6.8 Enthalpy of Formation

Enthalpy of Formation

Pick a reaction and watch stacked ΔH°f bars build for reactants and products on a shared kJ axis. Signed subtraction gives ΔH°rxn — element species pin to zero.

ΔH°f Table6 ReactionsProduct − ReactantElements = 0
Topic 6.8

Enthalpy of Formation

Calculate the enthalpy change for a chemical or physical process based on the standard enthalpies of formation.

Tables of standard enthalpies of formation can be used to calculate the standard enthalpies of reactions.

Definition. ΔH°f is the enthalpy change when one mole of a compound is formed from its elements in their standard states at 298 K and 1 bar (1 atm is also used).

The key consequence: the standard enthalpy of formation of any element in its standard state is zero. O₂(g), N₂(g), Br₂(l), Hg(l), C(graphite), and Na(s) all have ΔH°f = 0. Note that this is state-specific: O₃(g) is not zero, and neither is C(diamond).

The calculation — one of the equations printed on the AP sheet:

ΔH°rxn = Σ ΔH°f(products) − Σ ΔH°f(reactants)

Each ΔH°f must be multiplied by its stoichiometric coefficient before summing. This "products minus reactants" pattern reappears for entropy (9.2) and free energy (9.3) — learn it once and it serves three units.

Writing a formation equation. The compound must be the sole product with a coefficient of exactly 1, which often forces fractional coefficients on the reactant side. For ethanol:

2 C(graphite) + 3 H₂(g) + ½ O₂(g) → C₂H₅OH(l)

Fractions are correct and expected here — do not clear them.

Why formation enthalpies are more accurate than bond energies: they are measured for specific substances in specific states, so they include intermolecular effects that a gas-phase bond-energy average ignores.

Key points

  • ΔH°_f of an element in its standard state is zero — but only in that state.
  • ΔH°_rxn = Σ(products) − Σ(reactants), each weighted by its coefficient.
  • A formation equation makes exactly one mole of the compound, fractional coefficients allowed.
  • Formation enthalpies give exact values; bond energies give estimates.

Equations

  • on the exam sheetPrinted on the AP equation sheet.
    • standard enthalpy of formation (kJ/mol)

Common mistakes

  • Forgetting to multiply by coefficients. 2 H₂O(l) contributes 2 × (−285.8).
  • Reversing products and reactants.
  • Assuming every element is zero. Only in the standard state — O₃ and diamond are not.
  • Ignoring physical state. ΔH°_f for H₂O(l) and H₂O(g) differ by the enthalpy of vaporization.

Worked example

Calculate ΔH° for 2 C₂H₆(g) + 7 O₂(g) → 4 CO₂(g) + 6 H₂O(l). ΔH°_f (kJ/mol): C₂H₆(g) = −84.7, CO₂(g) = −393.5, H₂O(l) = −285.8.

Products:
4 × CO₂: 4(−393.5) = −1574.0 kJ
6 × H₂O(l): 6(−285.8) = −1714.8 kJ
Σ products = −3288.8 kJ

Reactants:
2 × C₂H₆: 2(−84.7) = −169.4 kJ
7 × O₂: 7(0) = 0 kJ (O₂ is an element in its standard state)
Σ reactants = −169.4 kJ

ΔH°rxn = −3288.8 − (−169.4) = −3119.4 kJ

That is for 2 mol of ethane, so per mole of C₂H₆ the value is −1559.7 kJ/mol. Strongly exothermic, as expected for a combustion.

Full notes for topic 6.8 →