Chemistry • Year 11 • Module 4 • Lesson 6
Bond Energy & Enthalpy Change
Apply the bond energy method to real reactions; interpret a bond energy bar chart and a multi-fuel comparison data table; reason about sign, magnitude, and sources of error.
1. Sequence the calculation steps
The five steps of a bond energy ΔH calculation are listed below in shuffled order. Write the correct order (1–5) in the “Order” column. 5 marks
| Step (shuffled) | Order (1–5) |
|---|---|
| Apply ΔH = ΣB(reactants) − ΣB(products) and state whether the result is exothermic or endothermic. | |
| Identify all bonds broken in the reactants; multiply each bond energy by its stoichiometric coefficient and sum. | |
| Draw structural formulas (Lewis structures) for every reactant and product molecule. | |
| Write and balance the chemical equation for the reaction. | |
| Identify all bonds formed in the products; multiply each bond energy by its stoichiometric coefficient and sum. |
2. Interpret bond energy data — fuel comparison
A researcher calculated ΔH values using the bond energy method for the complete combustion of five common fuels (gaseous water assumed as product throughout). The table below shows the bond energy data and calculated results. 8 marks
| Fuel | Balanced equation (gaseous products) | ΣB(reactants) (kJ mol−1) |
ΣB(products) (kJ mol−1) |
ΔH (kJ mol−1) |
|---|---|---|---|---|
| Hydrogen (H2) CSIRO Hydrogen Research fuel |
H2 + ½O2 → H2O | 436 + 249 = 685 | 2 × 460 = 920 | −235 |
| Methane (CH4) natural gas, main residential fuel |
CH4 + 2O2 → CO2 + 2H2O | 4(414) + 2(498) = 2652 | 2(743) + 4(460) = 3326 | −674 |
| Ethanol (C2H5OH) biofuel, E10 blend |
C2H5OH + 3O2 → 2CO2 + 3H2O | 5(414) + 347 + 360 + 460 + 3(498) = 4731 | 4(743) + 6(460) = 5732 | −1001 |
| Propane (C3H8) LPG, common in regional NSW |
C3H8 + 5O2 → 3CO2 + 4H2O | 8(414) + 2(347) + 5(498) = 6496 | 6(743) + 8(460) = 8138 | −1642 |
| Ammonia (NH3) proposed zero-carbon fuel |
4NH3 + 3O2 → 2N2 + 6H2O | 12(391) + 3(498) = 6186 | 2(945) + 12(460) = 7410 | −1224 |
Bond energies used (kJ mol−1): H–H 436; O=O 498; O–H 460; C–H 414; C=O 743; C–C 347; C–O 360; N–H 391; N≡N 945. Note: ½O2 uses half the O=O value; values from this lesson’s bond energy data table.
2.1 Identify which fuel releases the most energy per mole of fuel combusted according to these bond energy calculations. 1 mark
2.2 All five ΔH values are negative. Explain what this tells you about the relative magnitudes of ΣB(reactants) and ΣB(products) for each fuel. 2 marks
2.3 The CSIRO has investigated H2 as a clean transport fuel (CSIRO Hydrogen Research Program). Using the data in the table, explain one advantage and one limitation of H2 compared to methane as a fuel, based solely on these bond energy results. 2 marks
2.4 A student claims the ΔH values in this table are the actual heats of combustion. Identify two reasons why this claim is incorrect. 2 marks
2.5 Predict whether the true (experimental) ΔH for hydrogen combustion (with liquid H2O as product) would be more or less negative than −235 kJ mol−1. Justify your prediction. 1 mark
3. Read and interpret a bond energy bar chart
The chart below shows the bond dissociation energies of six bonds that appear regularly in HSC Chemistry calculations. Study the chart and answer the questions. 6 marks
3.1 Which bond requires the most energy to break? Suggest one chemical or biological reason why this makes that molecule especially stable. 2 marks
3.2 Consider the combustion of hydrogen: H2 + ½O2 → H2O. Using values read from the chart (accept H–H = 436, O=O = 498, O–H = 460 kJ mol−1), calculate ΔH. Show full working. 2 marks
3.3 The C=O bond energy (743 kJ mol−1) is much larger than the C–H bond energy (414 kJ mol−1). Explain what this difference tells you about the relative stability of CO2 compared to a hydrocarbon with only C–H bonds. 2 marks
4. Predict and justify
Read the scenario below and answer the question. 4 marks
Scenario: A chemist is comparing two industrial synthesis routes for producing ammonia. Route A is the Haber process: N2(g) + 3H2(g) → 2NH3(g) using bond energies N≡N = 945, H–H = 436, N–H = 391 kJ mol−1. Route B is a hypothetical reaction: 2NO(g) + 5H2(g) → 2NH3(g) + 2½O2(g) — note this is endothermic. The Australian government’s National Hydrogen Strategy classifies Route A (with green H2) as “green ammonia”.
