Enthalpy and Bond Strength

Enthalpies of Formation

Enthalpy of formation is the enthalpy of the reaction in which a compound is formed from its elements.

Hess's law states that the total enthalpy change for a reaction is the sum of all changes, regardless of the number of steps or stages of the reaction. This allows scientists to calculate the enthalpy of various reactions using measured enthalpies of reaction. Various databases listing various enthalpies exist. Enthalpies can be calculated for both chemical and physical changes. For example, in combustion reactions a substance reacts with oxygen. During vaporization a substance changes phase from liquid to gas form. During fusion a substance changes phase from a solid to liquid.

Enthalpy of formation (ΔHf\Delta H_f) is the enthalpy of the reaction in which a compound is formed from its elements. Enthalpies of formation are often shown with all reactants and products in standard state, a reference point used to calculate a substance's properties. In this case, the enthalpy of formation is called the standard enthalpy of formation and is shown as ΔHf\Delta H_f^\circ.

Standard Enthalpies of Formation for Common Substances

Substance Formula and Standard State Standard Enthalpy of Formation (ΔHf\Delta H_f^\circ) at 25°C (kJ/mol)
Ammonia NH3(g) −46.2
Benzene C6H6(l) 49.0
Calcium carbonate CaCO3(s) −1207.1
Carbon dioxide CO2(g) −393.5
Carbon monoxide CO(g) −110.5
Ethane C2H6(g) −84.7
Ethene C2H4(g) 52.3
Ethyne (acetylene) C2H2(g) 226.7
Glucose C6H12O6(s) −1273
Methane CH4(g) −74.8
Sodium carbonate Na2CO3(s) −1130.9
Sodium chloride NaCl(s) −411.2
Water H2O(l) −285.8

The change of enthalpy is when one substance is made from its constituent elements and either absorbs or releases heat. When the enthalpy of formation is exothermic (releases heat), it has a negative value. But when the enthalpy of formation is endothermic (absorbs heat), it has a positive value.

In determining the enthalpies of formation, keep in mind that some elements in a compound can appear in more than one form. For example, oxygen is stable as a diatomic molecule (O2) or ozone (O3). Single oxygen atoms can also exist for short periods of time. When enthalpy of formation is determined, diatomic oxygen (O2) is used in the calculation because it is more stable than ozone at 1 atm pressure and 25ºC. Similarly, carbon can occur as graphite or diamond under these conditions. Graphite is more common and is used in the calculation.

It is nearly impossible to measure absolute enthalpies. All enthalpy values are changes in enthalpy. This means enthalpies of formation are calculated in terms of the enthalpies of formation of the elements, such as diatomic oxygen (O2). The enthalpy of formation of the most stable form of an element at 1 atm pressure and 25°C is reported as 0. Diatomic oxygen (O2), hydrogen (H2), nitrogen (N2), and graphite all have an enthalpy of formation of 0. Other forms of these elements do have enthalpies of formation. For example, ozone has an ΔHf\Delta H_{f}^\circ of 142.7 kJ/mol. Diamond has an enthalpy of formation of 1.88 kJ/mol.

Enthalpies of formation can be used to calculate the enthalpies of more complex reactions.

Step-By-Step Example
Calculation of Enthalpy Change for Photosynthesis
Plants and other organisms use photosynthesis to produce glucose from atmospheric carbon dioxide. The balanced reaction for photosynthesis is:
Use the ΔHf\Delta H_{f}^\circ values in the table to calculate ΔH\Delta H for this reaction.
Step 1
Set up an equation for the change in enthalpy in terms of the enthalpies of formation of each compound in the reaction.
ΔH=ΔHproductsΔHreactants=ΔHf,glucose[6ΔHf,CO2+6ΔHf,water]\begin{aligned}\Delta {H}&=\Delta{H_{\rm{products}}}-\Delta H_{\rm{reactants}}\\&=\Delta{H_{{f},\;{\rm{glucose}}}^\circ}-\left[6\Delta{H_{{f},\;{\rm{CO}_{2}}}^\circ}+6\Delta{H_{{f},\;{\rm{water}}}^\circ}\right]\end{aligned}
Remember that oxygen gas consists of its elements already in standard state, so it is not included in the calculation.
Step 2
Substitute values from the table.
ΔH=1273kJ/mol[(6)(393.5kJ/mol)+(6)(285.8kJ/mol)]=2803kJ/mol\begin{aligned}\Delta {H}&=-1273\;{\rm{kJ}}/\rm{mol}-\left[(6)(-393.5\;{\rm{kJ}}/\rm{mol})+(6)(-285.8\;{\rm{kJ}}/\rm{mol})\right]\\&=2803\;{\rm{kJ}}/\rm{mol}\end{aligned}
The enthalpy change for the reaction is 2,803 kJ/mol. Note that the answer is a positive number, indicating an endothermic reaction.