Chemistry • Year 11 • Module 3 • Lesson 4
Combustion Reactions
Apply combustion chemistry to real fuel data, balancing equations, cause-and-effect reasoning, and a scenario involving Australian energy sources.
1. Interpret energy data — heats of combustion of Australian fuels
The bar chart below shows the heat of combustion (energy released per gram) for five fuels used or exported by Australia. Data are approximate standard values. 6 marks
Source: Approximate standard heats of combustion. Methane and propane as LNG/LPG; octane as petrol surrogate; black coal (NSW/QLD export grade); ethanol (E10 transport fuel blend). Data from NIST and IEA.
1.1 Identify the fuel with the highest heat of combustion per gram and the fuel with the lowest. State the difference in kJ g⁻¹. 2 marks
1.2 Methane has the highest heat of combustion per gram in the chart, and it is the main component of natural gas used in Australian gas heaters. Using the graph data and the lesson definition of heat of combustion, explain what advantage methane has over octane (petrol) in terms of energy released per gram of fuel burned. 2 marks
1.3 The graph shows that ethanol releases approximately 29.7 kJ per gram when burned completely, compared with 47.9 kJ per gram for octane. Using the heat of combustion data from the graph, explain why a vehicle would need to burn more ethanol than octane to travel the same distance. In your answer, use the lesson definition of heat of combustion. 2 marks
2. Interpret combustion data — products and combustion type
A laboratory investigation analysed the gas-phase products from burning four fuels under different oxygen conditions. The table below shows the molar ratio of CO₂ to CO in the exhaust gases. 7 marks
| Fuel | Oxygen condition | CO₂ : CO ratio (molar) | Soot observed? |
|---|---|---|---|
| Methane (CH₄) | Excess O₂ | All CO₂, no CO | No |
| Methane (CH₄) | Limited O₂ | 1 : 1 | No |
| Propane (C₃H₈) | Excess O₂ | All CO₂, no CO | No |
| Propane (C₃H₈) | Limited O₂ | 2 : 1 | Yes |
2.1 Identify which row(s) represent complete combustion and which represent incomplete combustion. Give the evidence from the table. 2 marks
2.2 Propane in limited O₂ shows a CO₂ : CO ratio of 2:1. Write a balanced equation for the incomplete combustion of propane (C₃H₈) that produces both CO₂ and CO in a 2:1 molar ratio, along with water. Show your atom check. 3 marks
2.3 The lesson states that the products of incomplete combustion include CO gas and/or solid carbon soot (C(s)), depending on how limited the oxygen supply is. Using this information, describe the two possible outcomes for carbon atoms when oxygen is severely limited during incomplete combustion, and identify which product (CO or C soot) requires less oxygen per carbon atom. 2 marks
3. Cause-and-effect chain — coal seam gas well with restricted air
A coal seam gas (CSG) facility in Queensland burns methane from a well head. A maintenance fault reduces the air supply to the burner stack. Trace the chain of consequences using the cause boxes below. 5 marks (1 per effect)
Cause: Air supply to the methane burner is reduced by 50%.
Effect 1: The oxygen-to-fuel ratio ...
Effect 2: Combustion shifts from complete to ...
Effect 3: Products formed include ...
Effect 4: Workers nearby are at risk of ...
Overall outcome (so…): The facility must ...
4. Predict and justify — prescribed burns and combustion chemistry
Land managers conducting a prescribed burn in southern Victoria wait for conditions of low humidity, moderate temperature, and a steady 15 km/h breeze before igniting. A colleague questions why wind conditions matter so much for a burn that will “just produce smoke and ash anyway.” 4 marks
4.1 Predict how the steady wind affects the type of combustion occurring in the burn, and justify this using combustion chemistry. 3 marks
4.2 Predict how the combustion type during a windy prescribed burn affects the safety of smoke for communities downwind. 1 mark
Q1.1 — Highest and lowest energy fuel (2 marks)
Highest: methane at 55.6 kJ g⁻¹. Lowest: ethanol at 29.7 kJ g⁻¹. Difference: 55.6 − 29.7 = 25.9 kJ g⁻¹. [1 mark both identified; 1 mark difference with units]
Q1.2 — Methane energy advantage over octane (2 marks)
Methane has a heat of combustion of 55.6 kJ g⁻¹ compared with 47.9 kJ g⁻¹ for octane [1]. This means that each gram of methane releases more energy when burned completely, so less mass of methane is needed to produce the same energy output — an advantage in terms of energy released per gram of fuel [1].
