THERMO CHEMISTRYCHAPTER 11.5 MOLAR ENTHALPIES OF FORMATION AND HESS’ LAW

When chemists measure energy changes, they must use a starting point. When looking at energy states of molecules, the elemental state is considered to be the point with zero potential energy:the reference energy state. This means all elements (O2, Mg, Ag, S8, etc…) in their elemental forms have an enthalpy of formation of zero kJ/mol.

When a compound is built from elements, the reaction either requires or releases energy which is called the standard enthalpy of formation. ΔfHo. The middle of your data booklet consists of a large table of these values. Combined with Hess’ Law, we can use these to determine the energy produced or absorbed in a reaction. (enthalpy of reaction).

Combustion of Methanol:

Combustion of Propane

C3H8(g) +5O2(g) 4H2O(g) + 3CO2(g)

C3H8(g) = -103.8 kJ/mol

O2(g) = 0 kJ/mol (element)

H2O(g) = -241.8 kJ/mol

CO2(g) = -393.5 kJ/mol

ΔrHo = [products – reactants] = [(4mol x -241.8kJ/mol) – (3mol x -393.5kJ/mol)] - [1mol x -103.8kJ/mol]

ΔrHo = (-2147.7kJ) – (-103.8kJ) = -2043.9kJ (if one mole of propane is combusted)

The molar enthalpies of formation can also be used to describe the thermal stability of a compound (which is its resistance to decomposition when heated).

Rule: the lower the enthalpy of formation (including negatives!!!!), the more stable it is

Ex: What is more stable: water or carbon dioxide?

Water: -285.8kJ/mol

Carbon dioxide: --393.5kJ/mol

Since CO2 is more negative (lower) it is more thermally stable.

Practice: page 514 #1-6