Coefficient in a Real gas can be easily Calculated with the help of the
Vander Waals equation
Since both a and b are small, the term ab/V2
in the vander waals equation can be neglected
provided the pressure not too high
Thus, The van der Waals equation may be written as
PV = RT –
a
V
+ bP ...(8)
Since, V =
RT
P

116
 PV = RT – aP/RT + bP ...(9)
or V = RT/P - a/RT + b ...(10)
Differentiating with respect to temperature at constant pressure. We get
2
P
V
R /P a /RT
T
 
   
...(11)
Rearranging Eq-11,
or RT = P(V – b) + aP/RT ...(12)
Dividing both sides by PT
2
R V b a
P T RT

  ...(13)
Substituting the value of R/P from Eq. 13 in Eq. 14 we have -
2 2
P
V V b a a
T T RT RT
  
    
2
P
V V b 2a
T T RT
  
   
P
V 2a
T V b
T RT
 
    
...(14)
Using the well known thermodynamic relation
P P
V H
V T
T P
    
        
...(15)
Eq.14, may be written as -
T
H 2a
b
P RT
 
   
...(16)
from Eq. (4) and Eq.(16) we have
pH
T 1 2a
b
P C RT
   
        ...(17)
Thus, The Joule - thomson coefficient is positive for real gas.
Inversion Temperature
Temperature below which gas show cooling effect in Joule thomson experiment is called Inversion
temperature
i
b
2a
T
R

for H2
Ti
= –80°C
for He Ti
= –242°C
Boyale Temperature
Temperature range in which real gases obey ideal gas law is called Boyale Temperature.
Tb
=
b
a
R
i bT 2T

117
3. THERMOCHEMISTRY
The branch of Physical Chemistry which deals with the energy changes accompanying chemical
transformation.
The energy change in a Chemical reaction is due to change of bond energy, i.e, it result from the
breaking of bonds and formation of new bonds in products.
Exothermic Reactions : Reactions which are accompanied by evolution of heat, are called
exothermic reactions.
for exothermic reactions
H = Vee
H = HP
- HR
HP
= enthalpy of Produce
HR
= enthalpy of reactant
for negative value of H P RH H
Energy
given out
Product
E
Reactant
Fig. : Exothermic reactions
Endothermic Reactions
Such reactions which are accompanied by absorption of heat, are called endothermic reaction
for endothermic reactions.
H = Positive
for positive value of H
P RH H
Product
Energy
taken in
Reactant
E
Fig. Endothermic reaction
Enthalpy of Neutralization or Heat of Neutralization
Enthalpy of neutralization is defined as the Heat evolved when one mole of a strong base such
as NaOH and KOH is neutralized by one mole of strong acid such as HCl, H2
SO4
, and HNO3
.
dilute solution.
Some examples are -
Strong acid + Strong base = Salt + water; H = –13.7 Kcal.mol–1
HCl (aq) + NaOH (aq) = NaCl (aq) + H2
O()
HNO3
(aq) + NaOH (aq) = NaNO3
(aq) + H2
O()
It is observed that heat of neutralization of a strong acid against a strong base is always nearly
the same, i.e, 13.7 Kcal mol–1
or 57 KJ mol-1

118
 because It simply involves the combination of H
ions and OH–
ions to form unionised H2
O
for example - 2H Cl Na OH Na Cl H O( )     
      
or 2H OH H O( ) 
  
If however, the acid or alkali, is weak, the enthalpy of neutralisation is different because the
reaction now involves dissociation of weak acid or the weak alkali as well.
for example:
(i) 3 3CH COOH(aq) CH COO (aq) H (aq) 

(ii) 2H (aq) OH (aq) H O( ); H 13.7 Kcal/ mol 
    
Enthalpy of Formation : (Hf)
The amount of heat absorbed or evolved when 1 mole of the substance is formed its element
is called enthalpy of formation.
for example;
2 2C(amorphous) O (g) CO (g); H 97.6 Kcal/ mol    
Enthalpy of formation of CO2
is - 97.6 Kcal mol–1
Thus, enthalpy of formation of HCl is –22 Kcal mol–1
Enthalpy of Combustion
The amount of heal evolved when one mole of substance is Completely oxidized
for example,
4 2 2 2CH (g) 2O (g) CO (g) 2H O / )   
1
combH 890.3KJmol
  
(g)2 2 2
1
H O (g) H O( )
2
  
1
combH 285.29KJmol
  
Enthalpy of Solution
The amount of heat evolved when one mole solute is dissolved in excess of solvent is called heat
of solution.
for example -
1
2 4KCl(s) aq. H SO (aq.) ; H 20.2kcalmol
    
1
KOH(S) aq. KOH(aq.) ; H 13.3kcalmol
    
Enthalpy changes during phase transformations :
(i) Enthalpy of Fusion, fus.
H : It is the enthalpy change that accompanies melting of one mole
of solid substance at constant temperature (melting point of solid) and pressure.
for example,
H2
O(s)  H2
O(l) ; fus
H = + 6.01 kJ mol–1
(ii) Enthalpy of Vaporisation, vap
H : It is the enthalpy change required to vapourise one mole
of a liquid substance at constant temperature (boiling point of liquid) and pressure for example :
H2
O(l)  H2
O(g) ; vap
H = + 40.79 kJ mol–1

119
(iii) Enthalpy of Sublimation, sub
H : It is the enthalpy change required to sublime one mole
of a solid substance at constant temperature and pressure. For example
CO2
(s)  CO2
(g) ; sub
H = + 25.2 kJ mol–1
(iv) Enthalpy of fransition : It is the enthalpy change when one mole of one allotropic form
changes to another under conditions of constant temperature and pressure. For example
C(graphite)  C(diamond) trs
H = 1.90 kJ mol–1
Lattice Enthalpy (lattice
H)
The lattice enthalpy of an ionic compound is the enthalpy change which occurs when one mole
of an ionic compound dissociates into its in gaseous state under conditions of constant temperature and
pressure.
Na+
Cl–
(s)  Na+
(g) + Cl–
(g) ;
lattice
H = + 788 kJ mol–1
Lattice enthalpy can also be defined for the reverse process. In that case the value of HLE
will
be negative.
BORN-HABER CYCLE FOR NaCl :
This cycle is based on thermochemical changes taking place in the formation of a lattice. This
cycle can be used to determine lattice energy which cannot be directly measured. It is defined as that
energy released when one mole of the ionic compound (lattice) is formed its isolated ions in the gaseours
state under standard condition.
nAm+
(g) + mBn–
(g)  An
Bm
(s)
H = –U (lattice energy)
Formation of NaCl(s) lattice involves thus.
S + I +
Cl Cl
2
 
– E – U = q
hence, U can be calculated.
here, S = enthalpy of sublimation on Na(s) = Hsublimation
I = ionisation of energy of Na(g) = Hionization
= bond energy of Cl2
U = lattice energy
q = enthalpy of formation of NaCl(s) = Hformation
If lattice is MgX2
(s) then
S + (I1
+ I2
) + – 2E – U = q
where, (I1
+ I2
) = total ionisation energy of form Mg2+
(g).
Enthalpy of Atomisation, a
H :
It is the enthalpy change when one mole of a substances is dissociated into atoms in the
gaseous state, under constant pressure and temperature condition.
For example
H2
(g)  2H(g) ; a
H = 435.0 kJ mol–1
CH4
(g) C(g) + 4H(g); a
H = 1665 kJ mol–1
Enthalpy of Hydration, hyd
H :
It is the enthalpy change when one mole of an anhydrous (or partly hydrated) compound combines
with the required number of moles of water to form a specific hydrate at the specified temperature and
pressure. For example :
CuSO4
(s) + 5H2
O(I)  CuSO4
· 5H2
O(s) ; (hyd)
H = –78.20 kJ mol–1