Q1.3 — Ethanol vs octane heat of combustion (2 marks)
Ethanol releases 29.7 kJ per gram when burned completely, while octane releases 47.9 kJ per gram — ethanol provides approximately 38% less energy per gram [1]. Therefore, to release the same total energy (needed to travel the same distance), a vehicle must burn a greater mass of ethanol than octane. The heat of combustion is the energy released per gram in complete combustion, so the lower value for ethanol directly means more fuel mass is consumed per kilometre [1].
Q2.1 — Complete vs incomplete combustion identification (2 marks)
Rows 1 and 3 (excess O₂ conditions for methane and propane) represent complete combustion — evidence: CO₂:CO ratio shows only CO₂ produced, no CO and no soot [1]. Rows 2 and 4 (limited O₂) represent incomplete combustion — evidence: CO is present in the exhaust gases, and soot is observed for propane [1].
Q2.2 — Balanced incomplete combustion of propane 2:1 CO₂:CO (3 marks)
We need 2 CO₂ and 1 CO from 3 carbon atoms → one C₃H₈ molecule works directly.
C₃H₈(g) + 4O₂(g) → 2CO₂(g) + CO(g) + 4H₂O(g) [1 for correctly written equation]
Atom check: Left → 3C, 8H, 8O. Right → 3C, 8H, 4+1+4=9O. — that’s not balanced yet.
Recalculate O on right: 2×2 + 1 + 4×1 = 4+1+4 = 9O atoms. Left: need 4.5 O₂ → multiply through by 2:
2C₃H₈(g) + 9O₂(g) → 4CO₂(g) + 2CO(g) + 8H₂O(g) [1 for correct balanced equation]
Atom check: Left → 6C, 16H, 18O. Right → 6C, 16H, 8+2+8=18O. Correct. CO₂:CO = 4:2 = 2:1. [1 for atom check confirmation]
Q2.3 — Carbon outcomes under severely limited oxygen (2 marks)
When oxygen is limited, carbon in the fuel can be only partially oxidised to CO(g) rather than fully oxidised to CO₂(g); or, with extreme oxygen limitation, carbon may remain as solid soot C(s) instead [1]. Forming C(s) soot requires less oxygen per carbon atom than forming CO (no oxygen is consumed for a carbon atom that stays as solid carbon), so soot is the product under the most severe oxygen restriction [1].
Q3 — Cause-and-effect chain (5 marks, 1 each)
Effect 1: The oxygen-to-fuel ratio falls below the amount required for complete combustion [1]. Effect 2: Combustion shifts from complete to incomplete combustion [1]. Effect 3: Products formed include carbon monoxide (CO) and possibly soot (C), as well as H₂O and some CO₂ [1]. Effect 4: Workers nearby are at risk of CO poisoning — CO binds haemoglobin with ∼200× the affinity of O₂, causing hypoxia [1]. Overall outcome: The facility must shut down the burner and restore adequate air supply (or ventilate the area and evacuate workers) to prevent fatalities [1].
Q4.1 — Wind and combustion type (3 marks)
A steady 15 km/h breeze continuously replenishes oxygen at the burning fuel surface, maintaining a high oxygen-to-fuel ratio [1]. This promotes complete combustion, where all carbon is oxidised to CO₂ and all hydrogen to H≶O [1]. Without wind, oxygen would be depleted in the enclosed space around smouldering material, causing incomplete combustion and accumulation of toxic CO gas [1].
Q4.2 — Smoke safety downwind (1 mark)
Complete combustion produces CO₂ and H₂O rather than CO and particulate carbon soot, so smoke from a windy prescribed burn contains far lower concentrations of toxic gases and fine particles, making it significantly safer for downwind communities and less likely to trigger CO poisoning or severe respiratory effects [1].