Lecture 4: Phase diagrams in
Material Science. Equilibrium
3-11-2009
• Lecture plan:
– phase diagrams in material science:
• microstructures
i
t t
in
i isomorphous
i
h
bi
binary
systems
t
• microstructures in eutectic alloys
• liquid crystals
– equilibrium
• equilibrium and Gibbs free energy
• description of equilibrium
• response of equilibrium to conditions (P, T, pH)
– problems
Phase diagrams and
Microstructure
Binary phase diagrams
• Ph
Phase di
diagram with
ith total
t t l solubility
l bilit iin b
both
th liliquid
id and
d solid
lid
state: isomorphous system
homogeneous liquid solution of Cu and Ni.
T(°C)
• 2 phases:
1600
1500
L (liquid)
(li id)
α (FCC solid solution)
L (liquid)
1400
• 3 phase fields:
L
L+α
α
s
u
d
i
il qu + α s
L lidu
so
1300
α
(FCC solid
solution)
l ti ) homogeneous solid solution of Cu and Ni.
1200
1100
1000
0
20
40
60
80
Cu-Ni phase diagram
100
wt% Ni
Cu-Ni phase diagram
Information we can extract from the diagram:
¾ the phases present;
¾ composition
p
of the p
phases
¾ percentage of fraction of the phases
C0=35 wt% Ni
at TA: Only liquid, composition of liquid
is given by the overall composition
(C0=35 wt% Ni)
at TD: Only liquid, composition of liquid
is given by the overall composition
(C0=35 wt% Ni)
at TB: Both L and α are present
Composition at TB:
• Liquid phase (L) of 32% Ni
• Solid phase (α) of 43% Ni
g ratio:
• Weight
T(°C)
A
TA
1300
L (liquid)
TB
1200
TD
WL S
R
(43 − 35)
= ;Wα =
=
= 73%
Wα R
S + R (43 − 32)
20
α
+
L
tie line dus
i
il qu
B
R S
D
α
+
L
s
u
d
i
l
so
α
( l d)
(solid)
3032 35 4043
CLCo
50
Cα
wt% Ni
Development of microstructure in a Cu-Ni alloy
Equilibrium case
(very slow cooling)
Development of microstructure in a Cu-Ni alloy
Non-Equilibrium case
(real)
• Fast cooling:
Cored structure
• Slow cooling:
Equilibrium structure
First α to solidfy:
46wt%Ni
Last α to solidfy:
< 35wt%Ni
Uniform Cα:
35wt%Ni
How we can prevent coring
and get equilibrium structure?
Binary Eutectic Systems: Sn-Pb
T(°C)
Sn-Pb system:
¾ limited solubility in solid state 300
¾ 3 single phase regions (L, a, b);
q
below TE.
¾ TE=183 0C, no liquid
200 α
¾ Eutectic composition 61.9%
At the eutectic temperature:
L(CE ) R α (Cα E ) + β (Cβ E )
150
100
Pb
•
L (liquid)
L+α
18.3
R
0 11 20
For a 40wt%Sn
40wt%Sn-60wt%Pb
60wt%Pb alloy at 150C,
150C find...
find
--the compositions of
the phases:
Ca = 11wt%Sn
Cb = 99wt%Sn
183°C
61.9
L+β β
97.8
S
α+β
40
Co
60
99100
Sn
Co, wt% Sn
59
= 67wt %
88
29
= 33wt %
Wβ =
88
Wα =
80
Microstructures in binary systems
T(°C)
400
L: Cowt%Sn
• Co < 2wt%Sn
• Result:
L
α
L
--polycrystal
polycrystal of α grains.
grains 300
200
TE
100
α
L+α
α: Cowt%Sn
(Pb-Sn
System)
α+β
0
10
20
30
Co
Co, wt%
2
(room T solubility limit)
Sn
Microstructures in binary systems
L: Cowt%Sn
T(°C)
400
• 2wt%Sn < Co < 18.3wt%Sn
• Result:
--α polycrystal with fine
β crystals.