120
Hess’s Law of Constant Heat Summation
According to this law, the amount of heat evolved or absorbed (total heat change) is always
constant, whether the process take place in one or in several steps.
Let a substance A be changed to B in three step & involving a change from A to C, C to D, and
finally from D to B
If H1
, H2
, H3
are the heat exhanged in the first step, second and third step, respectively, then
according to Hess’s law
Total enthalpy change (H), from A to D will be equal to the sum of enthalpies involved in various
steps.
HA B
C D
H1
H2
H3
1 2 3H H H H      
Applications of Hess’s Law
1. Calculation of Enthalpies of Reactions :
Hess’s law makes it Possible to calculate enthalpies of many reactions which cannot be determined
experimentally.
for example,
It is difficult to measure the heat evolved when carbon burns in oxygen to form Carbon monoxide
2
1
C(S) O (g) CO(g)
2
  ; H = ?
from Hess’s law, however, it is known that the heat evolved in the Combustion of one mole of
Carbondioxide is the same, viz, 393.5 kJ, whether the reaction takes place in a single step as
2 2C(S) O (g) CO (g)  ; H = –393.5 kJ
or in two steps are
2
1
C(S) O (g) CO(g)
2
  ; H = x kJ (say)
2 2
1
CO(g) O (g) CO (g)
2
  ; H = y kJ (say)
according to Hess’s law
H = x + y
or x + y = –393.5 kJ
The heat change involve in the Combustion of Carbon monoxide to give carbon dioxide can be
measured and has been found to be – 282.0 kJ
So, x + (–282) = –395.5
x = 395.5 + 282
= –111.5 kJ
Thus, H for the Combustion of carbon to give carbon monoxide is - 11.5 KJ

121
2. Calculation of Enthalpies of Formation
The enthalpies of formation of Compounds can calculated by the application of Hers’s law
for example:
Calculation of enthalpies of formation of benzene from its elements : carbon and hydrogen.
Step-I : The thermochemical equations for the known values are written as
(i) 6 6 2 2 2
15
C H ( ) O (g) 6CO (g) 3H O( );
2
    H = –3267.7 kJ
(ii) C(S) + O2
(g)  CO2
(g); H = –393.5 kJ
(iii) H2
(g) +
1
2
O2
(g)  H2
O() H = –285.9 kJ
Step-II : Equation (ii) is multiplied by 6 and equation (iii) is multiplied by 3
(iv) 6C(S) + 6O2
(g)  6 CO2
(g) ; H = –2361.0 kJ
(v) 3H2
(g) +
3
2
O2
(g)  3 H2
O(); H = –857.7 kJ
Adding (iv) and (v) and substracting (i), we get
6C(S) + 3H2
(g)  C6
H6
(); H = 49.0 kJ
Thus enthalpy of formation of benzene is + 49 kJ
KIRCHHOFF’s EQUATION
Variation of heat of the reaction with temperature is expressed mathematically by kirchhoff and
is known ask kirchhoff’s equation.
(A) Kirchhoff’s Equation at Constant Pressure
at constant P
d (H) = Cp
dT
on integrating -
2 2
1 1
T T
P
T T
d H C dT   
2 1 P 2 1HT HT C (T T )     
2 1
P
2 1
HT HT
C
T T
  
 

HT2
= Heat of reaction at T2
temperature
HT1
= Heat of reaction at T1
temperature
(B) Kirchhoff’s Equation at Constant Volume
at constant volume
d(E) = CV
dT
on integrating
2 2
1 1
T T
V
T T
d E C dT   
2 1T T V 2 1E E C (T T )     

122
or
2 1T T
V
2 1
E E
C
T T
  
 

2TE = Heat of reaction at T2
temperature
1TE = Heat reaction at T1
temperature
Bond Energies or Bond Enthalpies
The average amount of energy required to break one mole of bonds of that type present in
Compound.
Bond energy is also called, the heat of formation of the bond.
Consider the dissociation of H2
O molecule which Consist of two O-H bonds, The dissociation
occur in two stages.
2H O(g) H(g) OH(g);  H = 497.8 kJ mol–1
OH(g) H(g) O(g);  H = 428.5 kJ mol–1
The average of these two bonds dissociation energies gives the value of bond energies of
O – H bond =
497 8 428 5
2
  
= 463.15 kJ mol–1
Application of Bond Energy
i. Heat of Reaction
Heat of reaction =  Bond energy or reactant
–  Bond energy of products
for example -
Ex. Calculate the enthalpy of the following reaction,
2 2 2 3 3H C CH (g) H (g) CH CH (g)   
The bond energies of C–H, C–C, C=C and H–H are 99, 83, 147 and 104 k cal respectively
Sol. The reaction is :
H H H H
| | | |
(g) H H(g)C C H— C— C—H(g) ; H ?
| | | |
H H H H
    
H = Sum of bond energies of reactants
Sum of bond energies of Products
=    C C C H H H C C C HH 4 H H H 6 H              
= (147 + 4 × 99 + 104) – (83 + 6 × 99) = – 30 kcal
ii. Determination of Resonance Energy
When a compound show resonance, there is Considerable difference between the heat of formation
as calculated from bond energies and that determined experimentally.
Resonance energy can be calculated using the formula

123
H(Actual) – H(theoretical) = resonance energy of products – Resonance energy of reactants.
The proof of above formula is given by following diagram.
Consider a reaction
A(g) + B*(g)  C(g) + D*(g)
where (*) showing that molecules exhibit phenomena of resonance. Remember where ever
resonance take place, bond breaking become difficult.
Actual energy required to break a bond is equal to actual
= theoretical
– resonance energy
A(g) B*(g) C(g) D*(g)
Gaseous elements
A CB B– RE D D– RE
++
Hactual
Hactual
= A
+ B
– R.EB
– {C
+ D
– R.ED
}
Hactual
= (A
+ B
– C
– D
) + R.EB
– R.ED
[Hactual
– HTheoretical
= R.EProducts
– R.Ereactants
]
CALORIMETRY-MEASURING HEATS OF REACTIONS
All calorimetric techniques are based on the measurement of heat that may be generated
(exothermic process), consumed (endothermic process) or simply dissipated by a sample. There are
numerous methods to measure such heat. Any process that results in heat being generated and exchanged
with the environment is a candidate for a calorimetric study.
A calorimeter is a device used to measure heat of reaction. In order to measure heats of
reactions, we often enclose reactants in a calorimeter, initiate the reaction, and measure the temperature
difference before and after the reaction. The temperature difference enables us to evaluate the heat
released in the reaction.
Two basic types of calorimetry are discussed :
(a) measurement based on constant volume.
(b) measurement based on constant pressure.
A calorimeter may be operated under constant volume which measures internal energy change
U by bomb calorimeter or constant (atmosphere) pressure, which measures enthalpy H by calorimeter
Whichever kind to use, heat capacity of the calorimeter is required.
The heat capacity is the amount of heat required to raise the temperature of the entire calorimeter
by 1 K, and it is usually determined experimentally before or after the actual measurements of heat of
reaction. The heat capacity of the calorimeter is determined by transferring a known amount of heat into
it and measuring its temperature increase.
(i) Bomb calorimeter (U measurement) : For chemical reactions, heat absorbed at constant
volume, is measured in a bomb Calorimeter. In this Calorimeter, a steel vessel (the bomb) is immersed
in a water bath. A combustible substance is burnt in pure oxygen supplied in the bomb. Heat evolved
during the reaction is transferred to the water around the bomb and its temperature is monitored. Since
the bomb Calorimeter is sealed, its volume does not change, i.e., the energy changes associated with
reactions are measured at constant volume.

