Engineers designing a car to run on ethanol vs octane need to know which fuel delivers more energy — per mole, per gram, and per litre. They calculate these values entirely on paper using the three methods you have built across Lessons 6–9. This lesson puts them all together: when to use bond energies, when to use ΔH°f values, and when Hess's Law is the only option.
Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
A fuel with a larger ΔHc (enthalpy of combustion in kJ mol⁻¹) always gives you more energy for a given mass of fuel. True or false?
Think about ethanol (C₂H₅OH, ΔHc = −1367 kJ mol⁻¹, M = 46.07 g mol⁻¹) vs methanol (CH₃OH, ΔHc = −726 kJ mol⁻¹, M = 32.04 g mol⁻¹). Ethanol releases almost twice as much energy per mole. But if you fill a fuel tank by mass — which fuel gives you more energy per kilogram?
Before this lesson: Calculate (or estimate) energy per gram for each fuel. Does the fuel with higher ΔHc per mole also have higher energy per gram? Write your prediction with any working.
Type your prediction below — you will revisit this at the end of the lesson.
Write your prediction in your book. You will revisit it at the end of the lesson.
📚 Core Content
In HSC, the method is almost always signalled by the data provided. The question tells you which method to use — your job is to recognise the signal and apply the correct procedure.
Why do the methods give different numerical results?
Molar enthalpy of combustion (kJ mol⁻¹) is the chemist's comparison. Energy per gram (kJ g⁻¹) is the engineer's comparison. Neither is universally "better" — you must specify which comparison you are making.
Fuel efficiency data — selected fuels:
| Fuel | Formula | M (g mol⁻¹) | ΔHc (kJ mol⁻¹) | Energy per gram (kJ g⁻¹) |
|---|---|---|---|---|
| Hydrogen | H₂(g) | 2.02 | −286 | 141.6 |
| Methane (natural gas) | CH₄(g) | 16.04 | −890 | 55.5 |
| Octane (petrol) | C₈H₁₈(l) | 114.23 | −5471 | 47.9 |
| Ethanol | C₂H₅OH(l) | 46.07 | −1367 | 29.7 |
| Methanol | CH₃OH(l) | 32.04 | −726 | 22.7 |
Key observations from the data:
Higher ΔHc per mole always means more energy per gram.
Why students think this: ΔHc is the most visible number and larger magnitude seems "better."
What is actually true: Energy per gram = |ΔHc| ÷ M. A fuel can have large ΔHc per mole but low energy per gram if its molar mass is large. Always calculate both before comparing fuels.
The bond energy method and ΔH°f method always give the same ΔH for a reaction.
Why students think this: Both methods calculate ΔH for the same reaction, so the answer should be the same.
What is actually true: Bond energies are averages for bond types across many molecules; ΔH°f values are experimentally measured for specific substances in actual states. They give different numerical results, with ΔH°f being more accurate. Only under ideal conditions (all gas-phase species, exact bond energies) would they agree.
Hess's Law only works when you have exactly two equations to add together.
Why students think this: The NESA prototype (L08) uses two equations, and students generalise from this.
What is actually true: Hess's Law works with any number of equations. The worked example in L08 used three equations. The ΔH°f method is implicitly a Hess's Law cycle using as many formation equations as there are species in the balanced equation. The principle (path independence of enthalpy) is not limited by the number of steps.
🔬 Worked Examples
Calculate the standard enthalpy of combustion of propan-1-ol (C₃H₇OH) using ΔH°f data:
C₃H₇OH(l) + 9/2 O₂(g) → 3CO₂(g) + 4H₂O(l)
ΔH°f values: C₃H₇OH(l) = −303 kJ mol⁻¹ | CO₂(g) = −393.5 kJ mol⁻¹ | H₂O(l) = −285.8 kJ mol⁻¹ | O₂(g) = 0
Calculate the energy per gram released by propan-1-ol. (M = 60.10 g mol⁻¹, ΔHc = −2020.7 kJ mol⁻¹.) Then compare to ethanol (29.7 kJ g⁻¹). Which releases more energy per gram?
