Respuesta :
Answer:
The heat of the reaction or standard enthalpy of the reaction CO(g) + H₂O(g) → CO₂(g) + H₂(g) is ΔH(rxn) = -41.2 kJ
Explanation:
The reaction whose standard enthalpy is required, as obtained from the internet is
CO(g) + H₂O(g) → CO₂(g) + H₂(g)
The formation reaction for some of the reactants and products are given as
C(s) + O₂(g) → CO₂(g) ΔH = -393.5 kJ
2CO(g) + O₂(g) → 2CO₂(g) ΔH = -566.0 kJ
2H₂(g) + O₂(g) → 2H₂O(g) ΔH = -483.6 kJ
To find the standard enthalpies of the given reaction, we need the heat of formation of each of the species involved in the reaction
ΔH (CO₂(g)) = -393.5 kJ
ΔH (H₂O(g)) = -483.6 kJ ÷ 2 = -241.8 kJ (Because 2 moles of H₂O(g) are formed in the given formation reaction)
ΔH (O₂(g)) = ΔH (H₂(g)) = 0 kJ (No heat of formation for elements)
ΔH (CO(g)) = ? (This isn't given)
But it can be calculated from the second given reaction
2CO(g) + O₂(g) → 2CO₂(g) ΔH = -566.0 kJ
Heat of reaction = ΔH(products) - ΔH(reactants)
Heat of reaction = -566.0 kJ
ΔH (products) = 2 × ΔH (CO₂(g)) = 2 × -393.5 = -787 kJ
ΔH (reactants) = [2 × ΔH (CO(g))] + [1 × ΔH (O₂(g))] = 2 × ΔH (CO₂(g))
Hence, we have
-566 = -787 - [2 × ΔH (CO₂(g))]
2 × ΔH (CO₂(g)) = -787 + 566 = -221 kJ
ΔH (CO₂(g)) = -221 ÷ 2 = -110.5 kJ
CO(g) + H₂O(g) → CO₂(g) + H₂(g)
Heat of reaction = ΔH(products) - ΔH(reactants)
ΔH (products) = [ΔH (CO₂(g))] + [ΔH (H₂(g))]
= -393.5 + 0 = -393.5 kJ
ΔH (reactants) = [1 × ΔH (CO(g))] + [1 × ΔH (H₂O(g))] = -110.5 - 241.8 = -352.3 kJ
Heat of reaction = -393.5 - (-352.3) = -41.2 kJ
Hope this Helps!!!
