The document discusses phase diagrams and the phase rule. It begins by defining key terms like phase, components, and degrees of freedom. It then explains Gibbs phase rule which relates the number of phases (P), components (C), and degrees of freedom (F) as F=C-P+2. Several examples of 1-component and 2-component phase diagrams are discussed. The document also covers concepts like true equilibrium, metastable equilibrium, and phase transitions. Phase diagrams show the conditions under which different phases can exist in equilibrium and include lines separating single phase and two phase regions.

1. SHILPA JOY
ASSISTANT PROFESSOR
DEPARTMENT OF CHEMISTRY
KURIAKOSE ELIAS COLLEGE MANNANAMSJ KEC 1

2. PHASE RULE : to study heterogeneous system at equilibrium .
J. Willard Gibbs: 1876, Yale University
“the sum of the number of degrees of freedom and the no. of phases exceeds
the no. of components by two”
F+P=C+2
SJ KEC 2

3. PHASE DIAGRAM
application of phase rule
The effect of P, T & Conc. on the
equilibrium system can be predicted.
SJ KEC 3

4. PHASE
A phase is defined as any homogeneous and physically distinct part of the system which is
separated from the other parts of the system by definite boundary surfaces.
EXAMPLES No. of Phases
1. Single gas like oxygen or hydrogen (homogeneous) 1
2. Mixture of Gaseous (no boundary) 1
3. Single liquid (homogeneous) 1
4.Miscible liquids 1
5. Immiscible liquids as many phases as the no.
of liquids
6. Single solid 1
7. Mixture of solids as many phases as the no.
of solids
SJ KEC 4

5. CaCO3 (s) CaO (S) + CO2
(g)
Heterogeneous system
3- Phases
No transfer of energy and mass from one phase to another:- no further
change in P, T &Composition of various phases at eulibrium
SJ KEC 5

6. no system No of phases
1 ice& water 2 (1S+1L)
2 Water & water vapour 2 (1L+1G)
3 Ice, liquid water & water vapour 3 (1L+1G+1S)
4 Water+ ethanol 1 (1L)
5 Benzene+water 2 (1L+1L V)
6 CaCO3,CaO&CO2 3 (2S+1G)
7 Monoclinic and rhombic sulphur 2 (1S+1S)
SJ KEC 6

7. Freezing point of water: ice, liquid & water vapour-equilibrium: composition of each
phase expressed in terms of the component, water
Sulphur system: rhombic sulphur, monoclinic sulphur liquid sulphur vapour…. ??????
: 2 components
Any two of the three substance can be chosen as independently variable constituents
to express the composition of each phase by means of a equation- 2 component
system.
CaCO3 (s) CaO (S) + CO2
(g)
COMPONENTS
The no. of components of a system at equilibrium is defined as the smallest no
independently variable constituents by means of which the composition of each phase can
be expressed by means of a chemical equilibrium
Case 1: CaO & CO2 are 2 components
phase components
CaCO3 CaO + CO2
CaO CaO +0 CO2
CO2 0 CaO + CO2
SJ KEC 7

8. Case 2: CaCO3 & CO2 are 2 components
phase components
CaCO3 CaCO3 + CO2
CaO CaCO3  CO2
CO2 0CaCO3 + CO2
Case 3: CaCO3 & CaO are 2 components
phase components
CaCO3 CaCO3 +0 CaO
CaO 0CaCO3 + CaO
CO2 CaCO3 - CaO
SJ KEC 8

9. +Fe(s) H2(g)
FeO(s) +H2O (S)
3 components
+NH4CI(s) NH3 (g) HCI(g)
Phase composition components
Solid NH4CI 2 ???????????
Gaseous phase 1
+NH3 HCI
+xNH3 yHCI
SJ KEC 9

10. Degrees of freedom
The no. of degrees of freedom or variables of a system is the no of variables such as T,P
or Conc. which must be specified in order to define a system completely.
C P F Eg
1 1 2
( bivariant)
pure gas T & P
1 2 1
(monovariant)
Water in contact
with vapour
vapour pressure has a fixed
value at a particular
temperature
1 3 0
(non-variant/
invariant
triple point of
water(0.0098 0 C
273.1675 K &4.58
mm Hg)
lowering T: vapour condense
Raising T : Ice melts
SJ KEC 10