Methanol (CH₃OH, M = 32.04 g mol⁻¹, ΔHc = −726 kJ mol⁻¹) and ethanol (C₂H₅OH, M = 46.07 g mol⁻¹, ΔHc = −1367 kJ mol⁻¹) are compared as alternative automotive fuels.
(a) Which releases more energy per mole? (b) Which releases more energy per gram? (c) Suggest two reasons other than energy content why ethanol is often preferred as a fuel additive over methanol.
🔬 Activities
Given thermochemical equations:
a Write the target equation for the enthalpy of formation of ethane: the equation in which exactly 1 mol C₂H₆(g) is formed from its elements in standard states. Include state symbols.
b Using Hess's Law with the three given equations, calculate ΔH°f[C₂H₆(g)]. Show each manipulation step (which equations to use, whether to reverse or scale, the modified ΔH values), cancel intermediates, and verify the target equation before summing.
c Now write the balanced combustion equation for ethane and use your ΔH°f result to calculate ΔHc[C₂H₆(g)] using the ΔH°f formula. Use ΔH°f[C₂H₆(g)] = −83.7 kJ mol⁻¹.
Type your working below:
Answer in your workbook — show every manipulation step.
ΔH°f data (kJ mol⁻¹):
| Substance | ΔH°f (kJ mol⁻¹) | Substance | ΔH°f (kJ mol⁻¹) |
|---|---|---|---|
| CO₂(g) | −393.5 | H₂O(l) | −285.8 |
| CH₃OH(l) methanol | −239 | C₂H₅OH(l) ethanol | −277.7 |
| C₃H₇OH(l) propan-1-ol | −303 | C₄H₉OH(l) butan-1-ol | −327 |
| O₂(g) | 0 |
a Write the balanced combustion equation for butan-1-ol (C₄H₉OH) and calculate ΔHc using the ΔH°f data above.
b Complete the table below (calculate energy per gram for each alcohol). Then answer: does energy per gram consistently increase or decrease as the alcohol chain gets longer?
Molar masses: methanol = 32.04; ethanol = 46.07; propan-1-ol = 60.10; butan-1-ol = 74.12 g mol⁻¹
ΔHc values: methanol = −726; ethanol = −1367; propan-1-ol = −2021; butan-1-ol = your answer from (a).
Type your working below:
Answer in your workbook.
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
5 random questions from a replayable lesson bank — feedback shown immediately
✍️ Short Answer
6. A student uses two different methods to calculate ΔH for the combustion of methane (CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)).
Method 1 (Bond energy): Bond energies: C–H = 413; O=O = 498; C=O = 743 (in CO₂); O–H = 463 kJ mol⁻¹. ΔH = −674 kJ mol⁻¹
Method 2 (ΔH°f): ΔH°f: CH₄(g) = −74.8; CO₂(g) = −393.5; H₂O(l) = −285.8 kJ mol⁻¹. ΔH = −890.3 kJ mol⁻¹
(a) Which result is more accurate? Justify in terms of the data used. (2 marks)
(b) The bond energy method gave −674 kJ mol⁻¹ while ΔH°f gave −890.3 kJ mol⁻¹ — a difference of 216 kJ mol⁻¹. Identify and explain TWO specific reasons for this large discrepancy. (4 marks)
6 MARKS
Type your full answer below:
Answer in your workbook.
7. Propan-1-ol (C₃H₇OH, M = 60.10 g mol⁻¹) and butan-1-ol (C₄H₉OH, M = 74.12 g mol⁻¹) are being evaluated as biofuel alternatives to octane.