124
Since volume does not change, a bomb calorimeter measures the heat evolved under constant
volume, qv
,
qv
= C T,
where T is the temperature increase. The qv
so measured is also called the change in internal
energy, E.
E = qv
= C × T
Illustration
1. A calorimeter with heat capacity equivalent to having 13.3 moles of water is used to measure the
heat of combustion from 0.303 g of sugar (C12
H22
O11
). The temperature increase was found to
be 5.0 K. Calculate the heat released, the amount of heat released by 1.0 g, and 1.0 mole of
sugar.
Sol. Heat released, qv
,
qv
= 13.3 × 75.2 × 5.0 K
= 5000 J
The amount of heat released by 1.0 g would be,
5000 J/0.303 g = 16.5 kJ / g
Since the molecular weight of sugar is 342g/mol, the amount of heat released by 1.0 mole would
be 16.5 × 342 = 56431 kJ/mol.
Illustration
2. The temperature of a calorimeter increases 0.10 K when 7.52J of electric energy is used to heat
it. What is the heat capacity of the calorimeter?
Sol. Dividing the amount of energy by the temperature increase yields the heat capacity, C,
C = 7.52 / 0.10 = 75.2 J/K.
Note :
More heat is giving of if the reaction is carried out at constant pressure, since the P-V work (1.5
R T) due to the compression of 1.5 moles of gases in the reactants would contribute to dH. If

125
1.0 mole water is decomposed by electrolysis at constant pressure, we must supply an amount
of energy equivalent to enthalpy change, dH, a little more than internal energy, dE. More energy
must be supplied to perform the P-V work to be done by the products (H2
and O2
).
(ii) H measurement : Measurement of heat change at constant pressure (generally at atmospheric
pressure) can be done in a Calorimeter shown in the figure. In this case, the Calorimeter is left open
to atmosphere. As the reaction occurs in the Calorimeter, the temperature change is noticed and then
heat of reaction is measured with the knowledge of heat capacity of Calorimeter system.
The heat capacity of the calorimeter can also be determined by burning an exactly known amount
of a standard substance, whose enthalpy of combustion has been determined. Benzoic acid, C7
H6
O2
,
is one such standard. The problem below illustrates the calculations.
Illustraion
3. When 0.1025 g of benzoic acid was burnt in a bomb calorimeter the temperature of the calorimeter
increased by 2.165° C. For benzoic acid H°comb
= –3227 kJ mol–1
. Calculate the heat capacity
of the calorimeter.
Sol. The equation for the combustion is,
C7
H6
O2
(s) + 7.5 O2
(g)  7CO2
(g) + 3H2
O(1), H° = 3227 kJ
Since 7.5 moles of O2
gas is needed, and 7 moles of CO2
is produced, some pressure-volume
work is done, to the calorimeter.
PV = ng
R T, where n = (7 – 7.5) = –0.5 mol
E = H – ng
RT
= –3227 – (–0.5 × 8.314298 × 298)
= –3226 kJ/mol (a small correction)
The amount of heat produced by 0.1025 g benzoic acid is
q = 0.1025/122.13 × 3226 = 2.680 kJ
Thus, the heat capacity is
C = qv
/ T = 2.680 / 2.165 = 1.238 kJ / K.
After the heat capacity is determined, the calorimeter is ready to be used to measure the enthalpy
of combustion of other substances.

126
4. THE SECOND LAW OF THERMODYNAMICS
The second law of thermodynamics help us to determine the direction in which energy can be
transferred.
It also helps us to predict whether a given process or a Chemical reaction can occurs
spontaneously.
It also helps us to know the equilibrium Condition.
It states that It is impossible to use a cyclic process to extract heat from a reservoir and to
Convert it into work without transferring at the same time a certain amount of heat from a hotter to colder
part of the body.
Heat Engines : The Carnot Cycle
The conversion of heat into work required an energy. Any device that can convert heat into
mechanical work is called a heat engine. S Carnot (1824) devised a heat engine consisting of one mole
of an ideal gas enclosed in a cylinder fitted with a weightless, frictionless piston, and determined its
efficiency, that is, the ratio of work done to heat absorbed in the following way.
Carnot completed a cycle consisting of four reversible operations such that after the fourth
operation the system reverts back unchanged to its initial starting position. The operations are (I) reversible
isothermal expansion, (II) reversible adiabatic expansion, (III) reversible isothermal compression, and (IV)
reversible adiabatic compression.
P V T1 1 2
A
B
C
D
P V T2 2 2
P V T3 3 1
P V T4 4 1
The temperatures in (I) and (III) are T2
and T1
respectively and T2
> T1
because after the second
operation the temperature must fall. The parameters at different points of operation are shown in the
indicator diagram (Fig). The adiabatic curves are steeper because during these operations PV
= constant,
and not PV = constant.
Two large reservoirs containing liquids at temperatures T2
and T1
respectively are necessary for
per forming the isothermal processes. The first one is called the source as it supplies heat to the system
necessary for doing the expansion work and the other is called a sink as it takes up some heat, given
up by the system, as a result of compression.
Heat is taken up by the system only during the first operation and let this heat be Q2
. The work
done by the system at this stage is then equal to Q2
. The work in different stages are given as,
2
I 2 2
1
V
W RT ln Q
V
 
II V 1 2W C (T T ) 
4
III 1
3
V
W RT ln
V

and IV V 2 1W C (T T )  
WII
and WIV
are equal and opposite, and hence the total work done by the engine in the complete
cyclic operation is,

127
W = WI
+ WIII
(WII
and WIV
cancel out)
I III II IVW = W + W (W and W cancel out)
2 4
2 1
1 5
V V
= RT ln RT ln
V V
 ....(1)
As the operations II and IV are adiabatic,
1 1
1 3 2 2T V T V 
 or
1
1 2
2 3
T V
T V

 
  
 
and 1 1
1 4 2 1T V T V 
 or
1
1 1
4
T V
T V

 
  
 
Therefore,
2 1 4 1
3 4 3 2
V V V V
or
V V V V
 
For (1), and (2)
W = RT2
ln V2
/V1
+ RT1
ln V1
/V2
= RT2
ln V2
/V1
– RT1
ln V2
/V1
= R(T2
– T1
) ln V2
/V1
The efficiency , of the engine is given by
2 1 2 1
2 2 2 1
R(T T )ln V / VW
Q RT ln V / V

  
2 1
2
T T
T

 
The efficiency of a Carnot engine thus depends only on the temperature of the source and the
sink, and not on any other factor example, say the nature of, or the amount of the working substance.
Since, (T2
–T1
) < T2
therefore the efficiency is always less than unity, i.e., an engine operating in
cycles cannot convert heat completely into work. Only if T1
= 0 K, then and then only  = 1 and
consequently W = Q2
. So complete conversion of heat into work can be effected only if the sink can be
placed at 0 K. The temperature of 0 K cannot be achieved and hence in practice We cannot have any
heat engine of unit efficiency.
If Q1
is the amount of heat given up by the engine to the sink then the amount of heat converted
into work is Q2
–Q1
and hence,
2 1 2 1
2 2
Q Q T T
Q T
 
  
Carnot engine is a hypothetical one with all the stages of the cycle performed reversibly, and
since reversible processes yield maximum work, therefore this engine has maximum efficiency. The
engine being reversible can also be made to operate in the opposite direction example, in refrigeration.
Example : A refrigeration engine operates between 0 and 25°C heat being withdrawn from the
lower temperature and given up at higher temperature. What would be the work done in freezing 1 g of
water the latent heat of fusion of water being 80 kcal/kg?
Solution : Here T2
= 0°C = 273 K and T1
= 273 + 25 = 298 K
Q2
= 80 kcal/kg = 80 cal/g. Therefore,

128
2 1
2
2
T T (273 298)K
W Q 80cal
T 273K
  
   
 
= –7.324 cal
= –7.324 × 4.2 J = –30.76 J
Heat given up to the surroundings = Heat taken up by the engine – work done = Q2
– W
= 80 + 7.324
= 87.324 cal
= 366.7 J
Carnot’s theorem : This is a consequence of the second law of thermodynamics. It is usually
stated in two parts: (a) Of all the cyclic heat engines operating between the same two temperatures
(temperatures of source and sink), it is the reversible engine that has the maximum efficiency. (b) All
cyclic reversible engines operating between the same two temperatures have the same efficiency.
Entropy
Second law of thermodynamics gives a new thermodynamics term which is a state function is
called entropy :
 Entropy is denoted by ‘s’
 Entropy stands for two greek words
En - identity it with energy
trope - Change
 Entropy is measure of randomness or disorderness of a molecular system.
 It is difficult to define the actual entropy of a system
It is more convenient to define the change of entropy during a change of state.
Thus change of entropy of a system may be define as the summation of all the terms each
involving heat exchange (q), divided by the absolute temperature (T) during each infinitesimally small
change of the process carried out reversibly
Thus entropy change for a finite change of state of a system at Constant temperature is given by
revq
S
T
  ...(1)
For a measurable change between two states 1 and 2 then
2
rev
2 1
1
dq
S S S
T
     ...(2)
where S2
& S2
are final and initial state entropy of system.
Unit of Entropy : Since entropy Change is expressed by a heat term divided by the absolute
temperature, entropy is expressed in term of Jk–1
or cal k–1
entropy is an extensive property i.e, it depends on the amount of the substrate ; hence unit is of
entropy are expressed as calk–1
or Jk–1
mol–1
.
Entropy Change in Isothermal System
for any isothermal change in state T being Constant can be removed from the integral in equation
(2) which then reduced immediately to –
Qrev
S
T
  ...(3)

