Calculate the enthalpy change for the reaction Mn3O4 (s) + CO (g) ⟶3MnO (s) + CO2 (g) from the following:

Mn3O4 (s) + 4CO (g) ⟶3Mn (s) + 4CO2 (g) ΔH=255.6kJ

MnO (s) + CO (g) ⟶Mn (s) + CO2 (g) ΔH=102.1kJ

Express your answer using one decimal place and include the appropriate units.

- 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.