Hess’s law type 2

The second way to apply Hess’s law uses standard enthalpies of formation.

formation reaction

a reaction in which a compound is formed from its elements

C(s) + O2(g) → CO2(g), ΔH=−393.5 kJ\displaystyle \Delta{} H = - 393.5 \, \text{kJ}

standard enthalpy of formation (ΔHf\displaystyle \Delta{} H_{\text{f}})

the molar enthalpy for a formation reaction (expressed per mole of product produced) where the elements are in their standard states and the reaction occurs at SATP (25 ºC and 101.3 kPa)

ΔHf=−393.5 kJ/mol\displaystyle \Delta{} H_{\text{f}} = - 393.5 \, \text{kJ/mol} CO2(g)

We can calculate the enthalpy change of a reaction using the individual ΔHf\displaystyle \Delta{} H_{\text{f}} values of the products and reactants:

ΔH=∑nΔHf(products)−∑nΔHf(reactants)\displaystyle \Delta{} H = \sum n \Delta{} H_{\text{f(products)}} - \sum n \Delta{} H_{\text{f(reactants)}}.

Example

What is the enthalpy change for CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)?

Begin with the formula given above. Substitute moles and standard enthalpies of formation for each product and reactant:

ΔH=[(1 mol)(−393.5 kJ/mol)+(2 mol)(−285.8 kJ/mol)]\displaystyle \Delta{} H = \left [ \left ( 1 \, \text{mol} \right ) \left ( - 393.5 \, \text{kJ/mol} \right ) + \left ( 2 \, \text{mol} \right ) \left ( - 285.8 \, \text{kJ/mol} \right ) \right ]
−[(1 mol)(−74.4 kJ/mol)+0]\displaystyle - \left [ \left ( 1 \, \text{mol} \right ) \left ( - 74.4 \, \text{kJ/mol} \right ) + 0 \right ].

This evaluates to −890.7 kJ. Notice that the answer is in kJ, not kJ/mol.