Question

How do you calculate standard molar enthalpy of formation?

Answer 1

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