HSC Exam Practice

Sample response. The standard enthalpy of combustion is the enthalpy change when exactly one mole of a substance undergoes complete combustion in excess oxygen at standard conditions (25°C, 100 kPa), with all reactants and products in their standard states.

Marking notes. 1 mark for “one mole burns completely in O₂”; 1 mark for “standard conditions” or “25°C, 100 kPa / 1 atm”.

Sample response. The bond energy method uses tabulated average bond enthalpies for specific bond types (e.g. C–H, O=O) and applies ΔH = ΣB(reactants) − ΣB(products). It is approximate (accuracy ±5–20%) because average values do not reflect the exact environment of bonds in a specific molecule, and because it assumes all species are gaseous. The ΔH°_f method uses experimentally measured standard enthalpies of formation for each specific substance in its actual physical state at 25°C, applying ΔH°_rxn = ΣΔH°_f(products) − ΣΔH°_f(reactants). It is more accurate because no averaging is involved and actual physical states are used.

Marking notes. Bond energy: 1 mark for data type (average bond enthalpies) + 1 mark for source of error/accuracy limitation. ΔH°_f: 1 mark for data type (experimental, substance-specific) + 1 mark for why it is more accurate (no averaging, actual states).

Sample response. Bond energy calculations count only the energy required to break bonds in gaseous reactants and form bonds in gaseous products — this is because bond enthalpies are defined for bonds between isolated gaseous atoms. At standard conditions for combustion, however, the fuel may be liquid (e.g. ethanol(l)) and water is produced as a liquid (H₂O(l)). The bond energy calculation does not account for the energy required to vaporise the liquid fuel or the energy released when gaseous water condenses to liquid. For each mole of H₂O(l) vs H₂O(g), this is approximately 44 kJ mol⁻¹. For a combustion producing 3 mol H₂O, this alone introduces ∼132 kJ mol⁻¹ of error.

Marking notes. 1 mark for “bond enthalpies are defined for gaseous species only”; 1 mark for identifying the specific state problem (liquid fuel / liquid water product); 1 mark for quantifying the error or identifying it as significant and cumulative.

Sample response. Method: ΔH°_f method, because a table of ΔH°_f values is provided for all species. ΣΔH°_f(products) = 3(−393.5) + 4(−285.8) = −1180.5 + (−1143.2) = −2323.7 kJ mol⁻¹. ΣΔH°_f(reactants) = 1(−303) + (9/2)(0) = −303 kJ mol⁻¹. ΔH_c = −2323.7 − (−303) = −2020.7 kJ mol⁻¹.

Marking notes. 1 mark method identification + 1 mark Σproducts calculation + 1 mark Σreactants calculation + 1 mark correct final answer (accept −2021).

Sample response. E_g (kJ g⁻¹) = |ΔH_c| ÷ M, where |ΔH_c| is the magnitude of the molar enthalpy of combustion (kJ mol⁻¹) and M is the molar mass (g mol⁻¹). Ethanol: 1367 ÷ 46.07 = 29.7 kJ g⁻¹. Methane: 890 ÷ 16.04 = 55.5 kJ g⁻¹. Methane delivers 87% more energy per gram than ethanol. This difference arises because methane’s very low molar mass (only CH₄, 16.04 g mol⁻¹) concentrates a large fraction of its energy per unit mass despite having a much lower |ΔH_c| per mole.

Marking notes. 1 mark formula stated; 1 mark each correct calculated value (ethanol 29.7, methane 55.5); 1 mark explicit comparison naming the higher value.

Sample response. Step 1: Write and identify the target equation (the reaction whose ΔH is unknown). Step 2: Write the available known thermochemical equations. Step 3: Reverse and/or scale each known equation so that when they are added, all intermediate species cancel and the target equation is obtained (reversing changes the sign of ΔH; scaling changes the magnitude proportionally). Step 4: Verify that all intermediates have cancelled to confirm the target equation is correct, then sum the modified ΔH values to obtain ΔH_target.

Marking notes. 1 mark per step, 4 marks total. Must include: identify target / write known equations / reverse and scale with sign/magnitude rules / cancel and verify intermediates. Award 1 mark if step 3 addresses sign AND scale separately.

Sample response (a). Per mole, octane ranks first by a large margin (5471 kJ mol⁻¹); the alcohols rank in order of chain length below it. Per gram, octane still ranks first (47.9 kJ g⁻¹), but the relative gap between the alcohols is much smaller (22.7–36.1 kJ g⁻¹). Methanol shows the most dramatic change: it ranks last per mole among the four alcohols listed, but its per-gram value (22.7) is only slightly below ethanol (29.7) relative to the per-mole difference — its lower molar mass partially compensates its lower ΔH_c. The per-mole difference between methanol and ethanol is a factor of 1.88×, but the per-gram difference is only 1.31×.

Sample response (b). Octane ranks first on both comparisons because (1) it has the highest |ΔH_c| per mole (5471 kJ mol⁻¹) due to its 8 carbon atoms, 18 hydrogen atoms and absence of an oxygen atom — burning octane releases energy from 25 C–H bonds and 7 C–C bonds; (2) its high carbon and hydrogen content relative to molar mass gives it a favourable energy-per-gram ratio (47.9 kJ g⁻¹); the oxygen atoms in alcohols are “dead weight” — they add molar mass but contribute no energy, reducing energy density per gram.

