Answer:
- 50.7 kJ.
Explanation:
- To get the enthalpy change for the reaction:
Mn₃O₄(s) + CO(g) ⟶ 3MnO(s) + CO₂(g).
- We must orient the given reactions in a way that its sum give the required reaction.
The first reaction be as it is:
Mn₃O₄(s) + 4CO(g) ⟶ 3Mn(s) + 4CO₂(g), ΔH₁ = 255.6 kJ.
The second reaction should be reversed and multiplied by 3 and also the value of its ΔH must multiplied by (- 3):
3Mn(s) + 3CO₂(g) ⟶ 3MnO(s) + 3CO(g), ΔH₂ = (- 3)(102.1 kJ) = - 306.3 kJ.
- By summing the two reactions after the modification, we get the required reaction:
Mn₃O₄(s) + CO(g) ⟶ 3MnO(s) + CO₂(g).
∴ ΔH rxn = ΔH₁ + ΔH₂ = (255.6 kJ) + (- 306.3 kJ) = - 50.7 kJ.
