Chemistry • Year 11 • Module 4 • Lesson 7
Enthalpy of Formation
Synthesise, evaluate and justify — Band 5–6 extended-response questions requiring multi-step calculation, data interpretation and critical comparison.
Question 1 — Enthalpy of combustion: methane, ethanol and the hydrogen economy 8 marks
Australia is exploring a transition from natural gas (CH4) as the primary domestic heating fuel towards blended natural gas–hydrogen and eventually pure hydrogen (H2). A government report compares the enthalpy of combustion per mole of CO2 emitted and the enthalpy of combustion per gram of fuel for three candidate fuels. Use the data table below and the ΔH°f formula to complete this analysis.
| Substance | Formula | State | ΔH°f (kJ mol−1) | Molar mass (g mol−1) |
|---|---|---|---|---|
| Methane | CH4 | (g) | −74.8 | 16.0 |
| Ethanol | C2H5OH | (l) | −277.7 | 46.1 |
| Hydrogen | H2 | (g) | 0 | 2.02 |
| Carbon dioxide | CO2 | (g) | −393.5 | 44.0 |
| Water | H2O | (l) | −285.8 | 18.0 |
| Oxygen | O2 | (g) | 0 | 32.0 |
Combustion equations at standard conditions:
- CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
- C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)
- H2(g) + ½O2(g) → H2O(l)
(a) Calculate ΔH°rxn for the combustion of each of the three fuels. Show complete working for each, including ΣΔH°f(products) and ΣΔH°f(reactants) as separate steps. 3 marks
(b) Calculate the enthalpy of combustion per gram of fuel for each of the three fuels. Present your answers in a small table (columns: fuel | ΔH°rxn per mol | molar mass | ΔH°rxn per gram). 2 marks
(c) A domestic gas heater currently burning CH4 is retrofitted to burn pure H2. A government report states: “Switching to hydrogen eliminates CO2 emissions and delivers essentially the same energy per gram of fuel as methane.” Use your calculated values to evaluate this claim. State clearly which part(s) of the claim are supported by the ΔH°f data and which are not. 3 marks
(d) The ΔH°f method was used here rather than the bond energy method. State one specific reason why the ΔH°f method is more appropriate for this comparison. (bonus context)
Question 2 — Evaluating fuel selection: Apollo mission and modern rocket propellants 8 marks
In the Apollo programme, NASA engineers calculated propellant performance entirely from thermodynamic data tables before any test fire. The table below lists ΔH°f data for species relevant to three rocket propellant systems. Study the data and the scenario, then construct a response as directed.
| Substance | Formula | State | ΔH°f (kJ mol−1) |
|---|---|---|---|
| Hydrazine | N2H4 | (l) | +50.6 |
| Ammonia | NH3 | (g) | −46.1 |
| Methane | CH4 | (g) | −74.8 |
| Water | H2O | (l) | −285.8 |
| Carbon dioxide | CO2 | (g) | −393.5 |
| Nitrogen | N2 | (g) | 0 |
| Oxygen | O2 | (g) | 0 |
| Hydrogen | H2 | (g) | 0 |
Relevant combustion/oxidation equations at standard conditions:
- Propellant A (hydrazine): N2H4(l) + O2(g) → N2(g) + 2H2O(l)
- Propellant B (ammonia): 4NH3(g) + 3O2(g) → 2N2(g) + 6H2O(l)
- Propellant C (methane): CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
In your response you must:
- Calculate ΔH°rxn per mole of fuel burned for each of the three propellants, showing full working.
- Calculate the enthalpy released per gram of fuel for each propellant (molar masses: N2H4 = 32.0; NH3 = 17.0; CH4 = 16.0 g mol−1).
- Explain, with reference to the sign and magnitude of ΔH°f[N2H4(l)], why hydrazine releases more energy per mole than a compound with a negative ΔH°f of similar molecular weight would.
- Evaluate which propellant delivers the greatest energy per gram and whether this alone is sufficient to select the best propellant for a rocket, identifying at least one non-thermodynamic factor.
- Reach an evidence-based judgement: was NASA justified in using the ΔH°f method rather than the bond energy method when designing the Apollo Lunar Module engine?
Use this space for all working and your written response. Continue on the back if needed.
Q1(a) — Combustion enthalpies (3 marks, 1 per fuel)
Methane: ΣΔH°f(p) = 1(−393.5) + 2(−285.8) = −965.1; ΣΔH°f(r) = 1(−74.8) + 2(0) = −74.8; ΔH°rxn = −965.1 − (−74.8) = −890.3 kJ mol−1.