L
L
α
300
L+α
α
200
TE
α: C owt%Sn
α
β
100
0
α+β
10
20
30
Co Co, wt%
2
(sol limit at Troom)
(sol.
18 3
18.3
(sol. limit at TE)
Sn
Microstructures in binary systems
• Eutectic composition
Microstructures in binary systems:
eutectic and around
T(°C)
L
300
200
TE
α
L+α
L+β
α+β
100
Co
Co
0
0
hypoeutectic hypereutectic
20
40
hypoeutectic: Co=50wt%Sn
α
α
60
80
eutectic
18.3
α
61.9
100
Co, wt% Sn
97.8
hypereutectic: (illustration only)
eutectic: Co=61.9wt%Sn
β
α α
β
α
175μm
(Pb-Sn
System)
β
160μm
eutectic micro-constituent
From: W.D. Callister, “Materials Science and Engineering: An Introduction”, 6e.
β
β β
β
IRON-CARBON (Fe-C) PHASE DIAGRAM
L ⇒ γ + Fe3C
Eutectoid (B):
-Eutectoid
γ ⇒ α + Fe3C
T(°C)
1600
δ
L
1400
1200
γ+L
γ
(austenite)
γ
α+
800
600
B
400
0
(Fe)
Result: Pearlite =
alternating
lt
ti layers
l
off
α and Fe3C phases.
(Adapted from Fig. 9.24, Callister 6e.
(Fig. 9.24 from Metals Handbook, 9th ed.,
V l 9,
Vol.
9 Metallography
M t ll
h and
d
Microstructures, American Society for
Metals, Materials Park, OH, 1985.)
L+Fe3C
S
R
γ+Fe3C
727°C = Teutectoid
R
S
α+Fe3C
1
0.77
Ceutectoid
α
120μm
γ γ
γ γ
1000
A
1148°C
Fe3C (ce
ementite
e)
• 2 important
points
-Eutectic
E
i (A)
(A):
2
3
4
5
6
6.7
4.30
Co, wt% C
Fe3C (cementite-hard)
α (ferrite-soft)
Adapted from Fig. 9.21,Callister 6e. (Fig. 9.21
adapted from Binary Alloy Phase Diagrams, 2nd ed.,
V l 1,
Vol.
1 T.B.
T B Massalski
M
l ki (Ed
(Ed.-in-Chief),
i Chi f) ASM
International, Materials Park, OH, 1990.)
21
HYPOEUTECTOID STEEL
T(°C)
1600
δ
L
γ γ
γ γ
1200
γ γ
γ γ
1000
α
γ
γ
α
γ αγ
800
γ+L
γ
(austenite)
i )
γ+Fe3C
r s
727°C
αRS
400
0 Co
pearlite
0.77
wα =s/(r+s) 600
wγ =(1-w α)
α
α
α
wpearlite = wγ
wα =S/(R+S)
wFe3C =(1-w α)
L+Fe
L
Fe3C
1148°C
1148
C
α+Fe
α
Fe3C
1
2
3
4
5
(Fe-C
System)
Fe
e3C (cem
mentite))
1400
6
6.7
Adapted from Figs.
9.21 and 9.26,Callister
6e. (Fig
(Fig. 9
9.21
21 adapted
from Binary Alloy
Phase Diagrams, 2nd
ed., Vol. 1, T.B.
Massalski (Ed.-inChi f) ASM
Chief),
International, Materials
Park, OH, 1990.)
Co, wt% C
100μm Hypoeutectoid
steel
Adapted from
Fig. 9.27,Callister
6e. (Fig. 9.27 courtesy Republic Steel Corporation.)
