Standard enthalpy of formation

For many substances, the formation reaction may be considered as the sum of a number of simpler reactions, either real or fictitious. The enthalpy of reaction can then be analyzed by applying Hess's Law, which states that the sum of the enthalpy changes for a number of individual reaction steps equals the enthalpy change of the overall reaction. This is true because enthalpy is a state function, whose value for an overall process depends only on the initial and final states and not on any intermediate states. Examples are given in the following sections.

Standard enthalpy change of formation in Born–Haber diagram for lithium fluoride

For ionic compounds, the standard enthalpy of formation is equivalent to the sum of several terms included in the Born–Haber cycle. For example, the formation of lithium fluoride,

Li(s) + ​¹⁄₂ F₂(g) → LiF(s)

may be considered as the sum of several steps, each with its own enthalpy (or energy, approximately):

1. The standard enthalpy of atomization (or sublimation) of solid lithium.
2. The first ionization energy of gaseous lithium.
3. The standard enthalpy of atomization (or bond energy) of fluorine gas.
4. The electron affinity of a fluorine atom.
5. The lattice energy of lithium fluoride.

The sum of all these enthalpies will give the standard enthalpy of formation of lithium fluoride.

${\displaystyle \Delta H_{\text{f}}=\Delta H_{\text{sub}}+{\text{IE}}_{\text{Li}}+{\frac {1}{2}}{\text{B(F–F)}}-{\text{EA}}_{\text{F}}-{\text{U}}_{\text{L}}.}$

In practice, the enthalpy of formation of lithium fluoride can be determined experimentally, but the lattice energy cannot be measured directly. The equation is therefore rearranged in order to evaluate the lattice energy.[3]

${\displaystyle U_{\text{L}}=\Delta H_{\text{sub}}+{\text{IE}}_{\text{Li}}+{\frac {1}{2}}{\text{B(F–F)}}-{\text{EA}}_{\text{F}}+\Delta H_{\text{f}}.}$

The formation reactions for most organic compounds are hypothetical. For instance, carbon and hydrogen will not directly react to form methane (CH₄), so that the standard enthalpy of formation cannot be measured directly. However the standard enthalpy of combustion is readily measurable using bomb calorimetry. The standard enthalpy of formation is then determined using Hess's law. The combustion of methane (CH₄ + 2 O₂ → CO₂ + 2 H₂O) is equivalent to the sum of the hypothetical decomposition into elements followed by the combustion of the elements to form carbon dioxide and water:

CH₄ → C + 2 H₂

C + O₂ → CO₂

2 H₂ + O₂ → 2 H₂O

Applying Hess's law,

Δ_combH^⦵(CH₄) = [Δ_fH^⦵(CO₂) + 2 Δ_fH^⦵(H₂O)] − Δ_fH^⦵(CH₄).

Solving for the standard of enthalpy of formation,

Δ_fH^⦵(CH₄) = [Δ_fH^⦵(CO₂) + 2 Δ_fH^⦵(H₂O)] − Δ_combH^⦵(CH₄).

The value of Δ_fH^⦵(CH₄) is determined to be −74.8 kJ/mol. The negative sign shows that the reaction, if it were to proceed, would be exothermic; that is, methane is enthalpically more stable than hydrogen gas and carbon.

It is possible to predict heats of formation for simple unstrained organic compounds with the heat of formation group additivity method.

The standard enthalpy change of any reaction can be calculated from the standard enthalpies of formation of reactants and products using Hess's law. A given reaction is considered as the decomposition of all reactants into elements in their standard states, followed by the formation of all products. The heat of reaction is then minus the sum of the standard enthalpies of formation of the reactants (each being multiplied by its respective stoichiometric coefficient, ν) plus the sum of the standard enthalpies of formation of the products (each also multiplied by its respective stoichiometric coefficient), as shown in the equation below:[4]

Δ_rH^⦵ = Σν Δ_fH^⦵(products) − Σν Δ_fH^⦵(reactants).

If the standard enthalpy of the products is less than the standard enthalpy of the reactants, the standard enthalpy of reaction is negative. This implies that the reaction is exothermic. The converse is also true; the standard enthalpy of reaction is positive for an endothermic reaction. This calculation has a tacit assumption of ideal solution between reactants and products where the enthalpy of mixing is zero.

For example, for the combustion of methane, CH₄ + 2 O₂ → CO₂ + 2 H₂O:

Δ_rH^⦵ = [Δ_fH^⦵(CO₂) + 2 Δ_fH^⦵(H₂O)] − [Δ_fH^⦵(CH₄) + 2 Δ_fH^⦵(O₂)].

However O₂ is an element in its standard state, so that Δ_fH^⦵(O₂) = 0, and the heat of reaction is simplified to

Δ_rH^⦵ = [Δ_fH^⦵(CO₂) + 2 Δ_fH^⦵(H₂O)] − Δ_fH^⦵(CH₄),

which is the equation in the previous section for the enthalpy of combustion Δ_combH^⦵.

1. When a reaction is reversed, the magnitude of ΔH stays the same, but the sign changes.
2. When the balanced equation for a reaction is multiplied by an integer, the corresponding value of ΔH must be multiplied by that integer as well.
3. The change in enthalpy for a reaction can be calculated from the enthalpies of formation of the reactants and the products
4. Elements in their standard states make no contribution to the enthalpy calculations for the reaction, since the enthalpy of an element in its standard state is zero. Allotropes of an element other than the standard state generally have non-zero standard enthalpies of formation.

Thermochemical properties of selected substances at 298 K and 1 atm

• Calorimetry
• Enthalpy
• Heat of combustion
• Thermochemistry

• Zumdahl, Steven (2009). Chemical Principles (6th ed.). Boston. New York: Houghton Mifflin. pp. 384–387. ISBN 978-0-547-19626-8.

• NIST Chemistry WebBook