Heats of Combustion from Standard Enthalpies of Formation: 21-Mat-A1 Thermodynamics, May 2014, Q4

Question 4: Heats of Combustion from Standard Enthalpies of Formation (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Standard enthalpies of formation at 25 °C (all products of complete combustion are CO₂(g) and H₂O(l), and O₂ is a pure element with $\Delta H^\circ_f=0$).

Species	ΔH°_f (kJ/mol)
C₂H₆ (ethane)	−85
C₂H₅OH (ethanol)	−277
C₃H₈ (propane)	−105
C₄H₁₀ (butane)	−126
CO₂	−394
H₂O	−286

Find. $\Delta H_c$ per mole for each of the four fuels, and which one releases the most heat per unit mass.

Approach. Write the balanced combustion equation for each fuel, then apply Hess's law, $\Delta H_c=\sum\Delta H^\circ_{f,\text{products}}-\sum\Delta H^\circ_{f,\text{reactants}}$, with $\Delta H^\circ_f(\text{O}_2)=0$. Convert each molar heat of combustion to a per-gram basis using the molar mass to answer part (e).

(a) Ethane. $\text{C}_2\text{H}_6+\tfrac72\text{O}_2\rightarrow2\text{CO}_2+3\text{H}_2\text{O}$: $$\Delta H_c=[2(-394)+3(-286)]-(-85)=-1646+85=\boxed{-1561\ \text{kJ/mol}}.$$
(b) Ethanol. $\text{C}_2\text{H}_5\text{OH}+3\text{O}_2\rightarrow2\text{CO}_2+3\text{H}_2\text{O}$: $$\Delta H_c=[2(-394)+3(-286)]-(-277)=-1646+277=\boxed{-1369\ \text{kJ/mol}}.$$
(c) Propane. $\text{C}_3\text{H}_8+5\text{O}_2\rightarrow3\text{CO}_2+4\text{H}_2\text{O}$: $$\Delta H_c=[3(-394)+4(-286)]-(-105)=-2326+105=\boxed{-2221\ \text{kJ/mol}}.$$
(d) Butane. $\text{C}_4\text{H}_{10}+\tfrac{13}{2}\text{O}_2\rightarrow4\text{CO}_2+5\text{H}_2\text{O}$: $$\Delta H_c=[4(-394)+5(-286)]-(-126)=-3006+126=\boxed{-2880\ \text{kJ/mol}}.$$
(e) Per unit mass. Dividing $-\Delta H_c$ by the molar mass ($M_{\text{C}_2\text{H}_6}=30.07$, $M_{\text{C}_2\text{H}_5\text{OH}}=46.07$, $M_{\text{C}_3\text{H}_8}=44.10$, $M_{\text{C}_4\text{H}_{10}}=58.12$ g/mol) gives the heat released per gram of fuel burned, listed in the results table. Ethane's short, hydrogen-rich, low-molar-mass molecule gives it the highest heat release per gram: $\boxed{\text{C}_2\text{H}_6,\ \approx51.9\ \text{kJ/g}}$.

Fuel	ΔH_c (kJ/mol)	Heat release (kJ/g)
(a) C₂H₆ ethane	−1561	51.9
(b) C₂H₅OH ethanol	−1369	29.7
(c) C₃H₈ propane	−2221	50.4
(d) C₄H₁₀ butane	−2880	49.6
(e) Highest per unit weight	Ethane (C₂H₆), ≈ 51.9 kJ/g

Concept & Theory

Topic. Hess's law applied to heats of combustion, and the distinction between molar and gravimetric fuel value.

Key relations. $\Delta H_{rxn}=\sum\nu_i\Delta H^\circ_{f,i}(\text{products})-\sum\nu_i\Delta H^\circ_{f,i}(\text{reactants})$; gravimetric heating value $=-\Delta H_c/M$.

Why this works. Enthalpy is a state function, so the heat of combustion depends only on the initial (fuel + O₂) and final (CO₂ + H₂O) states, never on the reaction mechanism — this is exactly why tabulated formation enthalpies can be recombined arbitrarily. The ethanol data point illustrates the effect of an already-oxidized carbon: ethanol's O-H and C-O bonds mean less oxygen has to be supplied and less energy is released per mole (and per gram) than for the corresponding hydrocarbon.

Common pitfall. Ranking the fuels by molar (not gravimetric) heat of combustion for part (e) — butane releases the most heat per mole (largest molecule) but not the most per gram, because its molar mass is also the largest of the four.

Source. Gaskell, Introduction to the Thermodynamics of Materials, Ch. 6 (heats of reaction, Hess's law and standard enthalpies of formation).

Question 4 of 7: Heats of Combustion from Standard Enthalpies of Formation

Question 4: Heats of Combustion from Standard Enthalpies of Formation (20 marks)