# D19.2 Standard Formation Enthalpy

Standard formation enthalpy, ΔfH°, is the enthalpy change for a reaction in which exactly one mole of a pure substance in a specified state (s, l, or g) is formed from free elements in their most stable states under standard-state conditions. ΔfH° is also referred to as the standard heat of formation.

For example, ΔfH° of CO2(g) at 25 °C is −393.5 kJ/mol. This is the enthalpy change for the exothermic reaction:

C(s, graphite) + O2(g) ⟶ CO2(g)          ΔfH° = -393.5 kJ/mol (25 °C)

The gaseous reactant and product are at a pressure of 1 bar, the carbon is present as solid graphite, which is the most stable form of carbon under standard-state conditions.

For nitrogen dioxide, NO2(g), ΔfH° is 33.2 kJ/mol at 25 °C:

½N2(g) + O2(g) ⟶ NO2(g)          ΔfH° = +33.2 kJ/mol (25 °C)

A reaction equation with ½ mole of N2 and 1 mole of O2 is appropriate in this case because the standard enthalpy of formation always refers to formation of 1 mole of the substance, here, it is NO2(g).

By definition, the ΔfH° of an element in its most stable form under standard conditions is 0 kJ/mol. A table of ΔfH° values for many common substances can be found in the Appendix.

Activity: Equations for Standard Formation Enthalpy

Hess’s law can be used to determine the ΔrH° of any reaction if the ΔfH° of the reactants and products are available. In other words, we can think of any reaction as occurring via step-wise decomposition of the reactants into their component elements followed by re-combination of the elements to give the products. (Almost no reaction would actually occur via such a mechanism, but we can imagine such a path for the sole purpose of using Hess’s law to calculate ΔrH°.)

The ΔrH° of the overall reaction is therefore equal to:

ΔrH° = ∑ΔfH°(products) – ∑ΔfH°(reactants)

Exercise: Using Standard Formation Enthalpies

Activity: Applying Hess’s Law to Standard Formation Enthalpies 