Marking notes. Part (a): 1 mark for identifying the ranking change / smaller relative gap per gram; 1 mark for naming a specific fuel where the change is notable. Part (b): 1 mark for large |ΔH_c| per mole explained (many C–H/C–C bonds); 1 mark for high energy per gram due to high C/H content and no oxygen; 1 mark for explicitly stating that oxygen in alcohols adds mass without contributing combustion energy.

Sample response (a). Scale (1) ×2: 2C(graphite) + 2O₂(g) → 2CO₂(g); ΔH = 2(−393.5) = −787.0 kJ mol⁻¹. Scale (2) ×3: 3H₂(g) + &frac32;O₂(g) → 3H₂O(l); ΔH = 3(−285.8) = −857.4 kJ mol⁻¹. Reverse (3): 2CO₂(g) + 3H₂O(l) → C₂H₅OH(l) + 3O₂(g); ΔH = +1367.3 kJ mol⁻¹. Add: 2C + 2O₂ + 3H₂ + &frac32;O₂ + 2CO₂ + 3H₂O → 2CO₂ + 3H₂O + C₂H₅OH + 3O₂. Cancel 2CO₂(g) both sides; cancel 3H₂O(l) both sides; 2O₂ + &frac32;O₂ − 3O₂ = ½O₂ remaining on left. Remaining: 2C(graphite) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l) — matches target. ΔH°_f = −787.0 + (−857.4) + 1367.3 = −277.1 kJ mol⁻¹. (Literature: −277.7 kJ mol⁻¹ — small rounding difference.)

Sample response (b). Assumption: all H₂O intermediates are in the liquid state (l) as used in equations (2) and (3). If gaseous H₂O(g) were used instead, ΔH₂ would be −241.8 and ΔH₃ would change correspondingly — the ΔH_f result would be different (less negative by 3 × 44 = 132 kJ mol⁻¹).

Marking notes. Part (a): 1 mark scaling (1)×2 correctly; 1 mark scaling (2)×3 correctly; 1 mark reversing (3) correctly (+sign); 1 mark correct cancellation shown; 1 mark final value −277 ± 2 kJ mol⁻¹. Part (b): 1 mark for identifying H₂O state assumption and explaining the consequence quantitatively or qualitatively.

Sample response. Three methods are available for calculating ΔH°_c: the bond energy method, the ΔH°_f method, and Hess’s Law applied to a multi-step equation pathway.

The bond energy method sums average bond enthalpies for bonds broken in reactants minus bonds formed in products (ΔH = ΣB(reactants) − ΣB(products)). It requires only a table of average bond enthalpies and is conceptually straightforward for estimating ΔH when no other data is available. Its accuracy is inherently limited to approximately ±5–20% because: (a) bond enthalpies are averages across many molecular contexts rather than exact values for a specific bond in a specific molecule; (b) the method assumes all species are gaseous, so it fails to account for enthalpy changes associated with physical state changes — notably the condensation of H₂O(g) to H₂O(l) in standard combustion, which releases approximately 44 kJ mol⁻¹ per mole of water. These limitations make this method unsuitable when precise ΔH values are required.

The ΔH°_f method uses experimentally measured standard enthalpies of formation for each specific substance in its actual physical state at 25°C, applying ΔH°_rxn = ΣΔH°_f(products) − ΣΔH°_f(reactants), with each value multiplied by its stoichiometric coefficient. This is the most accurate method for standard conditions when database values are available: no averaging is used, and the actual physical states of all species are respected. It is the preferred method when a ΔH°_f table is given in an HSC question. For example, determining ΔH_c of ethanol produced from Bundaberg sugarcane bagasse would use this method given tabulated ΔH°_f values for C₂H₅OH(l), CO₂(g), and H₂O(l).

Hess’s Law (multi-step method) combines a set of known thermochemical equations by reversing and scaling so that intermediate species cancel, leaving the target reaction. Its accuracy equals the accuracy of the input data: if based on precise experimental calorimetric values, it can be more accurate than the ΔH°_f method for molecules not well-represented in standard tables. This method is required when ΔH°_f data for the target compound is unavailable — for example, Shell Prelude FLNG’s syngas-forming reaction (C + H₂O → CO + H₂) can only be determined by combining combustion equations for C, CO, and H₂, because the direct reaction enthalpy cannot be measured cleanly.

All three methods derive from the same underlying principle — Hess’s Law and conservation of energy — but differ in data source and precision. The selection rule is driven by the data provided: bond energies → bond energy method; ΔH°_f table → ΔH°_f method; thermochemical equations → multi-step Hess’s Law. Never apply the bond energy method when ΔH°_f data is available, as this introduces unnecessary error.

Marking notes. 1 mark — bond energy method: data type (average bond enthalpies), approximate, gaseous assumption. 1 mark — bond energy error sources: averaging AND physical state assumption both identified. 1 mark — ΔH°_f method: data type (experimental, compound-specific), formula stated (products − reactants). 1 mark — why ΔH°_f is more accurate (no averaging, actual physical states). 1 mark — Hess’s Law multi-step: describes manipulation process (reverse, scale, cancel), notes accuracy depends on input data. 1 mark — names at least one specific Australian energy context correctly (e.g. Bundaberg bagasse, Shell Prelude, E10 ethanol blends). 1 mark — reaches an explicit evaluative statement selecting which method is most appropriate for standard conditions and why (not a description, but an evaluation with justification).