22
HYPEREUTECTOID STEEL
T(°C)
1600
δ
L
γ γ
γ γ
Fe3C
γ γ
γ γ
1200
γ+L
γ
(austenite)
L+Fe3C
1148°C
1000
γ+Fe3C
α
wFe3C =r/(r+s)
(
)600
wγ =(1-w Fe3C)
400
0
pearlite
R
wpearlite = wγ
wα =S/(R+S)
wFe3C =(1-w α)
s
r
800
0.77
γ γ
γ γ
S
1
Co
α+Fe3C
2
3
4
5
(Fe-C
System)
Fe
e3C (cem
mentite
e)
1400
6
6.7
Adapted from Figs.
9.21 and 9.29,Callister
6e. (Fig. 9.21 adapted
from Binary Alloy
Phase Diagrams, 2nd
ed., Vol. 1, T.B.
Massalski ((Ed.-inChief), ASM
International, Materials
Park, OH, 1990.)
Co, wt% C
60μm Hypereutectoid
steel
Adapted from
Fig. 9.30,Callister
6e. (Fig. 9.30
copyright 1971 by United States Steel Corporation.)
23
Liquid crystals
•
•
Mesophase – an intermedediate phase between
solid and liquid. Example: liquid crystal
Liquid crystal – substance having a liquid-like
i
imperfect
f t order
d in
i att lleastt one direction
di ti and
d llongrange positional or orientational order in at least one
another direction
Nematic
Smectic
calamitic
(rod-like)
discotic
Cholesteric
Nematic crystals in LCD
EQUILIBRIUM
Chemical Equilibrium
A+ B R C + D
• Chemical reaction tend to move towards a dynamic
equilibrium in which both reactants and products are
present but have no tendency to undergo net change
The question: How to predict the composition of mixture at various condition
The Gibbs energy minimum
•
•
Spontaneous change at const P and T happens towards lower values of
the Gibbs energy
Let’s
Let
s consider reaction
ZZX
If some amount
dξ
A YZZ B
of A changed into B: dnA = − d ξ
d B = + dξ
dn
⎛ ∂G ⎞
⎟
ξ
∂
⎝
⎠ P ,T
Reaction Gibbs energy (definition): Δ r G = ⎜
dG = μ A dnA + μ B dnB = − μ A dξ + μ B d ξ = ( μ B − μ A )d ξ
⎛ ∂G ⎞
ΔrG = ⎜
⎟ = μB − μ A
⎝ ∂ξ ⎠ P ,T
At equilibrium
Difference between chemical
potentials of the products and
the reactants at the composition
fo the reaction mixture
ΔrG = 0
extent of the reaction
The Gibbs energy minimum
• Spontaneity reaction at const P, T
ΔrG < 0
Forward reaction is spontaneous, reaction exergonic (work-producing)
ΔrG = 0
Reaction at equilibrium
ΔrG > 0
Reverse reaction is spontaneous, reaction endergonic i.e.
