Chemistry • Year 11 • Module 1 • Lesson 11
Polymers: Structure and Properties
Build HSC Band 5–6 extended-response technique: integrate polymerisation mechanisms, structural analysis, IMF reasoning, and real material data into sustained scientific argument.
1. Data + scenario: selecting a polymer for a medical suture (Band 5–6)
8 marks Band 5–6
Scenario. A biomedical engineer is selecting a polymer for an absorbable surgical suture — a stitch that must hold a wound closed for 4–6 weeks then break down inside the body. Two candidate polymers are under evaluation. Polymer X is a condensation polymer derived from lactic acid monomers (each monomer has two functional groups: —OH and —COOH). It has an average chain length of n = 3,000 and a melting point of 155°C. Polymer Y is an addition polymer made from a vinyl monomer with a non-polar substituent; chain length n = 5,000; melting point 130°C; insoluble in body fluids; estimated to persist in the body for over 50 years.
| Property | Polymer X | Polymer Y |
|---|---|---|
| Polymerisation type | Condensation | Addition |
| Monomer functional groups | —OH and —COOH (bifunctional) | C=C double bond |
| Backbone linkage | Ester (—COO—) | C–C single bonds only |
| Melting point (°C) | 155 | 130 |
| Degradability in body | Hydrolysed by body enzymes in 4–8 weeks | Not degraded; >50 years persistence |
| Tensile strength at n=3000–5000 | ~55 MPa | ~90 MPa |
Illustrative data based on properties of poly(lactic acid) and polyethylene. Not intended for clinical use.
Q1. Evaluate which polymer is more suitable for an absorbable surgical suture. In your response you must:
- Identify the polymerisation type of Polymer X and explain how the ester linkage in its backbone makes it susceptible to enzymatic hydrolysis (breakdown in body fluids).
- Explain why Polymer Y’s backbone structure makes it resistant to degradation in the body, referring to bond types.
- Compare the two polymers on at least three criteria relevant to suture performance (e.g. strength, degradability, melting point, biocompatibility).
- Use the data table to support your comparison with specific values.
- Reach a justified conclusion about which polymer should be selected and identify one limitation of your conclusion.
2. Design an experiment — effect of chain length on tensile strength
7 marks Band 4–6
Research question: Does increasing the degree of polymerisation (chain length) of polyethylene increase its tensile strength, and if so, is the relationship linear?
Constraints: You have access to five batches of polyethylene with different average chain lengths (average n = 1,000; 5,000; 20,000; 80,000; 150,000). Each batch can be moulded into flat test strips 10 cm × 1 cm × 0.2 cm. You have a materials testing machine that can apply a known stretching force and record the force at which the strip breaks. You may request any additional equipment at no cost.
(a) State the independent variable, dependent variable, and three variables that must be controlled. (4 marks)
Independent variable:
Dependent variable:
Controlled variables (three):
(b) Describe how you would use your data to determine whether the relationship between chain length and tensile strength is linear or non-linear. What would a non-linear result tell you about the IMF changes at high chain lengths? (2 marks)
(c) State one limitation of this experimental design that could affect the reliability of results, and suggest how it could be addressed. (1 mark)
3. Source critique — a flawed claim about polymer recycling
6 marks Band 5
“All polymers can be recycled by melting them down and reshaping them. The challenge of polymer recycling is simply one of collection and sorting infrastructure, not chemistry. Once we solve the logistics, every polymer — from nylon ropes to epoxy resin boat hulls to polythene bags — can be remelted and made into something new.”
— Adapted from a hypothetical opinion article in a general-science publication.
Q3. Identify the scientific flaw in this statement, explain the correct chemistry, and describe how you would demonstrate the flaw experimentally.
- Identify the specific claim that is scientifically incorrect and name the polymer class that contradicts it.
- Explain, using structural chemistry, why that polymer class cannot be remelted and remoulded.
- Describe a simple experiment that would distinguish a recyclable polymer from a non-recyclable one.
Q1 — Marking criteria (8 marks)
Polymerisation type and hydrolysis (2 marks): Polymer X is a condensation polymer (1 mark). Its backbone contains ester linkages (—COO—) formed between the —OH and —COOH groups of the bifunctional lactic-acid monomers. Ester bonds are susceptible to hydrolysis — enzymes in body fluids (esterases) catalyse the reaction of water with the ester bond, breaking the chain at each linkage and progressively degrading the polymer into smaller fragments and ultimately lactic acid, which can be safely metabolised (1 mark).
