The following table gives some reactivity ratios for various comonomer pairs. Fo
ID: 482759 • Letter: T
Question
The following table gives some reactivity ratios for various comonomer pairs. For each pair, calculate:
a) the composition of the copolymer formed at low conversions of an equimolar mixture of the two monomers. Show an example calculation. Note: I have given you the answer to the first pair so that you can check your approach.
b) the comonomer composition required to form a copolymer consisting of 50 mol.% styrene repeat units. Show an example calculation. Note: I have given you the answer to the first pair so that you can check your approach.
Monomer 1 Monomer 2 Equimolar Composition ri composition or 50 mol% styrene (b Styrene Butadiene 0.78 1.39 42.7 mol 57.2 mol Styrene Styrene 0.52 046 Styrene Methyl methacrylate Styrene Acrylonitrile 0.40 0.04 Styrene Vinyl chloride 17 0.02 DescriptionExplanation / Answer
F1 = [r1 f12 + f1 f2] / [r1 f12 + 2 f1 f2 + r2 f22]
where F1 = Mole fraction of monomer 1 in copolymer
f1 = Mole fraction of monomer 1 in feed = 0.5 (as equimolar composition)
f2 = Mole fraction of monomer 2 in feed = 0.5 (as equimolar composition)
r1 = Reactivity ratio of monomer 1
r2 = Reactivity ratio of monomer 2
1. Styrene and Butadiene
r1 = 0.78, r2 = 1.39
F1 = [r1 f12 + f1 f2] / [r1 f12 + 2 f1 f2 + r2 f22]
= [(0.78) (0.5)2 + (0.5) (0.5)] / [(0.78) (0.5)2 + (2) (0.5) (0.5) + (1.39) (0.5)2]
= 0.445 / 1.0425 = 0.427 = 0.427 x 100 = 42.7%
Equimolar composition = 42.7%
2. Styrene and Methyl methacrylate
r1 = 0.52, r2 = 0.46
F1 = [r1 f12 + f1 f2] / [r1 f12 + 2 f1 f2 + r2 f22]
= [(0.52) (0.5)2 + (0.5) (0.5)] / [(0.52) (0.5)2 + (2) (0.5) (0.5) + (0.46) (0.5)2]
= 0.38 / 0.745 = 0.510 = 0.510 x 100 = 51.0%
Equimolar composition = 51.0%
3. Styrene and Acrylonitrile
r1 = 0.40, r2 = 0.04
F1 = [r1 f12 + f1 f2] / [r1 f12 + 2 f1 f2 + r2 f22]
= [(0.40) (0.5)2 + (0.5) (0.5)] / [(0.40) (0.5)2 + (2) (0.5) (0.5) + (0.04) (0.5)2]
= 0.35 / 0.61 = 0.574 = 0.574 x 100 = 57.4%
Equimolar composition = 57.4%
4. Styrene and Vinyl chloride
r1 = 17, r2 = 0.02
F1 = [r1 f12 + f1 f2] / [r1 f12 + 2 f1 f2 + r2 f22]
= [(17) (0.5)2 + (0.5) (0.5)] / [(17) (0.5)2 + (2) (0.5) (0.5) + (0.02) (0.5)2]
= 4.50 / 4.755 = 0.946 = 0.946 x 100 = 94.6%
Equimolar composition = 94.6%
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.