B.Sc II Sem III Chemistry College Name :- Late Ku.Durga K.Banmeru Science College Lonar University Name:- Sant Gadgebaba Amravati University Amravati

1. MR.SHIVSHANKAR PURUSHOTTAM MORE
ASSISTANT PROFESSOR
DEPARTMENT OF CHEMISRTY
LATE KU.DURGA K.BANMERU SCIENCE
COLLEGE,LONAR DIST.BULDANA
B.Sc-II Sem-III
Topic :- Thermodynamic

3. A. THERMODYNAMIC
Gibbs’s and Helmholtz free energy
Here the concept of entropy is the fundamental consequence of the second law
of thermodynamic. There are two other function, Which utilized entropy in
their derivation. These two function are “free energy function and work
function”. These function like the internal energy, heat constant and entropy are
fundamental thermodynamic properties and depend upon the states of the
system only.
We know that energy can be converted into work, but it is not always
necessary that all the energy may be converted into work.so any kind of energy
which can be converted into useful work is called ‘available energy 'such as the
operation of an engine or a motor. But the energy which cannot be converted
into useful work is known as ‘unavailable energy’

5. Total Energy= Isothermal available energy + Isothermal unavailable energy
H+ G+TS
Or G= H-TS
Definition:- The fraction of the total energy, which is isothermally available for
converting into useful work is called free energy of system. It is denoted by the
symbol G. Free energy is also known as thermodynamic potential or Gibbs
function and is defined by the equation.
G= H-TS--------------------------------(1)
Where,H,T, and S are the heat content (enthalpy),Temperature and entropy of
the system respectively.
As H,T and S depend only on the thermodynamic state and not on its previous
history,it is clear that G may be regarded as single valued function of the state
of the system hence,dG will be a complete differential.

6. PHYSICAL SIGNIFICANCE OF FREE ENERGY
Consider an isothermal change from the initial state to the final state at constant temperature T.
The two states are represented by the subscripts 1 and 2.
G1= H1-TS1 ------ (2)
G2= H2-TS2------(3)
So that, G2-G1=(H2-TS2)-(H1-TS1)
=(H2-H1) –T( S2-S1)
Or ∆G= ∆H-T∆S
Where, ∆G= G2-G1 is the change in Gibbs free energy of the system------(4)
∆H= H2-H1 is the change in enthalpy of the system.
And ∆S=S2-S1 is the change in enthalpy to the system.

7. At constant Temperature (T)
∆S=
𝒒 𝒓𝒆𝒗
𝑻
by the definition of entropy change.
i.e T∆S= q rev
At constant pressure (P)
∆H= ∆E+P∆V (by the definition of enthalpy change)
Substituting the values of T∆S and ∆H form the equation (5) and (6) in equation
(4) we get, ∆G =(∆E+P∆V)- qrev = (∆E-q rev) + P∆V---------(7)
Now, according to first law of thermodynamic,
∆E= qrev - wmax
(- sign before w indicate that work is done by the system)
Or ∆E-qrev = -wmax.
Substituting this value is equation (7),we get,
∆G=-wmax+ P∆V
Or -∆G=wmax-P∆V=wnet

8. But, P∆V is the work of expansion done by the system
corresponding to the increase in volume ∆V. Hence (wmax-P∆V)
stand for the maximum work done other than the work of
expansion. This is called as ‘net work done’ available form the
process.
Hence, we conclude that for a process occurring at constant
temperature and constant pressure, the decrease in Gibbs’s free
energy is equal to the ‘maximum net work’ obtainable form the
process .i.e the total work minus the pressure-volume work or
work of expansion. It is because the change in G is a measure of
the useful work done that G has been called as the ‘free energy’

9. WORK FUNCTION (HELMHOLTZ FUNCTION/HELMHOLTZ FREE ENERGY
We know that a part of internal energy of a system can be utilized at constant temperature
to do useful work. So, this fraction of internal energy which is isothermally available is
called as ‘work function’
This is denoted as by symbol ‘A’ (German Arbeit) and defined following equation,
E=A+TS
Or A=E-TS
Where, E is Internal energy of the system. T is Temp. S is entropy
Definition:- The fraction of the total internal energy,which is isothermally available for
converting into useful work,is called as Helmholtz free energy of system. It is denoted by
the symbol A. It is also known as thermodynamic work function.
A=E-TS-----(1)
Where, E,T and S are the internal energy,Absolute temp.and entropy of the system

10. Since, E,T and S are the function of the state of the system only (can not depend upon previous
history),Therefore.’A’ also must be state function.
Physical significance of the work function.
Consider an isothermal change taking place at temp.T,Then,
A1=E1-TS1 for initial state ---(1)
And A1=E2-TS2 for final state – --(2)
Where, A1,E1 and S1 respectively are the value of work function,Internal energy and entropy of the
system in the initial state and A2, E2 and S2 are correspounding value in the final state.
Therefore, increase in the function A accompanying the process will be given by,
A2-A1= (E2-TS2) –(E1TS1)=(E2-E1) -T(S2-S1)
Or ∆A=∆E-T∆S ---(3)
Where, ∆E=E2-E1 is the change in the internal energy of the system.
∆S =S2-S1 is the change in entropy of the system.
Now,according to the definition of entropy change,we know thath

11. 𝒒 𝒓𝒆𝒗
𝑻
-------(4)
Further, according to first law of thermodynamics as applied to an isothermal
reversible process,
We have
∆E= q rev – w max ----(5)
(-sign before w has been used to indicate that the work has been by the system during the
change under the consideration).
Putting the value of ∆S and ∆E form eq (4)and (5) in Eq.(3) we get
∆A= q rev – w max-T.
𝑞 𝑟𝑒𝑣
𝑇
Or -∆A=w max
Thus for a process occurring at constant temp. the decrease in the work function A is
equal to the maximum work done by the system. In fact, it is for this reason that this
thermodynamic quantity has been termed as ‘Work function ‘although at one time it was
called ‘Helmholtz free energy.

