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12 December, 00:06

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.

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  1. 12 December, 02:27
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    - 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.
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