129
In isothermal expansion of an ideal gas carried out reversibly. There will be no change in internal
energy.
dE or dU = 0
hence from first law
dU = Q – w
in expansion w = –w (negative)
So dU = Q + w and dU = 0
Qrev
= –w ...(4)
In such cases work done in the expansion of n-moles of a gas from volume V1
to V2
at Constant
temperature T is given by
–W = nRT ln
2
1
V
V
 
 
 
Qrev = nRT In
2
1
V
V
 
 
 
rev
2 1
Q 1
S nRT ln(V / V )
T T
   
= nR In (V2
/V1
)
2 1S nR ln(V / V ) 
Spontaneity in Term of Entropy Change
A thermodynamically irreversible process (Spontaneous) is always accompanied by an increase
in the entropy of system and its surrounding taken together while in a thermodynamically reversible
process, the entropy of the system and its surrounding taken together remains unaltered
we may put the above statement in the form of the following mathematical expressions:
(Ssys
+ Ssur
) = 0 (For reversible Process)
(Ssys
+ Ssur
) > 0 (For irreversible Process)
Combining the two, we get
sys surS S 0   
where ‘equal to’ sign refers to a reversible Process while the ‘greater than’ sign refers to an
irreversible Process.
This Conclusion is of great importance as it help us to predict whether a given Process can take
Place spontaneously or not, i.e, whether it is thermodynamically feasible or not.
Since all Processes in nature occur spontaneously, i.e, irreversibly, it follows that the entropy of
the universe is increasing continuously. This is another statement of the second law.
Entropy Changes Accompanying Changes of Phase
(1) Entropy Change from Solid Phase to Liquid Phase
Consider a case when one mole of solid changes into liquid phase at its fusion temperature in
under a Constant pressure if HF
is the molar heat of fusion than the entropy change accompanying the
Process will be given by :
F
F
m
H
S
T

 
here 

130
HF
= molar heat of fusion
Tm
= Melting Point
(2) Entropy Change from Liquid Phase to Vapour Phase
The entropy change at constant pressure of vaporisation at boiling temperature (Tb
) is given by
V
V
b
H
S
T

 
here 
HV
= molar heat of vaporisation
Tb
= Boiling Point
 Since HV
and HF
is positive hence SV
and SF
is increase.
(3) Entropy Change from one Crystalline form to Another
It molar heat of transition of the substances is H+ and transition temperature T than St
is given
by :
t
t
H
S
T

 
Trouton’s Law : For many liquids, the entropy of vaporisation at the normal boiling Point has
approximately the same value
SVap
 90 J/k mol–1
So, Hvap
 90 Jk–1
mol–1
× Tb
 Troution’s law failed for associated liquid such as water, alcohol
 It also fail for substrate with boiling point of 150 k or below.
Entropy Change in Cyclic Reversible Process
In a Completer cycle the total entropy Change of a system must be zero, since it has returned
exactly its Original thermodynamic state
hence 
rev
cyclic
Q
0
T
  
Entropy Change of an Ideal Gas
Since entropy of a system varies with the state of the system its value for a Pure gaseous
substances will depends upon two or three Variables T, P & V. Since T is taken generally as one of the
variables, the second variable to be Considered may be V or P. Thus two variables to be Considered
are either T and V or T and P
(i) When T & V are Two Variables
according to first law of thermodynamics :
Qrev
= dU + W for compression (V)
Qrev
= dU – W for expansion (V)
If the work involved is due to expansion (V) of the gas, then for an infinitesimal increase in
volume dV again a Pressure P
–W = P.dV
So Qrev
= dU + P.dV ...(1)
The increase in entropy of the gas for an infinitesimally small change is given by the expression-
revdQ
dS
T
 ...(2)

131
So
dU P.dV
dS
T

 ...(3)
Substituting the value of dU = Cv
.dT and
RT
P
V
 , we have
vC .dT l RT
dS dV
T T V
   
v
dT dV
dS C . R
T V
   ...(4)
Making the assumption that Cv is independent of temperature generate integration of equation to
get :
2 2
v
1 1
T V
S C ln Rln
T V
   ...(5)
for n moles of the ideal gases, the above equation may be written as :
2 2
v
1 1
T V
S nc ln nRln
T V
   ...(6)
It is evident that the entropy change for the change of state of an ideal gas depends upon the
initial and final volumes as well as on the initial and final temperatures.
(ii) When T and P are the Two Variables
If P1
is the Pressure of the ideal gas in the initial state and P2
is the final state then -
P1
V1
= RT1
So V1
= RT1
/P1
P2
V2
= RT2
So V2
= RT2
/P2
and
2 1 2
1 2 1
V P T
V P T
  ...(7)
Substituting equation (7) in equation (5) for one mole of gas, we have
S = Cv
ln T2
/T1
+ Rln P1
/P2
+ Rln T2
/T1
or S = Cv
ln T2
/T1
+ R ln T2
/T1
+ R ln P1
/P2
we know that
Cp
– Cv
= R
So S = Cp
ln T2
/T1
+ R ln P1
/P2
...(8)
for n mole of ideal gas, we have
p 2 1 1 2S nC lnT / T nRln P /P   ...(9)
or
p 2 1 2 1S nC lnT / T – nRln P /P  ...(10)

132
Entropy Change of an Ideal Gas in Different Process
(1) Isothermal Process
In isothermal Process, there is no change in temperature,
hence
2 2 1
T
1 1 2
V P P
S Rln Rln Rln
V P P
     ...(11)
Case-I If V2
> V1
then ST
will be positive.
i.e, ln isothermal expansion entropy will be increase
Case-II If V1
> V2
then ST
will be negative
i.e, in isothermal compression entropy will be decrease.
(2) Isobaric Process
In Isobaric Process there is no change in pressure
So
P p 2 1S C ln T / T  ...(12)
At Constant Pressure, increase in temperature of an ideal gas is accompanied by increase in
entropy of the ideal gas.
(3) Isochoric Process
In Isochoric process, volume remain constant
So,
v v 2 1S C ln T / T 
Evidently, increase in temperature of an ideal gas tat constant volume is accompanied by increase
in entropy.
In a reversible adiabatic process, the entropy. Remains constant and hence s should be zero
then from equation (5) and (8)
v 2 1 2 1C ln T / T Rln V / V  ...(13)
or v 2 1 1 2 2 1C ln T / T RlnP /P Rln P /P   ...(14)
Entropy of Mixture of Ideal Gases
For one mole of ideal gas -
ds = Cv
dT/T + R dV/V ...(15)
Integrating equation (15)
S = Cv
lnT + RlnV + So
Where, So = integration constant
Remembering the Cp
-Cv
= R, and V = RT/P, we get
S = (Cp
– R) lnT + Rln RT/P + So
S = Cp
lnT – RlnT + RlnR + RlnT – RlnP + So
’
= Cp
lnT – RlnP + So
’ ...(16)
RlnR is a Constant
where So
’ = (R lnR + So
) is another constant.
Let mixture of ideal gases A & B in system which have nA mole of ideal A and nB male of ideal
gas B at Partial Pressure PA
& PB
. Then the entropy of mixture is given by

