# How do you calculate standard molar enthalpy of formation?

Mar 19, 2014

You use the standard enthalpy of the reaction and the enthalpies of formation of everything else.

For most chemistry problems involving ΔH_f^o, you need the following equation:

ΔH_(reaction)^o = ΣΔH_f^o(p) - ΣΔH_f^o(r),

where p = products and r = reactants.

EXAMPLE:

The ΔH_(reaction)^o for the oxidation of ammonia

4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)

is -905.2 kJ. Calculate ΔH_f^o for ammonia. The standard enthalpies of formation are: NO(g) = +90.3 kJ/mol and H₂O(g) = -241.8 kJ/mol.

Solution:

4NH₃(g)+ 5O₂(g) → 4NO(g) + 6H₂O(g)

ΔH_(reaction)^o = ΣΔH_f^o(p) - ΣΔH_f^o(r)

ΣΔH_f^o(p) = 4 mol NO×(+90.3 kJ)/(1 mol NO) + 6 mol H₂O×(-241.8 kJ)/(1 mol H₂O) = 361.2 kJ – 1450.8 kJ = -1089.6 kJ

ΣΔH_f^o(r) = 4 mol NH₃ × (x kJ)/(1 mol NH₃) + 5 mol O₂ × (0 kJ)/(1 mol O₂) = 4x kJ

ΔH_(reaction)^o = ΣΔH_f^o(p) - ΣΔH_f^o(r); so

-905.2 kJ = -1089.6 kJ – 4x kJ

4x = -184.4

x = -46.1

ΔH_f^o(NH₃) = x kJ/mol = -46.1 kJ/mol