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

standard enthalpy of formation (ΔHf)

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 CO2(g)

We can calculate the enthalpy change of a reaction using the individual ΔHf values of the products and 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:

ΔH=[(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.