Chemistry • Year 11 • Module 4 • Lesson 2
Calorimetry — Combustion
Synthesise, evaluate and justify at Band 5–6 level: extended multi-criteria responses requiring calculation, data interpretation and evidence-based judgement.
1. Comparing fuels for the Karratha LNG export industry — data-driven evaluation
Scenario. Engineers at a Karratha, Western Australia LNG facility are evaluating four candidate fuels to power on-site electricity generation. The site is remote and supply logistics favour fuels with higher energy density per kilogram. The engineers ran a series of bomb calorimeter tests and recorded the results below. They are also considering the viability of blending ethanol (sourced from Australian agricultural waste) as a partial substitute for methane in some generator turbines.
| Fuel | Formula | M (g mol−1) |
Accepted ΔHc (kJ mol−1) |
Energy density (kJ g−1) |
CO2 emitted per kJ released (g kJ−1) |
|---|---|---|---|---|---|
| Methane (LNG) | CH4 | 16.04 | −890 | 55.5 | 0.050 |
| Ethanol | C2H5OH | 46.07 | −1367 | 29.7 | 0.064 |
| Propane | C3H8 | 44.10 | −2220 | 50.3 | 0.060 |
| Octane (petrol) | C8H18 | 114.23 | −5471 | 47.9 | 0.069 |
Table data: standard ΔHc values from NESA data booklet / CRC Handbook. CO2 emission factors calculated from stoichiometry of complete combustion. Energy density = |ΔHc|/M.
In a well-structured extended response, answer the following prompt. 8 marks
Extended response prompt: Evaluate which of the four fuels is most suitable for powering on-site electricity generation at the Karratha LNG facility, taking into account energy density, CO2 emission intensity, and supply logistics in a remote site context.
In your response you must:
- Define molar enthalpy of combustion and explain the sign convention.
- Calculate the energy density of propane and ethanol to verify the values in the table (show working for both).
- Compare all four fuels on three criteria: energy density per gram, CO2 emission per kJ, and a third criterion of your choice.
- Use a specific named example or calculation to justify your recommendation.
- Reach an evidence-based judgement, acknowledging at least one trade-off or limitation.
2. Multi-step calorimetry calculation — ethanol vs octane: a quantitative comparison
Scenario. A Year 11 class investigates whether ethanol (C2H5OH, M = 46.07 g mol−1) or octane (C8H18, M = 114.23 g mol−1) delivers more heat to a fixed mass of water for the same mass of fuel burned. They run two spirit burner experiments side by side, each using 250.0 g of water, burning exactly 1.000 g of each fuel. Their measured ΔT values are recorded below alongside the bomb calorimeter accepted values.
| Fuel | Mass burned (g) | Water mass (g) | Measured ΔT (°C) | Accepted ΔHc (kJ mol−1) |
|---|---|---|---|---|
| Ethanol | 1.000 | 250.0 | 14.2 | −1367 |
| Octane | 1.000 | 250.0 | 16.8 | −5471 |
(a) For each fuel, calculate: (i) q in kJ; (ii) experimental ΔHc in kJ mol−1; (iii) percentage error relative to the accepted value. Show all working. 6 marks (3 per fuel)
(b) The octane experiment shows a larger ΔT per gram than ethanol (16.8 vs 14.2°C), yet the percentage error for octane is substantially larger. Explain why the larger ΔT does not mean octane’s experimental ΔHc is more accurate, and identify two specific physical reasons why octane would be expected to show a larger percentage error in a spirit burner experiment. 4 marks
(c) A student proposes that they could improve the accuracy of the octane result by using a much larger mass of water (e.g. 1000 g instead of 250 g) so the temperature rise is more easily read on the thermometer. Evaluate this proposal, identifying one way it would help and one way it introduces a new problem. 2 marks
Q1 — Karratha fuel evaluation (marking guidelines)
Mark allocation (8 marks):
1 mark — Definition + sign convention: Molar enthalpy of combustion is the enthalpy change when one mole of a substance undergoes complete combustion in oxygen under standard conditions; it is negative (exothermic) because energy is released to the surroundings, making the system’s enthalpy lower after combustion than before.
