HSC Exam Practice

Sample response. Bond dissociation energy is the average energy required to break one mole of a specific covalent bond in a gaseous molecule, measured in kJ mol⁻¹. It is always positive because breaking a bond requires an input of energy from the surroundings; bond breaking is an endothermic process by definition.

Marking notes. 1 mark for “average energy to break 1 mol of a specific bond in the gaseous state”; 1 mark for “always positive because bond breaking is endothermic (requires energy input)”.

Sample response. ΔH = ΣB(reactants) − ΣB(products). ΣB(reactants) is the sum of bond energies of all bonds broken in the reactant molecules; ΣB(products) is the sum of bond energies of all bonds formed in the product molecules. Reactants appear first in the formula.

Marking notes. 1 mark for the correct formula with reactants first; 1 mark for correctly defining both ΣB(reactants) and ΣB(products).

Sample response. Exothermic: ΔH < 0 (negative); more energy is released forming bonds in products than is absorbed breaking bonds in reactants, so ΣB(products) > ΣB(reactants). Endothermic: ΔH > 0 (positive); more energy is absorbed breaking bonds in reactants than is released forming bonds in products, so ΣB(reactants) > ΣB(products).

Marking notes. 1 mark for exothermic ΔH < 0; 1 mark for endothermic ΔH > 0; 1 mark for correctly stating the relative magnitude of ΣB(reactants) vs ΣB(products) for at least one case.

Sample response. Reason 1: Bond energies used in calculations are average values across many different molecules. The actual bond energy of a C–H bond in CH₄ differs slightly from C–H in C₂H₅OH; using the same average introduces cumulative error. Reason 2: Bond energies are defined for species in the gaseous state. When products such as water are actually formed as liquids, the energy released during condensation is not included in the bond energy calculation, making the calculated ΔH less negative than the true experimental value.

Marking notes. 1 mark per reason clearly stated (max 2). 1 additional mark if one reason includes a specific example or quantification (e.g. “condensation of H₂O releases approximately 44 kJ mol⁻¹ not accounted for”).

Sample response. Thermodynamic factor: ΔH = −93 kJ mol⁻¹ for the Haber process, meaning the reaction is thermodynamically favourable — the products (NH₃) are at lower energy than the reactants. Thermodynamics predicts that the reaction should proceed. Kinetic factor: the activation energy is extremely high because the N≡N triple bond (945 kJ mol⁻¹) must be substantially disrupted to reach the transition state. At room temperature, virtually no N₂ or H₂ molecules have sufficient kinetic energy to overcome this barrier. These two factors are independent: ΔH describes the energy difference between start and finish (thermodynamics), while activation energy describes the energy cost to get there (kinetics). A reaction can be thermodynamically favourable yet kinetically inert.

Marking notes. 1 mark for correctly stating ΔH is negative / reaction is thermodynamically favourable; 1 mark for identifying the high activation energy as the kinetic barrier; 1 mark for linking the kinetic barrier to the N≡N bond energy specifically; 1 mark for explicitly distinguishing between thermodynamic (ΔH) and kinetic (activation energy / rate) factors as independent.

Sample response. ΣB(reactants) = 1(436) + 1(243) = 679 kJ mol⁻¹. ΣB(products) = 2(432) = 864 kJ mol⁻¹. ΔH = 679 − 864 = −185 kJ mol⁻¹ (exothermic).

Marking notes. 1 mark for correct ΣB(reactants) = 679; 1 mark for correct ΣB(products) = 864; 1 mark for correct ΔH = −185 kJ mol⁻¹ with sign. Consequential error permitted: if ΣB values are wrong but formula is correctly applied, award the method mark.

Sample response (a). Bonds broken (reactants, endothermic): 4 × C–H in CH₄; 2 × O=O in 2O₂. Bonds formed (products, exothermic): 2 × C=O in CO₂; 4 × O–H in 2H₂O. Bond breaking is always endothermic (absorbs energy); bond forming is always exothermic (releases energy). [2 marks: 1 for correctly listing bonds broken/formed; 1 for correctly labelling each process as endo/exothermic]

Sample response (b). ΣB(reactants) = 4(414) + 2(498) = 1656 + 996 = 2652 kJ mol⁻¹. ΣB(products) = 2(743) + 4(460) = 1486 + 1840 = 3326 kJ mol⁻¹. ΔH = 2652 − 3326 = −674 kJ mol⁻¹ (exothermic). [2 marks: 1 for ΣB values; 1 for correct ΔH with sign]