4.1 Using the bond energy data provided for Route A, calculate ΔH for the Haber process. Show full working. Then predict which route a chemical engineer would prefer on energy grounds and justify your reasoning with reference to sign of ΔH. 4 marks
Q1 — Step sequence
Correct order: 4, 3, 2, 5, 1
1 → Write and balance the equation (4th in list = order 1).
2 → Draw structural formulas (3rd in list = order 2).
3 → Identify bonds broken / ΣB(reactants) (2nd in list = order 3).
4 → Identify bonds formed / ΣB(products) (5th in list = order 4).
5 → Apply ΔH = ΣB(r) − ΣB(p) and state sign (1st in list = order 5).
Q2 — Fuel comparison data
2.1 Propane (C3H8) releases the most energy per mole: ΔH = −1642 kJ mol−1.
2.2 All ΔH values are negative, which means ΣB(products) > ΣB(reactants) in every case. More energy is released forming bonds in the products (CO2, H2O, N2) than is absorbed breaking bonds in the reactants (fuel + O2). The strong C=O and O–H bonds in combustion products account for the consistently high ΣB(products). [1 for ΣB(products) > ΣB(reactants); 1 for linking to which bonds are strong]
2.3 Advantage: H2 combustion produces only water vapour (no CO2), making it a zero-carbon fuel — consistent with the CSIRO Hydrogen Research Program’s goal of replacing fossil fuels. Limitation: H2 releases only 235 kJ mol−1, far less than methane (−674) per mole. However, on a per-gram basis H2 is actually energy-dense (molar mass 2 g mol−1) so a full analysis needs to compare per-kg values. Accept any valid advantage/limitation supported by data. [1 each]
2.4 Reason 1: Bond energy values are averages across many molecular environments, not the exact values in these specific molecules. Reason 2: The gaseous state assumption means liquid water products have been modelled as H2O(g); the latent heat of condensation of water is not accounted for, so the true ΔHc° (with H2O(l)) is more negative. [1 each]
2.5 More negative than −235 kJ mol−1. When H2O(l) forms, additional energy is released as the water vapour condenses. This condensation enthalpy (~44 kJ mol−1 of H2O) is not included in the bond energy calculation, so the actual ΔHc° (using H2O(l)) is approximately −286 kJ mol−1.
Q3 — Bar chart interpretation
3.1 N≡N (945 kJ mol−1) requires the most energy to break. N2 makes up 78% of the atmosphere and is essentially chemically inert to most organisms; the enormous energy cost to break N≡N means the activation energy for most N2 reactions is prohibitively high at ambient temperatures, explaining why atmospheric nitrogen is biologically inaccessible without specialised enzymes or the Haber process. [1 bond identified; 1 valid reason]
3.2 ΣB(reactants) = 1(436) + ½(498) = 436 + 249 = 685 kJ mol−1 [½O2 means one O=O counts as 249]. ΣB(products) = 2(460) = 920 kJ mol−1. ΔH = 685 − 920 = −235 kJ mol−1 (exothermic). [1 for correct ΣB values; 1 for correct ΔH and sign]
3.3 A higher bond energy means more energy is released when those bonds form. CO2 contains two C=O double bonds (743 kJ mol−1 each), releasing ~1486 kJ mol−1 per CO2 formed. Hydrocarbons bonded only by C–H (414) and C–C (347) release far less on a per-bond basis. This is why combustion of hydrocarbons is highly exothermic — the formation of strong C=O bonds in CO2 releases substantially more energy than was absorbed in breaking the weaker C–H and C–C bonds. [1 for linking higher BDE to more energy released on forming; 1 for comparing CO2 vs hydrocarbon context]
Q4 — Predict and justify
4.1 Route A — Haber process:
ΣB(reactants) = 1(945) + 3(436) = 945 + 1308 = 2253 kJ mol−1 [1]
ΣB(products) = 6(391) = 2346 kJ mol−1
ΔH = 2253 − 2346 = −93 kJ mol−1 [1]
Route A is exothermic (ΔH < 0); Route B is endothermic (ΔH > 0). A chemical engineer would strongly prefer Route A [1]: exothermic reactions release energy to surroundings, reducing the external energy input required to drive the process. An endothermic route like B requires continuous energy input to proceed, making it more expensive and energy-inefficient at industrial scale [1].