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Fuels and enthalpy of combustion (SACE Stage 2 Chemistry)

Syllabus dot point

“Define and calculate the enthalpy of combustion of fuels, and compare fuels by their energy content per gram and per mole.”

9 min read

Quick answer

The enthalpy of combustion is the heat released when one mole of a fuel burns completely in oxygen. It is found by calorimetry: q=mcΔTq = mc\Delta T for the water heated, then ΔHc=−q/n\Delta H_c = -q/n for the fuel. Fuels are compared per mole (theoretical) or, more usefully for real use, per gram, by dividing the molar value by the molar mass. Energy per gram favours fuels with a high hydrogen-to-carbon ratio, such as methane and hydrogen.

Jump to a section
  1. What this dot point is asking
  2. Lead worked calculation
  3. Defining enthalpy of combustion
  4. Comparing fuels: per mole versus per gram
  5. Complete versus incomplete combustion
  6. Renewable and alternative fuels
  7. Why it matters for managing resources
  8. Exam-style questions

What this dot point is asking

SACE expects you to define and calculate enthalpy of combustion, compare fuels per mole and per gram, and discuss complete versus incomplete combustion and renewable fuels.

Lead worked calculation

Enthalpy of combustion from calorimetry

Burning 0.860 g0.860\ \text{g} of propan-1-ol heats 200.0 g200.0\ \text{g} of water by 30.5 ∘C30.5\ ^\circ\text{C}. Find the experimental molar enthalpy of combustion. (c=4.18 J g−1 ∘C−1c = 4.18\ \text{J g}^{-1}\ ^\circ\text{C}^{-1}; M(C3H7OH)=60.1 g mol−1M(\text{C}_3\text{H}_7\text{OH}) = 60.1\ \text{g mol}^{-1}.)

Heat gained by the water

q=mcΔT=200.0×4.18×30.5=2.550×104 J=25.50 kJq = mc\Delta T = 200.0 \times 4.18 \times 30.5 = 2.550 \times 10^{4}\ \text{J} = 25.50\ \text{kJ}

Moles of fuel burned

n=mM=0.86060.1=1.431×10−2 moln = \frac{m}{M} = \frac{0.860}{60.1} = 1.431 \times 10^{-2}\ \text{mol}

Molar enthalpy of combustion

ΔHc=−qn=−25.501.431×10−2=−1.78×103 kJ mol−1\Delta H_c = -\frac{q}{n} = -\frac{25.50}{1.431 \times 10^{-2}} = -1.78 \times 10^{3}\ \text{kJ mol}^{-1}

The negative sign shows combustion is exothermic. The experimental value is lower in magnitude than the data-book figure because heat is lost to the surroundings and the apparatus.

Defining enthalpy of combustion

Enthalpy of combustion

The molar enthalpy of combustion (ΔHc\Delta H_c) is the energy released when one mole of a substance is completely burned in oxygen under standard conditions. It is always negative because combustion is exothermic. For a fuel, the more energy released per mole (and per gram), the better the fuel for energy supply.

Comparing fuels: per mole versus per gram

A fair comparison depends on the basis chosen.

Energy per gram is the practical measure

Comparing fuels per mole can mislead, because large molecules release more energy per mole simply by being bigger. The fairer practical measure is energy per gram (or per litre for liquids), found by dividing the molar enthalpy by the molar mass: ∣ΔHc∣M\dfrac{|\Delta H_c|}{M}. On a per-gram basis, fuels with a high hydrogen-to-carbon ratio (methane, hydrogen) release the most energy, because forming water from hydrogen is very exothermic.

For transport and portable use, energy per unit mass and per unit volume both matter, which is why hydrogen (very high energy per gram) still poses storage challenges (low energy per litre as a gas).

Complete versus incomplete combustion

  • Complete combustion (plenty of oxygen) gives carbon dioxide and water and releases the maximum energy: CH4+2O2→CO2+2H2O\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}.
  • Incomplete combustion (limited oxygen) produces carbon monoxide and/or soot (carbon) and releases less energy: 2CH4+3O2→2CO+4H2O2\text{CH}_4 + 3\text{O}_2 \rightarrow 2\text{CO} + 4\text{H}_2\text{O}. The toxic CO\text{CO} and particulates are also pollutants.

Incomplete combustion both wastes energy and creates hazards, so well-designed burners ensure a good air supply.

