1
KE-100.3410 Polymer properties
Exercise 3:
DSC, NMR, IR
The general Avrami equation is applicable to any type of crystallization.
X t  1  exp( kt n )
where k and n are constants typical of the nucleation and growth mechanisms.
Derived Avrami exponents for different nucleation and growth mechanisms (Sperling,
Introduction to physical polymer science, 4th ed., Wiley & Sons, 2006, pp. 277.):
Degree of crystallinity of a polymer can be calculated when melting enthalpy of polymer Hf
is measured with DSC and compared with the theoretical fully crystalline melting enthalpy
H 0f characteristic for each polymer:
X
H f
H 0f
For example for PP H 0f  209 J / g .
Exercise 3.1
The crystallinity of isotactic polypropylene was studied with DSC. Heating rate for melting
the sample was 10 oC/min up till 200 oC temperature. The sample was kept at 200 oC for five
minutes after which it was cooled to 121oC very quickly and kept there for 10 minutes.
Melting enthalpies (Hf) were used to obtain the values for relative degree of crystallinity Xt.
Calculate the degree of crystallinity for isotactic PP and determine the constants n and k for
Avrami equation. What can be said about geometry of the crystals based on the Avrami
constant n?
2
t (s)
0
20
60
120
180
240
300
420
600
Xt
0
0.0026
0.0383
0.1947
0.4454
0.6988
0.8753
0.9916
1.0000
Heat of fusion for fully crystalline
polypropylene 209 J/g.
Avrami equation:

X t   1  exp  kt n

DSC curve for melting of iPP
Endothermal heat flow W/g
26
Tm=166,2oC
24
Hf=92,95 J/g
22
Heating
10oC/min -->
20
18
100
120
140
T/oC
160
180
200
Figure 1. DSC curve of second heating scan after 10 min crystallization at 121 oC.
Solution 3.1
Degree of crystallinity can be calculated from the measured enthalpy data of isotactic PP Hf
and melting enthalpy of completely crystalline PP H 0f :
J
g
X

 0.44
0
J
H f
209
g
H f
92.95
Solving the constants from Avrami equation it needs to be manipulated into the form:
1  X t  exp(kt n )
 ln(1  X t )  kt n
 ln( ln(1  X t ))  ln k  n ln t
3
When ln(-ln(1-Xt)) is plotted as a function of ln t, constant n is obtained from the slope and ln
k is the cross point with y-axis.
t (s)
0
20
60
120
180
240
300
420
600
Xt
0
0.0026
0.0383
0.1947
0.4454
0.6988
0.8753
0.9916
0.9999
ln t
2.9957
4.0943
4.7875
5.1930
5.4806
5.7038
6.0403
6.3969
ln(-ln(1-Xt))
-5.9555
-3.2420
-1.5299
-0.5284
0.1822
0.7333
1.5644
2.2203
3
1
ln(-ln(1-Xt))
y = 2.4358x - 13.209
-1
-3
-5
-7
3
3.5
4
4.5
5
5.5
6
6.5
ln t
From the figure n = 2.44 and ln k =-13.209  k =1.8 10-6 .
Based on the Avrami constant n =2.44, the sample would contain two types of crystalline
structures, depending on nucleation at least disc-like crystallites.
Exercise 3.2
In quality control, analysis has to be quick and easily repeatable. A small amount of
antioxidant Irganox 1010 was added to PE in during production. The amount of the additive
was followed with UV spectroscopy. For the reference, a set of samples with the known
amount of antioxidant (0-0.1 mol-%) was measured:
t / cm
0.1
AO 0%
A
0.042
AO 0.01%
A
0.081
AO 0.05%
A
0.29
AO 0.1%
A
0.57
4
Use the measurements to formulate a correlation in the form A(c,t) = (kc+a)t, according to
Beer-Lambert law that can be used to determine the amount (c) of antioxidant (AO) in the PE
sample. The thickness (t) of the sample as well as the UV absorbance (A) is measured. Use
the equation then to determine the amount of antioxidant in a PE sample with a thickness of
0.4 cm and UV absorbance A = 0.61.
Solution 3.2
Beer Lambert: A  kct
Use the measured data to construct an equation following the Beer-Lambert law in the form:
A(c,t) = (kc+a)t
When there is no antioxidant in PE, cAO = 0, t = 0.1 cm, and A = 0.042, we get:
A(0, 0.1)  (k  0  a)  0.1cm  0.042  a 
0.042
 0.42cm 1
0.1cm
In order to get the constant k we must plot A/t as a function of c so that the starting point for
the plot is c = 0, A/t = 0.42 cm-1.
A/t
0.42
0.81
2.90
5.70
Determination of constants
6
y = 5205.6x + 0.42
R² = 0.9979
5
A/t (cm-1)
c
0
0.0001
0.0005
0.0010
4
3
2
1
0
0
0.0002
0.0004
0.0006
0.0008
0.001
c
The slope of the line gives as the constant k = 5206 cm-1. Thus the absorption equation gets
the form A(c,t) = (5206c+0.42)t.
Solve the equation for c and substitute the values:
5
A
0.61
a
 0.42cm1
t
0.4
cm
A(c, t )  (kc  a)t  c 

