STRUCTURAL HYERARCHY IN SEMICRYSTALLINE POLYMERS
atomic scale - crystal unit cell
-several Angstroms (<1 nm to a few nm)
- detailed crystal structure, conformation of individual bonds - wide-angle XRD
scale of crystal thickness -100-500 Å (10 - 50 nm)
- electron microscopy, small-angle XRD scale of crystal aggregates (spherulites etc.)
-μm scale - optical microscopy
Semicrystalline Polymers
detector
2θ diffracted
PHILIPS X-RAY DIFFRACTOMETER
Proportional counter Graphite single crystal monochromator Sample
holder X-Ray tube
Local radiation enclosure Divergence
Receiving slit
2θ
Semicrystalline Polymers
diffracted intensity
2θ
crystalline
diffracted intensity
2θ
amorphous
diffracted intensity
2θ
semicrystalline detector
2θ diffracted
Degree of crystallinity
a c
c a c
c
I I
I v v X v
= +
= +
X X
X v v
v v
a c a a
c a c
a a c c
) ( ) 1
( ρ ρ ρ
ρ ρ
ρ ρ ρ
− +
=
− +
= +
= +
density:
crystallinity:
polyethylene: ρc= 1.00 gcm-3, ρ= 0.85 gcm-3
Problem: Crystallinities of HDPE and LDPE are typically 0.5 and 0.8.
Calculate their densities.
Determination of X. Primary method: XRD
Secondary: DSC, density, (NMR, IR)
How are the two phases arranged?
“Fringed micelle” model
Polymer Single Crystals
9.5 nm
Polymer Single Crystals
polyethylene crystal showing screw dislocations (transmission electron microscopy - TEM)
A multilayer crystal of polyethylene (TEM)
screw dislocation
Structure of lamellar polymer crystals
• lamellar crystals thickness 10 nm (typically from solution)
• polymer chains normal to lamellar surface
• typical length of extended polymer chain:
1μm = 1000 nm
• What is the structure?
chain-folded lamellar crystal
Surface Decoration by Polymer Vapour Deposition:
Comparison of extended-chain n-alkane and folded-chain polyethylene single crystals
n-C36
H
74PE
Decorating PE lamellae crystallize epitaxially on top of chain folds
n-C36H74 polyethylene
What about melt-crystallizedpolymers?
• Also lamellar crystals?
• Small-angle X-ray scattering (SAXS) suggests so.
• Diffraction peak at small 2θcorresponds to d-spacing 20 – 50 nm.
= “long period”l Îstacking of lamellar crystals
Bragg equation relating diffraction angle θ to interplanar spacing d
θ
d θ
2dsinθ incident
diffracted
electron density
nλ= 2dsinθ Small d - large θ (wide angle XRD when layers are layers of atoms)
Large d - small θ (small-angle XRD when layers are thin crystals)
SAXS of oriented i-PP
Stacked melt-crystallized lamellae
X = lc/l• folds less regular
• amorphous phase more liquid-like
• tie-molecules – give mechanical integrity lc
la l
Atomic Force Micrograph (AFM) of melt-crystallized PE
- thin layer on graphite - lamellae edge-on - chains tilted
Lamellar structure in melt-crystallized PE
SEM of surface of PE from which low MW fraction was extracted
Why are polymers semicrystalline?
+ =
amorphous crystalline Binary alkane mixture
extended-chain alkane
once-folded alkane Monodisperse polymer (pure long chain alkane)
fully crystalline
chain length up to 390 C-atoms (M.W. = 5462)
Part of MAT215 and MAT6102 modules Prof. G. Ungar, © University of Sheffield
Polydisperse polymers – molecular weight distribution
7 5
3
2 1
0 1 2 3 4 5
0.3 6 7
10 9
8 5 7
1.8 weighte
d distribution
w (000)i
<Mn><Mw>
num ber distri
bution
< >=∑
M ∑
w n
n n
ii i
i i
< >=∑
M ∑
w n
w w n
i i
i ii i
2
P M
M
w n
=< >
< >
polydispersity:
P = 1.33
averages:
Polymer Crystallization
∑
Aσi i=minimumWhy thin chain-folded crystals?
σe
σ Why not the more stable extended-chain
crystals?
Most stable shapes of crystals are such that the total surfaces energy is minimised.
σ σe≅10
However, instead of being the smallest, the high energy surface (fold surface) is the largestin lamellar polymer crystals.
