Chemistry • Year 11 • Module 4 • Lesson 6

Bond Energy & Enthalpy Change

Build HSC Band 5–6 extended-response technique: evaluate bond energy calculations using real data, quantitative reasoning, and multi-criteria judgements.

Master • Extended Response

1. Extended response — evaluate bond energy data for a real clean-energy decision (Band 5–6)

8 marks   Band 5–6

Scenario: The Australian Government’s National Hydrogen Strategy (2019) identified hydrogen (H2) as a key clean-energy export. CSIRO researchers have compared H2 with methane (CH4) as fuels. The bond energy data table and chart below summarise key values.

Fuel reaction (gaseous products)Bond energies used (kJ mol−1)ΔH (bond energy method)ΔHc° experimental (liquid H2O)
H2(g) + ½O2(g) → H2O(g) H–H = 436; O=O = 498; O–H = 460 −235 kJ mol−1 −286 kJ mol−1
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) C–H = 414; O=O = 498; C=O = 743; O–H = 460 −674 kJ mol−1 −890 kJ mol−1

Additional data: Molar masses: H2 = 2 g mol−1; CH4 = 16 g mol−1. Energy per gram: H2 = 143 kJ g−1 (experimental); CH4 = 55 kJ g−1 (experimental).

Q1. Using the data provided, evaluate the accuracy of the bond energy method and the suitability of H2 as a replacement for CH4 as a fuel. In your response you must:

  • Show the full bond energy calculation for H2 combustion to verify the −235 kJ mol−1 value.
  • Compare the bond energy ΔH with the experimental ΔHc° for both fuels; calculate the percentage discrepancy for each fuel.
  • Explain, using two specific reasons, why the bond energy values differ from experimental values.
  • Using both per-mole and per-gram energy data, evaluate whether H2 is a suitable replacement for CH4 on energy grounds.
  • Reach an evidence-based judgement that explicitly weighs at least two criteria (energy density, carbon emissions, or accuracy of bond energy data).
Strategy: (1) Show working for H2. (2) Calculate % discrepancy = |bond energy − experimental| / |experimental| × 100 for each fuel. (3) Name average bond enthalpies AND gaseous state assumption as reasons. (4) Note H2 has lower ΔH per mole but higher per gram. (5) Carbon emissions: H2 produces only H2O; CH4 produces CO2.

2. Multi-step calculation and interpretation — ethanol as a biofuel (Band 5–6)

7 marks   Band 5–6

Context: Ethanol (C2H5OH) is used in Australian biofuel blends such as E10 (10% ethanol in petrol). Its structural formula is CH3CH2OH, which contains: 5 × C–H bonds, 1 × C–C bond, 1 × C–O bond, 1 × O–H bond.

Bond energies (kJ mol−1): C–H = 414; C–C = 347; C–O = 360; O–H = 460; O=O = 498; C=O = 743.
Balanced equation: C2H5OH(g) + 3O2(g) → 2CO2(g) + 3H2O(g).

(a) Using the structural formula information provided, calculate ΣB(reactants) for the combustion of ethanol. Show every bond and every multiplication. 2 marks

(b) Calculate ΣB(products) and hence determine ΔH. State whether the reaction is exothermic or endothermic. 2 marks

(c) The experimental ΔHc° for ethanol with liquid water as product is −1367 kJ mol−1. Calculate the percentage discrepancy between your bond energy answer and this experimental value. Then explain, using two specific reasons, why these values differ. 3 marks

For (c): % discrepancy = |your ΔH − experimental| / |experimental| × 100. For the explanation: one reason relates to average bond enthalpies; the other to the state of water in the product.
Answers — Do not peek before attempting

Q1 — Evaluate bond energy accuracy and H2 as a clean fuel

Bond energy calculation for H2 (verification):
ΣB(reactants) = 1(436) + ½(498) = 436 + 249 = 685 kJ mol−1
ΣB(products) = 2(460) = 920 kJ mol−1
ΔH = 685 − 920 = −235 kJ mol−1 ✓ (matches table value) [1]

Percentage discrepancies:
H2: |−235 − (−286)| / 286 × 100 = 51/286 × 100 = 17.8% [1]
CH4: |−674 − (−890)| / 890 × 100 = 216/890 × 100 = 24.3% [1]

Two reasons for discrepancy:
Reason 1: Average bond enthalpies — tabulated bond energies (e.g. C–H = 414 kJ mol−1) are averages across many different molecules; the actual C–H bond energy in CH4 differs slightly from this average, and the error accumulates across all bonds in the calculation. [1]
Reason 2: Gaseous state assumption — bond energies are defined for species in the gaseous state, so the calculation uses H2O(g) as the product. The experimental ΔHc° uses H2O(l); the condensation of water vapour to liquid releases additional energy (~44 kJ per mole H2O not captured in the bond energy calculation). For H2 producing 1 mol H2O, this accounts for ~44 of the 51 kJ discrepancy; for CH4 producing 2 mol H2O, condensation contributes ~88 kJ of the 216 kJ discrepancy (the remainder comes from average bond enthalpies). [1]

Energy comparison per mole and per gram:
Per mole: CH4 (−890) releases significantly more energy than H2 (−286). On a per-mole basis, H2 is a lower-energy fuel. However, per gram: H2 = 143 kJ g−1 vs CH4 = 55 kJ g−1. H2 has nearly 3× the energy density per gram. [1]

Evidence-based judgement: H2 is a suitable replacement for CH4 on two key criteria. First, per-gram energy density: H2 delivers 143 kJ g−1 vs CH4’s 55 kJ g−1, making H2 a mass-efficient fuel for vehicles where weight matters. Second, carbon emissions: combustion of H2 produces only H2O with no CO2, directly addressing climate targets — a clear advantage over CH4. The bond energy data overestimates ΔH by 9–16% for both fuels, so neither fuel’s energy advantage is an artefact of the calculation method; the conclusion holds when experimental values are used. The suitability of H2 is therefore well-supported by the evidence. [1]

Q2 — Ethanol combustion calculation and interpretation

(a) ΣB(reactants):
C2H5OH: 5(414) + 347 + 360 + 460 = 2070 + 347 + 360 + 460 = 3237 kJ mol−1
3O2: 3(498) = 1494 kJ mol−1
ΣB(reactants) = 3237 + 1494 = 4731 kJ mol−1 [1 for correct bond identification; 1 for correct arithmetic total]

(b) ΣB(products) and ΔH:
2CO2: 2 × 2(743) = 4(743) = 2972 kJ mol−1
3H2O: 3 × 2(460) = 6(460) = 2760 kJ mol−1
ΣB(products) = 2972 + 2760 = 5732 kJ mol−1
ΔH = 4731 − 5732 = −1001 kJ mol−1 [1 for ΣB(products); 1 for correct ΔH and sign, exothermic]

(c) Percentage discrepancy and explanation:
% discrepancy = |−1001 − (−1367)| / 1367 × 100 = 366 / 1367 × 100 = 26.8% [1]
Reason 1: Average bond enthalpies — the C–H, C–C, and C–O bond energies in ethanol differ slightly from the averaged values used; with many bonds summed, these small deviations accumulate into a noticeable error [1].
Reason 2: Gaseous state assumption — the calculation used H2O(g) as the product; the experimental value assumes H2O(l). Condensation of 3 mol of water vapour releases approximately 3 × 44 = 132 kJ mol−1 of additional energy not captured by the bond energy method. Combined with errors from average bond enthalpies, this accounts for most of the 366 kJ discrepancy observed [1].