Question

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.

Answer 1

- 50.7 kJ.

Step-by-step 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.