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2015-09-04
1
1Soft Matter Physics
Phase Transitions: Kinetics
2015-09-04
Andreas DahlinLecture 3/11Jones: 3.3-3.4
http://www.adahlin.com/
2015-09-04 Soft Matter Physics 2
Outline
We know from the previous lecture what phases to expect for a mixture of compounds
under given circumstances.
We also want to know how fast and by what mechanism a phase transition occurs!
• Kinetics of freezing (homogeneous or heterogeneous nucleation).
• Metastable liquid mixtures (nucleation again).
• Unstable liquid mixtures (spinodal decomposition).
Emphasis on the difference between equilibrium and kinetics!
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2015-09-04 Soft Matter Physics 3
Repetition: Reaction Kinetics
There is an “activated state”, corresponding to an activation energy, after which the
energy change is just “downhill”.
The probability that a reaction occurs is an exponential function of the activation energy
with rate constant k usually following Arrhenius kinetics:
reaction progression
free
en
ergy
ΔG*
ΔG
Tk
Gk
B
exp
The “reaction progression”
can for instance be the size
of the crystal formed
during a freezing event!
There will be an energy cost associated with creating the interface (γSL for solid-liquid)
which scales with the area and an energy release associated with the freezing that must
scale with volume.
The energy change upon freezing a sphere (ΔHm given per mass) of radius r is then:
2015-09-04 Soft Matter Physics 4
Nucleation in Freezing
SL
2
m
m
3
nuc π43
π4 r
T
THrrG
Clearly, this function has a
maximum for a certain r = r*.
Crystals with r < r* probably melt
again…
Crystals with r > r* probably
continue to grow…r < r*
r > r*
freezing
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Freezing Kinetics
We can find r* by taking the derivative:
The activation energy is then:
Since we also take the exponential of ΔG* to get the rate constant this becomes an
extremely temperature sensitive function!
In practice one usually gets homogenous nucleation for ΔT on the order of tens of
degrees…
TH
Trr
T
THr
r
rG
m
mSLSL
m
m2nuc 20π8π4
22
m
2
2
m
3
SL
SL
2
m
mSL
m
m
3
m
mSLnuc
3
π16
2π4
2
3
π4
TH
T
TH
T
T
TH
TH
TrGG
Often nucleation is instead initiated by a boundary (the walls of a vessel) or some kind
of impurity. The interfacial energy can then be much lower.
2015-09-04 Soft Matter Physics 6
Heterogeneous Nucleation
Icehotel
http://www.icehotel.se
Why bother so much with the
mechanism of freezing? One
reason is that the quality in
terms of crystallinity of the
final solid is strongly affected
by the nucleation process!
For instance, grain boundaries
are formed where crystals
meet if they cannot coalesce.
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Assume we have a planar surface where nucleation occurs. This “catalyst” can be a wall
or the surface of a piece of dirt. Young’s equation (force balance) gives:
If the solid is a spherical cap its volume is by geometry:
The relevant areas are:
“catalyst” (C)
solid (S)
liquid (L)
θ
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Nucleation on Surfaces
CLCSSL cos
cos2cos13
π 23
r
V
cos12π 2
SL rA
22
CS sinπrA
Sidenote: The use of Young’s
equation here is not obvious to
me since it is for liquid
droplets in equilibrium…
θ
r
2015-09-04 Soft Matter Physics 8
Reduced Activation Energy
Activation energy can be derived as for homogenous nucleation. The energy of forming
a crystal on the surface with radius of curvature r is:
The activation energy becomes quite similar to that of homogenous nucleation
(exercise). Amazingly, we get a factor containing only the contact angle:
CLCS
22
SL
2
m
m23
nuc sinπcos1π2cos2cos13
π
rr
T
THrG
4
cos2cos12
hom
het
G
G
Now we can plot the relative
reduction in activation energy for
different contact angles.
The effect from heterogeneous
nucleation is strong for the “right”
type of surface!
θ
ΔG
*het/Δ
G*
ho
m
0 60 120 1800
0.2
0.4
0.6
0.8
1
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A perfectly clean glass is filled with perfectly clean water and put into a freezer at -5 °C.
The enthalpy of melting is ΔHm = 335 Jg-1, the interfacial tension between water and ice
0.03 Jm-2 and the density of ice 0.93 gcm-3.
(A) Show that homogenous nucleation is unlikely to occur.
(B) Show that ice is likely to form on the walls of the container.
2015-09-04 Soft Matter Physics 9
Exercise 3.1
2015-09-04 Soft Matter Physics 10
Exercise 3.1
(A) The activation energy for homogeneous nucleation is:
We have:
ΔHm = 3.35×105 Jkg-1
γSL = 0.03 Jm-2
ρ = 930 kgm-3
T = 268 K
Tm = 273 K
ΔT = 5 K
Inserting all values gives ΔG* = 1.38…×10-17 J. At 268 K this is equal to 3757kBT which
clearly is a very high activation energy.