11. ????? If a one component system has two phases with each other ,the degree
of freedom will be ----------------
???? Triple point of water system at----- temperature
??? When the degree of freedom is zero the system is called…….
???? Explain the degrees of freedom with regard to phase rule
SJ KEC 11

12. CONDITION FOR PHASE EQUILIBRIUM
A multiphase system is in thermodynamic equilibrium if exist simultaneously three
kinds of equilibrium
Thermal equilibrium ( constant temp)
Mechanical equilibrium ( constant pressure)
Chemical equilibrium (constant composition or chemical potential)
Temperature, pressure and chemical potential must be same throughout out
the system at equilibrium.
SJ KEC 12

13. 1. Condition of temperature
Equilibrium between two phases in an isolated system(constant vol. and energy)
Phase 1 Phase 2
( dS)V,E = 0 dS1+ dS2 = 0
phase 1 absorb an infinitesimal amount of heat from phase 2
dSS 1 =
𝜕𝑞
𝑇1
dS2 = -
𝜕𝑞
𝑇2
𝜕𝑞
𝑇1
-
𝜕𝑞
𝑇2
= 0
T1= T2
SJ KEC 13

14. 2. Conditions of pressure
above equilibrium at constant total volume and constant temperature
an infinitesimal increase in the volume (dV) of phase 1 , & decrease in volume
of phase 2
( dA)V,T = 0 dA1+ dA2 = 0
( dA) T = - w rev
dA 1 = - P1 dV dA2 = + P2 dV
- P1 dV + P2 dV = 0
P1 = P2
SJ KEC 14

15. 3. Chemical equilibrium
consider a closed system of P phases designated as , , ……P which contain C
components Shown as 1,2,3,…C At constant T & P
Gibbs’s free energy of each phase
G  = f(T,P,ni ) 
G  = f(T,P,ni ) 
G P = f(T,P,ni ) P
dG = G  +G  +……… + G P
dG = -SdT+VdP+idni
At constant T & P (dG) T,P = idni P
Infinitesimal transfer of matter (dG) = i  dni  + i dni + i  dni 
Closed system i  dni + i dni + i  dni  = 0
At equilibrium
dn1 + dn1 + dn1  +…….+ dn1p= 0 1  = 1 = 1 = 1 P
dn2 + dn2 + dn2  +…….+ dn2p= 0 2  = 2 = 2 = 2 P
.
.
dnc + dnc + dnc  +…….+ dncp= 0 C  = C = C = C PSJ KEC 15

16. TRUE EQUILIBRIUM
A system is said to be in a state of true equilibrium under a given set of conditions if the
same state can be realised from either direction by following any possible procedure.
Ex: equilibrium between ice and water at 1atm &273 K
 partial melting of ice or partial freezing of water
METASTABLE EQUILIBRIUM
A system is said to be in a state of metastable equilibrium under a given set of conditions if
the same state can be realised only from one direction and too very careful to change of
conditions.
Ex: possible to cool water slowly and carefully at 271K or even lower without the
appearance of ice.
&
Impossible to have water at 271K by melting of ice.
271K : metastable equilibrium
SJ KEC 16

17. CaCO3 (s) CaO (S) + CO2
(g)
?????COMPONENT 2
PHASE 3
NO OF DEGREES OF FREEDOM F= C-P+2 = 2-3+2= 1
SJ KEC 17

18. GIBBS PHASE RULE
if the equilibrium in a heterogeneous system is not influenced by electrical, magnetic
or gravitational forces, the no. of degrees of freedom(F) of a system , the no. of
components (c) and the no. of phases (P) of the system are related by the equation
F= C-P+2
SJ KEC 18

19. Phase diagram
A diagram giving the conditions of
equilibrium between various phases
of a substance
Various phase of a substance present
in different regions of T&P
G- low P& high T
S- high P & low T
Two phases in equilibrium are represented on line.
No of lines = no of phase equ. in system
SJ KEC 19

