Enthalpy of Formation

Q1 — Sequence the steps (correct order)

1 Write the balanced equation with state symbols. → 2 List the ΔH°_f value for every species from the data table. → 3 Include elements; write ΔH°_f = 0 explicitly. → 4 Multiply each ΔH°_f(product) by its coefficient and sum: ΣΔH°_f(products). → 5 Multiply each ΔH°_f(reactant) by its coefficient and sum: ΣΔH°_f(reactants). → 6 Subtract ΣΔH°_f(reactants) from ΣΔH°_f(products) to give ΔH°_rxn. → 7 Check the sign (negative = exothermic; positive = endothermic). → 8 Report with units and sign.

Q2 — Graph interpretation

(a) CO₂(g) at −393.5 kJ mol⁻¹. This indicates CO₂(g) is the most thermodynamically stable substance shown — it sits lowest on the energy scale relative to its elements, meaning the most energy was released when it formed. It is very difficult to decompose back to carbon and oxygen.

(b) C₂H₄(g) (+52.4) and N₂H₄(l) (+50.6). A positive ΔH°_f means these compounds sit above the element baseline — they store more energy than their constituent elements. When they combust, this stored energy is released in addition to the energy from forming the products, amplifying the total enthalpy released per mole burned. This makes them energy-dense fuels/propellants.

(c) Difference = −285.8 − (−241.8) = −44.0 kJ mol⁻¹. Water vapour has 44 kJ mol⁻¹ less energy than liquid water — this is approximately the enthalpy of condensation (vapourisation). The bond energy method assumes water is produced as a gas, so for reactions producing liquid water it misses ~44 kJ per mole of H₂O — this systematically under-estimates the exothermicity. For CH₄ combustion (2 mol H₂O), the error is ~88 kJ mol⁻¹.

(d) Estimated from chart: CO(g) ≈ −110 kJ mol⁻¹ (exact: −110.5). CO₂ is more stable by −393.5 − (−110.5) = −283.0 kJ mol⁻¹. CO₂ is 283 kJ mol⁻¹ more thermodynamically stable than CO.

Q3 — Australian context: E10 and natural gas

(a) (i) C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l)    (ii) CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)

(b) Methane: ΣΔH°_f(products) = 1(−393.5) + 2(−285.8) = −393.5 − 571.6 = −965.1; ΣΔH°_f(reactants) = 1(−74.8) + 2(0) = −74.8; ΔH°_rxn = −965.1 − (−74.8) = −890.3 kJ mol⁻¹.
Ethanol: ΣΔH°_f(products) = 2(−393.5) + 3(−285.8) = −787.0 − 857.4 = −1644.4; ΣΔH°_f(reactants) = 1(−277.7) + 3(0) = −277.7; ΔH°_rxn = −1644.4 − (−277.7) = −1366.7 kJ mol⁻¹.

(c) Methane: 890.3 ÷ 1 mol CO₂ = 890.3 kJ per mol CO₂. Ethanol: 1366.7 ÷ 2 mol CO₂ = 683.4 kJ per mol CO₂. On this basis, methane releases more energy per mole of CO₂ produced. The student’s claim that ethanol is “worse” has some support from this metric, but the claim based solely on the number of carbons is an oversimplification — the ΔH°_f of each fuel and the state of products must be considered.

Q4 — Carbon monoxide vs carbon dioxide

(a) Reaction 1: ΔH° = ΔH°_f[CO(g)] − [ΔH°_f(C) + ½ΔH°_f(O₂)] = −110.5 − 0 = −110.5 kJ mol⁻¹. Reaction 2: ΔH° = ΔH°_f[CO₂(g)] − [ΔH°_f(C) + ΔH°_f(O₂)] = −393.5 − 0 = −393.5 kJ mol⁻¹.

(b) Reaction 2 (full oxidation to CO₂) releases 393.5 kJ mol⁻¹ compared to 110.5 kJ mol⁻¹ for Reaction 1. Full oxidation to CO₂ releases 3.56× more energy per mole of carbon.

(c) Trade-off 1: deliberately producing CO means losing 283 kJ mol⁻¹ of useful energy per carbon — the plant generates significantly less electricity. Trade-off 2: CO is a highly toxic gas that poses serious workplace safety hazards during capture/transport, whereas CO₂ is relatively inert at low concentrations.

Q5 — Comparison table

Feature	Bond energy method	ΔH°f method
Data source	Average bond enthalpies (tabulated)	Experimentally measured ΔH°f values
State assumed	All species assumed gaseous	Actual states at standard conditions
Accuracy	Approximate (±5–20%)	More accurate (precise measured data)
Formula direction	Reactants first: ΣBE(reactants) − ΣBE(products)	Products first: ΣΔH°f(products) − ΣΔH°f(reactants)