12. VARIATION OF HELMHOLTZ FUNCTION/WORK FUNCTION WITH
TEMPERATURE AND VOLUME
The work function is given by the equation
A=E-TS------(1)
Complete differential of this equation gives
dA=dE-TdS-SdT-----(2)
by the definition of entropy change, we know that
dS=
𝑞 𝑟𝑒𝑣
𝑇
----(3)
Further the first law of thermodynamic, we have
dqrev=dE-dw---------(4)
And if the work is restricted to work of expansion only,
-dw=PdV------------(5)
Putting this value in eq,no. (4)
dqrev-dE-PdV----(6)

13. VARIATION OF HELMHOLTZ FUNCTION/WORK FUNCTION WITH
TEMPERATURE AND VOLUME
putting this value in equation (3), we get
=
𝒅𝑬+𝑷𝒅𝑽
𝑻
Or TdS=dE+PdV---------(7)
Putting this value in equation no. (2) to get,
dA=dE-dE-PdV-SdT=-PdV-SdT-----(8)
If temperature is constant,dT=0 then eq. (8) become
(dA)T=-(PdV)T
Or 𝝏𝑨 𝝏𝑽 𝑻 = −𝑷-------(9)
If volume is kept constant,dV=0 equation (8)become
(dA)T= -(SdT)V
Or 𝝏𝑨 𝝏𝑻 𝑽 = −𝑺----(10)
Equation (9) and (10) gives the variation of the work function with temperature and
volume.

14. VARIATION OF GIBB’S FUNCTION/FREE ENERGY WITH TEMPERATURE
AND PRESSURE.
The free energy is given by the equation
G=H-TS------(1)
By definition enthalpy
H=E+PV------(2)
Putting this value in Eq.(1) we get G=E+PV-TS---------(3)
Complete differentiation of this equation gives
dG= dE+PdV+VdP-TdS-SdT--------(4)
but =
𝒒 𝒓𝒆𝒗
𝑻
( by the definition of entropy change)
=
𝒅𝑬+𝑷𝒅𝑽
𝑻
( dq rev= dE+PdV, by first law of
thermodynamics)
Or T∆S=dE+PdV--------(5)

15. VARIATION OF GIBB’S FUNCTION/FREE ENERGY WITH TEMPERATURE
AND PRESSURE.
Substituting this value in equation (4), we obtain
dG=dE+PdV+VdP-(dE+PdV)-SdT
=VdP-SdT
dG=VdP-SdT----------(6)
This is an important expression, which gives the change in free energy with change in pressure and
change in temperature in a reversible process. Equation(6) is called total differential equation.
a) If temperature is constant,dT=0 equation (6) takes the form
(dG)T=(VdP)T
Or 𝝏𝑮 𝝏𝑷 𝑻 = 𝑽-------(8)
(b) If pressure is kept constant, dP=0 equation (6) takes the form
Or 𝝏𝑮 𝝏𝑻 𝑷 = −𝑺-------(9)
Equation (8) & (9) gives the variation of free energy with pressure and temperature.

16. SPONTANEITY IN TERMS OF FREE ENERGY
i) Spontaneous change is uni-directional or irreversible. No work
has to be done for spontaneous change. It occur by itself. On the
other hand, reversible change is bi-directional. For a reversible
change to occur, work has to be done.
ii) There is no time factor for spontaneous change,it may occur
rapidly or very slowely.

17. CONDITION IN TERMS OF GIBBS FREE ENERGY CHANGE
We know that,
G=H-TS(By definition)
And H=E+PV
Therefore, G=E+PV-TS------------(1)
The differential of this equation is
dG=dE+PdV+VdP-TdS-SdT
Or TdS=-dG+dE+PdV+VdP-SdT--------(2)
When any change is brought about reversiblely,the entropy change is obtained by
TdS=dE+PdV--------(3)

18. CONDITION IN TERMS OF GIBBS FREE ENERGY CHANGE
For irreversible process entropy change become
TdS>dE+PdV----------(4)
By combining equation (3) and (4)
TdS≥dE+PdV
Substitute value of TdS from equation (2)
-dG+dE+PdV+VdP-SdT≥dE+PdV
Or -dG≥ 𝑺𝒅𝑻-VdP
Or dG≤VdP-SdT
At constant temperature and pressure dT=0 and dP=0
∴ dG≤0--------(5)
The equation (5) represent that, if in a chemical reaction change in free energy is negative i.e less than zero than the
reaction will be spontaneous or feasible and at equilibrium, free energy change will be zero.

19. GIBBS-HELMHOLTZ EQUATION.
The variation of free energy with change in temperature and change in pressure is given by the equation.
dG=VdP-SdT
So, at constant pressure the variation of free energy change with temperature is given by the equation
𝝏𝑮
𝝏𝑻 P= -S-------------(1)
i.e dG=-SdT
Let G1 and G2 be initial and final free energies of a system at constant and temperature T
∴ dG1=-S1dT ----------(2)
dG2=-S2dT-------------(3)
where, S1 and S2 are entropies in initial and final state of system.
Consider equation (3) & (4)
dG2-dG1=(S2-S1) dT
d(∆G)=-∆SdT
At constant pressure
𝒅(∆𝑮)
𝒅𝑻 P=-∆S

