Thermochemical principles and the properties of particles and substances

Practice

Recall it first, then use it — testing yourself builds the memory, where rereading the notes above only feels easier.

Recall and application shuffled together — you have to work out what each item wants before you can answer it, which is the part the exam actually tests. Both kinds feed the same spacing schedule, so anything you miss comes back sooner.

Say the answer out loud before flipping — that attempt is where the learning happens, even when you get it wrong.

Exam-style only. Instant feedback, shuffled each attempt, best score saved on this device.

What actually comes up

Three years of real questions, part by part, with the grade band each part sat at according to that year's marking schedule. Start with the pattern table — it tells you both halves of the paper: particles & substances (Question 1 / the start of Question 2) and thermochemistry (the rest).

Across 3 years

The pattern — what to prepare

The 2025 paper ↗ Marking schedule ↗

Question 1 — shapes, polarity & spontaneity

Sulfur forms a range of fluoride-containing ions and molecules. Part (a) is a table to complete for SF₃⁻ and SBrF₅ (Lewis structure and shape name). Part (b) gives the Lewis structure and "seesaw" shape of SF₄. Part (c) concerns S(s) + 3F₂(g) → SF₆(g), an exothermic reaction of finely powdered sulfur with fluorine gas.

• a Complete the table: Lewis structure and shape of SF₃⁻ and SBrF₅. Achieved
• b i Identify and explain the polarity of SF₄ (seesaw). Merit
• b ii Justify why the F–S–F bond angles in SF₄ differ from those in SBrF₅. Excellence
• c Justify, in terms of the entropy of system and surroundings, why S(s) + 3F₂(g) → SF₆(g) is spontaneous. Excellence

Question 2 — electron config, ionic radius & attractive forces

Part (a) is an s,p,d electron-configuration table for Cl, Cu and Fe³⁺, then asks why Fe²⁺ (92 pm) has a larger radius than Fe³⁺ (79 pm). Part (b) gives ΔvapH° values and asks for the attractive forces in each liquid. Part (c) is a Hess's Law calculation on 4CH₃NHNH₂(ℓ) + 5N₂O₄(ℓ) → 4CO₂(g) + 12H₂O(g) + 9N₂(g).

• Fe²⁺ 92 pm · Fe³⁺ 79 pm
• ΔvapH°: HBr 17.3 · HF 25.2 · Br₂ 29.6 kJ mol⁻¹

• a i Electron configurations (s, p, d) for Cl, Cu and Fe³⁺. Achieved
• a ii Explain why the Fe²⁺ ion (92 pm) is larger than Fe³⁺ (79 pm). Merit
• b i Identify all attractive forces in liquid HBr, HF and Br₂. Achieved
• b ii Explain why ΔvapH° of HF is higher than that of HBr. Merit
• b iii Justify why Br₂ has the highest ΔvapH° of the three. Excellence
• c i Calculate ΔrH° for 4CH₃NHNH₂ + 5N₂O₄ (Hess's law from formation/vaporisation data). Merit
• c ii Explain the effect on ΔrH° if the water were produced as a liquid. Merit

Question 3 — formation ΔH, trends & calorimetry

Nitric acid manufacture from ammonia supplies a table of formation equations to complete and use, then two periodic-trend explanations (Mg against Sr, and atomic radius across Period 3), and a calorimetry calculation working backwards from a temperature rise to a mass.

• 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(ℓ), ΔrH° = −1172 kJ mol⁻¹
• ΔfH°(NO) = +90.3 · ΔfH°(H₂O(ℓ)) = −286 kJ mol⁻¹
• CaCl₂: 20.4 → 23.2 °C, 68.0 g solution, c = 4.18 J g⁻¹ °C⁻¹, ΔH = −81.3 kJ mol⁻¹, M = 111 g mol⁻¹

• a i Write the balanced equation for the standard enthalpy of formation of NH₃. Achieved
• a ii Calculate ΔfH°(NH₃) from the data given. Merit
• b i Justify, using periodic trends, why Mg has a higher first ionisation energy than Sr. Merit
• b ii Explain the trend in atomic radius across Period 3 (Na→P). Merit
• c Dissolving CaCl₂ raises 68.0 g of solution 20.4→23.2 °C; find the mass of CaCl₂ (q = mcΔT, then ΔH). Merit

The 2024 paper ↗ Marking schedule ↗

Question 1 — shapes, Hess & spontaneity

A Lewis-structure and shape table for PF₅ and SeCl₄²⁻, then the explosive reaction of chlorine trifluoride with hydrazine — once investigated as a rocket fuel — first as a Hess calculation and then as a question about why it is spontaneous.

• 4ClF₃(g) + 3N₂H₄(ℓ) → 12HF(g) + 3N₂(g) + 2Cl₂(g)
• N₂H₄(ℓ) + O₂(g) → N₂(g) + 2H₂O(ℓ), ΔrH° = −623 kJ mol⁻¹
• 2ClF₃(g) + 2NH₃(g) → 6HF(g) + N₂(g) + Cl₂(g), ΔrH° = −1200 kJ mol⁻¹

• a Complete the table: Lewis structure and shape of PF₅ and SeCl₄²⁻. Achieved
• b i Calculate ΔrH° for 4ClF₃ + 3N₂H₄ → 12HF + 3N₂ + 2Cl₂ using the given data. Merit
• b ii Justify, in terms of the entropy of system and surroundings, why this explosive reaction is spontaneous. Excellence

Question 2 — electron config, ionisation trend & polarity

Reading an electron configuration symbol by symbol, ordering three elements by first ionisation energy from periodic trends, and then deciding which of two possible shapes for BrCl₃ — T-shaped or trigonal planar, both from a trigonal bipyramidal arrangement — makes the molecule polar.

