Given the following reactions Fe2O3 (s) + 3CO (s) → 2Fe (s) + 3CO2 (g) ΔH = -28.0 kJ 3Fe (s) + 4CO2(s) → 4CO (g) + Fe3O4(s) ΔH = +12.5 kJ the enthalpy of the reaction of Fe2O3 with CO 3Fe2O3 (s) + CO (g) → CO2 (g) + 2Fe3O4 (s) is ________ kJ.

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

[tex]Fe_2O_3(s)+3CO(s)\rightarrow 2Fe(s)+3CO_2(g)[/tex] [tex]\Delta H=-28.0kJ[/tex] [tex]\times 3[/tex]

[tex]3Fe_2O_3(s)+9CO(s)\rightarrow 6Fe(s)+9CO_2(g)[/tex] [tex]\Delta H=-84.0kJ[/tex] (1)

[tex]3Fe(s)+4CO_2(s)\rightarrow 4CO(g)+Fe_3O_4(s)[/tex] [tex]\Delta H=+12.5kJ[/tex] [tex]\times 2[/tex]

[tex]6Fe(s)+8CO_2(s)\rightarrow 8CO(g)+2Fe_3O_4(s)[/tex] [tex]\Delta H=+25.0kJ[/tex] (2)

The final reaction is:

Subtracting (2) from (1):

[tex]3Fe_2O_3(s)+CO(g)\rightarrow CO_2(g)+2Fe_3O_4(s)[/tex] [tex]\Delta H=-84.0-(+25.0)=-109kJ[/tex]

Thus the enthalpy of the reaction is -109 kJ