Standard heat of formation, reaction and combustion

Why it matters

Whether a reactor needs cooling coils or a fired heater, how hot a furnace runs and how much a fuel is worth all depend on the heat of reaction. Tabulated standard heats of formation and combustion, combined through Hess's law and corrected to process temperature, give reaction enthalpies for any reaction without a calorimeter.

Key ideas

Standard state. Pure substance at 1 bar (older tables: 1 atm – the difference is negligible here) in a specified phase, usually at 25 °C (298.15 K). The phase must always be stated: H₂O(l) and H₂O(g) have different values.

Standard heat of reaction ΔH_r°. The enthalpy change when the stoichiometric amounts of reactants, each in its standard state, react completely to products in their standard states at 25 °C. It is quoted per mole of reaction as written: doubling the equation doubles ΔH_r°. Negative ΔH_r° = exothermic; positive = endothermic.

Standard heat of formation ΔH_f°. ΔH_r° for forming 1 mol of a compound from its elements in their most stable forms (C as graphite, O₂(g), H₂(g), N₂(g), S rhombic). By definition ΔH_f° = 0 for such elements. A large negative ΔH_f° (CO₂, H₂O) means a stable compound.

Standard heat of combustion ΔH_c°. ΔH_r° for burning 1 mol of a substance completely in O₂ to specified products: CO₂(g), H₂O (liquid or vapour – must be stated), SO₂(g), N₂(g). Always negative.

• Higher heating value (HHV, gross): product water is liquid.
• Lower heating value (LHV, net): product water is vapour. HHV − LHV = (moles of water) × latent heat of water at 25 °C (44.0 kJ/mol). Heats of combustion are the practical route to ΔH_f° of organic compounds, which are hard to form directly from their elements.

Hess's law. Because enthalpy is a state function, equations can be added and subtracted like algebra, and so can their ΔH values. This gives the two working formulas: from formation data (products minus reactants) and from combustion data (reactants minus products).

Heat of reaction at another temperature (Kirchhoff). Take reactants from T down to 25 °C, react at 25 °C, take products back up to T. The result is ΔH_r(T) = ΔH_r° + ∫ΔC_p dT, where ΔC_p = Σ ν_i·C_p,i (products positive). Phase changes along the path must be included.

ΔH vs ΔU. At constant volume (bomb calorimeter) the measured heat is ΔU. For ideal gases ΔH = ΔU + Δn_gas·R·T.

Formulas

• ΔH_r° = Σ ν_i·ΔH_f,i°(products) − Σ |ν_i|·ΔH_f,i°(reactants) – kJ per mol of reaction as written.
• ΔH_r° = Σ |ν_i|·ΔH_c,i°(reactants) − Σ ν_i·ΔH_c,i°(products) – from combustion data.
• ΔH_r(T) = ΔH_r° + ∫ ΔC_p dT from 298.15 K to T ; ΔC_p = Σ ν_i·C_p,i – with mean values: ΔH_r(T) = ΔH_r° + ΔC_p,mean·(T − 298.15).
• HHV − LHV = n_H₂O × 44.0 kJ/mol – per mol of fuel, water latent heat at 25 °C.
• ΔH = ΔU + Δn_gas·R·T – ideal gases.
• Q = ξ·ΔH_r° – heat released or absorbed at 25 °C for extent ξ.

Worked examples

Example 1 (standard). (a) Using ΔH_f° (kJ/mol): CH₄(g) −74.8, CO₂(g) −393.5, H₂O(l) −285.8, H₂O(g) −241.8, find the HHV and LHV of methane. (b) From ΔH_c° of propane(g) = −2220.0 kJ/mol (liquid water), find ΔH_f° of propane.

1. CH₄ + 2O₂ → CO₂ + 2H₂O.
2. HHV basis (liquid water): ΔH_c° = [−393.5 + 2(−285.8)] − [−74.8] = −890.3 kJ/mol.
3. LHV basis (vapour): ΔH_c° = [−393.5 + 2(−241.8)] − [−74.8] = −802.3 kJ/mol. Check: difference 88.0 = 2 × 44.0 ✓.
4. (b) C₃H₈ + 5O₂ → 3CO₂ + 4H₂O(l): −2220.0 = 3(−393.5) + 4(−285.8) − ΔH_f°(C₃H₈).
5. ΔH_f°(C₃H₈) = −1180.5 − 1143.2 + 2220.0 = −103.7 kJ/mol.

Answer: HHV = 890.3 kJ/mol, LHV = 802.3 kJ/mol; ΔH_f°(propane) = −103.7 kJ/mol

Example 2 (GATE level). Methanol synthesis CO(g) + 2H₂(g) → CH₃OH(g) runs at 250 °C. ΔH_f° (kJ/mol): CO −110.5, CH₃OH(g) −200.7. Mean C_p (25–250 °C, J/(mol·K)): CO 29.5, H₂ 29.2, CH₃OH(g) 52.0. Find ΔH_r° and ΔH_r at 250 °C, and the heat removed for 100 mol/s of methanol formed at 250 °C.

1. ΔH_r° = −200.7 − (−110.5 + 0) = −90.2 kJ/mol.
2. ΔC_p = 52.0 − 29.5 − 2 × 29.2 = −35.9 J/(mol·K).
3. ΔH_r(523.15 K) = −90.2 + (−35.9 × 10⁻³) × (523.15 − 298.15) = −90.2 − 8.08 = −98.3 kJ/mol.
4. Heat released at 250 °C: ξ = 100 mol/s, Q = 100 × (−98.3) = −9.83 × 10³ kW, i.e. about 9.8 MW must be removed for an isothermal reactor.

Answer: ΔH_r° = −90.2 kJ/mol; ΔH_r(250 °C) ≈ −98.3 kJ/mol; ≈ 9.8 MW removed

Common mistakes

• Ignoring the phase of water (liquid vs vapour) – an 88 kJ/mol difference for methane.
• Reversing the sign convention: from formation data it is products − reactants; from combustion data it is reactants − products.
• Forgetting stoichiometric coefficients, or quoting ΔH_r° without saying which equation it refers to.
• Giving elements in non-standard forms (O atoms, diamond) a zero ΔH_f°.
• Using ΔH_r° at 25 °C for a high-temperature reactor without the ΔC_p correction.

For GATE CH

Expect NAT questions on ΔH_r° from formation or combustion data, HHV/LHV conversions, ΔH_f° from a heat of combustion, and ΔH_r at elevated temperature with mean or polynomial C_p. Practise writing the Hess path explicitly and checking the sign of the answer against chemical intuition.

Quick check

1. What is ΔH_f° of O₂(g) at 25 °C?
2. ΔH_r° for N₂ + 3H₂ → 2NH₃ given ΔH_f°(NH₃) = −45.9 kJ/mol?
3. What is the HHV − LHV of methane per mol?
4. If ΔC_p < 0, does an exothermic reaction become more or less exothermic at higher T?
5. Which heat-of-combustion value includes the latent heat of water?

Answers: 1. 0; 2. −91.8 kJ per mol of reaction; 3. 88.0 kJ/mol; 4. more exothermic; 5. the higher heating value.