Heats of Combustion of Four Hydrocarbon Fuels: 21-Mat-A1 Thermodynamics, December 2018, Q5

Question 5: Heats of Combustion of Four Hydrocarbon Fuels (20 marks)

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Given.

Standard enthalpies of formation at 25°C (kJ/mol)
Species	$\Delta H_f^{\circ}$
CH₄	−75
C₃H₈	−105
C₆H₁₄	−199
C₈H₁₈	−259
CO₂	−394
H₂O(l)	−286

Find. (a)–(d) $\Delta H_c$ per mole for each fuel; (e) which fuel releases the most heat per unit mass.

Approach. Write the balanced complete-combustion reaction for each fuel to CO₂ and liquid H₂O, then apply Hess's law, $\Delta H_c=\sum\Delta H_f^{\circ}(\text{products})-\sum\Delta H_f^{\circ}(\text{reactants})$, with $\Delta H_f^{\circ}(\text{O}_2)=0$.

(a) CH₄. CH₄$+2$O₂$\rightarrow$CO₂$+2$H₂O: $$\Delta H_c=[(-394)+2(-286)]-(-75)=-966+75=\boxed{-891\ \text{kJ/mol}}.$$
(b) C₃H₈. C₃H₈$+5$O₂$\rightarrow3$CO₂$+4$H₂O: $$\Delta H_c=[3(-394)+4(-286)]-(-105)=-2326+105=\boxed{-2221\ \text{kJ/mol}}.$$
(c) C₆H₁₄. C₆H₁₄$+\tfrac{19}{2}$O₂$\rightarrow6$CO₂$+7$H₂O: $$\Delta H_c=[6(-394)+7(-286)]-(-199)=-4366+199=\boxed{-4167\ \text{kJ/mol}}.$$
(d) C₈H₁₈. C₈H₁₈$+\tfrac{25}{2}$O₂$\rightarrow8$CO₂$+9$H₂O: $$\Delta H_c=[8(-394)+9(-286)]-(-259)=-5726+259=\boxed{-5467\ \text{kJ/mol}}.$$
(e) Heat per unit mass. Divide each $|\Delta H_c|$ by the fuel's molar mass: $$\frac{891}{16.04}=\boxed{55.5\ \text{kJ/g}},\quad \frac{2221}{44.10}=\boxed{50.4\ \text{kJ/g}},\quad \frac{4167}{86.18}=\boxed{48.4\ \text{kJ/g}},\quad \frac{5467}{114.23}=\boxed{47.9\ \text{kJ/g}}$$ for CH₄, C₃H₈, C₆H₁₄ and C₈H₁₈ respectively. CH₄ (methane) releases the most heat per unit weight, because it has the highest hydrogen-to-carbon ratio of the four and H₂O formation contributes disproportionately more heat per unit mass than CO₂ formation.

Fuel	$\Delta H_c$ (kJ/mol)	Per unit mass (kJ/g)
CH₄	−891	55.5
C₃H₈	−2221	50.4
C₆H₁₄	−4167	48.4
C₈H₁₈	−5467	47.9
(e) Highest heat per unit weight	CH₄ (methane), 55.5 kJ/g

Concept & Theory

Topic. Hess's law applied to standard heats of combustion of hydrocarbon fuels.

Key relations. $\Delta H_c=\sum n_i\Delta H_{f,i}^{\circ}(\text{products})-\sum n_j\Delta H_{f,j}^{\circ}(\text{reactants})$, $\Delta H_f^{\circ}(\text{O}_2)=0$; combustion products are CO₂(g) and H₂O(l) for the higher (gross) heating value used here.

Why this works. Enthalpy is a state function, so the heat released by complete combustion depends only on the initial fuel/O₂ and final CO₂/H₂O states, not on the reaction mechanism — Hess's law lets the reaction be built as products-minus-reactants from tabulated formation enthalpies.

Common pitfall. Ranking fuels by $|\Delta H_c|$ per MOLE instead of per unit MASS for part (e) — C₈H₁₈ releases the most heat per mole (it is the largest molecule) but the LEAST per unit mass, because gravimetric heating value is what actually matters for fuel selection and it favors the lightest, most hydrogen-rich molecule.

Source. Gaskell, Introduction to the Thermodynamics of Materials, Ch. 5 (Hess's law and enthalpies of formation/combustion).

Question 5 of 7: Heats of Combustion of Four Hydrocarbon Fuels

Question 5: Heats of Combustion of Four Hydrocarbon Fuels (20 marks)