20. GIBBS-HELMHOLTZ EQUATION.
We know that,
∆G=∆H-T∆S
- T∆S=∆G-∆H
∴ ∆𝑺 =
∆𝑮−∆𝑯
𝑻
------(5)
Substituting value of ∆S from equation (5) in equation (4)
𝒅(∆𝑮)
𝒅𝑻 P=
∆𝑮−∆𝑯
𝑻
Or 𝑻
𝒅(∆𝑮)
𝒅𝑻 P= ∆G-∆H
Or, ∆G=∆H+T
𝒅(∆𝑮)
𝒅𝑻 P----------------------(6)
This is known as Gibb’s Helmholtz equation in terms of Gibb’s free energy and enthalpy. By analogy,Gibb’s-
Helmholtz equation in terms of work function and internal energy can be derived at constant volume which can
be written as
∆A = ∆E+T
𝒅(∆𝑨)
𝒅𝑻 V

21. APPLICATION OF GIBBS-HELMHOLTZ EQUATION
1) Calculate of E.M.F of a reversible cell:
The decrease in free energy produced by the passage of nF coulombs of electricity thorough a
reversible cell is given by
-∆G=nEF
Where, n= number of electron lost or gained
E= E.M.F of the reversible cell
F= faraday of electricity (1F=96500 coulomb)
Putting the value of ∆G in Gibb’s Helmholtz equation we get,
=∆H+
𝒅(∆𝑮)
𝒅𝑻 P
EF=∆H-nFT
𝒅𝑬
𝒅𝑻 P
∴
−∆𝑯
𝒏𝑭
+ 𝑻
𝒅𝑬
𝒅𝑻 P
Here E gives E.M.F of a reversible cell

22. APPLICATION OF GIBBS-HELMHOLTZ EQUATION
2)If the standard potential of an electrode is known at two or more
temperature, the entropy of the reversible ion can be determined by using
suitable form of Gibb’s Helmholtz equation.
3) Calculation of enthalpy changes: The Gibbs-Helmholtz can be used for
calculating entropy changes and the enthalpy changes occurring in
isothermal reaction.

23. PROBLEM
The free energy changes(∆G) for the reaction.
2Ag+Hg2Br2 2AgBr + 2Hg is found to be -13129.39 J
300K and d(∆G)dTP for it is -60.207 JK-1 .Find (∆H) for the
reaction at 300K?
Solution:-We know that Gibb's-Helmholtz equation
=∆H+Td(∆G)dTP
=∆G-Td(∆G)dTP
=-13129.39-300(-60.307)
=-13129.39+18062.1
=4932.71Joule

24. CHEMICAL POTENTIAL OR PARTIAL MOLAR FREE ENERGY
Consider the extensive properties, free energy. Let it be
represented by G. Suppose that the system consist of a n constituents
(for homogenous system),the amount of which present in the system
are n1,n2,n3…….. moles. Then the properties G is a function not only
of temperature and pressure but of the amounts of the different
constituents as well ,so that we can write.
G=f(T,P,n1,n2,n3……..)
Now, if there is a small change in the temperature, pressure and
the amounts of the constituents, then the change in the properties G
is given by,
=
𝜕𝐺
𝜕𝑇 P,n1,n2 dT+
𝜕𝐺
𝜕𝑃 T,n1,n2 dP+
𝜕𝐺
𝜕𝑛1 T,P,n2.n3 dn1+
𝜕𝐺
𝜕𝑛2 T,P,n1,n3dn2------(2)

25. CHEMICAL POTENTIAL OR PARTIAL MOLAR FREE ENERGY
On the right hand side, the quantity gives the change in the value of G with
temperature when pressure and composition are kept constant, the second
quantity gives the change in the value G with pressure when temperature and
composition kept constant, the remining quantities gives the change in the G
with a change in the amount in the amount of a constituent,when
temperature,pressure and amount of other constituents are kept constant.
If the temperature and pressure of the system kept constant,then
dT=0 and dP=0 so that eq.(2) become
(dG)T,P= +
𝜕𝐺
𝜕𝑛1 T,P,n2.n3 dn1+
𝜕𝐺
𝜕𝑛2 T,P,n1.n3 dn2+--------(3)
Each derivative on the right hand side is called partial molar property and is
represented by putting a bar over the symbol of the particular properties i.e
𝐺1,𝐺2for 1st .2nd compound etc. Respectively. Thus

26. CHEMICAL POTENTIAL OR PARTIAL MOLAR FREE ENERGY
In general,for ith component
𝝏𝑮
𝝏𝒏𝟏 T,P,n2.n3 =𝐺i------------(4)
This quantity is called partial molar free energy or chemical potential is usually
represented by the symbol 𝜇,Thus,
𝜇 = 𝐺I =
𝜕𝐺
𝜕𝑛𝑖 T,P,n1.n2------------------(5)
If dni -1mole,𝜇=dG T,P,n1,n2
Hence the chemical potential may be defined as follows:-
The chemical potential of a constituent is a mixture is the increase in the free energy
which takes place at constant temperature and pressure when 1 mole of the
constituent is added to the system, keeping the amounts of all other constituent
constant i.e when 1 mole of the constituent is added to such a large quantity of the
system that its composition remains almost unchanged.

27. CHEMICAL POTENTIAL OR PARTIAL MOLAR FREE ENERGY
(dG)T,P=𝜇1dn1+ 𝜇2dn2 +------(6)
For a system of defined composition represented by the number of moles
n1,n2,n3—etc
Equation (6) on integration gives
G=T,P,N= n1𝜇1+ n2𝜇2
Where, the subscript N stand for constant composition
On the right hand side, the first terms gives contribution of the first
constituent to the total free energy of the system. The second term gives
the contribution of the second constituent and so on. Obviously n1,n2 etc
gives the contribution per mole to the total free energy.

28. CHEMICAL POTENTIAL OR PARTIAL MOLAR FREE ENERGY
Hence chemical potential may also be defined under:-The chemical
potential of a constituent in a mixture is its contribution per mole to the
total free energy of the system of a constant composition at constant
temperature and pressure.
It may be noted that whereas free energy is an extensive property i.e
depend on number of moles the chemical potential is an intensive
property because it refer to one mole of the substance.

