INTRODUCTION At sea level At top of a mountain Water boils at 100 C Water boils at lt 100 C bP fP P 10133 bar bp 100 C P 026 bar bp 69 C elevation 8848 m ID: 276829
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Slide1
PHASE EQUILIBRIASlide2
INTRODUCTION
At sea level
At top of a mountain
Water boils at 100
C
Water boils at < 100C
b.P
= f(P)Slide3
P = 1.0133 bar
bp
= 100CSlide4
P = 0.26 bar
bp
= 69C
elevation = 8848 mSlide5
P = 2
atm , b.p = 124C+ EG 50/50 (v/v), b.p =129.5 C =>
b.p = f(P, c)Slide6
stability map
stability of the state of aggregation
Depending on the circumstances, molecules aggregate to form solid, liquid, or gas
If I specify P, T, and c, what are the stable phases?
Phase diagramSlide7
Examples of single-phase system (
= 1):Pure water (l)White gold (alloy of Au-Ag-Ni) => Ag and Ni substitute Au in its FCC lattice => homogeneous composition throughout => solution
Air (N
2, O
2,
Ar, CO2
)
Phase = region of a substance that is:
Uniform in chemical composition
physically distinct
mechanically separable
Examples of dual-phase system (
= 2)
:
Ice cubes in liquid water
Milk (fat globules in aqueous solution) Slide8
Equilibrium =
the condition which represents the lowest energy level the properties are invariant with time
Component = the measure of chemical complexity
N
1
2
1
> 1
Water (l)
diamond
Ice cube in liquid water (slush)
White gold
(Au-Ag-Ni)
CCl
4
– H
2
OSlide9
ONE-COMPONENT SYSTEM (N = 1)
Phase diagram of water
P (
atm
)
T (
C
)
1
100
Solid
= 1
Gas
= 1
Liquid
= 1
l = v coexistence curve
= 2
s = v coexistence curve
= 2
s = l coexistence curve
= 2
0
0.01
Triple point
= 3
4.58 mm Hg
Critical point
(374
C, 218
atm
)Slide10
POLYMORPHS
Phase diagram of sulfur
B : 95,5
C and 0,51 PaC :
115C and 2,4 PaE :
151C and 1,31108
Pa
Different atomic arrangement at constant composition Slide11
The transformation from one polymorph to another can be :
Reversible : two crystalline forms are said to be enantiotropicIrreversibel : two crystalline forms are said to be
monotropicSlide12
Pressure-temperature diagram for
dimorphous substances: (a) enantiotropy, (b) monotropySlide13
TWO-COMPONENT SYSTEM (N = 2)
There are tree variables that can affect the phase
equilibria
of a binary system: T, P, and C
T
T
P
P
cSlide14
Two-component phase diagram
3 dimensionT-P-c
2 dimension
T-P
P-c
T-c
Crystallization: liquid and solid
The effect of P can be ignored
T-c Diagram Slide15
BINARY SYSTEM– TYPE 1
Complete solubility in solid and liquid states
Change of state (s = l)
Metal (
Hume-
Rothery
Rule)
Similar crystal structure
Similar atomic volumes
Small values of
electronegativity
For
interstitial
solid solutions, the Hume-
Rothery
rules are:
Solute
atoms must be smaller than the pores in the solvent
lattice.
The
solute and solvent should have similar
electronegativity
.Slide16
The
atomic radii of the solute and solvent atoms must differ by no more than 15
%:
For
substitutional solid solutions, the Hume-
Rothery rules are:
The
crystal structures
of solute and solvent must match.
Maximum
solubility
occurs when the solvent and solute have the same
valency
. Metals with lower
valency
will tend to dissolve in metals with higher
valency
.
The
solute and solvent should have similar
electronegativity
. If the
electronegativity
difference is too great, the metals will tend to form
intermetallic
compounds
instead of solid solutions.Slide17
T
c
A
B
Liquid
= 1
Solid
= 1
Liquid + solid
= 2
Liquidus
SolidusSlide18
Liquidus
: l s + l : the lowest T for only liquid (at any c)Solidus : s s + l : the highest T for only solid (at any c)
The diagram represents the phase behavior of two components having many similarities (type of bonding, atomic size, crystal structure, etc), we call it
ISOMORPHOUS DIAGRAM
.
The shape of the diagram looks like a lens lens-shape diagram / lenticular
diagram.
Example: naphthalene - -
naphtholSlide19
Solid solution of naphthalene -
-
naphthol
Similar
Slide20
Solid solution of naphthalene -
-naphthylamine
>Slide21
BINARY SYSTEM– TYPE 2
Partial or limited solubility
miscibility gap
No change of state
always liquid or always solid
T
c
A
B
= 2
l
= 1
s
l
1
+
l
2
+
Coexistence curveSlide22
Hexane - nitrobenzene
upper critical solution temperature (upper consolute temperature)Slide23
lower critical solution temperature (lower consolute temperature)Slide24Slide25
BINARY SYSTEM– TYPE 3
Partial or limited solubility
Change of state
T
c
A
B
l
l
+
l
+
Solid solution (A rich)
Solid solution (B rich)
Liquid solution
= 1
= 1
= 1
= 2
= 2
= 2
Eutectic
l
EUTECTIC DIAGRAMSlide26
Freezing point depression of both components (A and B)
Eutectic point is equilibrium of , , and l
Eutectic point is unique; it only happens at one T, P, and cSlide27
Phase diagram for the simple eutectic system naphthalene - benzeneSlide28
Phase diagram for system
NaCl – H2O Slide29
ENTHALPY-COMPOSITION DIAGRAMSlide30
Solution A having composition of
xA and enthalpy of HA
is mixed adiabatically with solution A having composition of x
B
and enthalpy of HB
.The composition and the enthalpy of the mixture is calculated using material balance:
Total balance:
Component balance:
Combining both equations yields:Slide31
Similarly, if mixture A were to be removed adiabatically from mixture C, the enthalpy and composition of residue B can be located on the straight line through points A and C by means of the equation:Slide32
EXAMPLE
Calculate (a) the quantity of heat to be removed and (b) the theoretical crystal yield when 5000 lb of a 30 per cent solution of MgSO4
by mass at 110F is cooled to
70F. Evaporation and radiation losses may be neglected.
SOLUTION
(a)
Initial solution (A)
x
A
= 0.3 H
A
= - 31 Btu/lb
Cooled system (B)
x
B
= 0.3 H
B
= - 75 Btu/lb
Enthalpy change
H = - 44 Btu/lb
Heat to be removed = (- 44) (5000) = - 220000 Btu
(b)
The cooled system B is located in the region where MgSO
4
.7H
2
O is in equilibrium with solution.Slide33Slide34
The crystal MgSO
4.7H2O contains:
x
C
= 0.49
From the graph:
x
A
= 0.26
The MgSO
4
.7H
2
O crystal yield is 869.6 lb