You use the standard enthalpy of the reaction and the enthalpies of formation of everything else.
For most chemistry problems involving ΔH_f^oΔHof, you need the following equation:
ΔH_(reaction)^o = ΣΔH_f^o(p) - ΣΔH_f^o(r)ΔHoreaction=ΣΔHof(p)−ΣΔHof(r),
where p = products and r = reactants.
EXAMPLE:
The ΔH_(reaction)^oΔHoreaction for the oxidation of ammonia
4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)
is -905.2 kJ. Calculate ΔH_f^oΔHof 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)ΔHoreaction=ΣΔHof(p)−ΣΔHof(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