20. Slope
𝑑𝑃
𝑑𝑇
=
Δ𝐻
TΔ𝑉𝑚
Clapeyron eqn.
Sublimation & vapourisation - +ve slope
Fusion - -ve slpoe
Triple point
the lines representing S V,SL and LV
equilibria meet one another at this point.
This point is called the triple point of a
system.
SJ KEC 20

21. C P F Eg
1 1 2
( bivariant)
pure gas T & P
1 2 1
(monovariant)
Water in contact
with vapour
vapour pressure has a fixed
value at a particular
temperature (T/P)
1 3 0
(non-variant/
invariant
triple point of
water(0.0098 0 C
273.1675 K &4.58
mm Hg)
lowering T: vapour condense
Raising T : Ice melts
One component system
SJ KEC 21

22. PHASE DIAGRAM OF WATER : 1 Component system
 1C
 3P
 3 possible combination of two phase in
equilibrium
 S V,SL & LV F=1
 one three phase equ. SL V F=0
SL V
S V
LV
SL
3 SINGLE PHASE F=2
SJ KEC 22

23. 𝑑𝑃
𝑑𝑇
=
Δ𝐻
T(𝑉𝑔−𝑉𝑙)
=
+ 𝑣𝑒
+𝑣𝑒
Molar heat of vapourisation
Boiling point of water
Molar volume of water vapours
Molar volume of liquid water
OA
OC
OB
OA: vapour pressure curve of water
At a given temperature only one
vapour pressure F=1 (T/P)
Slope +ve VP T
SJ KEC 23

24. 𝑑𝑃
𝑑𝑇
=
Δ𝐻
T(𝑉𝑔−𝑉𝑠)
=
+ 𝑣𝑒
+𝑣𝑒
OA
OC
OB
OB: SUBLIMATION CURVE OF ICE
At a given temperature only one vapour
pressure F=1(T/P)
Slope +ve VP T
OC: fusion CURVE OF ICE
At a given temperature only one vapour
pressure F=1 (T/P)
melting pont of ice or freezing point of
water lowered by increase of pressure
Slope -ve VP 1/T
𝑑𝑃
𝑑𝑇
=
Δ𝐻
T(𝑉𝑙−𝑉𝑠)
=
+ 𝑣𝑒
−𝑣𝑒
SJ KEC 24

25. OA’
METASTABLE CURVE OA’
 liquid water &water vapour
 possible to cool water below its freezing
Point or even lower without the appearance of ice
 Supercooled state
 Unstable
 Spontaneously converted in to stable state by adding
 Small crystals of ice or stirring
 OA’ above OB – High VP
TRIPLE POINT O
The phase co-exist.
F=0
SJ KEC 25

26. AREA BETWEEN THE LINES
1. OA
x equ .L&V
xb constant T-L
xz constant T-V
xc constant P-L
xd constant P –V
2.OB
SJ KEC 26

27. EFFECT OF TEMPERATURE &PRESSURE
TEMPERATURE
Solid ice heated L-L’
L’ – FUSION START
M- VAPOURISATION START
SJ KEC 27

28. VERY HIGH PRESSURE
The existence of a substance
in different crystalline forms
is called polymorphism.
5 triple point
SJ KEC 28

29. SJ KEC 29

30. SJ KEC 30

31. SULPHUR SYSTEM
SR stable at ordinary T&P
SM stable 95 .6 o C to 119. 6o C
95 .6o C is the transition
temperature
1 component
Two solid form of sulphur can exist in equilibrium with each other and can under
go reversible transformation at the transition temperature. such a change is called
ENANTIOTROPY .
Sulphur can exist in 4 different forms
1) Solid sulphur SR 2) solid sulphur SM 3) liquid sulphur SL 4) sulphur vapour SVSJ KEC 31

32. 2 Phase equilibria - 6 curves 6
F=1 MONOVARIANT (P=2, C=1 , F=C-P+2)
3 Phase Equilibrium – 4 TRIPLE point F=0 (P=3,C=1) NONVARIANT
(O) (B) (C) (A’)
SJ KEC 32