29. GIBBS DUHEM EQUATION
Free energy (G), being an extensive property, depends not only upon the
temperature and pressure of the system but also the composition of the system. If
the system consists of number of constituents, the amounts of which are n1, n2, n3....
moles respectively, then we
can write :
G=f(T, P, n1, n2, n3....)---------(1)
For small changes in temperature, pressure and the quantity of different
constituents, the small change in free energy can be obtained by differentiation of
Eq. (1), this gives
=
𝜕𝐺
𝜕𝑇 P,n1,n2 dT+
𝜕𝐺
𝜕𝑃 T,n1,n2,dP+
=
𝜕𝐺
𝜕𝑛1 T,P,n2,n3 dn1+
𝜕𝐺
𝜕𝑛2 T,P,n1,n3 dn2+…….------------(2)

30. GIBBS DUHEM EQUATION
If temperature and pressure are kept constant dT=0 and dP=0
So that Eq.(2) become
(dG)T,P = =
𝜕𝐺
𝜕𝑛1 P,n2,n3 dn1+
𝜕𝐺
𝜕𝑛2 T,P,n1,n1,n2 dP+……--------(3)
Putting
𝜕𝐺
𝜕𝑛1 P,n2,n3 = 𝜇1
Putting
𝜕𝐺
𝜕𝑛2 P,n1,n3= 𝜇2 and so on
We get, (dG)T,P = 𝜇1dn1+𝜇2dn2+……(4)
For a system of definite composition represented by the number of moles n1,n2,n3
etc
Equation (4) on integration gives;
(G)T,P,N = n1𝜇1+ n2𝜇2 +….

31. GIBBS DUHEM EQUATION
Differentiating this equation under condition of constant temperature
and pressure but varying composition,we get,
(dG)T,P=(n1d𝜇1+𝜇1dn1)+(n2d𝜇2+𝜇2dn2)+……
= (𝜇1dn1+ 𝜇2dn2+…….)+ (n1d 𝜇1+ n2d𝜇2)---------(5)
Comparing equation (4) and (5), we get,
n1d𝜇1+ 𝜇2dn2+……) =0
𝑛id𝜇𝑖=0
This equation which is applicable to a system under constant
temperature and pressure is called as Gibbs-Duhem equation.

32. VAN’T HAFF REACTION ISOTHERM
Let us consider following reaction at temperature T(K) aA(g)+bG(g) lL(g)+mM(g)
free energy change (∆G) for this reaction is give by
∆G= 𝐺Product- 𝐺Reactant
∆G=(l𝜇L+ m𝜇M)-(a𝜇A+ b𝜇B)--------------------(1)
Where 𝜇L,𝜇M , 𝜇A , 𝜇B, are the chemical potential of various species in the reaction mixture at
temperature T and given by,
𝜇L= 𝜇L
o RT ln PL
𝜇M= 𝜇M
o RT ln PM -------------------------(2)
𝜇A= 𝜇A
o RT ln PA
𝜇B= μB
o RT ln PB
In the above equation 𝜇L
o,𝜇M
o,𝜇A
o,𝜇B
o are chemical potential is standard state(partial pressure
chemical potential form equation (2) in equation (1) we get,
∆G=[l(𝜇L
o RT ln PL) + m(𝜇M
o RT ln PM )]-[a( 𝜇A
o RT ln PA )+b( 𝜇B
o RT ln PB)
On rearranging

33. VAN’T HAFF REACTION ISOTHERM
∆G=[l 𝜇L
o+ 𝜇M
o)-(a𝜇A
ob 𝜇B
o)]+RT [l ln PL+m ln PM)-(a ln PA+ b ln PB)
Or ∆Go+RT[ln (PL)lx (PM)m]-[ln(PA)ax(PB)b
∆Go+RT ln
𝑃𝐿 𝑙
(𝑃𝐴𝑎)
× 𝑃𝑀 𝑚
× 𝑃𝐵 𝑏 =reaction quotient--------------(3)
Thus,Vant’s Hoff reaction isotherms gives free energy change (∆G) of a chemical reaction in terms of
standard free (∆Go) and the partial pressure of reactants and products at a given temperature(T).
∆Go+RT ln
𝑃𝐿 𝑙
(𝑃𝐴𝑎)
× 𝑃𝑀 𝑚
× 𝑃𝐵 𝑏 equilibrium
Or
∆Go=-RT ln
𝑃𝐿 𝑙
(𝑃𝐴𝑎)
× 𝑃𝑀 𝑚
× 𝑃𝐵 𝑏 equilibrium----------------------(4)

34. VAN’T HAFF REACTION ISOTHERM
Since∆Go represent standard free energy change of the reaction, it must be constant at a given
temperature. Hence, the expression on the right hand eq.(4) must be also constant. Since R is gas
constant, it follows that at constant temperature.
𝑃𝐿 𝑙
(𝑃𝐴𝑎)
× 𝑃𝑀 𝑚
× 𝑃𝐵 𝑏 equilibrium must be constant
Thus,
𝑃𝐿 𝑙
(𝑃𝐴𝑎)
× 𝑃𝑀 𝑚
× 𝑃𝐵 𝑏 equilibrium=Kp-----------------(5)
Where, Kp is the equilibrium constant of the reaction at a given temperature
Therefore, equation (4) may be also be written as
∆Go=-RT lnKp
∆Go=-2.303RTlogKp-------(6)
Derivation of equation (5) and (6) is thus a thermodynamic derivation of the law chemical
equilibrium