22
m
2
2
m
3
SL
3
π16
TH
TG
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2015-09-04 Soft Matter Physics 11
Exercise 3.1
(B) Glass is a hydrophilic surface. We can expect a low contact angle of θ = 30° or less.
The activation energy of heterogeneous nucleation compared with homogenous
nucleation is reduced by the factor:
Inserting θ = 30° shows that ΔG*het is almost down to 1% of ΔG*hom, which means that it
is tens of kBT. Thus the nucleation can be expected to occur on “everyday timescales”.
4
cos2cos12
2015-09-04 Soft Matter Physics 12
Crystal Branching
Why do ice crystals become
branched? The same effect
contributes to the complicated
shapes of snowflakes.
It appears that after nucleation and
some growth the solid generally
becomes “pointy” in shape.
The growth of sharp perturbations
in the forming crystal amplified!
We will just make a qualitative
argument here (Jones goes into
more details if you are interested).
Wikipedia: Ice Crystals
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Dendritic Growth
When the solid is formed, heat corresponding to ΔHm is released. This heat needs to
diffuse away for the surrounding liquid to remain undercooled!
On pointy structures, the heat flux is greater and the undercooling remains higher.
Sidenote: Faster growth releases more heat, so the argument is not obvious. The thermal
conductivities of the solid and the liquid must also come into play…
heat fluxcoldest spot
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Reflections and Questions
?
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8
In the unstable region, any small fluctuation in Φ will be amplified directly. The liquids
separate through spinodal decomposition.
But let us first look at the metastable region. Here small fluctuations are stable. Phase
separation can only occur through nucleation: A fluctuation must cause a reasonably
large volume with concentration similar to the opposite coexisting phase.
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Liquid-Liquid Phase Separation
ΔFmix (Φ)
Φ0
r
Φ1 Φ2
Φ0
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The Critical Size
radius (r)
free
en
ergy c
han
ge
(ΔF
nuc)
ΔFnuc(r)
critical radius r*
For describing nucleation in the metastable region, the theory is very similar to freezing.
Free energy change for spherical droplet of radius r of the coexisting phase scales with
volume. Extra energy cost associated with creating the interface (scales with area).
Again there must be a certain critical droplet size corresponding to an activation barrier.
nuc
23
nuc π43
π4rf
rrF
Here Δf is the free energy change
per volume inside the formed
nucleus. (Must be negative!)
Importantly, γnuc is now the
interfacial tension of the coexisting
compositions!
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2015-09-04 Soft Matter Physics 17
Nucleation Activation Energy
If we take the derivative as in the treatment of freezing kinetics we get:
Thus we expect that nucleation events (with r > r*) occurs with a frequency proportional
to exp(-ΔF*/[kBT]).
It should be possible to relate Δf to ΔFsep (Jones does not but I give it a try). If we know
how many “lattice sites” there are in the nucleated droplet we can use ΔFsep:
Here v is a characteristic molecular volume (of a lattice site). Thus we connect the theory
of phase separation from an equilibrium perspective with the kinetics!
nuc
2nuc π8π4 rfrr
F
fr
nuc2
2
3
nuc
1
3
π16
fF
v
ΦΦΦFf
210sep ,,
A binary mixture follows the regular solution model with χ = 2.5.
(A) Show that the mixture is metastable at Φ = 0.2.
(B) Estimate the activation energy in of homogenous nucleation at 300 K. The molecules
have a volume of 7 nm3 and the interfacial tension between the coexisting compounds is
2×10-5 Nm-1.
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Exercise 3.2
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(A) First condition: Curvature of ΔFmix should be positive.
Also, Φ = 0.2 must be larger than the lower value of Φ for the coexisting phases!
Must be solved numerically. Testing a few values close to Φ = 0.2 gives Φ1 ≈ 0.15 as one
solution. Opposite coexisting phase is then at Φ2 ≈ 0.85.
2015-09-04 Soft Matter Physics 19
Exercise 3.2
ΦΦΦΦΦΦTkF 15.21log1logBmix
5.251loglogBmix
ΦΦΦTk
Φ
F
5
1
11B2
mix
2
ΦΦTk
Φ
F
04
5B
2.0
2
mix
2
TkΦ
F
Φ
05.251loglog0mix
ΦΦΦ
Φ
F
(B) We calculate ΔFsep for formation of the coexisting compositions.
We have Φ0 = 0.2, Φ1 = 0.15 and Φ2 = 0.85. Inserting these values into ΔFmix gives
ΔFmix(Φ0) = -0.1004kBT and ΔFmix(Φ1) = ΔFmix(Φ2) = -0.104kBT.