133
S = nA (Cp
lnT – Rln PA
+ So
’) + nB (Cp
lnT – Rln PB
+ So
’]
S = n [Cp
lnT – RlnP + So
’] ...(17)
From Dalton’s law of partial pressure
Partial Pressure
P  xp
P = Partial Pressure
x = mole fraction
p = Total Pressure
So we have -
n p oS [C ln T R lnp R ln x S ']     ...(18)
Entropy of Mixing
Entropy of mixing is defined as the difference between the entropy of mixture and the sum of the
entropies of separate gases each at Pressure P
Thus
Smix
= n
(Cp
lnT – RlnP – Rlnx + So
’)
n
(Cp
ln T – RlnP + So
’)
= –Rnlnx
= –R (n1
lnx, + n2
lnx2
+ ....)
Where - mix i iS R n ln x   
ni and xi represent the number of moles and the mole fraction, respectively equation (19) represent
the total entropy of mixing
Free Energy
Free Energy Change and Spontaneity
In the free energy change of a chemical reaction is negative, the reaction can take place
spontaneously If the free energy change is zero, the reaction is in a state of equilibrium and if the free
energy change is positive, the reaction would not be proceed
Free energy G = H – T.S
at constant temperature
G = H – TS ...(1)
There are two factors which contribute to the value of G. There are the energy factor H and
TS neither H nor TS alone can determine the spontaneity of a reaction. The spontaneity of a reaction
would be decide by the over all value of two factors i.e., G for a reaction to be spontaneous G must
have a negative value for this H should be negative and TS should be positive when both factors are
favourable i.e, H is negative and TS is positive, then result in a large negative value of G the reaction
would therefore be highly feasible.
If H is negative TS is also negative. In such a case the feasibility of the reaction will be
determine by the factors which predominates.
If the numerical value of H(energy factor) is more than TS (entropy factor), the reaction would
be feasible.
If the numerical value of TS is greater than of H, the reaction would not be feasible, because
now the value of G would be positive not let us Consider that

134
H – is positive sign
TS – is also positive sign
If H > TS then reaction would not be feasible
If TS > H then reaction would be feasible
H TS Condition G
1. –Ve +Ve (any) ve, spontaneous
(favorable) (favorable)
2. –Ve –Ve –Ve, spontaneous
favorable unfavorable H > TS
3. +Ve +Ve –Ve, spontaneous
unfavorable favorable TS > H
Role of Temperature on Spontaneity
Temperature being a multiplying Parameter for entropy factor (TS) of a system. Plays an important
role in controlling the spontaneity of reaction.
1. Exothermic Process
In the Case of exothermic reaction
H = –Ve (Favorable Condition)
Case-I : When TS is +Ve (favorable), then G will be negative and the process will be spontaneous
at all temperature
Case-II : When TS is –Ve (unfavorable), then G will be negative when H > TS. To reduce
the magnitude of TS, the temperature should be low.
Thus, exothermic reactions can be made favourable (When TS = –Ve) by lowering the temperature.
2. Endothermic Process
In the case of endothermic process
H = +Ve (unfavorable Condition)
In the Case of endothermic process, G will be negative when H < TS. To increase the
magnitude of TS, temperature should be increase.
Thus, endothermic reactions can be made favorable by increasing the temperature.
Helmholtz Free Energy
The function that is Particularly applicable to Constant Volume Process is introduced, as called
helmholtz free energy. It is denoted by ‘A’
A E TS  ...(1)
Consider for thermodynamic change from initial state to final state at constant temperature.
for the initial state
A1
= E1
– TS1
...(2)
for the final state
A2
= E2
– TS2
...(3)
From equation (2) and (3)
A2
– A1
= E2
– E1
– T [S2
– S1
]
So A = E – TS ...(4)
we know that
revQ
S
T
  ...(5)

135
revQ
A E T.
T
  
revA E Q    ...(6)
according to first law of thermodynamics.
assuming reversible isothermal Process -
E = Qrev
– Wrev
...(7)
from equation (6) and (7)
A = Qrev
– Wrev
– Qrev
or
rev
rev
A W
or
A W
  
 
...(8)
So that, the decreasing in the helmholtz free energy in any Process at Constant temperature in
equal to the reversible work.
Gibbs Free Energy
The function that is Particularly applicable to Constant pressure is called Gibbs free energy and
it is denoted by ‘G’
It is given by -
G = H – TS ...(9)
But H = E + PV
or G = E + PV – TS
or G = A + PV ...(10)
 A = E – TS
and G = A + (PV)
or G = A + PV + VP ...(11)
When pressure is constant then
A is equal to – Wmax
So equation (11) gives -
G = –Wmax
+ PV
or –G = Wmax
– PV ...(12)
The quantity PV is the work done by the gas on expansion against the constant external
Pressure P. Therefore, –G gives the maximum work obtainable from a system other than that due to
change of volume , at constant temperature and Pressure. The work other than that due to change of
volume is called the net work. Thus,
max expNet Work G W W    ...(13)
The quantity G is also called free energy, thermodynamic Potential and Available energy.
Variation of free energy change with temperature and Pressure :
G = H – TS ...(1)
H = U + PV ...(2)
So G = U + PV – TS ...(3)

136
Upon differentiation,
dG = dU + PdV + VdP – TdS – SdT ...(4)
The first law of thermodynamics -
dQ = dU – dW (For expansion)
and work done is only due to expansion
–dW = PdV
So dQ = dU + PdV
For reversible Process -
dQ
dS
T

dQ = TdS = dU + PdV ...(5)
dG = dU + PdV + VdP – dU – PdV – SdT
dG VdP SdT  ...(6)
This equation gives change of free energy when a system undergoes reversibly process, a
change of pressure well as change of temperature.
Case - I
If pressure remains constant :
dP = 0, then eq. (6)
dG = –SdT
P
dG
S
dT
 
  
 
...(7)
G = G2
– G1
= –
2
1
T
T
S.dT ...(7A)
Case - II
If temperature is constant :
dT = 0, then equation (6)
dG = VdP
T
dG
V
dP
 
 
 
...(8)
G = G2
– G1
=
2
1
P
P
VdP
But V = RT/P
So
2
1
P
P
dP
G RT
P
  
2 1 1 2G RTln P /P RTln V / V   ...(9)
Variation of work function with Temperature and volumes
A = U – TS ...(10)
According to first law of thermodynamics -

137
dQ = dU + PdV (for expansion)
but
dQ
dS
T

So TdS = dQ = dU + PdV ...(11)
upon differentiation of equation (10)
dA = dU – TdS – SdT
From equation (11)
dA = dU – dU – PdV – SdT
dA PdV SdT   ...(12)
At constant volume dV = 0
[dA]V
= –SdT
V
A
S
T
 
   
...(13)
At Constant temperature
dT = 0
dA = – PdV
t
A
P
V
 
   
...(14)
these relationship gives the variation of the work function with temperature & volume.
Maxwell Relationship
The various expressions Connecting internal energy (U), enthalpy (H), Helmhotz free energy (A)
and Gibb’s free energy (G), With relevant Parameters such as entropy, Pressure, temperature and
volume may be put as :
(i) dU = TdS - PdV ...(1)
(ii) dH = TdS + VdP ...(2)
(iii) dA = –SdT – PdV ...(3)
(iv) dG = –SdT + VdP ...(4)
If V is Constant then equation (i) yields the result -
V
U
T
S
 
  
...(5)
If S is constant then equation (1) yields the result -
S
U
P
V
 
   
...(6)
differentiate equation (5) with respect To V keeping (S) Constant and equation (6) with respect
to S keeping (V) Constant, we get -
2
S
U T
( S)( V) V
  
     
...(7)
2
V
U P
( V)( S) S
  
      
...(8)
S V
T P
V S
    
        
...(9)

138
Following some mathematical Procedures as the equation (ii), (iii) & (iv) —
S P
T V
P S
    
        
...(10)
V V
S P
V T
    
       
...(11)
T P
S V
P T
    
        
...(12)
equation (9), (10), (11) and (12) are known as Maxwell’s relationship.
The Gibbs - Helmholtz Equation
Gibbs - Helmholtz equation relates the free energy to the internal energy or enthalpy change and
the rate of Changes of free energy with temperature. This equation may be applied to any change
whether physical or Chemical that take place at Constant volume and Pressure These equation given
by Gibb’s in 1875 and Helmholtz in 1882.
We know that -
G = H – TS at Constant Temp.
For initial state -
G1
= H1
– TS1
For final state -
G2
= H2
– TS2
G = H2
– H1
– T(S2
– S1
)
G = H – TS ...(1)
So –S =
G H
T
  
