KE-100.3410 Polymer properties
Exercise 3:
DSC, NMR, IR
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 is 209 J/g.
Avrami equation:

X t   1  exp  kt n
DSC curve for melting of iPP
26
Endothermal heat flow W/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?

Tm=166,2oC
24
Hf=92,95 J/g
22
Heating
10oC/min -->
20
18
100
120
140
T/oC
160
Figure 1. DSC curve of second heating scan after 10 min crystallization
at 121 oC.
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
180
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
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
Use the measurements to formulate a correlation in the form A(c,t) =
where Ai is the area of each absorbance peak and ai is the absorptivity
(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.
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
Exercise 3.3
to 841 cm-1 absorbance peak and similarly A841 is the area of the
Syndiotactic polystyrene (sPS) crystallizes in various crystal forms
amorphous fraction of a partially -crystalline sample that corresponds
() depending on crystallization conditions. The formation of -
to 841 cm-1 absorbance peak.
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 cm1
).
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
-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