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A scientist measures the standard enthalpy change for the following reaction to be 67.9 kJ:
Fe2O3(s) + 3 H2(g) →2Fe(s) + 3 H2O(9)
Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation
of H2O(g) is kJ/mol.

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Answer:

-252.5 kJ/mol = ΔH H2O(g)

Explanation:

ΔH Fe2O3 = -825.5kJ/mol

ΔH H2 = 0kJ/mol

ΔH Fe = 0kJ/mol

Based on Hess's law, ΔH of a reaction is the sum of ΔH of products - ΔH of reactants. For the reaction:

Fe2O3(s) + 3 H2(g) →2Fe(s) + 3 H2O(g)

ΔHr = 67.9kJ/mol = 3*ΔH H2O + 2*ΔHFe - (ΔH Fe2O3 + 3*Δ H2)

67.9kJ/mol = 3*ΔH H2O + 2*0kJ/mol - (ΔH -825.5kJ/mol + 3*Δ H2)

67.9 = 3*ΔH H2O(g) + 825.5kJ/mol

-757.6kJ/mol = 3*ΔH H2O(g)

-252.5 kJ/mol = ΔH H2O(g)

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