Ethanol: ΣΔH°f(p) = 2(−393.5) + 3(−285.8) = −787.0 + (−857.4) = −1644.4; ΣΔH°f(r) = 1(−277.7) + 3(0) = −277.7; ΔH°rxn = −1644.4 − (−277.7) = −1366.7 kJ mol−1.
Hydrogen: ΣΔH°f(p) = 1(−285.8) = −285.8; ΣΔH°f(r) = 1(0) + ½(0) = 0; ΔH°rxn = −285.8 − 0 = −285.8 kJ mol−1.
Q1(b) — Energy per gram (2 marks)
| Fuel | ΔH°rxn (kJ mol−1) | Molar mass (g mol−1) | ΔH°rxn per gram (kJ g−1) |
|---|---|---|---|
| CH4 | −890.3 | 16.0 | −55.6 |
| C2H5OH | −1366.7 | 46.1 | −29.6 |
| H2 | −285.8 | 2.02 | −141.5 |
Marking: 1 mark for correct CH4 and H2 per-gram values; 1 mark for correct ethanol per-gram value.
Q1(c) — Evaluating the government claim (3 marks)
Part supported: H2 combustion produces only H2O(l) — no CO2 is in the balanced equation. The claim that switching to hydrogen eliminates CO2 emissions at the point of combustion is supported by the data. [1 mark]
Part not supported: H2 delivers −141.5 kJ g−1 while CH4 delivers −55.6 kJ g−1. Per gram, hydrogen delivers approximately 2.5× more energy — the claim that it delivers “essentially the same energy per gram” is incorrect. [1 mark]
Judgement: The claim is partially correct regarding CO2 elimination but significantly wrong about equal energy density. A better summary would be: hydrogen provides far more energy per gram but requires specialised high-pressure or cryogenic storage because H2 is an extremely low-density gas. The energy-per-gram advantage does not translate to energy-per-litre advantage at ambient pressure. [1 mark]
Q2 — Marking criteria (8 marks)
Point 1 — ΔH°rxn per mole (3 marks, 1 per correct calculation with working):
- Propellant A (N2H4): 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.
- Propellant B (NH3): ΣΔH°f(p) = 2(0) + 6(−285.8) = −1714.8; ΣΔH°f(r) = 4(−46.1) + 3(0) = −184.4; ΔH°rxn = −1714.8 − (−184.4) = −1530.4 kJ per 4 mol NH3, or −382.6 kJ mol−1 NH3.
- Propellant C (CH4): ΔH°rxn = −890.3 kJ mol−1 (see Q1).
Point 2 — kJ per gram (1 mark, all three correct): N2H4: 622.2 ÷ 32.0 = 19.4 kJ g−1; NH3: 382.6 ÷ 17.0 = 22.5 kJ g−1; CH4: 890.3 ÷ 16.0 = 55.6 kJ g−1.
Point 3 — Positive ΔH°f and propellant advantage (2 marks): ΔH°f[N2H4(l)] = +50.6 kJ mol−1 means hydrazine sits above the element baseline (N2 and H2) on the enthalpy scale. [1] When hydrazine burns, the −622.2 kJ mol−1 released comes from two sources: (i) the energy from forming the stable products (N2, H2O) and (ii) the extra +50.6 kJ stored in the hydrazine structure above the element baseline. This amplifies the combustion enthalpy compared to a fuel with a negative ΔH°f. Compare to NH3 (ΔH°f = −46.1): when NH3 forms, energy is already released, so some of the “downhill energy” from combustion was pre-released during formation — the net combustion enthalpy per mole is smaller. [1]
Point 4 — Greatest energy per gram + non-thermodynamic factor (1 mark): CH4 delivers the greatest energy per gram (55.6 kJ g−1). However, energy density per gram alone is insufficient: a rocket propellant must also ignite reliably in a vacuum (or be hypergolic — spontaneously igniting on contact with the oxidiser), have manageable storage properties, and not produce solid residues. Hydrazine is hypergolic with N2O4 — no ignition system is needed — which is critical for a restart-capable lunar descent engine. CH4 is not hypergolic and requires an ignition system.
Point 5 — Justified use of ΔH°f method (1 mark): Yes. The ΔH°f method uses experimentally measured values for actual substances in their real physical states, providing an accurate ΔH. The bond energy method uses average bond enthalpies and assumes all species are gaseous — for condensed-phase propellants such as N2H4(l), the gaseous assumption introduces significant error. When lives depend on the calculation, the more accurate method is required.