# 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), $\displaystyle \Delta{} H = - 393.5 \, \text{kJ}$

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

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)

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

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

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

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