35. PROBLEM
3
2
O(g) O(g) at 298 K, AG' for the reaction is 163430 J mole
Solution
∆Go=-2.303 RT log Kp
log Kp=
∆Go
−2.303 RT
=
−163430
303𝑋8.315𝑋298
Kp= 2.0𝑋10−20
 Problem The value of Kp for the water gas reaction
CO (g)+ H2O (g) CO2(g) +H2(g) is 1.06× 105
𝑎𝑡 250
Calculate the standard free energy change (∆Go) of the reaction at 25°C. (R = 8.314
JK mol-1).
is 1.06 x 10 at 25°C
Solution: AG = - RT In Kp
=-2.303 RT log Kp
-(2.303 x 8.314 x 298 x log (1.06 x 10)
28673.61 J = 28.674 kJ

36. VAN'T HOFF EQUATION (TEMPERATURE DEPENDENCE OF EQUILIBRIUM CONSTANT)
Let us consider following reaction at equilibrium
aA(g)+ bB(g) lL(g)+mM(g)-----------------------(1)
Applying Van't Hoff reaction at equilibrium, we get
∆Go = -RT In Kp
On differentiating with respect to temperature, we have,
𝑑(∆Go
)
𝑑𝑇
=-R 𝑇
𝑑 𝑙𝑛𝐾𝑝
𝑑𝑇
+ 𝑙𝑛𝐾𝑝
𝑑(∆Go
)
𝑑𝑇
=-RT
𝑑 𝑙𝑛𝐾𝑝
𝑑𝑇
− 𝑅𝑙𝑛𝐾𝑝 ----------------(2)
T
𝑑(∆Go
)
𝑑𝑇
=-RT2 𝑑 𝑙𝑛𝐾𝑝
𝑑𝑇
− 𝑅𝑙𝑛𝐾𝑝
But,
∆Go=-RT lnKp
Hence, T T
𝑑(∆Go
)
𝑑𝑇
=-RT2 𝑑 𝑙𝑛𝐾𝑝
𝑑𝑇
+ ∆Go
Or RT2 𝑑 𝑙𝑛𝐾𝑝
𝑑𝑇
= ∆Go- T
𝑑(∆Go
)
𝑑𝑇

37. VAN'T HOFF EQUATION (TEMPERATURE DEPENDENCE OF EQUILIBRIUM CONSTANT)
According to Gibb’s Helmholtz equation,
∆G0=∆H0+T
𝑑(∆G0
)
𝑑𝑇 P
∆H0=∆G0-T
𝑑(∆G0
)
𝑑𝑇 P
Comparing equation (3) and (4) we get,
RT2 𝑑(𝑙𝑛𝐾𝑝)
𝑑𝑇
=∆H0---------------------(5)
Or
𝑑(𝑑𝑙𝑛𝐾𝑃)
𝑑𝑇
=
∆𝐻0
𝑅𝑇2-------------------------(6)
This is known as Van’t Hoff Equation
It has exprimentaly observed that the enthalpy change
Integrated from of Von’t Haff equation ∆H0 accompanying a chemical rection does not change
appreciably with the change in partial pressure of reactant or product.Thus we may replce ∆H0 by ∆H
𝑑(𝑑𝑙𝑛𝐾𝑃)
𝑑𝑇
=
∆𝐻
𝑅𝑇2-------------------------(7)
Hence, Van’t Hoff Equation takes the form
Equation (7) may be written as,

38. VAN'T HOFF EQUATION (TEMPERATURE DEPENDENCE OF EQUILIBRIUM CONSTANT)
d(lnKp)=
∆𝐻
𝑅𝑇2 𝑑𝑇-------------------------(8)
Let,Kp1 is the equlibrium constant at T1K and Kp2 is the equlibrium constant at T2K further,let us
assume that ∆H remain constant between temperature T1K and T2K.On integrating equation (8) we
get,
d(lnKp)=
∆𝐻
𝑅 𝑇1
𝑇2 1
𝑇2dT
lnKp2-lnKp1=
∆𝐻
𝑅
−
1
𝑇 𝑇1
𝑇2
lnKp2-lnKp1=
∆𝐻
𝑅
1
𝑇1
−
1
𝑇2
lnKp2-lnKp1=
∆𝐻
𝑅
𝑇2−𝑇1
𝑇1𝑇2
Or log10
𝐾𝑝2
𝐾𝑝1
=
∆
𝐻
2
.
303
𝑇2−𝑇1
𝑇1𝑇2

39. APPLICATION OF VAN’T HOFF EQUATION
 i)Calculation of equilibrium constant(Kp) at any desired temperature form the
knowledge of equilibrium constant at a known temperature and heat of reaction
(∆H).
 ii) Calculation of enthalpy change (∆H) accompanying the chemical reaction from
the knowledge of equilibrium constant at two different temperature.

40. PROBLEM
The equilibrium constant Kp for the reaction,
H2(g) + S(g) H2S(g) is 20.2 atm at 9450C and 9.21 at 10650C
Calculate the heat of reaction.
Applying Van’t Haff equation
logKp2-Kp1= ∆𝐻
2.303
𝑇2−𝑇1
𝑇1𝑇2
T1= 9450C+273=1218K
T2-10650C+273=1338K
Kp1=20.2atm
Kp2=9.21atm
R= 8.314JK-1
∆H=?
log9.21-log20.2= ∆
𝐻
(
2
.
303
)(
8
.
314𝐽
𝐾−
1
1338−1218𝐾
(1338)(1218)
∆H=-88126.3J=-88.826kJ.

