Heats of Combustion of Four Fuels, Per Mole and Per Unit Mass: 21-Mat-A1 Thermodynamics, December 2015, Q5

Question 5: Heats of Combustion of Four Fuels, Per Mole and Per Unit Mass (20 marks)

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Compound	Formula	$\Delta H^\circ_f$ (kJ/mol)	Molar mass (g/mol)
Methane	CH₄	−75	16.04
Methanol	CH₄O	−201	32.04
Ethanol	C₂H₆O	−235	46.07
Hexane	C₆H₁₄	−199	86.18
Carbon dioxide	CO₂	−394	—
Water (liquid)	H₂O	−286	—

Find. $\Delta H_c$ per mole for each fuel, and which fuel releases the most heat per gram burned.

Approach. Write each combustion reaction to CO₂(g) and liquid H₂O, then $\Delta H_c=\sum n_i\Delta H^\circ_{f,products}-\Delta H^\circ_{f,fuel}$ (O₂ has $\Delta H^\circ_f=0$). Divide each $|\Delta H_c|$ by the fuel's molar mass to compare heat release per unit weight.

(a) CH₄ + 2O₂ → CO₂ + 2H₂O(l). $$\Delta H_c=\left[-394+2(-286)\right]-(-75)=-966+75=\boxed{-891\ \text{kJ/mol}}.$$
(b) CH₄O + ⅓/2O₂ → CO₂ + 2H₂O(l). $$\Delta H_c=\left[-394+2(-286)\right]-(-201)=-966+201=\boxed{-765\ \text{kJ/mol}}.$$
(c) C₂H₆O + 3O₂ → 2CO₂ + 3H₂O(l). $$\Delta H_c=\left[2(-394)+3(-286)\right]-(-235)=-1646+235=\boxed{-1411\ \text{kJ/mol}}.$$
(d) C₆H₁₄ + 19/2 O₂ → 6CO₂ + 7H₂O(l). $$\Delta H_c=\left[6(-394)+7(-286)\right]-(-199)=-4366+199=\boxed{-4167\ \text{kJ/mol}}.$$
(e) Heat released per gram of fuel. Divide each $|\Delta H_c|$ by molar mass: $$\frac{891}{16.04}=55.5,\quad \frac{765}{32.04}=23.9,\quad \frac{1411}{46.07}=30.6,\quad \frac{4167}{86.18}=48.4\ \ \text{(all kJ/g)}.$$ $$\boxed{\text{Methane (CH}_4\text{) generates the most heat per gram: } 55.5\ \text{kJ/g}.}$$ Although hexane and ethanol release far more heat per mole, methane's low molecular weight (mostly hydrogen by atom count, and no oxygen already "spent" in its own structure) gives it the highest heating value on a mass basis — the same reason natural gas outperforms liquid hydrocarbons per kilogram.

Fuel	$\Delta H_c$ (kJ/mol)	$\Delta H_c$ (kJ/g)
(a) CH₄	−891	55.5
(b) CH₄O	−765	23.9
(c) C₂H₆O	−1411	30.6
(d) C₆H₁₄	−4167	48.4
(e) Highest per gram	CH₄ (methane), 55.5 kJ/g

Concept & Theory

Topic. Standard heat of combustion from tabulated formation enthalpies, and the distinction between molar and mass-specific heating value.

Key relations. $\Delta H_c=\sum n_i\Delta H^\circ_{f,products}-\sum n_i\Delta H^\circ_{f,reactants}$ (elements and O₂ have $\Delta H^\circ_f=0$); mass-specific value $=|\Delta H_c|/M$.

Why this works. Enthalpy is a state function, so Hess's law lets $\Delta H_c$ be built purely from tabulated formation enthalpies without needing to know the reaction mechanism. Ranking fuels "per mole" and "per gram" can disagree because molar mass varies far more between these four fuels than their combustion enthalpy per mole of carbon — the practical (economic, tank-volume) comparison is almost always the mass or volume basis, not the molar one.

Common pitfall. Answering part (e) using the molar $\Delta H_c$ ranking (hexane highest) instead of converting to a per-gram basis as the question actually asks for "per unit weight."

Source. Gaskell, Introduction to the Thermodynamics of Materials, Ch. 6 (heats of formation and combustion).

Question 5 of 7: Heats of Combustion of Four Fuels, Per Mole and Per Unit Mass

Question 5: Heats of Combustion of Four Fuels, Per Mole and Per Unit Mass (20 marks)