33. CURVES
1.OA
2.OB
3.BL
4.OC
5.BC
6.CD
POINTS
1.O
2.B
3.C
4.A’
SJ KEC 33

34. METASTABLE EQUILIBRIA
(dotted lines)
Conversion of SR to SM at the transition
temp. can occur only if heating done
extremely slowly( involves molecular
rearrangements).
Hence SR & Sv are in metastable equ.
1) OA’
SR heated quickly above 95.6 SR SV
The vapour pressure at each T is higher
than V.P of SM
2) A’ B SL is cooled along L, SM will not
separate at B. A’ is the MP of SR
SL SV
3) A’C effect of P on the MP of SR
SR SL
Fusion curve of metastable rhombic
sulphur.SJ KEC 34

35. AREAS BETWEEN THE LINES
OA&OC- rhombic sulphur
BC&BL- liquid phase
AO,BO& BL- vapour phase
OC, OB & BC- monoclinic sulphur
SJ KEC 35

36. Is it possible to have a quadrupole point in a phase diagram of one component
system?
P=4
C=1
F=C-P+2= 1-4+2= -1
SJ KEC 36

37. TWO COMPONENT SYSTEM
F=C-P + 2 = 4 - P
P=1
C=2
F=3 ( T, P& Composition)
To represent three variables graphically , necessary to have three
coordinate axes at right angles to one another. (3D on paper difficult).
Two of the three variables are chosen for graphic representation while
the third one is assumed constant (PRESSURE – CONSTANT)
Such a system pressure variable is kept constant is called CONDESD
SYSTEM
Degree of freedom for a condensed system is reduced by one,
reduced phase rule is
F=C-P+1
SJ KEC 37

38. based on miscibility of the two components in the molted state
 Two components are completely miscible
 Two components are partially miscible
 Two components are immiscible
SOLID – LIQUID EQUILIBRIA
SJ KEC 38

39. Based on nature of solid phase that separate out during cooling of
the system of type 1
A. Two components do not form any compound and solidification
they simply form an initimate mixture known as EUTECTIC. Eg ..
Pb-Ag, Pb-Sb, Cd-Bi
B. The components enter into chemical composition forming cmpd
with CONGRUENT MPs , the two components form a solid
compound stable upto its MP
C. The components enter into chemical composition forming cmpd
with INCONGRUENT MPs , the two components form a solid which
decompose before attaining its MP.
SJ KEC 39

40. EUTETIC SYSTEM
A binary system consisting of two
substances which do not chemically react
but miscible in all proportions in liquid
phase is called Eutectic system.
EUTETIC MIXTURE
A eutectic mixture is a solid solution of two
or more substances having the lowest
freezing point of all the possible mixtures
of the component .
SJ KEC 40

41. SJ KEC 41

42. 4. the 2 components are completely miscible in the solid
state and yield thereby a complete serious of solid
solution.
5. the 2 components are partially miscible in the solid
state and form stable solid solution.
6. the 2 components form solid solution with PERIECTIC
ie 2 components form solid solution which are stable
upto transistion temperature.
SJ KEC 42

43. 1. Formation of Simple EUTECTIC ( thermal analysis –COOLING CURVE)
 A& B completely miscible in liquid
stat
 Soln. gives pure A or B as the solid
phase
 T –composition curve
point A- MP of Pure A
point B- MP of Pure B
If B is gradually added the FP of A lowered
along AC
If A is gradually added the FP of B lowered
along BC
Thermal analysis involves the study of the cooling curves of
various compositions of a system during solidification.SJ KEC 43

44. AC – various compositions of solutions saturated with solid A at
temp b/w A &AC
Along curve 2 phases
1) Solid A
2) Solutions of B in A
BC – various compositions of solutions saturated with solid B at
temp b/w B &BC
Along curve 2 phases
1) Solid B
2) Solutions of A in B
F= C-P+1
2-2+1= 1
MONOVARIANT
SJ KEC 44

45. SJ KEC 45

46. C – two curves intersect – solid A&B in equilibrium with liquid phase
F=c-p+1= 2-3+1 =0 NON VARIANT
C- t & comp.constant - 3 phase
coexist
Point C – lowest T @ LIQUID
mixture exist.
pointC
Lowest MP form mixture of
solid A&B is called EUTECTIC
POINT
T&COMPOSITION------ EUTECTIC
T&COMP. SJ KEC 46

