Standard Enthalpies of Formation

A formation reaction makes exactly one mole of a compound from its elements in the form they naturally take at 1 bar and the stated temperature. Oxygen is O₂(g), carbon is graphite, bromine is Br₂(l). Fractional coefficients on the element side are expected, because the product side is fixed at one mole.

Construct a formation equation by fixing the product coefficient at one, writing each element in its reference form at the stated temperature, and balancing only the elemental reactants. For Al₂O₃(s): 2 Al(s) + 3/2 O₂(g) → Al₂O₃(s). For liquid ethanol: 2 C(graphite) + 3 H₂(g) + 1/2 O₂(g) → C₂H₅OH(l). A reaction from CO₂ and H₂ to ethanol is a synthesis reaction but is not its defining formation reaction because CO₂ is a compound.

Forming an element from itself involves no change, so the enthalpy change is zero by definition. This is a chosen reference point rather than a claim that elements contain no energy, and it works because only differences in enthalpy are ever measured.

The zero is assigned to the element's reference form, conventionally its stable form at the chosen pressure and temperature. At 298 K, graphite is the carbon reference, not diamond; O₂(g) is the oxygen reference, not O(g) or O₃(g). Standard pressure is 1 bar, while temperature must still be specified, commonly 298.15 K. “Standard” does not mean 0 °C. Solute standard states are defined through unit activity; 1 M is the ideal-dilute concentration convention used in introductory calculations, not a promise that every real 1 M solution is ideal.

Imagine breaking every reactant down into its elements and rebuilding the products from them. The first stage is the negative of the reactant formation enthalpies, the second is the product formation enthalpies. Adding the two stages gives the standard relationship.

Make two explicit sums before subtracting. For CH₄ + 2 O₂ → CO₂ + 2 H₂O(l), the product sum is −393.5 + 2(−285.8) = −965.1 kJ per reaction amount. The reactant sum is −74.6 + 2(0) = −74.6 kJ. Products minus reactants gives −890.5 kJ. The minus sign before the negative reactant sum is essential. If the desired unknown is a formation enthalpy, insert it as x into this same relationship and solve algebraically; do not switch subtraction rules.

A negative ΔH_f° means formation from the reference elements is exothermic. It does not establish stability against every possible decomposition, nor reaction speed, because those questions involve other products, entropy, composition and kinetic barriers. A positive value similarly does not prove that a compound rapidly decomposes. Compare the Gibbs-energy change of a specified reaction to assess thermodynamic direction, and kinetic evidence to assess lifetime.

Phase labels remain essential. ΔH_f°[H₂O(g)] − ΔH_f°[H₂O(l)] = +44.0 kJ/mol at 298 K gives the vaporization enthalpy at that temperature. This is not automatically the value at the boiling point, because enthalpies vary with temperature.