If the molecular volume is 7 nm3 we have the free energy density as:
We can then use the formula for ΔF*:
2015-09-04 Soft Matter Physics 20
Exercise 3.2
0mix2mix
12
101mix
12
02sep ΦFΦF
ΦΦ
ΦΦΦF
ΦΦ
ΦΦF
TkTkF BBsep 0036.01004.0104.015.085.0
15.02.0104.0
15.085.0
2.085.0
323
27B27kJm 1.23001038.1
107
0036.0
107
0036.0
Tkf
J103101.2
1102
3
π161
3
π16 20
23
35
2
3
nuc
f
F
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2015-09-04 Soft Matter Physics 21
Spinodal Decomposition
We now move to the kinetics of liquid-liquid phase separation for unstable mixtures, i.e.
the region which has negative curvature for ΔFmix(Φ).
What we want is Φ(x, y, z, t) to
describe the phase separation process
in space and time.
We can first make a qualitative
argument: If the new phases are formed
as very small units here and there will
be a lot of interfacial tension. If they
are formed as large units separated by
long distances the diffusion of material
will take a very long time.
There should be a characteristic length
for the phase separation!
optimum length
slow diffusion
high interfacial energy
vo
lum
e fr
acti
on
(Φ
)
position (x)
Fick’s law of diffusion usually says that the flux J (e.g. molecules per area and time) is
proportional to the concentration gradient:
Here D is the diffusion constant (m2s-1) for a particular molecule in a particular
environment (depends on almost anything). During spinodal decomposition we have
uphill diffusion against the concentration gradient. We are separating the mixture! How?
2015-09-04 Soft Matter Physics 22
Uphill Diffusion
CDJ
JJ
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2015-09-04 Soft Matter Physics 23
Repetition: Chemical Potential
Strictly speaking it is the gradient in chemical potential which determines diffusive flux.
N is some measure of number of molecules. Boltzmann statistics makes it possible to
relate concentration to chemical potential:
Here μ° is the chemical potential at a standard state and C° is a standard concentration.
Note that we must work with relative concentrations with this formula, but Φ is indeed
dimensionless so:
PTVT N
G
N
F
,,
TkC
C
B
exp
ΦTk logB
2015-09-04 Soft Matter Physics 24
Rewriting Fick’s Law
Note that the gradient of chemical potential is:
Fick’s first law in terms of chemical potential should thus be:
Note that the concentration C = Φ/v where v is the molecular (lattice site) volume.
Here comes the trick: The chemical potential must per definition also be possible to write
as a derivative of free energy density (f) with respect to concentration:
Sidenote: In our binary incompressible mixture, if we add A we must also remove B.
Strictly speaking we have an exchange chemical potential: μ = μA – μB
zyxΦzyxΦ
TkzyxΦTk ,,
,,,,log B
B
Tvk
DΦΦ
v
D
v
ΦDCDJ
B
vΦ
f
C
zyxfzyx
,,,,
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2015-09-04 Soft Matter Physics 25
Continuity Equation
We still miss the time dependence! This we get from the continuity equation.
For a gradient in direction x and a control volume dxdydz the material in per unit time is
J(x)dydz and out J(x + dx)dydz. In the absence of chemical reactions one thus has:
If we take the gradient of our flux, which is based on chemical potential, we get:
x
J
x
xxJxJ
zyx
zyxxJzyxJ
t
C
d
d
ddd
ddddd
xx + dx
C(x)
J(x)J(x + dx)
dydz
Φ
fΦ
Tvk
Dv
Φ
f
Tvk
DΦ
Tvk
DΦ
t
C
BBB
Φ
fΦ
Tk
D
t
Φ
B
The derivation in Jones looks quite different but is
essentially the same…
2015-09-04 Soft Matter Physics 26
Model of Free Energy Density
So far things are quite logical. Let us summarize:
• Rewrite Fick’s law in terms of chemical potential.
• Define chemical potential in terms of free energy density and concentration.
• Get time dependence from continuity equation.
What we need now is basically an expression for the free energy density. One such
expression (no details here) is:
Here we have f0(Φ) which is the free energy density for a uniform mixture. The other
term with a coefficient κ introduces an energy penalty for gradients in concentration,
which essentially models interfacial energy.
