D8.4 Standard Formation Enthalpy
Data derived from calorimetric measurements for many different chemical reactions, that is from many thermochemical expressions, can be summarized in a table of standard formation enthalpies and used to calculate accurate enthalpy changes for a variety of chemical reactions. A 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:
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:
A reaction equation with ½ mol of N2 and 1 mol of O2 is appropriate in this case because the standard enthalpy of formation always refers to formation of 1 mol of the substance; here, it is 1 mol 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
Write the standard enthalpy of formation reaction equations for:
- C2H5OH(ℓ)
- Ca3(PO4)2(s)
Write in your notebook, then left-click here for an explanation.
The standard enthalpy of formation reaction equations are for forming one mole of the compound from its constituent elements in their most stable states under standard-state conditions. If you are unsure what the most stable state of an element is, look in the thermodynamics table and see which form of the element has a 0 kJ/mol standard enthalpy of formation.
- 2 C(s, graphite) + 3 H2(g) + ½ O2(g) ⟶ C2H5OH(ℓ)
- 3 Ca(s) + ½ P4(s, white) + 4 O2(g) ⟶ Ca3(PO4)2(s)
A simple calculation can be used to determine the ΔrH° of any reaction if the ΔfH° of the reactants and products are available. The ΔrH° of the overall reaction is equal to the sum of all the standard formation enthalpies of the products minus the sum of all the standard formation enthalpies of the reactants:
Exercise: Using Standard Formation Enthalpies
Activity: Hess’s Law and Standard Formation Enthalpies
Consider this reaction:
3 NO2(g) + H2O(ℓ) ⟶ 2 HNO3(aq) + NO(g)
Write a multi-step process that sums to this reaction involving the decomposition of each reactant into its constituent elements, and the formation of each product from its constituent elements. Then, relate each step in the multi-step process to the standard enthalpy of formation reaction for each compound, and derive the enthalpy change for the overall reaction in terms of standard formation enthalpies.
Write in your notebook, then left-click here for an explanation.
Start with the first reactant, NO2, and write an equation in which the elements N2 and O2 form. Multiply the equation by 3 because there is a coefficient of 3 for NO2 in the equation. Then do the same for water. Then write equations in which the product molecules (with appropriate coefficients) are formed from their elements. Make certain that each element is in its most stable form at 25 °C (in this case, the elements are all gases).
3 NO2(g) ⟶  N2(g) + 3 O2(g) |
ΔrH°1 = −3×ΔfH°(NO2) = −99.54 kJ/mol |
| H2O(ℓ) ⟶ H2(g) + ½ O2(g) | ΔrH°2 = −1×ΔfH°(H2O) = +285.83 kJ/mol |
| H2(g) + N2(g) + 3 O2(g) ⟶ 2 HNO3(aq) | ΔrH°3 = 2×ΔfH°(HNO3) = −414.72 kJ/mol |
| ½ N2(g) + ½ O2(g) ⟶ NO(g) | ΔrH°4 = 1×ΔfH°(NO) = +90.25 kJ/mol |
Summing these reaction equations gives the reaction we are interested in:
3 NO2(g) + H2O(ℓ) ⟶ 2 HNO3(aq) + NO(g)
Summing the enthalpy changes of the reactions gives the value we want to determine:
ΔrH° = ΔrH°1 + ΔrH°2 + ΔrH°3 + ΔrH°4 = −99.54 kJ/mol + 285.83 kJ/mol − 414.72 kJ/mol + 90.25 kJ/mol = −138.18kJ/mol
Note that the enthalpy change for each thermochemical expression must always have the same value. If the thermochemical expressions are summed to give an overall reaction, the enthalpy changes can also be summed to give the enthalpy change for the overall reaction.
Note that this result was obtained by (1) multiplying the ΔfH° of each product by its stoichiometric coefficient and summing those values, (2) multiplying the ΔfH° of each reactant by its stoichiometric coefficient and summing those values, and then (3) subtracting the result found in (2) from the result found in (1). In other words:
ΔrH° = ∑ΔfH°(products) − ∑ΔfH°(reactants)
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