Answer:
ΔH∘ = - 41.2 KJ
Explanation:
We want to obtain the change in enthalpy for the reaction
CO(g) + H₂O(g) → CO₂(g) + H₂(g) (Main reaction)
And we're given the heat of formation of the reactants and products in the reaction
C(s) + O₂(g) → CO₂(g) ΔH∘=−393.5kJ (Reaction A)
2CO(g) → 2C(s) + O₂(g) ΔH∘=+221.0kJ (Reaction B)
2H₂(g) + O₂(g) → 2H₂O(g) ΔH∘=−483.6kJ (Reactions C)
To achieve this, we use the Born-Haber cycle.
The Born-Haber cycle entails writing the change in enthalpy of a reaction as a sum of change in enthalpies of a number of reactions that sum up to give the reaction whose enthalpy we needed from the start.
The main reaction is a sum of a sort of combination of Reactions A, B and C. We find this combination now.
From the reactions whose change in enthalpies are given,
C(s) + O₂(g) → CO₂(g) ΔH∘=−393.5kJ (Reaction A)
2CO(g) → 2C(s) + O₂(g) ΔH∘=+221.0kJ (Reaction B)
Dividing through by 2
CO(g) → C(s) + (1/2)O₂(g) ΔH∘=+110.5kJ (the enthalpy is divided by 2 too)
This reaction becomes (Reaction B)/2
2H₂(g) + O₂(g) → 2H₂O(g) ΔH∘=−483.6kJ (Reactions C)
Changing the direction of the reaction
2H₂O(g) → 2H₂(g) + O₂(g) ΔH∘=483.6kJ (the sign on the change in enthalpy changes)
Then, dividing by 2
2H₂O(g) → 2H₂(g) + O₂(g) ΔH∘=+483.6kJ
H₂O(g) → H₂(g) + (1/2)O₂(g) ΔH∘=241.8kJ (the change in enthalpy is divided by 2 too)
This reaction becomes (-Reaction C)/2
But, now, our main reaction can be written as a sum of these new Reactions,
Main Reaction = (Reaction A) + [(Reaction B)/2] + [(- Reaction C)/2]
C(s) + O₂(g) + CO(g) + H₂O(g) → CO₂(g) + C(s) + (1/2)O₂(g) + H₂(g) + (1/2)O₂(g)
Which gives the main reaction after eliminating the O2 that appears on both sides.
CO(g) + H₂O(g) → CO₂(g) + H₂(g)
Hence,
(ΔH∘ for the main reaction) = (ΔH∘ for reaction A) + [(ΔH∘/2) for reaction B) - [(ΔH∘/2) for reaction C)
ΔH∘ = - 393.5 + (221/2) - (-483.6/2) = - 41.2 KJ