47. Above AC&BC unsaturated soln or liquid melt
P=1
F=C-P+1=2 BIVARIANT (T&P)
SJ KEC 47

48. a- pure liquid cool b solid B separate& unsaturated solution more B
again cool solid B separates liquid compos. bC curve.
TO find the composition at any Temp Q
@ component B is in equilibrium with
Saturated solution of composition m
Apply LEVER rule Cm/CQ
d solid A begins to separate with B.
de mixed solid separates.
C’ same composition of eutectic ie C
C solid A&B separates till whole the liquid solidifies-
fine crystal A&B - mixture –not compound
Left CD LINE- large crystals A & intimate mixture of fine crystals of A&B
LARGE CRYSTALS are Primary cryatsals
Right CD – primary crystals of B
SJ KEC 48

49. Liquidus curve…. T-Composition curve of liquid phase –
AC&BC (BEGINNING OF FREEZING)
Solidus curve…. T -Composition curve of solid phase –EA, FB
& ECF(END OF FREEZING ON COOLING)
SJ KEC 49

50. Pb-Ag
Completely miscible
4 Phase in Equlibrium
1) Solid Pb
2) Solid Ag
3) Solution of Pb-Ag
4) Vapour
BP high- gas phase absent
F=C-P+1
C
AC&BC intersect . F=0
AC –freezing point curve of silver & MP is 961- monovariant
BC –freezing point curve of LEAD & MP is 327- monovariant
C – 303 eutectic temp
C- 2.6% Ag & 97.6 % Pb. eutectic composition
SJ KEC 50

51. EUTETIC POINT
The lowest temperature at which the liquid
mixture of two components can exist and if the
liquid is cooled below this temperature both the
components separate simultaneously in the
solid form having the same composition as in
the solution.
SJ KEC 51

52. Pattinson’s process – desilverisation of Pb - galena
0.1 to 2.6 %
the process of raising
the amount o f silver
SJ KEC 52

53. A. The components enter into chemical composition forming cmpd with
CONGRUENT MPs , the two components form a solid compound stable upto its
MP
or a compound said to congruent MP if it melts
3 solid phases
1) A
2) B
3) Compound AB
3 CURVES—FREEZING POINT
1) AC .. Solid A Liquid phase(EQUILIBRIUM)
2) BE .. Solid B Liquid phase
3) CDE … Solid AB Liquid phase
SJ KEC 53

54. D congruent MP( SOLID AND LIQUID have same composition)
At this temperature ,two component system becomes a single component as both solid and
liquid phase contains compound AB.
NON VARIANT
-congruent MP lies above MP of pure A&B
C & E – two Euetectic point
C – solid A& AB in equilibrium with liquid phase
E - solid B & AB in equilibrium with liquid phase
Any ‘’t” liquid phase has two compositions
x & x’ in equ. With solid AB.. So cmpd 2
solubilities at same T.
SJ KEC 54

55. Draw line DD’
1) Left halt- A& AB, DC- Depression in freezing point of cmpd AB on the addition of
A
2) RIGHT halt- B& AB, DE- Depression in freezing point of cmpd AB on the addition
of B
SJ KEC 55

56. FERRIC CHLORIDE –WATER
No of Hydrates with CMP
Sharply melts at constant T in to liquid of
the same composition
Fe2CI6.12H2O
Fe2CI6.7H2O
Fe2CI6.5H2O
Fe2CI6.4H2O
consider 100 MOLES OF WATER
POINT A- FREEZING POINT OF WATER
Ferric chloride added –AB (FeCI3 &H2O) –
MONOVARIANT
Addition FeCI3 , T LOWERS till eutectic T.
POINTB Fe2CI6.12H2O separates out . 3 phases- INVARIANT- 550C. -2.75
Moles FeCI3. – LOWEST T
FeCI3 & T , INCREASED- BCD(2 C- water melts) – monovariant-
BCD- solubility curve of Fe2CI6.12H2O
SJ KEC 56