So now we just insert this expression for f…
20 ΦΦff
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2015-09-04 Soft Matter Physics 27
Don’t Drink and Derive…
Looking at just one dimension we get something quite messy:
4
4
3
3
2
2
2
0
22
3
0
3
2
0
22
B
4
4
3
3
2
0
2
2
0
22
B
3
3
2
0
2
B
2
2
0
B
2
2
0
B
0
B
2
0
B
2
2
2
22
2
x
ΦΦ
x
Φ
x
Φ
x
Φ
Φ
f
x
Φ
Φ
fΦ
Φ
f
x
Φ
Tk
D
x
ΦΦ
x
Φ
x
Φ
x
Φ
Φ
f
xΦ
Φ
f
x
Φ
Tk
D
x
Φ
x
Φ
Φ
fΦ
xTk
D
x
Φ
Φ
f
xΦ
xTk
D
x
Φ
Φ
f
xΦ
xTk
D
x
Φ
Φx
Φ
Φ
f
xΦ
xTk
D
x
ΦΦf
ΦxΦ
xTk
D
t
Φ
2015-09-04 Soft Matter Physics 28
Cahn-Hilliard
By simplifying things a bit one can get an approximate description of the problem, the
Cahn-Hilliard equation (Jones 1D version):
Here M is a kind of transport coefficient. Note that without the fourth derivative term we
have Fick’s second law with an effective diffusion constant:
Note that the effective diffusion constant becomes negative (in an unstable mixture).
The derivatives of f0 and M are obviously concentration dependent. However, in order to
solve the equation one can assume that they are constant and get:
Here Φ0 is the initial state.
4
4
2
2
2
0
2
2x
ΦΦM
x
Φ
Φ
fΦM
t
Φ
2
0
2
effΦ
fMD
tq
Φ
fMqqxΦtxΦ 2
2
0
22
0 2expcosconstant,
negative
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2015-09-04 Soft Matter Physics 29
Amplification Wavevector
The cosine factor contains a certain wavevector q representing the characteristic
lengthscale of the phase separation. The exponential factor also contains q and amplifies
certain values of it by the factor in the exponential:
The function has a maximum for some q! So Cahn-Hilliard shows that a characteristic
lengthscale emerges.
2
2
0
22 2 qΦ
fMqqA
wavevector
amp
lifi
cati
on
maximum
amplification
3
2
0
2
82 qMΦ
fMq
q
A
2/1
2
0
2
max4
1
Φ
fq
2
2
0
2
max8
3
Φ
fMA
2015-09-04 Soft Matter Physics 30
Simulation of Phase Separation
Simulation of phase separation in 2D with Cahn Hilliard. Pattern is not entirely random!
Wikipedia: Cahn-Hilliard Equation
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2015-09-04 Soft Matter Physics 31
Imaging Spinodal Composition
Experiments with biopolymers. During different stages of phase separation, one liquid
phase solidifies (gelation) so that the pattern becomes fixed and can be imaged.
Wassen et al.
Soft Matter 2014
from low (a) to high
(e) temperature
from slow (a) to fast
(e) “freezing”
2015-09-04 Soft Matter Physics 32
Limits of Cahn-Hilliard
Initial phase: The Cahn-Hilliard solution shows a sinusoidal function, which means the
“borders” between the domains are as wide as the domains themselves.
For phase separation by nucleation the frequency of nucleation events determines the
initial domain sizes instead. (Cahn-Hilliard is not applicable.)
Intermediate phase: Eventually the borders become sharper as clearer interfaces between
coexisting compositions are formed. The domains are growing (coarsening). This is not
predicted by Cahn-Hilliard but must happen for full phase separation!
vo
lum
e fr
acti
on
(Φ
)
position (x)
coexisting
coexistinginitial
coarsened
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2015-09-04 Soft Matter Physics 33
Late Stage Growth
In the late stage of growth, smaller domains tend to be destroyed while only the larger
ones grow. This is due to a diffusive flux generated by the increased local concentration
around surfaces of higher curvature, i.e. smaller particles!
position (x)
vo
lum
e fr
acti
on
(Φ
)
J
The domain size grows
proportional to t1/3 if
diffusion is rate
limiting rather than
release of material from
the particle.
Jones gives a more
detailed explanation…
2015-09-04 Soft Matter Physics 34
Ostwald Ripening
The phenomenon where larger particles grow and smaller disappear is generally referred
to as Ostwald ripening. It occurs for emulsions but also for crystals.
Ostwald ripening explains the crunchyness of ice cream that has been in the freezer too
long. (After infinite time you get a single huge ice block separated from the cream!)
bulk (stable)
surface (less stable) pointy surface (much less stable)
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2015-09-04 Soft Matter Physics 35
Video: Supersaturated Solution
2015-09-04 Soft Matter Physics 36
Video: Mentos + Coke
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2015-09-04 Soft Matter Physics 37
Reflections and Questions
?
Derive the expression for the activation energy of heterogeneous nucleation in freezing
(see earlier in this lecture).
→
2015-09-04 Soft Matter Physics 38
Exercise 2.3
4
cos2cos1
3
π162
22
m
2
2
m
3
SL
TH
TG