...(2)
We also know that
dG = – SdT + Vdp
at Constant Pressure
dp = 0
So dG = – SdT
P
( G)
S
T
  
   
...(3)
P
G H G
T T
    
   
P
( G)
G H T
T
  
      
...(4)
This equation is known as Gibbs - Helmholtz equation similarly for a reaction at Constant Volume
the corresponding equation will be —
V
( A)
A V T
T
  
      
...(5)

139
5. CHEMICAL EQUILIBRIUM
Introduction
“Equilibrium is a state at which there are no observable changes as time goes by when a reaction
observable changes as time goes by. When a reaction has reached the equilibrium state. The concentration
of reactants and Products remain Constant over time and no visible change in the system is observed.”
A chemical reaction can be classified into types
Reaction
Reversible Irreversible
 A B  A B
The chemical reaction which take place The reaction proceed in such a way that
in both direction under similar Conditions reactants are completely converted into
i.e, the reaction proceeding from left to right products, i.e, the reaction move in one
is called the forward reaction and the oppose direction i.e, forward reaction only is called
reaction is called reverse or backword reaction. irreversible reaction.
For example : For example :
(1) CH3
COOH + NaOH  CH3
COONa + H2
O (1) NaOH + HCl NaCl + H2
O + 13.7 Kcal
(2) FeCl3
+ 3H2
O  Fe(OH)3
+ 3HCl (2) KClO3
(s) 2MnO  2KCl (s) + 3O2

(3) PCl5
(g)  PCl3
(g) + Cl2
(g) (3) BaCl2
(aq) + H2
SO4
(aq)
BaSO4
(s) + 2HCl (aq)
State of Chemical Equilibrium
Chemical equilibrium in a reversible reaction is the state at which both forward and backward
reactions or two opposing reaction occur at the same speed.
A + B
RF
Rb
 C + D
RF = Rate of forward reaction
Rb = Rate of backward reaction
At equilibrium -
RF Rb
Rf
Rb
Rate
Equilibrium State
Time
Characterstics :
 At equilibrium state, the Concentration of the reactants and Products do not change with
time.

140
 At equilibrium state Certain measurable properties such as Pressure, density, Colour,
Concentration etc. do not change with time.
 It is dynamic in nature i.e, both the reactions moves with the same speed.
 Equilibrium can only be achieved in closed vessel
 equilibrium state is reversible in nature.
 At equilibrium state
G = 0
So that H = TS
Law of Mass Action and The Equilibrium Constant
According to the low of Mass action, the rate of reaction, at given temperature is directly Proportional
to the Product of the molar Concentration of the reactants at a Constant temperature at any given time.
Law of mass action is applicable to elementary Process (one step reaction)
Consider the following reaction :
aA + bB  cC + dD
A/c to law of mass action -
Rate of forward reaction (RF
)  [A]a
[B]b
or kF
[A]a
[B]b
...(1)
Kf = Rate Constant for forward reaction
Rate of back ward reaction (Rb
)  [C]c
[D]d
or Kb [C]c
[D]d
...(2)
Kb
= Rate Constant for backward reaction
A equilibrium
kF
= kb
kF
[A]a
[B]b
= kB
[C]c
[D]d
or
c d
F
eq a b
b
K [C] [D]
K
K [A] [B]
  ...(3)
where keq
is called equilibrium Constant
The Concentration (active mass) of Pure solid and pure liquid are Constant.
So Pure solid and Pure liquid cannot appear in the expression of Kc.
Equilibrium Constant in term of Partial Pressure.
c d
c D
P a b
A B
P .P
k
P .P
 ...(4)
Equilibrium Constant in term of mole fraction -
c d
c D
x a b
p B
x .x
k
x x
 ...(5)
Thus the equilibrium Constant at a given temperature is the ratio of the rate Constant of forward
and backward reaction equilibrium Constant of a given reaction depends only on the temperature and
independent of initial amount of reactant, product, Pressure, Volume and catalyst.
Relation between kP
and kC
When the reactant and Product are in gaseous state, the Partial Pressure can be used instead
of Concentration
Consider the following equilibrium

141
aA(g) + bB(g)  cC(g) + dD(g)
For the above equilibrium kc is given by
Kc =
c d
a b
[C] .[D]
[A] .[B]
for gas,
Pv = nRT
P =
n
RT
v
 
 
 
P = CRT
P = Active Mass × RT
or a a b b c c d d
A A B B C C d DP C (RT) ,P C (RT) ,P C (RT) ,P C (RT)    ...(1)
then kc
=
c d
a b
[C] [D]
[A] [B]
or kc
=
c d
C D
a b
A B
C .C
C .C
...(2)
kP
is given by
kP
=
c d
C D
a b
A B
P .P
P .P
...(3)
Putting the value of c d a b
C D A BP ,P ,P and P from the equation (1)
kP
=
c c d d
C D
a a b b
A B
C (RT) C (RT)
C (RT) C (RT)


or kP
=
c d c d
C D
a b a b
A B
C .C (RT)
C .C (RT)


from the equation (2)
or kp
= kC
c d
a b
(RT)
(RT)


or kP
= kC
(RT)(c+d) – (a+b)
or kP
= kC
(RT)n
here  n = total number of molecules of gaseous products – total number of molecules of
gaseous reactants.
Unit of kP
= (atm)n
Hence, it is Constant for a given reaction, This Proves that Kp depends only on temperature (and
on nature of reaction) and nothing else.

142
Some Facts about Equilibrium Constant
(1) The equilibrium Constant of a forward reaction and that of its backward reaction are
reciprocal of each other.
if A + B  C + D k = x
Then C + D  A + B K1
=
1
x
(2) If a chemical reaction is multiplied by certain factor, its equilibrium Constant must be
raised to a power equal to that factor in order to obtain the equilibrium Constant for the
new reaction
If, for NO +
1
2
O2
 NO2
K =
2
1/2
2
[NO ]
[NO][O ]
then for 2NO + O2
 2NO2
K’ =
2
2
2
2
[NO ]
[NO] [O ]
k k '
(3) If k1
, k2
and K3
are step - wise equilibrium Constant for
A  B ...(1)
B  C ...(2)
C  D ...(3)
There fore the equilibrium Constant ‘k’ for A  D is given as follows:
1 2 3K K K k  
Effect of Temperature on K
According to Arrhenious equation -
K = Ae-Ea
/RT ...(i)
2
1 2 1
k Ea 1 1
log
k 2 303R T T
 
    
  
...(ii)
When T2
> T1
Ea = Activation energy
for forward reaction R = gas constant
log
2 f
1 2 1
Kf E 1 1
Kf 2.303R T T
   
     
   
...(iii)
for backword reaction -
log
b2
1 2 2
EKb 1 1
Kb 2.303R T T
   
     
   
...(iv)
Substraction equation (iv) from (iii), we get
log
 f b2 2
1 1 2 1
E Ekf / kb 1 1
kf / kb 2.303R T T
   
     
   

143
or log
2
1 2 3
k H 1 1
Rk 2 303 T T
 
   
  
...(v)
when H is the heat of reaction K1
and K2
= equlibrium Constant,
Case - I If H = 0
log k2
- log k1
= 0
log k2
= log k1
or 2 1k k
Case - II If H = (+)ve that is heat is absorbed the reaction is endothermic
thus
2 1
1 1
T T
 
 
 
= (-)ve
So log k2
– log k1
= ve
or log k2
> log k1
or 2 1k k
In endothermic reactions, increase in temperature increases the k value
Case - III If H = (-)ve. i.e, heat is released.
the reaction is exothermic
thus
2 1
1 1
T T
 
 
 
= ve
So log k2
– log k1
= (-)ve
or log k2
< log k1
or 2 1k k
In exothermic reactions, increase temperature, decreases the equilibrium constant K.