required work to go in forward reaction
Exergonic
E
i reaction,
ti
e.g. glucose oxidation Endergonic reaction,
e.g. protein synthesis
The description of equilibrium
•
Perfect gas equilibrium
Δ r G = μ B − μ A = ( μ B 0 + RT ln pB ) − ( μ A 0 + RT ln p A ) =
pB
= Δ r G + RT ln
pA
0
Q – reaction quotient
K- equilibrium constant
At equilibrium:
pB
Δ r G = 0 = Δ r G + RT ln
pA
0
RT ln K = −Δ r G 0
Wh reaction
Why
ti doesn’t
d
’t go till th
the end:
d
Δ mix G = nRT (κ A ln κ A + κ B ln κ B )
The description of equilibrium
2 A + B ⎯⎯
→ 3C + D
• General case of a reaction
0 = 3C + D − 2 A − B
Δ r G = Δ r G 0 + RT ln Q
ΔrG 0 =
∑
νΔ f G0 −
products
∑
νΔ f G0
reactants
activities of products
Q=
activities of reactants
For example, for the reaction above:
Q = ∏aj
j
aC 3 aD
Q= 2
a A aB
vj
The description of equilibrium
⎛
vj ⎞
K = ⎜∏aj ⎟
⎝ j
⎠equilibrium
At equilibrium:
q
RT ln K = −Δ
ΔrG 0
Example: Find degree of dissociation of water vapour at 2300K and 1 bar if
standard Gibbs energy for decomposition is 118 kJ/mol
ZZX H 2 ( g ) + 1 O2 ( g )
H 2O( g ) YZZ
2
ΔG 0
118*10
118
103
ln K = −
=
RT
8.3* 2300
K=
pH 2 pO2
pH 2O
1
2
α 3 2 p1 2
=
(1 − α )(2 + α )1 2
K = 2.08*103
α = 0.0205
The description of equilibrium
RT ln K = −Δ r G
0
K =e
−Δ r G 0 RT
=e
−Δ r H 0 / RT
e
Δr S 0 / R
Increase with reaction entropy
decrease with reaction enthalpy
Boltzmann distribution interpretation:
e − Ei / kT
pi = N
− Ei / kT
e
∑
i
Due to higher
density of
energy
gy levels
(i.e. higher S),
B is dominant
at equilibrium
The description of equilibrium
• Relation between equilibrium constants
aC aD γ C γ D bC bD
K=
=
×
= Kγ Kb
a A aB γ Aγ B bAbB
At low concentration:
K ≈ Kb
• Using biological standard state
If a biological reaction involves H+ ions, we have to take into account that
standard biological condition is at
pH = − log aH + = 7
NADH (aq ) + H + (aq ) R NAD + (aq ) + H 2 ( g )
Δ r G ⊕ = Δ r G 0 + 7 ln10 × RT =
= −21.8kJ / mol + 16.1× 8.3 ×10−3 kJ / K mol × 310 K = 19.7kJ / mol
The response of equilibria to the conditions
• E
Equilibria
ilib i will
ill respond
d tto ttemperature,
t
pressure and
d
concentration changes
RT ln K = −Δ r G 0
Pressure dependence:
Depends on standard Δ r G
(standard pressure)
0
⎛ ∂K ⎞
⎜
⎟ =0
⎝ ∂P ⎠T
ZZX 2 B( g )
A( g ) YZZ
pB 2
K=
pA p 0
• Pressure increase by injecting inert gas: no change as partial pressures of
reactants and products stay the same .
• Pressure increase by compression: system will adjust partial pressures so the
constant stays the same.
The response of equilibria to the conditions
• Le Chatelier principle:
A system at equilibrium, when subjected to disturbance
responds in a way that tends to minimize the effect of
disturbance
Extent of dissociation, α:
ZZX 2 B( g )
A( g ) YZZ
(1 − α )n
2α n
pB 2
K=
pA p 0
Mole fractions at equilibrium:
κA =
(1 − α )n
1−α
=
(1 − α )n + 2α n 1 + α
pB 2 κ B 2 p 2 4α 2 p
K=
=
=
κ A p 1−α 2
pA
⎛
½
⎞
1
0 ⎟
1
4
p
Kp
+
⎝
⎠
α =⎜
κB =
2α
1+ α
The response of equilibria to the conditions
• Temperature response
Equilibrium
q
will shift in endothermic direction if temperature
p
is increased and in
exothermic direction if temperature is lowered.
