4.2: Standard Enthalpies of Formation

In principle, we could use eq. 4.1.3 to calculate the reaction enthalpy associated with any reaction. However, to do so, the absolute enthalpies \(H_i\) of reactants and products would be required. Unfortunately, absolute enthalpies are not known—and theoretically unknowable, since this would require an absolute zero for the enthalpy scale, which does not exist.\(^{1}\) To prevent this problem, enthalpies relative to a defined reference state must be used. This reference state is defined at the constituent elements in their standard state, and the enthalpies of 1 mol of substance in this reference state are called standard enthalpies of formation.

Definition: Standard Enthalpy of Formation

The standard enthalpy of formation of compound \(i\), \(\Delta_{\mathrm{f}} H^{\; -\kern-7pt{\ominus}\kern-7pt-}_i \), is the change of enthalpy during the formation of 1 mol of \(i\) from its constituent elements, with all substances in their standard states.

The standard pressure is defined at \(P^{\; -\kern-7pt{\ominus}\kern-7pt-} = 100 \; \mathrm{kPa} = 1 \; \mathrm{bar}\).\(^2\) There is no standard temperature, but standard enthalpies of formation are usually reported at room temperature, \(T = 298.15 \; \mathrm{K}\). Standard states are indicated with the symbol \(\; -\kern-7pt{\ominus}\kern-7pt-\) and they are defined for elements as the form in which such element is most stable at standard pressure (for example, for hydrogen, carbon, and oxygen the standard states are \(\mathrm{H}_{2(g)}, \mathrm{C}_{(s,\text{graphite})}, \text{and }\mathrm{O}_{2(g)}\), respectively).\(^3\)

For example, the standard enthalpies of formation of some common compounds at \(T = 298.15 \; \mathrm{K}\) are calculated from the following reactions:

\[\begin{equation} \begin{aligned} \mathrm{C}_{(s,\text{graphite})}+\mathrm{O}_{2(g)} \rightarrow \mathrm{CO}_{2(g)} \qquad & \Delta_{\mathrm{f}} H_{\mathrm{CO}_{2(g)}}^{\; -\kern-7pt{\ominus}\kern-7pt-}= -394 \; \text{kJ/mol} \\ \mathrm{C}_{(s,\text{graphite})}+2 \mathrm{H}_{2(g)} \rightarrow \mathrm{CH}_{4(g)} \qquad & \Delta_{\mathrm{f}} H_{\mathrm{CH}_{4(g)}}^{\; -\kern-7pt{\ominus}\kern-7pt-}= -75 \; \text{kJ/mol} \\ \mathrm{H}_{2(g)}+\frac{1}{2} \mathrm{O}_{2(g)} \rightarrow \mathrm{H}_2 \mathrm{O}_{(l)} \qquad & \Delta_{\mathrm{f}} H_{\mathrm{H}_2 \mathrm{O}_{(l)}}^{\; -\kern-7pt{\ominus}\kern-7pt-}= -286 \; \text{kJ/mol} \end{aligned} \end{equation} \nonumber \]

A comprehensive list of standard enthalpies of formation of inorganic and organic compounds is also reported in appendix 16.

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1. An example of a known absolute zero for a scale is the zero of the temperature scale, a temperature that can be approached only as a limit from above. No such thing exists for the enthalpy.︎
2. prior to 1982 the value of \(P^{\; -\kern-7pt{\ominus}\kern-7pt-} = 1.0 \mathrm{ atm}\) was used. The two values of \(P^{\; -\kern-7pt{\ominus}\kern-7pt-}\) are within 1% of each other, since 1 atm = 101.325 kPa.︎
3. There are some exception, such as phosphorus, for which the most stable form at 1 bar is black phosphorus, but white phosphorus is chosen as the standard reference state for zero enthalpy of formation. For the purposes of this course, however, we can safely ignore them.