Chemistry » Thermochemistry » Enthalpy

# Standard Enthalpy of Formation

## Standard Enthalpy of Formation

A standard enthalpy of formation $$\text{Δ}{H}_{\text{f}}^{°}$$ is an enthalpy change for a reaction in which exactly 1 mole of a pure substance is formed from free elements in their most stable states under standard state conditions. These values are especially useful for computing or predicting enthalpy changes for chemical reactions that are impractical or dangerous to carry out, or for processes for which it is difficult to make measurements. If we have values for the appropriate standard enthalpies of formation, we can determine the enthalpy change for any reaction, which we will practice in the next section on Hess’s law.

The standard enthalpy of formation of CO2(g) is −393.5 kJ/mol. This is the enthalpy change for the exothermic reaction:

$$\text{C}\left(s\right)+{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{CO}}_{2}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}{H}_{\text{f}}^{°}=\text{Δ}{H}_{298}^{°}=-393.5\phantom{\rule{0.2em}{0ex}}\text{kJ}$$

starting with the reactants at a pressure of 1 atm and 25 °C (with the carbon present as graphite, the most stable form of carbon under these conditions) and ending with one mole of CO2, also at 1 atm and 25 °C. For nitrogen dioxide, NO2(g), $$\text{Δ}{H}_{\text{f}}^{°}$$ is 33.2 kJ/mol. This is the enthalpy change for the reaction:

$$\frac{1}{2}{\text{N}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{NO}}_{2}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}{H}_{\text{f}}^{°}=\text{Δ}{H}_{298}^{°}=\text{+33.2 kJ}$$

A reaction equation with $$\frac{1}{2}$$mole of N2 and 1 mole of O2 is correct in this case because the standard enthalpy of formation always refers to 1 mole of product, NO2(g).

You will find a table of standard enthalpies of formation of many common substances in this lesson. These values indicate that formation reactions range from highly exothermic (such as −2984 kJ/mol for the formation of P4O10) to strongly endothermic (such as +226.7 kJ/mol for the formation of acetylene, C2H2). By definition, the standard enthalpy of formation of an element in its most stable form is equal to zero under standard conditions, which is 1 atm for gases and 1 M for solutions.

## Example: Evaluating an Enthalpy of Formation

Ozone, O3(g), forms from oxygen, O2(g), by an endothermic process. Ultraviolet radiation is the source of the energy that drives this reaction in the upper atmosphere. Assuming that both the reactants and products of the reaction are in their standard states, determine the standard enthalpy of formation, $$\text{Δ}{H}_{\text{f}}^{°}$$ of ozone from the following information:

$$3{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}2{\text{O}}_{3}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}{H}_{298}^{°}=\text{+286 kJ}$$

### Solution

$$\text{Δ}{H}_{\text{f}}^{°}$$ is the enthalpy change for the formation of one mole of a substance in its standard state from the elements in their standard states. Thus, $$\text{Δ}{H}_{\text{f}}^{°}$$ for O3(g) is the enthalpy change for the reaction:

$$\frac{3}{2}{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{O}}_{3}\left(g\right)$$

For the formation of 2 mol of O3(g), $$\text{Δ}{H}_{298}^{°}=\text{+286 kJ.}$$ This ratio, $$\left(\frac{286\phantom{\rule{0.2em}{0ex}}\text{kJ}}{2\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{O}}_{3}}\right),$$ can be used as a conversion factor to find the heat produced when 1 mole of O3(g) is formed, which is the enthalpy of formation for O3(g):

$$\require{cancel}\text{Δ}\text{H}\text{° for}\phantom{\rule{0.2em}{0ex}}1\phantom{\rule{0.2em}{0ex}}\text{mole of}\phantom{\rule{0.2em}{0ex}}{\text{O}}_{3}\left(g\right)=1\phantom{\rule{0.2em}{0ex}}\cancel{\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{O}}_{3}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{286\phantom{\rule{0.2em}{0ex}}\text{kJ}}{2\phantom{\rule{0.2em}{0ex}}\cancel{\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{O}}_{3}}}\phantom{\rule{0.2em}{0ex}}=143\phantom{\rule{0.2em}{0ex}}\text{kJ}$$

Therefore, $$\text{Δ}{H}_{\text{f}}^{°}\left[{\text{O}}_{3}\left(g\right)\right]=\text{+143 kJ/mol}.$$

## Example: Writing Reaction Equations for $$\text{Δ}{H}_{\text{f}}^{°}$$

Write the heat of formation reaction equations for:

(a) C2H5OH(l)

(b) Ca3(PO4)2(s)

### Solution

Remembering that $$\text{Δ}{H}_{\text{f}}^{°}$$ reaction equations are for forming 1 mole of the compound from its constituent elements under standard conditions, we have:

(a) $$2\text{C}\left(s,\phantom{\rule{0.2em}{0ex}}\text{graphite}\right)+3{\text{H}}_{2}\left(g\right)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{C}}_{2}{\text{H}}_{5}\text{OH}\left(l\right)$$

(b) $$3\text{Ca}\left(s\right)+\frac{1}{2}{\text{P}}_{4}\left(s\right)+4{\text{O}}_{2}\left(g\right)⟶{\text{Ca}}_{3}\left({\text{PO}}_{4}\right)_{2}\left(s\right)$$

Note: The standard state of carbon is graphite, and phosphorus exists as P4.