(a) Using ΔH°f data from Activity 2, confirm ΔHc[propan-1-ol] = −2021 kJ mol⁻¹. (3 marks)
(b) Calculate the energy per gram for propan-1-ol and butan-1-ol (use ΔHc[butan-1-ol] = −2676 kJ mol⁻¹). Which has higher energy density per gram? (2 marks)
(c) Both values are lower than octane (47.9 kJ g⁻¹). Suggest one engineering advantage of alcohol fuels over octane that offsets their lower energy density. (1 mark)
6 MARKS
Type your full answer:
Answer in your workbook.
8. Australia's transport sector is exploring ethanol (E10 blend, 10% ethanol by volume) as a fuel additive to petrol.
(a) Calculate ΔHc[ethanol] from ΔH°f data: C₂H₅OH(l) = −277.7; CO₂(g) = −393.5; H₂O(l) = −285.8 kJ mol⁻¹. Balanced equation: C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l). (3 marks)
(b) A car fuel tank holds 50 L of fuel. Petrol (octane, density 0.703 g mL⁻¹, M = 114.23 g mol⁻¹, ΔHc = −5471 kJ mol⁻¹) is compared with E10 (assume the ethanol component contributes 10% of volume and has density 0.789 g mL⁻¹). Estimate the energy difference (in MJ) from the ethanol component alone in a 50 L tank. (3 marks)
(c) The Australian government promotes E10 despite its lower energy density. Evaluate whether energy density alone is a valid basis for fuel policy decisions. (2 marks)
8 MARKS
Type your answer below:
Answer in your workbook.
Go back to your Think First prediction. The statement was: "A fuel with a larger ΔHc per mole always gives more energy per gram." Let's resolve it precisely:
Type your reflection below:
Write your reflection in your book.
(a) 2C(graphite) + 3H₂(g) → C₂H₆(g) — formation equation producing exactly 1 mol C₂H₆ from elements in standard states.
(b) Scale (1) ×2: ΔH = −787.0; scale (2) ×3: ΔH = −857.4; reverse (3): ΔH = +1560.7. After cancellation: 2C + 3H₂ → C₂H₆ ✓. ΔH°f = −787.0 + (−857.4) + 1560.7 = −83.7 kJ mol⁻¹.
(c) ΣΔH°f(products) = 2(−393.5) + 3(−285.8) = −1644.4; ΣΔH°f(reactants) = −83.7. ΔHc = −1644.4 − (−83.7) = −1560.7 kJ mol⁻¹ — matches ΔH₃ ✓. Self-consistent cycle.
(a) C₄H₉OH(l) + 6O₂(g) → 4CO₂(g) + 5H₂O(l). ΣΔH°f(p) = 4(−393.5) + 5(−285.8) = −3003.0; ΣΔH°f(r) = −327. ΔHc = −3003.0 − (−327) = −2676 kJ mol⁻¹.
(b) Methanol: 22.7 | Ethanol: 29.7 | Propan-1-ol: 33.6 | Butan-1-ol: 36.1 kJ g⁻¹. Trend: energy per gram increases with chain length — each CH₂ addition increases ΔHc faster than molar mass, approaching but never reaching alkane energy density.
1. B — When ΔH°f data is provided, use the ΔH°f method — it is more accurate (experimental data, actual states) and directly applicable to the data given. The bond energy method is used only when ΔH°f data is unavailable.
2. A — Propane: 2220 ÷ 44.1 = 50.3 kJ g⁻¹; Butane: 2877 ÷ 58.1 = 49.5 kJ g⁻¹. Propane has slightly higher energy per gram. Option D demonstrates the exact misconception this lesson addresses.
3. B — The ΔH°f method uses experimentally measured values for each specific substance in its actual state at 25°C — the most accurate because no averaging is involved. Bond energy uses averages; Hess's Law accuracy depends on input quality.
4. C — ΣΔH°f(p) = 4(−393.5) + 5(−285.8) = −1574.0 − 1429.0 = −3003.0; ΣΔH°f(r) = 1(−126.2) + 0 = −126.2; ΔH = −3003.0 − (−126.2) = −3003.0 + 126.2 = −2876.8 ≈ −2877 kJ mol⁻¹. Option A uses the wrong formula direction (reactants − products).