41. B. PHASE EQUILIBRIUM
Immiscible liquids refer to liquids that cannot mix. These types of liquids
are completely insoluble in each other.
Example: Kerosene and water, oil and water, vanaspati ghee and water.
Nernst distribution law
If we take two immiscible solvents A and B in beaker, they form separate
layers. When a solute 'X' which is soluble in both solvents is added, it
gets distributed or partitioned between them. Molecules of X pass from
solvent A to B and vice-versa. Finally, a dynamic equilibrium is set up.
Nernst (1891) studied the distribution of several solutes between
different appropriates of solvents. The generalization given is known as
'Nernst-Distribution Law' or simply
Immiscible liquid oil & water

42. 'NERNST-DISTRIBUTION LAW'
Statement: "If a solute X' distributes itself between two immiscible solvents A and
B atconstant temperature and X is in the same molecular state in both solvents,
then the ratio of concentration of solute in two solvents remain constant".
∴
Concentration of X in A
Concentration of X in B
=Constant =K
If Concentration of X in A is CA and concentration of X in B is Cp then,
𝐶𝐴
𝐶𝐵
=KD
The constant ‘KD' (or simply K) is called the 'Distribution coefficient' or 'Partition co-
efficient' or 'Distribution ratio:

43. CONDITIONS FOR VALIDITY OF NERNST DISTRIBUTION LAW
i) Temperature should be constant.
ii) Solute should not associate or dissociate in any of the solvents.
iii) Molecular state of solute in both the solvents should be same.
iv) The solutions should be dilute. For concentrated solutions,
Nernst distribution law shows deviations.
v) The two solvents should be either immiscible or very
slightly miscible in each other.

44. PROBLEM
A Solid added to a mixture of benzene and water.After shaking well and allowing
to stand,10ml benzene layer was found to contain 0.13 g of solute and 100 ml of
water layer contained 0.22 g of solute, Calculate the value of distribution
coefficient.
Solution:-Concentration in benzene (CB)=
0.13
10
=0.013g/ml
Concentration in water (CW) =
0.22
100
0.0022g/ml
∴=Accoudring to Distribution law:-
𝐶𝐵
𝐶𝑊
=
0.013
0.0022
=5.9

45. PROBLEM
In the distribution of succinic acid between ether and water at 150C,20ml of the the ether layer
contain 0.092 g of the acid. Find out weight of the acid present in 50 ml of the aqueous solution in
equilibrium with it if the distribution co-efficient of succinic acid between water and ether is 5.2
Solution:- Let weight succinic acid in aqueous layer be x g.
∴ Concentration in aqueous layer=
𝑋
15
g-ml
∴ Concentration in ether layer=
0.092
20
g-ml
KD=
𝐶𝑊𝑎𝑡𝑒𝑟
𝐶𝐸𝑡ℎ𝑒𝑟
=
𝑋/50
0.092−20
5.2=
20𝑋
0.092𝑋50
5.2 × 0.09220 × 50
20
=1.196g

46. APPLICATION NERNST DISTRIBUTION LAW
Nernst distribution law is applicable when the solute exists as simple molecule in
the two solvents.
The ratios of two concentrations do not remain constant if solute undergoes
association or dissociation in one of the solvents. In this case law is applied with
some two solvents modifications, which are useful to determine degree of
association or dissociation.
Determination of degree of association
Let in solvent B, 'n' number of molecules of 'X' associates to form 'Xn’ molecule.In
solvent A, the solute molecules remain as normal molecules. This case is as
shown in figure

47. APPLICATION NERNST DISTRIBUTION LAW
Let C1 be concentration of X in solvent A. C3 be concentration of X in solvent B.
C2 be concentration of Xn, in solvent B. In this system there are two equilibria.
First equilibrium is- X in solvent A X in solvent B
∴ K1=
𝐶1
𝐶3
i.e C3=
𝐶1
𝐾1
------------------(1)
(No change)
(A) Solvent
nX Xn (B) Solvent
(Association)

48. Second equilibrium is
nX in solvent B Xn is solvent B
∴ form law of Mass action
Kc=
[𝑋𝑛]
[𝑋]𝑛 =
[𝐶2]
[𝐶3]𝑛
[C3]n=
𝐶2
𝐾𝑐
Taking nth root we get,
C3=
𝑛
𝐶2
𝑛
𝐾𝑐
--------------(2)
From equation (1) and (2) we get,
𝐶1
𝐾1
=
𝑛
𝐶2
𝑛
𝐾𝑐
∴
𝐶1
𝑛
𝐶2
=
𝐾1
𝑛
𝐾𝑐
=KD (Constant)
Thus when association occurs in one solvent the distribution equation is modified as;
𝐶1
𝑛
𝐶2
=KD

49. As the solute exists largely as associated molecule, the total concentration of ‘X’ determined experimentally in solvent B is taken
as the concentration of the associated molecule Xn.
Problem
When benzoic acid was shaken with mixture of benzene and water at constant temperature, the following results were obtained
Determine degree of association of benzoic acid in benzene.
Solution:-
𝐶1
𝑛
𝐶2
=KD
Consider, C1=0.24 C’1=0.55
C2=0.015 C’2=.022
∴
𝐶1′
𝑛
𝐶1
′
=∴
𝐶2
𝑛
𝐶2
′
=
𝑛 𝐶1
𝐶1
′
Taking log, log
𝐶2
𝐶2
′=
1
𝑛
=log
𝐶1
𝐶1
′
log
𝐶1
𝐶1
′= n log
𝐶2
𝐶2
′
log
0.24
0.55
=n log
0.015
0.022
∴ n=2
Concentration of acid in benzene (C1)
Concentration of acid in water(C2)
0.24
0.015
0.55
0.022
0.93
0.029

50. Determination of degree of dissociation
 Degree of dissociation is the fraction of total number of moles dissociated at equilibrium
 Let solute undergoes dissociation in solvent B and it normal in solvent A. The equilibrium set up in the two
solvent are shown fig
Let,C1 be the concentration of ‘X’ in solvent A.
C2 be the concentration of X (dissociation and undissociated) in solvent B
If degree of dissociation in solvent B is ′ ∝ ′ then,
(No Change)
X (Solvent A)
X (Solvent B)
Y+Z
(Dissociation)