144
6. IONIC EQUILIBRIUM
PH
and Buffer solutions;
Hydrolysis
Solubility Product
Introduction
In Chemical equilibrium we studied reactions involving molecules only but in ionic equilibrium we
will study reversible reactions involving formation of ion in water.
The Compounds which gives ions either in molten state or in solution are called electrolyte.
There are two types of electrolytes.
electrolytes
Strong electrolyte weak electrolyte
 
These electrolytes are almost These electrolytes are not completely
ionised when dissolved in a ionised in a polar solvent and they
Polar medium like water. behaves as poor Conductors of electricity
for examples - HNO3
, HCl, KOH, for examples - CH3
COOH, H3
PO4
, H3
BO3
,
NaOH, etc. NH4
OH etc.
Arrhenius Theory of Electrolytic Dissociation
(1) Postulates of Arrhenius Theory
(i) In aqueous solution, the molecules of an electrolyte undergo spontaneous dissociation to
form positive and negative ions
(ii) Degree of ionization ()
 =
No. of dissociated molecules
Total no. of molecules of electrolyte before dissociation
(iii) At moderate Concentration, there exist an equilibrium between the ion and undissociated
molecule such as,
NaOH  Na+
+ OH–
KCl  K+
+ Cl–
This equilibrium state is called ionic equilibrium
(2) Factors Affecting Degree of Ionization
(i) At normal dilution, value of  is nearly 1 for strong electrolyte, while its very less than 1
for weak electrolytes.
(ii) Higher the dielectric Constant of a solvent, more is its ionizing Power. Water is the most
Powerful ionising solvent as its dielectric Constant is highest.
(iii)  
1 1
Amount of Solution
Conc. of solution weight of solution dilution of solution
 

(iv) Degree of ionization of an electrolyte in solution increases with rise in temperature.
(v) The degree of ionization of an electrolyte decrease in the Presence of a strong electrolyte
having Common ion.

145
Ostwald’s Dilution Law
Consider a binary electrolyte AB, which dissociate into A+
and B-
ions and equilibrium state is
represented by the reaction :
A B (aq)  AA+
(aq) + B–
(aq)
Initially, C
At equilibrium, (c - x) x x
[A B]eq
= (c - x) [A+
] eq. = x [B–
] eq = x
we define degree of ionization (some time called degree of dissociation) as the fraction of the
electrolyte ionized in solution and it is represented as ‘’
ionised
taken
n x
or x c
n c
   
Hence
[A B] eq. = c – x = c – c = c (1 – x)
[B–
]eq = [A+
]eq = x = c
Kin
=
2
C C C
C(1 ) 1
   

   
The value of Kin
depends upon -
(a) Nature of electrolyte
(b) Nature of solvent
(c) Temperature.
Here, we always Consider aqueous solution, if the temperature is fixed, the Kin
value will depend
on the nature of electrolyte only.
for very weak electrolytes -
 < < < 1, (1 – )  1
k = C2
 = k / c
k c 
for weak electrolyte, degree of dissociation is inversely proportional to the square root of molar
concentration.
Hydrogen ion Concentration (pH) :
The negative logarithm of the Concentration of hydrogen ions is known as pH
i.e, pH = – log10
[H3
O+
] = – log10
[H+
]
Actually,
pH = – log aH
+ (where aH
+
is the activity of H+
ions)
Activity of H+
ions is the Concentration of free H+
ions in a solution.
In dilute solution, the activity of an ion is same as its Concentration.
Therefore, the earlier given expression of pH
can be modified for dilute solutions as, pH
= –log [H+
]
This assumption can only made when the solution is very much dilute, i.e, [H+
] < M for higher Concentration
of H+
ions, one needs to calculate the activity experimentally and then calculate the pH

146
Note
x  log ×  – log x  [H+
]  pH 
 If the value of x increase then log10
x also increases  - log10
x decreases  px
decreases
 [H+
]  PH

The greater the Concentration of hydrogen ions, lesser is the pH
.
for example : Let the [H+
] of an acid solution be its pH can be calculated as
pH
= - log [H+
]
= - log [10-3
]
= (-) (-3) log10
= 3  log 10 = 1
Substances PH
range substances PH
range
Gastric Contents 1.0 - 3.0 milk 6.3 - 6.6
Lemons 2.2 - 2.4 saliva 6.5 - 7.5
vinegar 2.4 - 3.4 Blood plasma 7.3 - 7.5
urine 4.8 - 8.4 Sea water 8.5
PH
of Mixture
Sample – 1 Sample – 2
PH
= 2 PH
= 3
Given [H+
] = 10–2
M [H+
] = 10–3
M
V = 1 ltr V = 2 ltr
m1
v1
+ m2
v2
= mR
(v1
+ v2
)
10–2
× 1 + 10–3
× 2 = MR
(1 + 2)
MR
=
3
12 10
3


4 × 10–3
= MR (Here MR = Resultant molarity)
pH
= – log (4 × 10–3
)
Buffer Solution
There are some solution which have definite pH i.e, their pH do not change by the addition of
small quantities of an acid or a base is called buffer solutions.
Buffer solution can be classified as
Single Buffer Mixed buffer
         
Salt of weak acid and Acidic Basic
weak base Buffer Buffer
(i) Single buffer : The solution of the salt of a weak acid and weak base
for example - CH3
COONH4
+ NH4
CN
(ii) Acidic Buffer : it is the solution of a mixture of a weak acid and a salt of this weak acid
with a strong base
for example:- CH3
COOH + CH3
COONa
(iii) Basic Buffer : It is the solution of a mixture of a weak base and a salt of this weak base
with a strong acid.
for example : NH4
OH + NH4
Cl

147
Buffer Action
Buffer action is the mechanism by which added H+
ions or OH–
ions are almost neutralised. So
that PH
practically remain Constant. Reverse base of Buffer neutralised the added H+
ion while the
reverse acid of buffer neutralise the added OH-
ions
Acidic Buffer Solution
Weak acid + salt of (WA + S.B)
for example : CH3
COOH + CH3
COONa
CH3
COOH  H+
+ CH3
COO (Feebly ionized)
CH3
COONa  Na+
+ CH3
COO (Completely ionized)
H2
O  H+
+ OH–
(very feebly ionized)
Case-I : on addition of acid 
CH3
COO + H
  CH3
COOH
thus ionization of CH3
COOH is suppressed by addition of an acid due to Common ion effect
Case-II : on addition of Base 
CH3
COOH + OH–
  CH3
COO–
+ H2
O
So OH–
ions furnished by a base are removed and PH
of the solution is Practically unaltered.
Basic Buffer Solution
Weak base + salt of (W.B + S.A)
For example, NH4
OH + NH4
Cl
NH4
OH  NH4
+
+ OH–
NH4
Cl  NH4
+
+ Cl–
H2
O  H+
+ OH–
Case-I : On addition of acid :
NH4
OH + H+
 NH4

+ H2
O
Case-II : On addition of Base :
NH4

+ OH–
 NH4
OH
Handerson’s Equation (PH
of a Buffer)
for acidic Buffer
HA  H+
+ AA–
Ka
=
 
H A
HA
 
      
...(1)
or [H+
] = Ka
 
–
HA
A  
...(2)
or [H+
] = Ka
 
 
Acid
Salt ...(3)

148
Taking logarithm and reversing sign :
– log [H+
] = – log Ka
– log
 
 
Acid
Salt
or pH = log
 
 
Salt
Acid – log Ka
 
 a
Salt
pH pk log
Acid
  ...(4)
This is known as handerson’s equation.
when
 
 
Salt
Acid = 10 then,
pH = 1 + pka
and when
 
 
Salt
Acid =
1
10
then
pH = pka
– 1
For Basic Buffer
 
 b
Salt
pOH pk log
Base
  ...(5)
knowing pOH, pH can be calculated by the application of the formula
pH pOH 14 
Buffer Capacity
It is the number of moles of H+
ion or OH–
ion that should be added to 1 litre of a buffer solution
so as to change the pH by unity
no. of moles of acid or base added to 1ltr
change inpH
 
It has been found that if the ratio of
 
 
Salt
Acid or
 
 
Salt
Base is unity. The pH of a particular buffer does
not change at all.
For buffer capacity to be maximum =
 
 
Conjugate base
1
acid

SALT HYDROLYSIS
It is the reaction of the cation or the anion or both the ions of the salt with water to produce either
acidic or basic solution.