Renewable and alternative fuels

Biofuels such as bioethanol (from fermenting sugars) and biodiesel (esters from plant oils) are renewable because the CO2\text{CO}_2 released on burning was recently absorbed from the atmosphere by the plant, giving them a smaller net carbon footprint than fossil fuels. Hydrogen burns to give only water (2H2+O2→2H2O2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}) with no CO2\text{CO}_2, but its overall sustainability depends on how it is produced. Evaluating a fuel means weighing energy content against emissions, renewability and practicality.

Using the fuel mass in q = mcΔT, or comparing per mole only

The most common SACE error is putting the mass of fuel into q=mcΔTq = mc\Delta T; that equation uses the mass of water being heated. The fuel mass is only used to find its moles for ΔHc=−q/n\Delta H_c = -q/n. A second error is comparing fuels per mole and concluding the bigger molecule is "better"; convert to energy per gram for a fair comparison. Always assign the negative sign to ΔHc\Delta H_c (combustion is exothermic) and convert qq to kJ\text{kJ} before dividing.

Why it matters for managing resources

Quantifying and comparing the energy content of fuels guides the choice of energy sources, balancing energy density against cost, supply and greenhouse emissions. This connects combustion chemistry directly to the resource and sustainability decisions at the heart of Topic 4.

Exam-style questions

Questions in the style of SACE Board exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

SACE 2022-style5 marks
In a calorimetry experiment, burning 1.15 g1.15\ \text{g} of ethanol raised the temperature of 250.0 g250.0\ \text{g} of water by 24.5 ∘C24.5\ ^\circ\text{C}. (a) Calculate the experimental molar enthalpy of combustion of ethanol. (b) The data-book value is −1367 kJ mol−1-1367\ \text{kJ mol}^{-1}; calculate the percentage of the energy transferred to the water. (c=4.18 J g−1 ∘C−1c = 4.18\ \text{J g}^{-1}\ ^\circ\text{C}^{-1}; M(C2H5OH)=46.07 g mol−1M(\text{C}_2\text{H}_5\text{OH}) = 46.07\ \text{g mol}^{-1}.)
Show worked answer →

(a) q=mcΔT=250.0×4.18×24.5=2.560×104 J=25.60 kJq = mc\Delta T = 250.0 \times 4.18 \times 24.5 = 2.560 \times 10^{4}\ \text{J} = 25.60\ \text{kJ}. n(ethanol)=1.1546.07=2.496×10−2 moln(\text{ethanol}) = \dfrac{1.15}{46.07} = 2.496 \times 10^{-2}\ \text{mol}. ΔHc=−25.602.496×10−2=−1.03×103 kJ mol−1\Delta H_c = -\dfrac{25.60}{2.496 \times 10^{-2}} = -1.03 \times 10^{3}\ \text{kJ mol}^{-1}. (3 marks)

(b) Percentage transferred =10261367×100=75.1%= \dfrac{1026}{1367} \times 100 = 75.1\%. (2 marks)

SACE 2019-style4 marks
Compare methane (CH4\text{CH}_4) and octane (C8H18\text{C}_8\text{H}_{18}) as fuels by calculating the energy released per gram, given molar enthalpies of combustion ΔHc(CH4)=−890 kJ mol−1\Delta H_c(\text{CH}_4) = -890\ \text{kJ mol}^{-1} and ΔHc(C8H18)=−5470 kJ mol−1\Delta H_c(\text{C}_8\text{H}_{18}) = -5470\ \text{kJ mol}^{-1}. (M(CH4)=16.0M(\text{CH}_4) = 16.0, M(C8H18)=114.0 g mol−1M(\text{C}_8\text{H}_{18}) = 114.0\ \text{g mol}^{-1}.) State which releases more energy per gram.
Show worked answer →

Methane: energy per gram =89016.0=55.6 kJ g−1= \dfrac{890}{16.0} = 55.6\ \text{kJ g}^{-1}. (2 marks)

Octane: energy per gram =5470114.0=48.0 kJ g−1= \dfrac{5470}{114.0} = 48.0\ \text{kJ g}^{-1}. (1 mark)

Methane releases more energy per gram (55.655.6 versus 48.0 kJ g−148.0\ \text{kJ g}^{-1}), because it has the higher hydrogen-to-carbon ratio. (1 mark)

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Sources & how we know this

Based on the official SACE Board subject outline. Not yet individually reviewed by a person; always check SACE Board and your school's materials for current rules.

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