5. A — The Hess's Law cycle via elements produces ΔHc = ΣΔH°f(products) − ΔH°f(fuel), which is exactly the ΔH°f formula. Both methods are mathematically identical — one is the cycle diagram representation and the other is the algebraic rearrangement of the same Hess's Law equation.
Q6 (6 marks):
(a) ΔH°f method gives −890.3 kJ mol⁻¹ — more accurate [½]. It uses experimentally measured values for each substance in its actual state (CO₂(g) and H₂O(l)) — not averages, and not assuming gaseous water [1½].
(b) Reason 1: The bond energy method assumed H₂O(g) as product (gaseous state assumption), but the actual product is H₂O(l) [1]. The condensation of 2 mol H₂O vapour → liquid releases an additional 2 × 44 ≈ 88 kJ mol⁻¹ of energy (latent heat) that is not counted in the bond energy result — this alone accounts for 88 of the 216 kJ mol⁻¹ discrepancy [1]. Reason 2: Bond energy values are averages across many molecular environments [1]. The C–H bond in methane, C=O bond in CO₂, and O–H bond in water each have slightly different actual values from the tabulated averages; these errors accumulate across 8 bond values in this calculation, introducing substantial total error [1].
Q7 (6 marks):
(a) ΣΔH°f(p) = 3(−393.5) + 4(−285.8) = −1180.5 + (−1143.2) = −2323.7 kJ mol⁻¹ [1]; ΣΔH°f(r) = 1(−303) + 9/2(0) = −303 kJ mol⁻¹ [1]; ΔHc = −2323.7 − (−303) = −2020.7 ≈ −2021 kJ mol⁻¹ ✓ [1].
(b) Propan-1-ol: 2021 ÷ 60.10 = 33.6 kJ g⁻¹ [1]; Butan-1-ol: 2676 ÷ 74.12 = 36.1 kJ g⁻¹ [1]. Butan-1-ol has higher energy density per gram [accept either in bold] [½]. Both are still below octane's 47.9 kJ g⁻¹ [½ credit for noting this].
(c) Any valid advantage — e.g. higher octane rating (reduces engine knock and allows higher compression ratios, improving efficiency) [1]; or renewable/carbon-neutral production cycle (produced from biomass); or reduced CO and particulate emissions; or improved oxygen content improves combustion completeness [1].
Q8 (8 marks):
(a) ΣΔH°f(p) = 2(−393.5) + 3(−285.8) = −787.0 + (−857.4) = −1644.4 kJ mol⁻¹ [1]; ΣΔH°f(r) = 1(−277.7) + 3(0) = −277.7 kJ mol⁻¹ [1]; ΔHc = −1644.4 − (−277.7) = −1366.7 ≈ −1367 kJ mol⁻¹ [1].
(b) Volume of ethanol = 10% × 50 L = 5 L = 5000 mL [½]. Mass(ethanol) = 5000 × 0.789 = 3945 g [½]. n(ethanol) = 3945 ÷ 46.07 = 85.6 mol [1]. Energy = 85.6 × 1367 = 117,000 kJ = 117 MJ [1].
(c) Energy density alone is not a valid sole basis for fuel policy [½]. Other critical factors include: (1) Renewability — ethanol from sugarcane is carbon-neutral over its lifecycle, unlike fossil fuels; reducing greenhouse gas emissions is a policy priority that outweighs the 3–4% reduction in fuel economy from E10 [1]. (2) Energy security — domestic fuel production reduces dependence on oil imports; for Australia, locally grown ethanol from sugarcane provides energy security benefits. (3) Environmental co-benefits — higher oxygen content in E10 reduces CO and hydrocarbon emissions, improving urban air quality [1]. Policy decisions require balancing multiple competing objectives, of which energy density is only one.
Put your knowledge of Hess's Law to the test. Answer correctly to deal damage — get it wrong and the boss hits back. Pool: lessons 1–10.
Tick when you've finished all activities and checked your answers.