149
Hydrolysis is the reverse of neutrolization.
Neutral  No hydrolysis
Salt 2H O
 aqueous Solution  Acidic  Cationic hydrolysis
Basic  Anionic hydrolysis
(i) Hydrolysis Constant : The general equation for the hydrolysis of a salt (BA)
BA + H2
O  HA + BOH
Salt acid base
  
  2
HA BOH
K
BA H O

here k is the equilibrium constant.
Since water is present in very large excess in the aqueous solution, its concentration [H2
O] may
be regarded as constant so,
  
 
 2 h
HA BOH
K H O K
BA
 
Kh
is called hydrolysis constant.
(ii) Degree of Hydrolysis : It is defined as the fraction (or percentage) of the total salt which is
hydrolysed at equilibrium for example if 90% of a salt which is hydrolysed at equilibrium its degree of
hydrolysis is 0.90 or as 90%.
h =
No. of moles of the salt hydrolysed
Total number of moles of salt taken
Type of salt Exp for Kh Exp for h Exp for pH
(i) Salt of weak acid Kh
= Kw
/Ka
h =
hK
c
 
 
 
PH
=
1
2
 [log Kw
+ log Ka
– log c]
and strong acid
(ii) Salt of strong acid Kh
= Kw
/Kb
h =
hK
c
 
 
 
PH
=
1
2
 [log Ka
+ log Kw
- log Kb]
and weak base
(iii) Salt of weak acid Kh
=
w
a b
K
K K h =  hK PH
=
1
2
 [ log Ka
+ log Kw
- log Kb
]
and weak base
(iv) Salt of strong acid and strong base do not undergo hydrolysis (they undergo only ionization)
hence the resulting aqueous solution is neutral.

150
Indicator : An indicator is a substance, which is used to determine the end point in a titration.
In acid - base titration organic substance (weak acid or weak base) are generally used as indicatrors.
They change their colour within a certain PH
range. The colour change and the PH
range of some
common indicators are tabulated below -
Indicator PH
range Colour change
Methyl orange 3.2-4.5 Pink to yellow
methyl red 4.4 - 6.5 Red to yellow
litmus 5.5 - 7.5 Red to blue
Phenolphthalein 8.3 - 10.5 Colourless to pink
Selection of suitable indicator or choice of indicator
(i) Strong acid Vs strong base  Phenolphtalein, methyl red, and methyl orange.
(ii) Weak acid Vs strong base  Phenoiphthalein
(iii) Strong acid Vs weak base  Methyl red and methyle orange
(iv) Weak acid Vs weak base No suitable indicator
Ex. Resaon for use of different indicator for different system.
Solubility Product
In a saturated solution of an electrolyte two equilibria exist and can be represented as :
AB  AB  ions
A B 


solid unionised
(dissolved)
Applying the law of mass action –
K =
 
A B
AB
 
      
Since the solution is saturated, the Concentration of unionised molecules of the electrolyte is
Constant at a particular temperature
i.e, [AB] = K!
= Constant
Hence; [A+
] [B–
] = K [AB] = k k!
= Ksp (Constant)
Ksp is termed as the solubility product. It is defined as the product of the Concentration of ions
in a saturated solution of an electrolyte at a given temperature.
for example, The electrolyte of type AxBy which is dissociated as
y x
x yA B xA yB 

Applying law of mass action -
x y
y x
x y
A B
A B
 
      
  
= k
When the solution is saturated
[Ax
By
] = k!
(Constant)
or [Ay+
]x
[Bx–
]y
= k [Ax
By
]
= k k!
= ksp (Constant)

151
Different Expressions for Solubility Products
(i) Electrolyte of the type AB 
AB  AA+
+ B–
x = solubility
Ks
= [A+
] [B–
]
Thus, AgCl  Ag+
+ Cl–
Ksp
= x2
x x or
x = spK
(ii) Electrolyte of the type AB2

PbCl2
 Pb2+
+ 2Cl–
x 2x
Ksp
= [Pb2+
] [Cl–
]2
: Ksp = [x] [2x]2
= 4x3
x = 3
spK / 4
(iii) Electrotyte of type A2
B 
Ex. H2
S  2H+
+ S–2
2x x
Ksp
= [H+
]2
[S–2
]
= [2x]2
[x]
Ksp
= 4x3
or x = 3
spK / 4
(iv) Electrolyte of type A2
B3
Ex As2
S3
, Sb2
S3
3 2
2 3
2x 3x
As S 2As 3s 

Ksp
= [AS3-1
]2
[S-2
]3
= (2x)2
(3x)3
= 4x2
× 27x3
Ksp
= 108x5
 x =
sp5
K
108
(v) Electrolyte of the type AB3
Ex AlCl3
, Fe(OH)3
AB3
 AA3+
+ 3B–
kS
= [A3+
] [B–
]3
3
3AlCl Al 3Cl
x 3x
 

Ksp
= [Al+3
] [Cl-
]3
= [x] [3x]3
= 27x4
x =
sp4
K
27

152
Creiteria of Precipitation of an Electrolyte
When ionic product of an electrolyte is greater than its solubility product. Precipitation occurs.
Case-I : When Kip
< Ksp
then solution is unsaturated in which more solute can be dissolved i.e,
no Precipitation.
Case-II : When Kip
= Ksp
than solution is saturated in which no more solute can be dissolved but
no ppt is formed.
Case-III : When Kip
> Ksp
then the solution is supersaturated and ppt is formed.
Calculation of Remaining Concentration After Precipitation
Sometimes an ion remains after precipitation if it is in excess. Remaining concentration can be
determined.
Ex. [A+
] left
=
spK [AB]
[B ]
[Ca2+
] left
=
sp 2
2
K [CaCOHl ]
[OH ]
In general
 m sp m nn
nleft m
K A B
A
B


   
  
% PPt of ion =
Initial Conc remaining Conc
100
Initial conc
 
 
 
Calculation of Simultaneous Solubility
Solubility of two electrolytes having Common ion ; when they are dissolved in the same solution,
is called simultaneous solubility.
Case-I : When the two electrolytes are almost equally strong (having close solubility product)
AgBr (Ksp
= 5 × 10–13
) ; AgSCN (Ksp
= 10–12
)
Case-II : When solubility products of two electrolytes are not close, i.e, they one not equally
strong e.g.
CaF2
· (Ksp
= 3.4 × 10–11
)
SrF2
(Ksp
= 2.9 × 10–9
)
Application of solubility product :
(i) Purification of Common salt 
NaCl  Na+
+ Cl–
HCl gas passed through the saturated solution of
NaCl
HCl  H+
+ Cl–
The concentration of Cl–
increase hence, the ionic product [Na+
] [Cl–
] exceeds the solubility
product of NaCl and therefore pure NaCl precipitate out from solution
(ii) Salting out of Soap 
Cn
H2n+1
COONa  Cn
H2n+1
COO + Na+
Soap +
NaCl  Na
+ Cl
Concentration of Na+

ionic product [Cn
H2n+1
COO ] [Na+
] exceeds the solubility products of soap and therefore, soap
precipitates out from the solution.

153
(iii) In quantitative analysis 
(a) Estimation of barium as barium sulphate 
H2
SO4
as precipitating agent is added to the aqueous solution of BaCl2
BaCl2
+ H2
SO4
 BaSO4
+ 2HCl
(b) Estimation of Silver as silver chloride 
AgNO3
+ NaCl AgCl + NaNO3
(c) Precipitation of the sulphides of group II and IV :
II group (Hg2+
, Pb+2
, Bi+3
Cu2+
, Cd+2
, As+3
, Sb+3
and Sn+2
)
IV group  (Co+2
, Ni+2
, Mn2+
, and Zn+2
)
The group reagent is H2
S
(d) Precipitation of III group hydroxides 
(Fe+3
, Al+3
and Cr+3
)
The group reagent is NaOH in presence of NH4
Cl
Stability Constant  Let us Consider dissociation of the ion FeBr+
FeBr+
 Fe2+
+ Br-
dissociation Constant for above equilibria may be given as
Kd
=
2
Fe Br
FeBr
 

      
  
Reciprocal of dissociation Constant is called stability constant
Ks
= 2
FeBr
Fe Br

 
  
      

chemical equilibrium and thermodynamics

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