# How do you calculate standard molar enthalpy of formation?

##### 1 Answer
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