Polymers
Level 2 (Recall & Standard Problems)
Time: 30 minutes | Total Marks: 40
Answer all questions. Use for chemical/mathematical expressions where needed.
Q1. Define the terms addition polymer and condensation polymer, giving one example of each. (4 marks)
Q2. Distinguish between thermoplastic and thermosetting polymers, giving one example of each. (4 marks)
Q3. Write the monomer(s) required for the preparation of the following polymers: (4 marks) (a) PVC (b) Teflon (PTFE) (c) Nylon-6,6 (d) Terylene (PET)
Q4. Name the type of polymerization mechanism involved in each of the following: (4 marks) (a) Polymerization of ethene using Ziegler–Natta catalyst (b) Formation of nylon-6,6 from adipic acid and hexamethylenediamine (c) Polymerization of styrene initiated by benzoyl peroxide (d) Polymerization of isobutylene with
Q5. Define number-average molecular weight and weight-average molecular weight . Write the expression for polydispersity index (PDI) and state its value for a monodisperse polymer. (5 marks)
Q6. A polymer sample contains 3 molecules of molar mass and 2 molecules of molar mass . Calculate , and the PDI. (6 marks)
Q7. Give one important use of each of the following aerospace/high-performance polymers: (4 marks) (a) Kevlar (b) Nomex (c) Epoxy resin (d) Polyimide
Q8. What are biodegradable polymers? Name two examples and state the monomer units of any one of them. (4 marks)
Q9. Classify the following as natural or synthetic polymers: (3 marks) (a) Cellulose (b) Nylon-6 (c) Rubber (natural)
Q10. Explain why bakelite is a thermosetting polymer while polyethene is a thermoplastic. (2 marks)
End of paper
Answer keyMark scheme & solutions
Q1. (4 marks)
- Addition polymer: formed by repeated addition of monomers (usually unsaturated, containing C=C) without loss of any small molecule. (1.5) Example: polythene (from ethene). (0.5)
- Condensation polymer: formed by repeated condensation between bi-/poly-functional monomers, usually with elimination of a small molecule like or . (1.5) Example: nylon-6,6. (0.5) Why: the defining criterion is loss/no-loss of a small molecule.
Q2. (4 marks)
- Thermoplastic: softens on heating and hardens on cooling; can be remoulded; has linear/branched chains held by weak forces. (1.5) Example: polythene / PVC. (0.5)
- Thermosetting: undergoes permanent hardening (cross-linking) on heating; cannot be remelted/remoulded. (1.5) Example: bakelite / melamine. (0.5)
Q3. (4 marks, 1 each) (a) PVC → vinyl chloride, (b) Teflon → tetrafluoroethene, (c) Nylon-6,6 → adipic acid + hexamethylenediamine (d) Terylene → ethylene glycol + terephthalic acid
Q4. (4 marks, 1 each) (a) Coordination (Ziegler–Natta) polymerization (b) Step-growth (condensation) polymerization (c) Free-radical polymerization (d) Cationic polymerization Why: generates carbocation; peroxide generates free radicals; diacid+diamine = step growth.
Q5. (5 marks)
- — averaged over number of molecules. (2)
- — averaged weighting by mass. (2)
- ; for a monodisperse polymer . (1)
Q6. (6 marks) Data: ; .
(2)
(3)
(1)
Q7. (4 marks, 1 each) (a) Kevlar → body armour / bullet-proof vests (also parachutes, tyres) (b) Nomex → fire-resistant clothing / thermal protection (c) Epoxy resin → adhesives / composite matrix (CFRP/GFRP) (d) Polyimide → high-temperature insulation / aerospace components
Q8. (4 marks)
- Biodegradable polymers: polymers that are decomposed by microorganisms/enzymes into simpler substances (e.g. ) in a reasonable time. (2)
- Examples: PHBV and nylon-2-nylon-6. (1)
- PHBV monomers: 3-hydroxybutanoic acid and 3-hydroxypentanoic (valeric) acid. (1)
Q9. (3 marks, 1 each) (a) Cellulose — natural (b) Nylon-6 — synthetic (c) Rubber (natural) — natural
Q10. (2 marks)
- Bakelite is heavily cross-linked (3-D network from phenol + formaldehyde) so it cannot be softened/remoulded on heating → thermosetting. (1)
- Polyethene has linear chains with only weak van der Waals forces, which loosen on heating allowing softening/remoulding → thermoplastic. (1)
[
{"claim":"Mn = 14000 g/mol","code":"N=[3,2]; M=[10000,20000]; Mn=sum(N[i]*M[i] for i in range(2))/sum(N); result = Mn==14000"},
{"claim":"Mw = 15714.2857... g/mol","code":"N=[3,2]; M=[10000,20000]; Mw=sum(N[i]*M[i]**2 for i in range(2))/sum(N[i]*M[i] for i in range(2)); result = abs(Mw-15714.2857142857)<1e-3"},
{"claim":"PDI ~ 1.122","code":"N=[3,2]; M=[10000,20000]; Mn=sum(N[i]*M[i] for i in range(2))/sum(N); Mw=sum(N[i]*M[i]**2 for i in range(2))/sum(N[i]*M[i] for i in range(2)); PDI=Mw/Mn; result = abs(PDI-1.1224489795918)<1e-4"},
{"claim":"PDI of monodisperse = 1 when Mw=Mn","code":"Mn=14000; Mw=14000; result = (Mw/Mn)==1"}
]