51. X Y + Z
At equilibrium (1-∝) ∝ ∝
Hence, the concentration of the undissociated (Normal) molecule in solvent B is C2(1-∝) Applying the
distribution law to normal molecule in the two solvents,
𝐶1
𝐶2(1 −∝)
This is modified equation in dissociation case.
iii) Extraction
The most important application of distribution law in industry as well as laboratory is the extraction
(removal by a solvent) of an organic substance from an aqueous solution. The is carried by shaking the
aqueous solution with a immiscible organic solvent, say ether in a separating funnel. The distribution
ratio of most of the organic compounds is very large in favour of organic solvents. On standing, the
aqueous and ether layers separate in the funnel. Etherlayer is separated and applied to distillation
The organic substance remains as residue in the

52. Types of extraction
There are main two types of extraction
1) Simple extraction:
The extraction process when carried out with the total amount of a flask.
2) Multiple extraction:
When the extraction is carried out in more number of steps (operations) by using
small amounts of organic solvent (extracting liquid) from total solvent from same
aqueous solution is known as multiple extraction or multistep extraction. Multiple
extraction gives more extraction of organic substance than single step extraction by
using same amount of extracting solvent

53. Suppose V ml of an aqueous solution contain 'W kg of an organic substance
Let V1 ml of given organic solvent is used for extraction in each step
First extraction: Consider X1' kg be the substance left unextracted in aqueous solution de the
first operation
∴ Concentration in aqueous layer=
𝑋1
𝑉
And concentration in organic solvent=
𝑊−𝑋1
𝑉1
From distribution law,
∴K=
𝐶𝑤𝑎𝑡𝑒𝑟
𝐶𝑆𝑜𝑙𝑣𝑎𝑛𝑡
𝑋1
𝑉
𝑊−𝑋1
𝑉1
=
𝑋1
𝑉
×
𝑉1
𝑊−𝑋1
∴ KVW-KVX1=X1V1
∴ KVX1+X1V1=KVW
X1=W
𝐾𝑉
𝐾𝑉+𝑉1
----------------(1)
Equation for the solute left unextracted after nth extractions:

54. Problem
Succinic acid is distributed in water and ether.In aqueous solution it undergoes dissociation. The
concentration of acid in aqueous layer was 42.5 × 10−3 kg dm-3 and ether layer was 7.1 × 10−3 kg dm-
3.If distribution co-efficient of acid is 0.25, calculate the degree of dissociation assuming the normal
molecule in ether.
Solution:-
Here. Concentration in aqueous layer (C3) =42.5× 10−3
Concentration in ether layer (C1) =7.1× 10−3 kg dm-3
Distribution coefficient =(KD)=0.25
We know,
KD=
𝐶1
𝐶2(1−∝)
0.25=
7.1×10−3
42.5×10−3(1−∝)
∴ 1-∝ = 0.6666

55. Suppose V ml of an aqueous solution contain 'W kg of an organic substance
Let V1 ml of given organic solvent is used for extraction in each step
First extraction: Consider X1' kg be the substance left unextracted in aqueous solution de the
first operation
∴ Concentration in aqueous layer=
𝑋1
𝑉
And concentration in organic solvent=
𝑊−𝑋1
𝑉1
From distribution law,
∴K=
𝐶𝑤𝑎𝑡𝑒𝑟
𝐶𝑆𝑜𝑙𝑣𝑎𝑛𝑡
𝑋1
𝑉
𝑊−𝑋1
𝑉1
=
𝑋1
𝑉
×
𝑉1
𝑊−𝑋1
∴ KVW-KVX1=X1V1
∴ KVX1+X1V1=KVW
X1=W
𝐾𝑉
𝐾𝑉+𝑉1
----------------(1)
Equation for the solute left unextracted after nth extractions:

56. 2) Second Extraction is carried by using same V1 ml of fresh solvent form aqueous solution remained after
first extraction. Let ‘X2’ kg be the substance unextracted in aqueous layer.
∴ Concentration in aqueous layer =
𝑋2
𝑉
And concentration in organic solvent=
𝑋2−𝑋1
𝑉1
∴ form distribution law, K=
𝐶𝑤𝑎𝑡𝑒𝑟
𝐶𝑆𝑜𝑙𝑣𝑎𝑛𝑡
𝑋1
𝑉
𝑊−𝑋1
𝑉1
=
𝑋2
𝑉
×
𝑉1
𝑋1−𝑋2
K(X1-X2) =
𝑉1
𝑉
X2
(KX1-KX2) =
𝑉1
𝑉
X2
KX1=KX2 +
𝑉1
𝑉
X2
KX1=
𝐾𝑋2𝑉+𝑉1𝑋2
𝑉
=
𝐾𝑉+𝑉1
𝑉
X2
X2=
𝐾𝑋1𝑉
𝐾𝑉+𝑉1
--------------(2)

57. Substitute value of x1 from Eq.(1)
X2= W
𝐾𝑉
𝐾𝑉+𝑉1
𝐾𝑉
𝐾𝑉+𝑉1
X2=W
𝐾𝑉
𝐾𝑉+𝑉1
2
3) nth extraction:- Similar to 2nd extraction,n- extraction are carried out,Let ‘xn’ kg be
the solute unextracted then we get
Xn=W
𝐾𝑉
𝐾𝑉+𝑉1
n -----------------(4)
Equation (4) in the general formula the calculation of the amount of substance left
unextracted after certain number of extractions using V1 ml of organic solvent in
every step.

