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Elan Coil [88]
3 years ago
10

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

Chemistry
1 answer:
faust18 [17]3 years ago
3 0

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), <em>ΔH₁ = 255.6 kJ.</em>

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), <em>ΔH₂ = (- 3)(102.1 kJ) = - 306.3 kJ.</em>

  • By summing the two reactions after the modification, we get the required reaction:

<em>Mn₃O₄(s) + CO(g) ⟶ 3MnO(s) + CO₂(g).</em>

<em></em>

<em>∴ ΔH rxn = ΔH₁ + ΔH₂ = (255.6 kJ) + (- 306.3 kJ) = - 50.7 kJ.</em>

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