Sample response (c). % error = |−674 − (−890)| / 890 × 100 = 216/890 × 100 = 24.3%. [1 mark] Discrepancy arises because: (1) the chart uses average bond enthalpies, not exact values for CH₄/CO₂/H₂O; (2) the calculation assumes H₂O(g) as product, but the experimental value uses H₂O(l) — condensation of 2 mol water releases ~88 kJ not counted in the bond energy method. [2 marks: 1 per reason]

Sample response (a). ΣB(reactants): propene = 1(614) + 1(347) + 6(414) = 614 + 347 + 2484 = 3445; H₂ = 1(436). Total = 3445 + 436 = 3881 kJ mol⁻¹. ΣB(products): propane = 2(347) + 8(414) = 694 + 3312 = 4006 kJ mol⁻¹. [2 marks: 1 per ΣB]

Sample response (b). ΔH = 3881 − 4006 = −125 kJ mol⁻¹ (exothermic). [1 mark]

Sample response (c). The bond energy values used are average enthalpies; the actual C=C and C–H bond energies in propene differ slightly from these averages, causing a small discrepancy from the accepted value. Accept: gaseous state assumption (though less relevant here since no H₂O is formed). [1 mark]

Sample response. The bond energy method calculates ΔH by subtracting ΣB(products) from ΣB(reactants): ΔH = ΣB(reactants) − ΣB(products). It provides a useful estimate of the sign and approximate magnitude of ΔH when no calorimetric data is available. Taking the Haber process as an example: ΣB(reactants) = 1(945) + 3(436) = 945 + 1308 = 2253 kJ mol⁻¹; ΣB(products) = 6(391) = 2346 kJ mol⁻¹; ΔH = 2253 − 2346 = −93 kJ mol⁻¹. The negative sign correctly predicts that the reaction is exothermic, which matches experimental measurements. The method therefore succeeds in predicting the correct sign of ΔH. However, two specific limitations affect the magnitude. First, bond energies are average values: the N≡N bond energy used (945 kJ mol⁻¹) is an average across molecules containing N≡N, not the precise value in N₂ specifically. Similarly, N–H = 391 kJ mol⁻¹ is averaged over different amines and amides; the true N–H energy in NH₃ differs slightly. Over six N–H bonds and one N≡N bond, this accumulates to meaningful error. The accepted ΔH for the Haber process is −92 kJ mol⁻¹ — very close to the calculated −93, but this agreement is coincidental for this particular reaction; other reactions show larger discrepancies. Second, the method assumes all species are in the gaseous state. This is appropriate for the Haber process (all reactants and products are gases) but introduces significant error for combustion reactions where water is produced as a liquid: the latent heat of condensation (~44 kJ mol⁻¹ per H₂O) is not included in the bond energy calculation, making the calculated ΔH less negative than the true ΔH_c°. For the combustion of methane producing 2 mol H₂O(l), this contributes ~88 kJ of additional energy not counted in the bond energy calculation; the total discrepancy between the calculated (−674 kJ mol⁻¹) and experimental (−890 kJ mol⁻¹) values (216 kJ) reflects both the state assumption and accumulated average-bond-energy errors. In summary, the bond energy method reliably predicts the sign of ΔH and gives reasonable magnitude estimates (error varies by reaction), making it valuable when bond energy tables are available and experimental data is absent. Its reliability is highest when all species are gaseous; it is less accurate for reactions producing liquid or solid products. More accurate methods such as Hess’s Law or the enthalpy of formation method should be preferred when precise ΔH values are required.

Marking notes. 1 mark — correct formula ΔH = ΣB(reactants) − ΣB(products) stated with reactants first. 1 mark — correct calculation of ΣB(reactants) = 2253 kJ mol⁻¹ for Haber process. 1 mark — correct calculation of ΣB(products) = 2346 kJ mol⁻¹. 1 mark — correct ΔH = −93 kJ mol⁻¹ with sign. 1 mark — correctly identifies sign prediction as a strength of the method. 1 mark — Limitation 1: average bond enthalpies explained with reference to why averages introduce error. 1 mark — Limitation 2: gaseous state assumption explained with specific example (H₂O(l) not accounted for in combustion). 1 mark — quantitative or comparative treatment: either quotes a % error, or compares the calculated vs experimental value for a named reaction. 1 mark — names a more accurate method (Hess’s Law or enthalpy of formation) and states why it is more accurate. 1 mark — reaches an explicit evaluative judgement that acknowledges both the strengths and limitations and gives a qualified recommendation for when to use the method.