57. POINT C - CONGRUENT MP of do-decahydrate
CURVE CB & CD- effect of adding water & ferric
chloride in lowering CMP of dodecahydrate.
POINT D-- Fe2CI6.7H2O SEPARATES OUT .- 2nd
eutectic point.
CURVE DEF– solubility curve ofFe2CI6.7H2O
POINT E —CMP Fe2CI6.7H2O
POINT F-- Fe2CI6.5H2O SEPARATES OUT .- 3rd
eutectic point.
CURVE FGH– solubility curve ofFe2CI6.5H2O
POINT G —CMP Fe2CI6.5H2O
POINT H-- Fe2CI6.4H2O SEPARATES OUT .- 4TH
eutectic point.
CURVE HJK– solubility curve ofFe2CI6.4H2O
POINT J —CMP Fe2CI6.4H2O
POINT K-- Fe2CI6SEPARATES OUT .- 5TH eutectic
point.
CURVE KL– solubility curve ofFe2CI6
NON VARIANT POINTS
eutectic &congruent MP POINTS
SJ KEC 57

58. A. The components enter into chemical composition forming cmpd with INCONGRUENT
MPs , the two components form a solid which decompose before attaining its MP.
The compound formed by the combination of two components ,decompose when
heated giving a new solid phase and solution
S1 S2 + solution
(original) (new solid)
Decomposition T- TRANSITION
TEMPERATURE
INCONGRUENT MPs are known as
TRANSITION or PERITECTIC TEMPERATURE
SJ KEC 58

59. Consider A&B pure components
POINT D-NM INCONGRUENT MPs OF AB2 / TRANSITION TEMPERATURE
CURVE AC- fusion curve A ,solid A in EQu. with liquid
CURVE BC- fusion curve B ,solid B in EQu. with liquid
CURVE CB- fusion curve AB2 ,solid B in EQu. with liquid
POINT C-- AB2 start forming (2s +1l)- invariant
POINT D-- AB2 start forming (2s+1l)- invariant
SJ KEC 59

60. SODIUM SULPHATE WATER SYSTEM
2 Enantiotropic crystalline form
Rhombic &monoclinic
2 hydrates
Na2SO4 10.H20 , Na2SO4 7.H20
SJ KEC 60

61. CURVE AB- freezing point curve – solid&l iquid equilibrium
POINT A . – ICE , ANHDROUS SODIUMSULPHATE ADDED MOVES ALONG AB,
(3 PHAESE, ICE,SOLN& VAPOUR) F=1
POINT B- Na2SO4 10.H20 F=O (ICE,SOLN
,VAPOUR, Na2SO4 10.H20)-QUADRUPOLE
POINT
CURVE BC- Na2SO4 7 .H20 , SEPARATES OUT
METASTABLE CURVE
POINT D- RHOMBIC SEPARTES OUT F=O
METASTABLE QUADRUPOLE POINT
lowering T ,E, Na2SO4 4..H20 separates
POINT C - Na2SO4 7H20 F=O (ICE,SOLN ,VAPOUR,
Na2SO4 7.H20)-QUADRUPOLE POINT
CURVE CD- METASTABLE CURVE OF
Na2SO4 7 .H20 , SEPARATES OUT
(3 PHAESE, heptahydarate ,SOLN& VAPOUR)
F=1 SJ KEC 61

62. POINT F - Na2SO4 10.H20,RHOMBIC ,
SOLUTION ,VAPOUR F=O
CURVE FG- SOLUBILITY CURVE OF
Na2SO4 , WARMING DECAHYDRATES
(3 PHAESE, decahydarate ,SOLN& VAPOUR)
F=1
POINT G – MONOCLINIC,RHOMBIC ,SOLUTION
VAPOUR F=O, QUADRUPOLE POINT,NO WATER
CURVE HG- SOLUBILITY CURVE OF
MONOCLINIC Na2SO4 ,
(3 PHAESE, MONOCLINIC ,SOLN& VAPOUR) F=1
CURVE BF- SOLUBILITY CURVE OF Na2SO4 10 .H20 , SEPARATES OUT
(3 PHAESE, decahydarate ,SOLN& VAPOUR) F=1
SJ KEC 62

63. 1) Point B,C,D, F- QUAFROPLE POINT
F=0
2) CURVE AB,BC,CD,BF,FG&GH
monovariant F=1
SJ KEC 63