58. Problem
The distribution co-efficient of an alkaloid between chloroform and water is 20 in favour of chloroform. Compare the
weight of the alkaloid remaining in 100 ml aqueous solution containing 1 gram when shaken with (a) 100 ml chloroform
and (b) two successive 50 ml portions.
Solution: -
K=
𝐶𝑤𝑎𝑡𝑒𝑟
𝐶𝑆𝑜𝑙𝑣𝑎𝑛𝑡
1
20
V=100 ml
V1=100 ml,
W=1 g and n=1
∴ X1=W
𝐾𝑋1𝑉
𝐾𝑉+𝑉1
= 1 ×
1
20
×100
1
20
×100×100
= 0.0476𝑔
(𝑏)V=100 ml
V1=50 ml
W= 1 g and n=2 Therefore X2=W
𝐾𝑉
𝐾𝑉+𝑉1
2 X2=1
1
20
×100
1
20
×100×50
2 =0.0083g
Hence, the solute remained unextracted is more in case (a) than case (b)

59. Problem :- An aqueous 0.1dm3 solution of organic compound contains 0.01Kg of compound
would be extracted in five instalments of 0.02 dm3 each of ether. If the partition coefficient is
5 in favor of ether, Calculate the amount extracted.
Solution:-
𝐾 =
Concentration of water
Concentration of ether layer
=
1
5
=0.2
Given: W= 0.01kg,V=0.1dm3,V1=0.02dm3,n=5
∴ Amount unextracted =W
𝐾𝑉
𝐾𝑉+𝑉1
= 0.01
0.2 × 0.1
0.2 × 0.1 + 0.02
= 3.125 × 10−4
Kg
Hence the amount extracted =0.01-3.125× 10−4Kg
=9.6874 × 10−3
Kg

60. Problem
A solid added to a mixture of benzene and water. After shaking well and allowing
to stand,10 ml of benzene layer was found to contain 0.13g of solute and 100ml of
water layer contained 0.22 g of solute, Calculate the value of distribution co-
efficient.
Solution: -
Concentration in benzene(CB)=
0.13
10
=0.013g/ml
Concentration in water (CW)=
0.22
100
=0.0022g/ml
∴ According to Distribution law,
𝐶𝑛
𝐶𝑊
=
0.013
0.0022
=5.9

61. B. PHASE TRANSITION
Phase transition is the transformation of a
thermodynamic
system from one phase to another. Phases of a
thermodynamic c system and the states of a matter
have uniform physical properties. During a phase
transition of a given medium certain properties of the
medium change, often discontinuously, as a result of
some external condition, such as temperature, pressure
and others. For example, a liquid may become gas upon
heating to the boiling point, resulting in an abrupt
change in volume.
The measurement of the external conditions at which
the Solid transition occurs is termed the phase
transition.

62. CLAUSIUS - CLAPEYRON EQUATION
 Clausius - Clapeyron equation is a way of characterizing a phase transition
between two phases of matter of a single constituent. On a pressure - temperature
diagram, the line separating the two phases is known as the coexistence curve. The
Clausius, Clapeyron relation gives the slope of the tangents to this curve.
Mathematically,

𝑑𝑃
𝑑𝑇
=
∆𝐻𝑣
𝑇(𝑉𝐵−𝑉𝐴)
𝑓𝑜𝑟 𝑙𝑖𝑞𝑢𝑖𝑑 𝑉𝑎𝑝𝑜𝑢𝑟
𝑑𝑃
𝑑𝑇
=
∆𝐻𝑉
𝑇(𝑉𝑣−𝑉𝑙)
∆𝐻𝑉 → Molar heat of vaporization of liquid
𝑓𝑜𝑟 𝑆𝑜𝑙𝑖𝑑 𝑉𝑎𝑝𝑜𝑢𝑟
𝑑𝑃
𝑑𝑇
=
∆𝐻𝑆
𝑇(𝑉𝑣−𝑉𝑠)
∆𝐻𝑆 →Molar heat of Sublimation of Substance

63. 𝑓𝑜𝑟 𝑆𝑜𝑙𝑖𝑑 𝑙𝑖𝑞𝑢𝑖𝑑
𝑑𝑃
𝑑𝑇
=
∆𝐻𝑓
𝑇(𝑉𝐿−𝑉𝑆)
∆𝐻𝑓 → Molar heat of Fusion of substance
Partially miscible liquid
There are number of liquid pairs which are only partially miscible with one another.
Their miscibility varies with temperature in a characteristics manner. Accordingly
three types are observed
i) Phenol- water system
ii)Triethylamine-water system
iii)Nicotine-water system

64.  1) Phenol - water system
 Phenol and water are only partially miscible
at ordinary temperatures. On shaking they
form two layers. The upper layer is a saturated
solution of phenol in water and the lower layer
is a saturated solution of water in phenol. The
mutual solubility of phenol and water in one
another increases with rise in temperature
Therefore at certain higher temperature there
is formation of homogeneous one phase solution
i.e, phenol and water becomes completely
miscible.

65. This temperature is known as the critical or consolute solution
temperature (CST) or mutual solubility temperature (MST). The MST
can be defined as the temperature at which a mixture of a definite
composition becomes homogenous and exists in a single layer. This CST
is different for different composition of phenol and water. There is one
highest temperature above which any composition remains completely
miscible. This temperature is known as Upper Critical Solution (UCST).
For phenol-Water system the UCST is 68.10C with a composition of 34%
phenol.

66. ii)Triethylamine-water system
There are some liquid pairs
in which mutual solubility
decrease with increase of
temperature. Triethylamine-
water is an exchange of this
type. This will have lower
CST below which the system
has only one homogenous
layer of completely miscible
liquids, The lower CST is
18.50C as shown in diagram.

67. iii)Nicotine-water system
 This is the case of closed mutual
solubility curve having upper as well
as lower CST following figure shows
the variation of mutual solubility with
temperature for this system. Within
the liquid are partially miscible while
outside the closed area, they are
completely miscible. The upper CST is
2080C and lower is 60.80C for 32% of
Nicotine