 0.00021
k
5206cm1
Thus the antioxidant content of the PE would be 0.021 mol-%.
Exercise 3.3
Syndiotactic polystyrene (sPS) crystallizes in various crystal forms () depending on
crystallization conditions. The formation of -form is kinetically favorable whereas the
formation of -form thermodynamically favorable. Fourier transform spectroscopy (FTIR) is
used for analysis of polymer crystallinity. In the FTIR spectrum of syndiotactic polystyrene,
characteristic absorbance peaks can be detected; for amorphous (841 cm-1), - (851 cm-1) and
-form (858 cm-1).
The proportion of - or -crystallinity (Xi) in syndiotactic polystyrene can be determined with
the following equations:
A851
a
X 
A
A
A841  851  858
a
a
A858
a
X 
A
A
A841  851  858
a
a
where Ai is the area of each absorbance peak and ai is the absorptivity ratio of each crystal
form:
a 
A851

A  A841
*
841
a 
A858

A  A841
*
841
A841* is the area of a fully amorphous sample, A841 is the area of the amorphous fraction of a
partially -crystalline sample that corresponds to 841 cm-1 absorbance peak and similarly
A841 is the area of the amorphous fraction of a partially -crystalline sample that corresponds
to 841 cm-1 absorbance peak.
6
The characteristic infrared band of neat
s-PS ranging from 940 to 820 cm−1: (a)
FTIR spectra of: (a) quenched; (b) coldcrystallized; (c) melt-crystallized s-PS
quenched s-PS; (b) melt-crystallized sPS at 240°C for 240 min; (c) sample in
thin film sample at 240°C for arbitrary
duration using the same s-PS sample.
(b) heating scan up to 264°C from 30°C
with 10°C/min, then quenched by liquid
nitrogen.
Wu, H.-D., Wu, S.-C., Wu, I.-D., Chang, F.-C., Novel determination
of the crystallinity of syndiotactic polystyrene using FTIR spectrum,
Polymer 42 (2001) 4719-25.
Three syndiotactic PS samples were analysed by FTIR to determine absorptivity ratios.
a)
Based on the measurement data calculate the averages for each absorptivity ratio (a,
a).
b)
Three samples of - and -forms were crystallized by using different isothermal
crystallization times 2-240 min. Calculate the crystallinity degrees of each sample
(Xor X) by using attained absorptivity ratios (a, a).
c)
Based on crystallinity degrees what can be said of the crystallization of syndiotactic
PS?
Sample
1
2
3
A841*
1.115
1.240
0.916
A841
0.706
0.842
0.642
A841
0.608
0.742
0.544



A851
0.072
0.071
0.050
A858
0.137
0.134
0.101
7
-crystalline samples:
Sample t (min)
1
2
2
10
3
240
A841
1.123
1.086
0.311
A851
0.153
0.225
0.077
A841
0.439
0.448
0.377
A858
0.043
0.066
0.068
-crystalline samples:
Sample t (min)
1
2
2
10
3
240
Solution 3.3
a)
Determine absorptivity ratios for both - and -forms in each sample. As an example,
absorptivity ratio was calculated for the first sample:
a ,1 
A851,1
A
0.072
0.137

 0.176 a ,1   858,1  
 0.270

A  A841,1 1.115  0.706
A841,1  A841,1 1.115  0.608

841,1
The other values are calculated in a similar manner and the average values are calculated as
well:
Näyte
1
2
3
A841*
1,115
1,240
0,916
A841
0,706
0,842
0,642
A841
0,608
0,742
0,544
A851
0,072
0,071
0,050
A858
0,137
0,134
0,101
ka
a
0,176
0,178
0,182
0,179
a
0,270
0,269
0,272
0,270
a= 0.179 ja a= 0.270.
b)
Taking into account that there is no -form crystalline structure in a sample of -form and
vice versa, the degree of crystallinity is calculated:
A851,1
0.153
a ,1
0.179
X  ,1 

 0.432
A851,1
0.153
1
.
123

A841,1 
0.179
a ,1
8
A858,1
0.043
a ,1
0.270
X  ,1 

 0.266
A858,1
0.043
0.439 
A841,1 
0.270
a ,1
The other samples of -form are calculated respectively:
Sample
1
2
3
t (min)
2
10
240
A841
1,123
1,086
0,311
A851
0,153
0,225
0,077
X
0,432
0,537
0,580
A841
0,439
0,448
0,377
A858
0,043
0,066
0,068
X
0,266
0,353
0,400
and samples of -form:
Sample
1
2
3
t (min)
2
10
240
From the result can be seen that crystallinity increases with extending isothermal
crystallization period in both crystalline forms. The crystallinity of samples of -form is
clearly bigger than the samples of -form. Therefore the crystallization rate of -form is
faster. As it was already mentioned the formation of -form is kinetically favorable whereas
the formation of -form thermodynamically favorable.