One of the challenges in thermochemistry is that absolute values of enthalpy cannot be measured directly. Only changes in enthalpy can be observed experimentally. To create a consistent thermodynamic scale, a reference point must be chosen.
The reference point used in chemistry is the standard enthalpy of formation, \(\Delta H_f^\circ\).
The standard enthalpy of formation is defined as the enthalpy change for the reaction that forms exactly one mole of a compound from its constituent elements in their standard states.
Examples include:
\[ C(gr) + O_2(g) \rightarrow CO_2(g) \]
\[ \Delta H_f^\circ = -393.5\ \mathrm{kJ\,mol^{-1}} \]
\[ H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l) \]
\[ \Delta H_f^\circ = -285.8\ \mathrm{kJ\,mol^{-1}} \]
\[ 3C(gr) + 4H_2(g) \rightarrow C_3H_8(g) \]
By definition, the standard enthalpy of formation of any element in its standard state is zero:
\[ \Delta H_f^\circ \bigl(O_2(g)\bigr) = 0 \]
\[ \Delta H_f^\circ \bigl(H_2(g)\bigr) = 0 \]
\[ \Delta H_f^\circ \bigl(C(gr)\bigr) = 0 \]
This convention provides a common reference point from which all other formation enthalpies can be measured.
Consider the combustion of methane:
\[ CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l) \]
The standard enthalpy of combustion can be determined from the following formation reactions:
\[ C(gr) + 2H_2(g) \rightarrow CH_4(g) \qquad \Delta H_f^\circ = -74.6\ \mathrm{kJ\,mol^{-1}} \]
\[ C(gr) + O_2(g) \rightarrow CO_2(g) \qquad \Delta H_f^\circ = -393.5\ \mathrm{kJ\,mol^{-1}} \]
\[ H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l) \qquad \Delta H_f^\circ = -285.8\ \mathrm{kJ\,mol^{-1}} \]
Reverse the methane formation reaction and multiply the water formation reaction by two:
\[ CH_4(g) \rightarrow C(gr) + 2H_2(g) \qquad \Delta H = +74.6\ \mathrm{kJ} \]
\[ 2H_2(g) + O_2(g) \rightarrow 2H_2O(l) \qquad \Delta H = -571.6\ \mathrm{kJ} \]
Adding the three equations yields the combustion reaction:
\[ CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l) \]
\[ \Delta H^\circ = +74.6\ kJ - 393.5\ kJ - 571.6\ kJ \]
\[ \Delta H^\circ = -890.5\ \mathrm{kJ} \]
The previous calculation can be generalized. Because formation enthalpies are simply a convenient application of Hess' Law, the standard reaction enthalpy can always be calculated using
\[ \Delta H^\circ_{\mathrm{rxn}} = \sum_{\mathrm{products}} \nu_i \Delta H_{f,i}^\circ - \sum_{\mathrm{reactants}} \nu_i \Delta H_{f,i}^\circ \]
Applying this expression to methane combustion:
\[ \Delta H^\circ_{\mathrm{rxn}} = \Bigl[ (1\ mol) \left(-393.5 \frac{kJ}{mol} \right) + (2\ mol) \left(-285.8 \frac{kJ}{mol} \right) \Bigr] - \Bigl[ (1\ mol) \left(-74.6 \frac{kJ}{mol} \right) + (2\ mol)(0) \Bigr] \]
\[ \Delta H^\circ_{\mathrm{rxn}} = -965.1\ kJ + 74.6\ kJ \]
\[ \Delta H^\circ_{\mathrm{rxn}} = -890.5\ \mathrm{kJ} \]
This is exactly the same result obtained using Hess' Law directly.
Big picture: Standard enthalpies of formation provide a convenient thermodynamic database that allows reaction enthalpies to be calculated without explicitly constructing a Hess' Law cycle. The familiar products minus reactants equation is simply Hess' Law written in a compact form.
Calculate the standard reaction enthalpy for
\[ PbBr_2(s) + Cl_2(g) \rightarrow PbCl_2(s) + Br_2(l) \]
Given the following standard enthalpies of formation:
| Substance | \(\Delta H_f^\circ\) (kJ/mol) |
|---|---|
| \(PbBr_2(s)\) | \(-278.7\) |
| \(PbCl_2(s)\) | \(-359.4\) |
| \(Cl_2(g)\) | \(0\) |
| \(Br_2(l)\) | \(0\) |
The standard reaction enthalpy is calculated using
\[ \Delta H^\circ_{\mathrm{rxn}} = \sum_{\mathrm{products}} \nu_i \Delta H_{f,i}^\circ - \sum_{\mathrm{reactants}} \nu_i \Delta H_{f,i}^\circ \]
Substituting the values:
\[ \Delta H^\circ_{\mathrm{rxn}} = \Big[ (1\ mol)(-359.4\ kJ/mol) + (1\ mol)(0) \Big] - \Big[ (1\ mol)(-278.7\ kJ/mol) + (1\ mol)(0) \Big] \]
\[ \Delta H^\circ_{\mathrm{rxn}} = -359.4\ kJ + 278.7\ kJ \]
\[ \Delta H^\circ_{\mathrm{rxn}} = -80.7\ \mathrm{kJ} \]
Therefore,
\[ \Delta H^\circ_{\mathrm{rxn}} = -80.7\ \mathrm{kJ} \]
Physical interpretation: The reaction is exothermic because \(PbCl_2(s)\) is thermodynamically more stable than \(PbBr_2(s)\). The formation of stronger Pb–Cl interactions releases more energy than is required to break the Pb–Br interactions, resulting in a negative reaction enthalpy.
Use the given standard enthalpies of formation to calculate \(\Delta H^\circ_{\mathrm{rxn}}\) for the reaction shown.
\[ \boxed{ \Delta H^\circ_{\mathrm{rxn}} = \text{Products} - \text{Reactants} } \]
Big picture: Standard enthalpies of formation are simply a standardized application of Hess' Law. By tabulating the enthalpies needed to form compounds from their elements, reaction enthalpies can be obtained quickly using the products-minus-reactants relationship rather than constructing a separate thermodynamic cycle for every reaction.