Answer:
- 416 kJ/mol
Step-by-step explanation:
The standard enthalpy of the reaction (Δ H ∘ rxn) is independent of the pathway, so it can be calculated by the enthalpy of formation of the reactants and the products:
Δ H ∘ rxn = ∑n*Δ H ∘f products - ∑n*Δ H ∘f reactants
Where n is the number of moles in the balanced reaction. So, for the reaction given:
Na₂O(s) + 1/2O₂(g) → Na₂O₂(s)
Because O₂ is formed by only one elements, its Δ H ∘f is 0 kJ/mol:
-89.0 = (1*(-505) - (1*Δ H ∘fNa₂O)
Δ H ∘fNa₂O = -505 + 89
Δ H ∘fNa₂O = - 416 kJ/mol