HSC Exam Practice

Sample response. The standard enthalpy of formation (ΔH°_f) is the enthalpy change when exactly one mole of a compound is formed from its constituent elements in their standard states at 25°C (298 K) and 100 kPa.

Marking notes. 1 mark for “one mole of compound formed from its elements”; 1 mark for standard conditions (25°C and 100 kPa, or standard states).

Sample response. ΔH°_f[O₂(g)] = 0 kJ mol⁻¹. This value applies by convention because O₂(g) is the standard state of elemental oxygen. No chemical change occurs when an element is “formed” from itself — there is no reaction and therefore no enthalpy change. All elements in their standard states are assigned ΔH°_f = 0 as the common reference baseline.

Marking notes. 1 mark for stating ΔH°_f[O₂(g)] = 0; 1 mark for explaining the convention (O₂ is the standard state of oxygen / forming an element from itself involves no chemical change).

Sample response. ½N₂(g) + &frac32;H₂(g) → NH₃(g)

Marking notes. 1 mark for correct balanced equation with state symbols; 1 mark for producing exactly 1 mol NH₃ (fractional coefficients accepted). If 2 mol NH₃ is produced, award 0.

Sample response. The ΔH°_f method uses experimentally measured formation enthalpies for each specific compound in its actual physical state; the bond energy method uses average bond enthalpies tabulated across many molecules. The ΔH°_f method gives more accurate results because the data is specific to the actual substance and state (e.g. H₂O(l) includes the condensation energy). The bond energy method assumes all species are gaseous and uses averages, introducing error of ±5–20%. Both methods require a balanced equation with state symbols.

Marking notes. 1 mark for identifying the data type difference (experimental vs average bond values); 1 mark for accuracy difference and linking to actual states; 1 mark for identifying the gaseous-state assumption as the source of inaccuracy in the bond energy method.

Sample response. ΣΔH°_f(products) = 1(−393.5) + 2(−285.8) = −393.5 − 571.6 = −965.1 kJ mol⁻¹. ΣΔH°_f(reactants) = 1(−74.8) + 2(0) = −74.8 kJ mol⁻¹. ΔH°_rxn = −965.1 − (−74.8) = −890.3 kJ mol⁻¹. Exothermic.

Marking notes. 1 mark for correct ΣΔH°_f(products); 1 mark for correct ΣΔH°_f(reactants) with O₂ = 0 shown; 1 mark for correct ΔH°_rxn with sign and units.

Sample response. The bond energy method gives a less negative value because: (i) it assumes all species are gaseous, so it uses ΔH°_f[H₂O(g)] rather than [H₂O(l)]. The difference is 44 kJ mol⁻¹ per mole of water; for 2 mol H₂O this is 88 kJ mol⁻¹ of condensation energy that is not counted. (ii) It uses average C–H, O=O, C=O and O–H bond enthalpies that are averages across many molecules and not specific to CH₄ or H₂O, introducing cumulative error.

Marking notes. 1 mark for identifying the gaseous water assumption as the main source; 1 mark for quantifying (~88 kJ mol⁻¹ for 2 mol H₂O) or explaining the condensation energy is missed; 1 mark for mentioning average bond enthalpies introduce additional approximation.

Sample response (a). The more negative the ΔH°_f value, the more thermodynamically stable the compound is relative to its elements. CO₂(g) (−393.5 kJ mol⁻¹) is the most stable; N₂H₄(l) (+50.6 kJ mol⁻¹) sits above its elements and is less stable (energy-rich). (2 marks: 1 for the relationship, 1 for using chart data.)

Sample response (b). A positive ΔH°_f means N₂H₄(l) stores +50.6 kJ mol⁻¹ more energy than its elements (N₂ and H₂). When hydrazine burns, energy is released from two sources simultaneously: the formation of the very stable products (H₂O(l), ΔH°_f = −285.8; N₂(g), ΔH°_f = 0) and the release of the extra energy stored in the hydrazine above the element baseline (+50.6). Together these give a large negative ΔH°_rxn. (3 marks.)

Sample response (c). ΣΔH°_f(p) = 1(0) + 2(−285.8) = −571.6; ΣΔH°_f(r) = 1(+50.6) + 1(0) = +50.6; ΔH°_rxn = −571.6 − (+50.6) = −622.2 kJ mol⁻¹. (2 marks: 1 for correct sums, 1 for correct final answer with sign.)

Sample response (a). ΣΔH°_f(products) = 2(−285.8) + 1(0) = −571.6 kJ mol⁻¹; ΣΔH°_f(reactants) = 2(−187.8) = −375.6 kJ mol⁻¹; ΔH°_rxn = −571.6 − (−375.6) = −196.0 kJ mol⁻¹. (3 marks: 1 for Σproducts, 1 for Σreactants, 1 for answer.)

Sample response (b). ΔH°_rxn = −196.0 kJ mol⁻¹ is negative, indicating the reaction is exothermic. Energy is released to the surroundings. This is consistent with the observable fact that concentrated H₂O₂ decomposes vigorously and can cause fires — the exothermic decomposition releases significant heat. (2 marks: 1 sign interpretation, 1 physical interpretation.)

Sample response (c). The calculation assumes all substances are at standard conditions (25°C, 100 kPa, in their standard states). In practice, the decomposition of H₂O₂ may occur under different temperatures and concentrations. (1 mark for any valid assumption.)

Sample response. The ΔH°_f method is a powerful and accurate approach to calculating the enthalpy change of any reaction for which formation data is available. The formula ΔH°_rxn = ΣΔH°_f(products) − ΣΔH°_f(reactants) uses experimentally measured values that account for each substance in its actual physical state at standard conditions. This is its key advantage over the bond energy method, which assumes all species are gaseous and uses average bond enthalpies that may not accurately represent the bonds in specific molecules. For the combustion of CH₄, the ΔH°_f method gives −890.3 kJ mol⁻¹ (matching the accepted value) while the bond energy method gives approximately −674 kJ mol⁻¹ — a 24% error arising partly from treating H₂O as a gas and missing the 88 kJ mol⁻¹ of condensation energy for 2 mol H₂O. However, the ΔH°_f method also has limitations. It requires access to a data table of measured ΔH°_f values; for novel or unstable compounds these values may not be available or may have large experimental uncertainties. The method is strictly valid only under standard conditions; at elevated temperatures or pressures, ΔH values change. Additionally, applying the formula correctly requires attention to stoichiometric coefficients and state symbols — common student errors (e.g. using reactants minus products, or forgetting to multiply by the coefficient) undermine the accuracy. In summary, the ΔH°_f method is more accurate and physically realistic than the bond energy method when tabulated data is available, but it depends on the quality of that data and is condition-specific.

Marking notes. 1 mark — defines or correctly applies the formula ΔH°_rxn = ΣΔH°_f(products) − ΣΔH°_f(reactants). 1 mark — advantage: uses experimentally measured values for actual states (not averages). 1 mark — advantage: more accurate; cites a specific example (e.g. CH₄ combustion result). 1 mark — contrast with bond energy method: identifies the gaseous state assumption / average bond enthalpy limitation. 1 mark — limitation of ΔH°_f method: data may be unavailable for novel/unstable compounds. 1 mark — limitation: valid only at standard conditions; may differ at other temperatures/pressures. 1 mark — reaches an explicit evaluative judgement that weighs advantages against limitations and reaches a qualified conclusion about when to prefer each method.