Minimum energycrystal In polymers:
• thick crystal – more stable (lower free energy) – less surface = less surface energy
–Æhigh thermodynamic driving forcefor crystallization DF = Fcrystal– Fmeltis highly negative
melt
crystal
free energy barrier F* = E*-TS*
driving force ΔF
Lamellar crystals are favoured bygrowth and nucleation.
Thinnest crystals grow fastest.
Crystals grow as thin as possible.
2 reasons for this:
1. if llarge - long chain segment must become straight
• all bonds must be in correct conformation (e.g. trans)
• probability low – chain extension is a rare event 2. there is no longitudinal growth
• chains deposit side-by-side
• growth is 2-dimensional
a l σe
Growth of σ
layer Growth of
crystal
Free energy barrier for chain attachment is mainly entropic
• Probability of 1 C-C bond being trans= 1/3
• Probability of 2 C-C bonds being trans= 1/3 × 1/3 = 1/32 = 1/9
• Probability of 100 C-C bonds being trans= 1/3100
• S* = k ln (1/3100) = - 100 ln3 k
• free energy barrier F* = (E) –TS* = kT × 100 ln3 F* = kT×L (L = crystal thickness)
• frequency of chain attachments Sexp(-F*/kT) = exp(-L)
melt crystal
F* = E*-TS*
melt ΔF
crystal
F* = E*-TS*
ΔF
thick thin
Rate
Barrier factor
Driving Force factor
kinetically favoured because barrier lower
How does lamellar thickness depend on T
c?
• For a given Tcthe minimum allowed lis lmin.
• lmin= thickness of crystals whoseTm= Tc.
• Growth rate of crystals with l = lminis zero.
(Driving force Gc-Gl= 0.)
• Hence:
l > lmin
• To find l(Tc) need findlmin(Tc).
• Find inverse: Tm(l)
MELTING TEMPERATURE OF LAMELLAR POLYMER CRYSTALS
Start from:
Change in Gibbs free energy dG = Vdp – SdT
At p=const all free energies decrease with T.
dG = -SdT
Slope = -S (entropy) Gc0= Gibbs free e. of ∞thick crystal Gl= Gibbs free e. of liquid (melt)
At the equilibrium melting pointTm0: Gc0=Gl
Free e. of thin crystal of thickness l (per unit volume):
Gc‘ = Gbulk+ Gsurface= GC0+ ΔG What is ΔG ?
ΔG = Aσe (A = surface area) (σe= fold surface free energy)
= (2/l) σe
Free e. of thin crystal of thickness l (per unit volume):
Gc‘ = Gbulk+ Gsurface= GC0+ ΔG What is ΔG ?
ΔG = Aσe (A = surface area) (σe= fold surface free energy)
= (2/l) σe
1
1 1
l
No. of layers = 1/l
Total area of layers (both surfaces) UNIT VOLUME
σe
Free e. of thin crystal of thickness l (per unit volume):
Gc‘ = Gbulk+ Gsurface= GC0+ ΔG What is ΔG ?
ΔG = Aσe (A = surface area) (σe= fold surface free energy)
= (2/l) σe
Now determine melting point depression ΔT = Tm0-Tm
Geometry: bc = ΔG ac = ΔT Sl
ab = ΔT Sc
bc = ac – ab
ΔG = ΔT Sl-ΔT Sc= ΔT (Sl-Sc) ΔG = ΔT ΔSf (ΔSf= entropy of fusion) Substitute ΔG:
(2/l)σe= ΔTΔSf
Thus melting point depression ΔTis:
since
where ΔHf= heat of fusion
f m e f e
H l 2 T S = l
=2
T Δ Δ
Δ σ σ
m f
f T
S ΔH
= Δ
• Melting point depression is inversely proportional to crystal thickness.
• Thin crystals: high free energy = low stability = low Tm f
m e m
m l H
2 T
= T T0− Δσ
• For a given supercooling ΔT the minimum allowed lis lmin.
• Hence:
l ≥lmin since:
Î
T H l T
f m e
Δ
=Δ2
σ
min
Îl∝1/ΔT
How does l depend on T
c?
f m e
H l 2 T
=
T Δ
Δ σ
Lamellar thickness l vs.
crystallization temperature for polyethylene
Crystallization rate vs. temperature
Close to Tmo:Only thick crystals can grow. Growth is slow.
Tm
Tg
T rate
With increasing ΔT - growth accelerates exponentially.
2 reasons:
1. Driving force ⏐Gc-Gl⏐increases.
2. Entropic barrier decreases since thinner crystals can grow.
At still lower temperatures, close to Tg, melt viscosity increases Îcrystallisation rate Æ0.
Typical crystallization rate vs. Tcfor polymers:
