Chemistry • Year 11 • Module 1 • Lesson 20

Module 1 Synthesis and Review

Build HSC advanced extended-response technique by integrating all four Module 1 inquiry questions in multi-criteria evaluation and experimental design tasks.

Master · Extended Response

1. Data + scenario: comparing four materials for an electrode application (Evaluate)

8 marks Evaluate

Scenario. A chemical engineer at a chlor-alkali plant (where brine is electrolysed to produce chlorine gas, sodium hydroxide and hydrogen) needs to select an electrode material for the cell. The electrode must: (i) conduct electricity, (ii) withstand temperatures up to 300 °C without melting, (iii) resist chemical attack by chlorine gas and sodium hydroxide, and (iv) be mechanically strong enough to maintain its shape under pressure. The table below summarises data for four candidate materials.

Material	Melting point (°C)	Electrical conductivity	Chemical resistance (Cl₂, NaOH)	Mechanical strength
Graphite (C, layered)	>3600 (sublimes)	High (in plane)	Moderate, slowly oxidised by Cl₂	Low tensile strength; brittle
Titanium metal (Ti)	1668	Moderate	Excellent, forms protective TiO₂ layer	High
Silicon dioxide (SiO₂, quartz)	1710	Very poor (insulator)	Good, inert to most acids and bases	High compressive; very brittle
Iron (Fe)	1538	High	Poor, corrodes rapidly in Cl₂ and NaOH	High

Illustrative data. Real chlor-alkali cells typically use dimensionally stable anodes (DSA) coated with RuO₂/TiO₂ on a Ti substrate.

Q1. Evaluate the four materials above as candidate electrodes for the chlor-alkali cell. In your response you must:

For each material, explain whether it satisfies or fails each of the four requirements (i)–(iv), using Module 1 IQ4 bonding and structure reasoning.
Recommend which single material best satisfies the four requirements overall, justifying your choice with evidence from the data table.
Explain why SiO₂ fails requirement (i), linking to its electron structure using IQ2 reasoning.
Explain why the metallic bonding model accounts for why titanium satisfies requirements (i), (ii), and (iv).
State one limitation of your recommended material and suggest one way it could be improved for industrial use.

Stuck? Plan: systematically evaluate each material vs. each criterion (table approach works well) → identify Ti as best (conductors that also resist corrosion) → link SiO₂ failure to covalent network + no free electrons (IQ2 electron config) → metallic bonding: delocalised electrons (conductivity), strong lattice (high MP, mechanical strength) → Ti’s limitation: oxidises to TiO₂ surface layer, reducing conductivity; improvement: coat with RuO₂.

2. Experimental design, investigating the effect of molecular size on boiling point in alkanes (Evaluate)

7 marks Evaluate

Research question. A student claims: “The boiling point of a straight-chain alkane is determined only by whether it is polar or non-polar, not by its molecular size.” Design a scientific investigation to test whether molecular size (number of carbon atoms) affects the boiling point of non-polar covalent molecular substances, using straight-chain alkanes.

Available data (safety): gaseous and volatile alkanes such as methane and butane, and boiling flammable liquids such as octane, dodecane and hexadecane, are not heated by students in a school laboratory. Instead you are provided with a reliable published dataset of boiling points for the straight-chain alkane homologous series, methane (CH₄), butane (C₄H₁₀), octane (C₈H₁₈), dodecane (C₁₂H₂₆) and hexadecane (C₁₆H₃₄), with their carbon numbers, plus graphing tools (a spreadsheet or data logger) to analyse the trend.

Q2. Design the investigation and present it using the format below.

State a hypothesis (a testable prediction) that addresses the student’s claim, including independent and dependent variables.
Identify the independent variable, dependent variable, and at least two controlled variables.
Describe, in at least four numbered steps, how you will analyse the provided boiling-point dataset (tabulate, plot BP vs carbon number, determine the trend), since these alkanes are not heated by students.
Explain what result would support the student’s claim and what would falsify it.
Link your expected results to Module 1 IQ4 reasoning about dispersion forces and molecular size.
State two limitations of your design and one way to improve reliability.

Stuck? Hypothesis: if the student’s claim is wrong, BP should increase with carbon chain length. IV = number of carbon atoms (molecular size/molar mass). DV = boiling point (°C). IQ4 link: longer alkane → more electrons → stronger dispersion forces → more energy to separate molecules → higher BP. Falsification: if BP stays constant across all alkanes, student’s claim is supported.

Answers, Do not peek before attempting

Q1, Sample Band 6 response (8 marks), annotated

Evaluation of each material against four criteria:

Graphite: (i) Satisfies, high in-plane conductivity due to delocalised π electrons; (ii) Satisfies, very high sublimation point, no melting issue; (iii) Partially fails, slowly oxidised by Cl₂, limiting electrode lifetime; (iv) Fails, brittle with low tensile strength (layers slide easily). [1, correct analysis of at least two criteria for graphite]

Titanium: (i) Satisfies, metallic bonding provides delocalised electrons for conductivity; (ii) Satisfies, MP 1668 °C well above 300 °C; (iii) Satisfies, forms protective TiO₂ passivation layer preventing further corrosion; (iv) Satisfies, metallic lattice provides high mechanical strength. [1, all four criteria correctly applied to Ti]

SiO₂: (i) Fails, covalent network insulator; (ii) Satisfies; (iii) Satisfies; (iv) Fails (brittle). [1, correct classification of at least two]

Iron: (i) Satisfies; (ii) Satisfies; (iii) Fails critically, rapid corrosion; (iv) Satisfies. [1, identifies critical failure]

