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), `Delta H=-393.5\ "kJ"`

standard enthalpy of formation (`Delta H_"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)

`Delta H_"f" = -393.5\ "kJ/mol"` CO2(g)

The standard enthalpy of formation for elements already in their standard states is zero.

We can calculate the enthalpy change of a reaction using the individual `Delta H_"f"` values of the products and reactants:

`Delta H = sum nDelta H_"f(products)" - sum nDelta H_"f(reactants)"`.


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:

`Delta H = [(1\ "mol")(-393.5\ "kJ/mol") + (2\ "mol")(-285.8\ "kJ/mol")]`
`- [(1\ "mol")(-74.4\ "kJ/mol") + 0]`.

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