Polymer Y backbone (1 mark): Polymer Y is an addition polymer; its backbone consists entirely of C–C single bonds (1 mark). Carbon–carbon single bonds in a non-polar chain are not susceptible to enzymatic hydrolysis because there is no polar bond or functional group for water or enzymes to attack. Hence Polymer Y persists indefinitely in the body.
Comparison on three criteria (3 marks, 1 per criterion):
- Degradability: Polymer X degrades in 4–8 weeks (required for absorbable suture); Polymer Y persists >50 years (unsuitable).
- Tensile strength: Polymer Y is stronger (90 MPa vs 55 MPa), but 55 MPa is typically sufficient to hold most wound types together for 4–6 weeks. If strength is marginal, chain length could be increased for Polymer X.
- Melting point: Both polymers have MPs well above body temperature (155°C and 130°C), so neither will soften in the body. Sterilisation at 121°C (autoclave) is feasible for Polymer X; Polymer Y also survives autoclave temperatures.
Justified conclusion and limitation (2 marks): Polymer X should be selected because it meets the essential criterion of being degradable within the required timeframe (4–8 weeks), whereas Polymer Y would require surgical removal or cause a chronic foreign-body response (1 mark). One limitation: the data do not include in-vivo degradation rates under actual physiological conditions (pH, enzyme concentration, temperature fluctuation), which may differ from laboratory estimates; clinical trials would be required to confirm safety and efficacy (1 mark).
Q2 — Marking criteria (7 marks)
(a) Variables (4 marks):
Independent variable (1 mark): Average degree of polymerisation (chain length, n) of the polyethylene samples — five levels: 1,000; 5,000; 20,000; 80,000; 150,000.
Dependent variable (1 mark): Tensile strength (force per unit cross-sectional area at break), measured in MPa.
Controlled variables (2 marks, 1 mark for any two of):
- Dimensions of each test strip (length, width, thickness — e.g. all strips 10 cm × 1 cm × 0.2 cm).
- Degree of branching / structural isomerism (all samples must be linear polyethylene with negligible branching).
- Temperature at which testing occurs (room temperature, e.g. 22°C; temperature affects chain mobility).
- Rate of applied force / crosshead speed of the testing machine (consistent loading rate).
(b) Linear vs non-linear (2 marks): Plot tensile strength (y-axis) against n (x-axis) and assess whether the points follow a straight line through the origin or a curve. A non-linear (curved/plateauing) result would indicate that the increase in total dispersion force per additional monomer unit diminishes at high chain lengths — consistent with chains becoming so long that new surface contact area added per extra monomer is negligible compared to already extensive entanglement (1 mark per valid point).
(c) Limitation (1 mark): Any one of: only five data points makes it difficult to distinguish linear from non-linear trend precisely — more intermediate values of n would improve resolution; or the strip moulding process may introduce variable degrees of crystallinity across samples affecting strength; or a single test per strip does not account for variation in individual samples (need multiple repeats per n value for reliability).
Q3 — Marking criteria (6 marks)
Identify the flaw (1 mark): The claim that “every polymer can be remelted and reshaped” is incorrect. Thermosetting polymers (e.g. epoxy resin, bakelite, vulcanised rubber) cannot be remelted or remoulded regardless of logistics.
Explain the chemistry (2 marks): Thermosetting polymers contain covalent cross-links between polymer chains (1 mark). These cross-links are covalent bonds, not intermolecular forces; they cannot be broken by heating. Heating a thermosetting polymer does not allow chains to slide past each other — instead, the material chars or decomposes at high temperatures (1 mark). This is fundamentally different from thermoplastics (e.g. polyethylene, nylon), where chains are held only by intermolecular forces that are overcome by heating, allowing the material to flow and be remoulded.
Experimental distinction (2 marks): Take two small samples — one thermoplastic (e.g. PE) and one thermoset (e.g. epoxy resin) (1 mark). Heat both samples with a heat gun or in an oven at approximately 200°C. The thermoplastic sample will soften and flow/deform; the thermoset sample will remain rigid and eventually char if heated further, but will not soften or flow. This observable difference confirms that thermosetting polymers cannot be recycled by remelting (1 mark).
Specific example named (1 mark): Epoxy resin boat hulls (mentioned in the quote) are thermosetting — the epoxy cross-links formed during curing are permanent covalent bonds; once set, the material cannot be remelted under any practical conditions.