Recommendation: Titanium best satisfies all four requirements. It is the only material that simultaneously conducts, withstands 300 °C, resists both Cl₂ and NaOH through passivation, and has high mechanical strength. Graphite fails on mechanical strength; SiO₂ fails on conductivity; iron fails on chemical resistance. [1, justified recommendation with comparison to at least two alternatives]

SiO₂ failure explained via IQ2: SiO₂ is a covalent network solid. Si (Z=14, [Ne]3s²3p²) uses all four valence electrons forming four Si–O covalent bonds throughout the 3D network. O (Z=8, [He]2s²2p&sup4;) uses both available bonding electrons. No electrons are left unbound or delocalised, every electron is localised in a covalent bond. Without mobile charge carriers, SiO₂ cannot conduct electricity. [1, IQ2 electron configuration reasoning correctly applied]

Metallic bonding and titanium requirements (i), (ii), (iv): In Ti, positively charged Ti cations occupy a regular lattice surrounded by a ‘sea’ of delocalised electrons. (i) Conductivity: the mobile electrons carry electrical charge under an applied potential. (ii) High MP: the strong electrostatic attraction between the cation lattice and delocalised electrons requires a large amount of energy to overcome; MP 1668 °C is far above the operating temperature. (iv) Mechanical strength: the non-directional nature of metallic bonding means layers of cations can shift without bond rupture, and the strength of the lattice gives high tensile strength. [1, metallic bonding model applied to all three criteria]

Limitation and improvement: Ti develops a TiO₂ surface oxide layer which, while protective, slightly reduces electrical conductivity over time, increasing cell resistance [1]. Improvement: coat the Ti substrate with a thin layer of ruthenium dioxide (RuO₂), which is highly conductive and chemically stable, this is the basis of the dimensionally stable anode (DSA) used industrially [1].

Marking criteria (8 marks): 1 = correct analysis of at least two criteria for graphite; 1 = all four criteria applied correctly to Ti; 1 = identifies correct failures for SiO₂ and Fe; 1 = justified recommendation comparing at least two alternatives; 1 = IQ2 electron config reasoning for SiO₂; 1 = metallic bonding model applied to all three Ti criteria; 1 = valid limitation with reasoning; 1 = specific improvement (accept any chemically valid coating/modification that improves conductivity or corrosion resistance).

Q2, Sample Band 6 response (7 marks), annotated

Hypothesis: If molecular size (number of carbon atoms) increases the strength of dispersion forces, then the boiling point of straight-chain alkanes will increase with increasing carbon chain length, contradicting the student’s claim that polarity is the only factor. Independent variable: number of carbon atoms per alkane molecule (5 levels: C1, C4, C8, C12, C16). Dependent variable: boiling point (°C). Controlled variables: (1) straight-chain (unbranched) alkane structure, all molecules must be n-alkanes; (2) atmospheric pressure at 1.0 atm during all measurements; (3) heating rate, measurement method, and apparatus type kept constant. [1, testable hypothesis with IV, DV, and both controlled variables]

Procedure (secondary-data analysis, no heating of alkanes): (1) Tabulate the provided boiling points against carbon number for the five n-alkanes (CH₄, C₄H₁₀, C₈H₁₈, C₁₂H₂₆, C₁₆H₃₄). (2) Check the data are comparable (all straight-chain alkanes, boiling points quoted at the same pressure, 1.0 atm). (3) Plot boiling point (°C) on the y-axis vs number of carbon atoms on the x-axis. (4) Describe the trend and calculate the approximate increase in boiling point per added carbon (per CH₂ unit) to quantify it. [1, four valid steps analysing the dataset]

What supports the student’s claim: If the boiling points of all five alkanes are approximately equal or show no systematic trend with carbon number, the student’s claim is supported (polarity may be the dominant factor). [1, correct identification of supporting result]

What falsifies the student’s claim: If boiling point increases systematically with carbon chain length (CH₄ < C₄H₁₀ < C₈H₁₈ < C₁₂H₂₆ < C₁₆H₃₄), the student’s claim is falsified, molecular size, not polarity, explains the trend within non-polar alkanes. [1, correct falsification criterion]

IQ4 link to expected results: All straight-chain alkanes are non-polar covalent molecular substances. The only IMFs between molecules are dispersion forces. Dispersion force strength increases with molecular size because larger molecules have more electrons, creating stronger and more frequent temporary dipoles. More energy is therefore required to separate larger molecules, giving a higher boiling point. Expected result: boiling point increases linearly (approximately) with carbon number. [1, dispersion force reasoning correctly linking molecular size to boiling point]

Limitations: (1) The analysis relies on the accuracy and comparability of the published dataset; values must be quoted at the same pressure and for pure straight-chain isomers, or the trend is distorted [1]. (2) Boiling point depends on more than carbon number alone, branching reduces dispersion-force strength, so the conclusion holds only for straight-chain (unbranched) alkanes [1].

Improvement: Repeat each alkane measurement three times, report mean ± standard deviation, and check measured values against published boiling points (literature values) to confirm instrument calibration. [1]

Marking criteria (7 marks): 1 = testable hypothesis naming IV, DV, and both controlled variables; 1 = four clear procedural steps with safe measurement protocol; 1 = correctly identifies the result that supports the student’s claim; 1 = correctly identifies the result that falsifies the claim; 1 = IQ4 dispersion force reasoning linking molecular size to BP; 1 = one valid, explained limitation; 1 = specific improvement with justification (can include second limitation if stated).