• Ar 1s²2s²2p⁶3s²3p⁶ · Ne 1s²2s²2p⁶ · P 1s²2s²2p⁶3s²3p³
• BrCl₃ is known to be polar

• a i What do the 3, p and 6 each represent in the 3p⁶ part of argon's configuration? Achieved
• a ii Arrange Ar, Ne and P in order of decreasing first ionisation energy, and justify the order. Excellence
• b Compare the T-shaped and trigonal planar shapes of BrCl₃ — which is polar, referring to bond polarity and dipole arrangement? Excellence

Question 3 — attractive forces & calorimetry

Attractive forces and boiling points for ammonia, sulfur dioxide and pentane, then an experimental determination of the enthalpy of combustion of ethanol using a burner under a beaker of water — including which measurement made the result unreliable.

• NH₃ −33 °C · SO₂ −10 °C · pentane 36 °C
• water 22.1 → 31.2 °C
• burner + ethanol 59.2 → 58.7 g
• ΔcH(C₂H₅OH) found to be −770 kJ mol⁻¹
• c = 4.18 J g⁻¹ °C⁻¹, M = 46.0 g mol⁻¹

• a i Identify all attractive forces between particles of liquid NH₃, SO₂ and pentane. Achieved
• a ii Explain the difference in boiling point between ammonia and sulfur dioxide. Merit
• a iii Explain why pentane boils higher than sulfur dioxide. Excellence
• b i Use q = mcΔT and ΔcH = −770 kJ mol⁻¹ to find the mass of water in the beaker. Merit
• b ii Which measured quantity was a source of error in the enthalpy value — and why? Merit

The 2023 paper ↗ Marking schedule ↗

Question 1 — shapes, polarity & spontaneity

Opens with a Lewis-structure and shape table for SF₄ and SF₆, then the shape and polarity of PF₃Cl₂ from a given Lewis structure, and closes on hydrazine: the equation for its enthalpy of vaporisation, whether that process is exothermic or endothermic, and the spontaneity of its exothermic reaction with oxygen.

• a Complete the table: Lewis structure and shape of SF₄ and SF₆. Achieved
• b The Lewis structure for phosphorus trifluorodichloride, PF₃Cl₂, is given in the paper. Identify and explain the shape and polarity of PF₃Cl₂. Excellence
• c i Write the equation that represents the enthalpy of vaporisation, ΔvapH°, for hydrazine, N₂H₄. Achieved
• c ii Explain whether the process in part (i) is exothermic or endothermic. Merit
• c iii The reaction N₂H₄(ℓ) + O₂(g) → N₂(g) + 2H₂O(g) is exothermic. Justify, in terms of the entropy changes of the system and surroundings, why the reaction is spontaneous. Excellence

Question 2 — ionisation energy & a formation enthalpy

Three periodic-trend questions on gallium — the equation for its first ionisation energy, how that compares with selenium, and why its atomic radius exceeds boron’s — then a Hess-style calculation of the standard enthalpy of formation of urea, and what changes if the water is produced as a liquid.

• 2NH₃(g) + CO₂(g) → CO(NH₂)₂(s) + H₂O(g), ΔrH° = −89.0 kJ mol⁻¹
• ΔfH°(NH₃) = −46.0
• ΔfH°(CO₂) = −394
• ΔfH°(H₂O(g)) = −242 kJ mol⁻¹

• a i Write the equation to show the reaction that has an enthalpy change equal to the first ionisation energy for the element gallium, Ga. Achieved
• a ii Use your knowledge of periodic trends to identify and justify the difference in the first ionisation energy of gallium, Ga, and selenium, Se. Merit
• a iii Use your knowledge of periodic trends to explain why gallium, Ga, has a larger atomic radius than boron, B. Merit
• b i Calculate the standard enthalpy of formation of urea, ΔfH°(CO(NH₂)₂(s)), using the data given. Merit
• b ii Explain what would happen to the ΔrH° provided in part (i) if the water was produced as a liquid, H₂O(ℓ). Excellence

Question 3 — configs, attractive forces & Hess

An electron-configuration table (S, Co, Cr³⁺) and the Ca / Ca²⁺ radius difference, then attractive forces and boiling points for three liquids of very different types, and finally the enthalpy change for the combustion of glucose from formation and combustion data.

• Ca atom 197 pm · Ca²⁺ 100 pm
• propan-1-amine 48.5 °C · chloroethane 12.3 °C · decane 174 °C
• ΔfH°(C₆H₁₂O₆) = −1270 · ΔcH°(C) = −394 · ΔcH°(H₂) = −286 · ΔvapH°(H₂O) = +6.01 kJ mol⁻¹

• a i Complete the table of electron configurations (s, p, d notation) for S, Co and Cr³⁺. Achieved
• a ii Explain why the radii of the Ca atom (197 pm) and the Ca²⁺ ion (100 pm) are different. Merit
• b i Identify all the types of attractive forces between particles of liquid propan-1-amine, chloroethane and decane. Achieved
• b ii Explain why the boiling point of propan-1-amine is higher than that of chloroethane. Merit
• b iii Explain why the boiling point of decane is higher than that of chloroethane. Excellence
• c Calculate the enthalpy change, ΔrH°, for C₆H₁₂O₆(s) + 6O₂(g) → 6CO₂(g) + 6H₂O(g), using the data given. Merit

Read the last column as a to-do list. Question 1 is nearly always "shapes + polarity + spontaneity", and the thermochemistry half (Hess, formation ΔH, calorimetry) always rolls in. No suggested slug — this is an existing topic.

Past exams & resources

⏱ Timed opens the paper and starts a countdown matching the exam length. Official NZQA PDFs. If a link 404s, that year’s paper may sit under a slightly different code: browse the NZQA subject page.