2 marks — Verification calculations:
Propane: |ΔHc|/M = 2220/44.10 = 50.3 kJ g−1 ✓ [1 mark]
Ethanol: 1367/46.07 = 29.7 kJ g−1 ✓ [1 mark]
3 marks — Three-criteria comparison:
(i) Energy density per gram: Methane (55.5) > propane (50.3) > octane (47.9) > ethanol (29.7) kJ g−1; methane delivers by far the most energy per kg, ideal for a remote site where fuel mass is a logistical constraint. [1 mark]
(ii) CO2 per kJ: Methane (0.050) < propane (0.060) < ethanol (0.064) < octane (0.069) g kJ−1; methane emits the least CO2 per kJ, partly because of its high H:C ratio (more energy comes from C–H bonds relative to C–C bonds). [1 mark]
(iii) Third criterion (accept any of): supply availability/infrastructure (LNG is already on-site at Karratha — no additional procurement); handling safety (LNG requires cryogenic storage vs ambient liquid fuels); renewability (ethanol can be produced from agricultural waste, relevant to long-term sustainability). [1 mark]
1 mark — Named evidence / calculation to justify recommendation: Methane releases 55.5 kJ g−1 compared to the next-best propane at 50.3 kJ g−1; for a remote site generating, say, 10,000 kJ of electricity (at 35% turbine efficiency, requiring ~28,600 kJ of fuel energy), methane requires 515 g versus 569 g of propane — a saving of 54 g per 10,000 kJ generated, compounding significantly at industrial scale.
1 mark — Evidence-based judgement with trade-off: Methane (LNG) is the most suitable fuel for the Karratha site, given its highest energy density and lowest CO2 intensity of the four fuels, compounded by the site-specific advantage that LNG infrastructure is already in place. The key trade-off is the capital and safety cost of cryogenic storage at −162°C; in the long term, partial ethanol blending could reduce net CO2 emissions if sourced from waste biomass, but at a significant penalty in energy density per kilogram of fuel shipped.
Q2 — Ethanol vs octane multi-step calculation
(a) Ethanol:
q = 250.0 × 4.18 × 14.2 = 14,839 J = 14.84 kJ [1]
n = 1.000/46.07 = 0.02170 mol; ΔHc = −14.84/0.02170 = −684 kJ mol−1 [1]
% error = |−684 − (−1367)|/1367 × 100 = 683/1367 × 100 = 50.0% [1]
Octane:
q = 250.0 × 4.18 × 16.8 = 17,556 J = 17.56 kJ [1]
n = 1.000/114.23 = 0.008755 mol; ΔHc = −17.56/0.008755 = −2006 kJ mol−1 [1]
% error = |−2006 − (−5471)|/5471 × 100 = 3465/5471 × 100 = 63.3% [1]
(b) A larger ΔT per gram only means more heat reached the water per gram burned — but percentage error measures how much of the total available energy was captured. Because octane’s accepted ΔHc is ~4× larger per mole than ethanol’s, the absolute heat gap between true and measured is enormous, giving a larger % error even with a bigger ΔT. [1 mark for this reasoning.] Two specific reasons for octane’s larger error: (1) Incomplete combustion: octane has a high C:H ratio (C8H18); burning 8 carbons per mole in an open-air flame demands more O2 than the diffuse flame can supply, leading to visible soot and CO, releasing less energy per mole. (2) Greater heat flux / radiation loss: octane’s much higher heat of combustion per mole means the flame is hotter and more energy radiates outward per unit time than during ethanol combustion, proportionally losing more heat to surroundings before it can reach the water. [2 marks, 1 per reason.] [Total 3 marks for part (b).]
(c) Benefit: Using more water gives a smaller ΔT for the same heat input, which makes the percentage uncertainty in reading ΔT smaller if the thermometer has a fixed precision (e.g. ±0.1°C). New problem: A larger mass of water also has a larger heat capacity, so the water takes longer to reach thermal equilibrium with the surroundings; this increases the proportion of combustion energy lost to the environment (larger absolute heat capacity means more energy is needed to raise temperature, but proportionally the water is also a larger “cold sink” competing with heat retention). In practice, the effect is neutral or makes accuracy worse because heat losses scale with time, not water mass alone. [1 mark each.]