Calculate the standard molar enthalpy of formation for na2o(s), given that the standard enthalpy of formation for na2o2(s) is -505 kj/mol and the enthalpy change for the following reaction is -89.0 kj/mol. na2o(s) + 1/2 o2(g) → na2o2(s)

- 416 kJ/mol

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