Enthalpy - Chemistry

Question

Calculate ΔH^o for the process Co₃O₄(s) → 3 Co(s) + 2 O₂(g)

from the following information

Co(s) + ½ O2(g) → CoO(s)	ΔHo = -108.1 kJ/mol
3 CoO(s) + ½ O2(g) → Co3O4(s)	ΔHo = -265.3 kJ/mol

Answer 1

Co₃O₄(s) → 3 Co(s) + 2 O₂(g) ... ∆Hº = ?

Using Hess' Law, we can approach the problem as follows:

GIVEN:

eq.1: Co(s) + ½ O₂(g) → CoO(s) ... ∆Hº = -108.1 kJ/mol

eq.2: 3 CoO(s) + ½ O₂(g) → Co₃O₄(s) ... ∆Hº = -265.3 kJ/mol

PROCEDURE:

reverse eq.2: Co₃O₄(s) ==> 3 CoO(s) + ½ O₂(g) ... ∆Hº = +265.3 kJ/mol

reverse & multiple eq.1 by 3: 3CoO(s) ==> 3Co(s) + 3/2 O₂(g) ... ∆Hº = 3 x +108.1 kJ/mol = +324.3 kJ/mol

Add these together to get :

Co₃O₄(s) + 3CoO(s) ==> 3 CoO(s) + ½ O₂(g) + 3Co(s) + 3/2 O₂(g)

Combine and/or cancel like terms to get the final target equation:

Co₃O₄(s) ==> 2 O2(g) + 3Co(s) ... target equation

∆Hº = 265.3 kJ/mol + 324.3 kJ/mol = 589.6 kJ/mol