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=−393.5kJ
- 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=−393.5kJ/mol
CO2(g)
We can calculate the enthalpy change of a reaction using the individual
ΔHf
values of the products and reactants:
ΔH=∑nΔHf(products)−∑nΔHf(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=[(1mol)(−393.5kJ/mol)+(2mol)(−285.8kJ/mol)]
−[(1mol)(−74.4kJ/mol)+0].
This evaluates to −890.7 kJ. Notice that the answer is in kJ, not kJ/mol.