1. Polymer Properties Exercise 3

2. 1. Crystallinity and DSC t (s)XtXt 00 200.0026 600.0383 1200.1947 1800.4454 2400.6988 3000.8753 4200.9916 6001.0000 Heat of fusion for fully crystalline polypropylene is 209 J/g. Avrami equation: Crystallinity of isotactic polypropylene was studied with DSC. Melting enthalpies (  H f ) were used to obtain the values for relative degree of crystallinity X t. 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?

3. 1) Degree of crystallinity can be calculated from the measured enthalpy data of isotactic PP  H f and melting enthalpy of completely crystalline PP: Solving the constants from Avrami equation it needs to be manipulated for the form:

4. 1) By plotting ln(-ln(1-X t )) as a function of ln t, constant n is obtained from the slope and ln k is the cross point with y-axis. From the figure n = 2.44 and t (s)XtXt ln tln(-ln(1-X t )) 00-- 200.00262.9957-5.9555 600.03834.0943-3.2420 1200.19474.7875-1.5299 1800.44545.1930-0.5284 2400.69885.48060.1822 3000.87535.70380.7333 4200.99166.04031.5644 6000.99996.39692.2203

5. 1) Avrami constant Derived Avrami exponents for different nucleation and growth mechanisms: 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.

6. 2. Characterization of a mixture In quality control the analysis has to be quick and easily repeatable. A small amount of antioxidant Irganox 1010 was added to PE in the 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: a)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 PE sample. Thickness (t) of the sample as well as the UV absorbance (A) is measured. b)Use the equation then to determine the amount of antioxidant in a PE sample with thickness of 0.4 cm and UV absorbance A = 0.61. AO 0%AO 0.01%AO 0.05%AO 0.1% t / cmAAAA 0.10.0420.0810.290.57

7. 2) Beer Lambert in general: 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, c AO = 0, t = 0.1 cm, and A = 0.042, we get: In order to get the constant k, A/t must be plotted as a function of c so that the starting point for the plot is c = 0, A/t = 0.42 cm -1. AO 0%AO 0.01%AO 0.05%AO 0.1% t / cmAAAA 0.10.0420.0810.290.57

8. 2) cA/t 00.42 0.00010.81 0.00052.90 0.00105.70 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: Thus the antioxidant content of the PE would be 0.021 mol-% with sample thickness 0.4 cm and UV absorbance 0.61.

9. Do you know, how to define the slope and y- cross point of a line without spreadsheet program or graph calculator?

10. 3. Crystallinity and IR 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. In the FTIR spectrum of syndiotactic polystyrene can be detected characteristic absorbance peaks; for amorphous (841 cm -1 ),  - (851 cm -1 ) and  -form (858 cm -1 ). Reminder: Atactic polystyrene is totally amorphous!

11. 3. IR spectrum of syndiotactic PS The characteristic infrared band of neat s-PS ranging from 940 to 820 cm −1 : a.quenched s-PS; b.melt-crystallized s-PS at 240°C for 240 min; c.sample in (b) heating scan up to 264°C from 30°C with 10°C/min, then quenched by liquid nitrogen.

12. 3a) Based on the measurement data calculate the averages for each absorptivity ratio (a , a  ). As an example, absorptivity ratio was calculated for the first sample: The other values are calculated in a similar manner and the average values are calculated as well: a  = 0.179 and a  = 0.270 SampleA 841 *A 841a A 841b A 851 A 858 aa aa 11.1150.7060.6080.0720.1370.1760.270 21.2400.8420.7420.0710.1340.1780.269 30.9160.6420.5440.0500.1010.1820.272 mean0.1790.270

13. 3b) Taking into account that there is no b-form crystalline structure in a sample of a-form and vice versa, the degree of crystallinity is calculated:

14. 3b,c) Samplet (min)A 841 A 851 XX 121.1230.1530.432 2101.0860.2250.537 32400.3110.0770.580 Based on crystallinity degrees what can be said of the crystallization of syndiotactic PS? 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. Samplet (min)A 841 A 858 XX 120.4390.0430.266 2100.4480.0660.353 32400.3770.0680.400 Crystallinity of  -form samples Crystallinity of  -form samples