Answer: The standard enthalpy of formation of [tex]H_2O(g)[/tex] is -252.1 kJ/mol.
Explanation:
The balanced chemical reaction is,
[tex]Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)[/tex]
The expression for enthalpy change is,
[tex]\Delta H=[n\times H_f_{products}]-[n\times H_f_{reactantss}][/tex]
Putting the values we get :
[tex]\Delta H=[2\times H_f{Fe}+3\times H_f{H_2O}]-[1\times H_f{Fe_2O_3}+3\times H_f{H_2}][/tex]
[tex]67.9kJ=[(2\times 0)+(3\times H_f{H_2O})]-[(1\times -824.2kJ/mol)+3\times 0kJ/mol)][/tex]
[tex]H_f{H_2O}=-252.1kJ/mol[/tex]
Thus standard enthalpy of formation of [tex]H_2O(g)[/tex] is -252.1 kJ/mol.