Van’t Hoff equation:
RT ln K = −Δ r G 0
Gibbs-Helmholtz equation
d ln K
1 d (Δ r G 0 / T )
=−
dT
R
dT
d ln
l K Δr H
=
dT
RT 2
0
Δr H 0
d (Δ r G 0 / T )
=−
dT
T2
Δr H
d ln
l K
=−
d (1/ T )
R
0
i.e. for exothermic
reaction:
ti
d ln K
Δr H 0 < 0
<0
dT
Δr H 0
So, we can predict the equilibrium constant at another temperature: ln K 2 − ln K1 = − R
⎛1 1⎞
⎜ − ⎟
⎝ T2 T1 ⎠
The response of equilibria to the conditions
• Boltzmann distribution interpretation
p
The response of equilibria to the conditions
•
Noncalorimetric measuring reaction enthalpy
d ln K Δ r H 0
=
d (1/ T )
R
Δr H 0
slope:
R
The response of equilibria to the conditions
•
Value of K at different temperatures
d ln K Δ r H 0
=
d (1/
( /T)
R
Δr H 0
1 2
0
ln K 2 − ln K1 =
Δ r H d (1/ T ) =
∫
R 1/ T1
R
1/ T
⎛1 1⎞
⎜ − ⎟
⎝ T2 T1 ⎠
Equilibria and pH
• Dissociation of water (autoprotolysis)
ZZX H 3O + (aq ) + OH − (aq )
2 H 2O(l ) Y
YZZ
Kw =
aH O+ aOH −
3
aH 2 O 2
= aH O+ aOH − = 10−14 at 298 K
3
For pure water:
Ionic dissociation
constant of water
aH O+ = aOH − = 10−7
3
pH = − log aH O+
3
at low concentration equal to molarity
The response of equilibria to pH
•
Arrhenius acid: increases concentration of H3O+ in solution
Arrhenius base: increases concentration of OH- in solution
Can be done via donation of OH- or removing of H+.
ZZX Na + (aq ) + OH − (aq )
NaOH (aq ) YZZ
ZZX NH 4 + (aq ) + OH − (aq )
NH 3 (aq ) + H 2O(l ) YZZ
Acidity constant, Ka:
ZZX H 3O (aq ) + A (aq ) K a =
HA(aq ) + H 2O(l ) YZZ
+
−
aH O+ a A−
3
aHA
pK a = − log K a
Conjugate base
Basicity constant, Kb:
Conjugate
j g
acid
ZZX HB + (aq ) + OH − (aq )
B (aq ) + H 2O(l ) YZZ
Kb =
aHB + aOH −
aB
pK b = − log K b
Example: dissociation of formic acid
•
Example: pK of formic acid is 3.77 at 298K. What is pH of 0.01M solution?
What would happen if it were strong acid?
ZZX HCOO − + H 3O +
HCOOH + H 2O(l ) YZZ
x± =
⎡⎣ H 3O + ⎤⎦ ⎡⎣ HCOO − ⎤⎦
= 1.695 ×10−4
Ka =
[ HCOOH ]
Answer:
−b ± b 2 − 4ac
2a
x 2 + 1.695 × 10−4 x − 1.695 × 10−6 = 0
[ H 3O + ] = 1.22 ×10−3
pH = 2.91
•
What would be dissociation degree at pH=4 and pH=10?
x = 1.22 × 10−3
Class problems: Last lecture
• Atkins 6.9b: sketch the phase diagram of
th system
the
t
NH3/N2H4 given
i
th
thatt the
th two
t
substances do not form a compound and
NH3 freezes at -78C,, N2H4 freezes at
+2C, eutectic formed with mole fraction
of N2H4 0.07 and melts at -80C.
• Atkins
Atki 6
6.10b
10b D
Describe
ib th
the di
diagram and
d
what is observed when a and b are
cooled down
Class problems:
• Atkins 7.2b: Molecular bromine is 24% dissociated at 1600K
and 1 bar.
bar Calculate K
K, ∆rG0 at 1600K and predict K at 20000C,
C
given ∆rH0=+112kJ/mol over the temperature range
• Atkins 7.4b: In the gas phase reaction A+B=C+2D it was found
th t when
that
h 2
2moll A
A, 1
1moll B and
d 3 moll D were mixed
i d and
d allowed
ll
d
to come to equilibrium at 25C, the mixture contained 0.79mol of
C at 1 bar. Calculate mol fraction of everyy species at
equilibrium, Kx, K and ∆rG0.