Given the following data \(\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{CO}(g) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{CO}_{2}(g) \quad \Delta H^{\circ}=-23 \mathrm{~kJ}\) \(3 \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{CO}(g) \longrightarrow 2 \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{CO}_{2}(g) \quad \Delta H^{\circ}=-39 \mathrm{~kJ}\) \(\mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{CO}(g) \longrightarrow 3 \mathrm{FeO}(s)+\mathrm{CO}_{2}(g) \quad \Delta H^{\circ}=18 \mathrm{~kJ}\) calculate \(\Delta H^{\circ}\) for the reaction $$ \mathrm{FeO}(s)+\mathrm{CO}(g) \longrightarrow \mathrm{Fe}(s)+\mathrm{CO}_{2}(g) $$

We are looking for the following reaction and its heat of reaction (enthalpy change): $$ \mathrm{FeO}(s)+\mathrm{CO}(g) \longrightarrow \mathrm{Fe}(s)+\mathrm{CO}_{2}(g) $$

To obtain the target reaction, we might need to reverse or multiply some of the given reactions. After analyzing the reactions, a good plan is: Reaction 1: Reverse it: $$ 2\mathrm{Fe}(s)+3\mathrm{CO}_{2}(g) \longrightarrow \mathrm{Fe}_{2}\mathrm{O}_{3}(s)+3\mathrm{CO}(g) \quad \Delta H^{\circ}=23 \mathrm{~kJ} $$ Reaction 2: No changes needed: $$ 3 \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{CO}(g) \longrightarrow 2 \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{CO}_{2}(g) \quad \Delta H^{\circ}=-39 \mathrm{~kJ} $$ Reaction 3: Reverse and multiply by 3: $$ 3\mathrm{FeO}(s)+3\mathrm{CO}_{2}(g) \longrightarrow \mathrm{Fe}_{3}\mathrm{O}_{4}(s)+3\mathrm{CO}(g) \quad \Delta H^{\circ}=-18 \mathrm{~kJ} \times 3=-54\mathrm{~kJ} $$ Now, by adding these manipulated reactions, we can obtain the target reaction.

Sum the manipulated reactions: $$ \begin{array}{ccccccc} 2\mathrm{Fe}(s)&+&3\mathrm{CO}_{2}(g)&+&3 \mathrm{Fe}_{2} \mathrm{O}_{3}(s)&+&\mathrm{CO}(g)& \longrightarrow &\mathrm{Fe}_{2}\mathrm{O}_{3}(s)&+&3\mathrm{CO}(g)&+&2 \mathrm{Fe}_{3} \mathrm{O}_{4}(s)&+&\mathrm{CO}_{2}(g) \\ & & &- &3\mathrm{FeO}(s)&-&3\mathrm{CO}_{2}(g)& \longrightarrow &-& & & -& & \mathrm{Fe}_{3}\mathrm{O}_{4}(s)& & 3\mathrm{CO}(g) \\ \cline{1-14} \mathrm{FeO}(s)&+&\mathrm{CO}(g)& & & & & \longrightarrow &\mathrm{Fe}(s)&+& & & & &\mathrm{CO}_{2}(g) \end{array} $$ So, our target reaction has been obtained from the given reactions.

To find \(\Delta H^{\circ}\) for the target reaction, add the heat of reactions corresponding to the manipulated reactions: $$ \Delta H^{\circ}=23\mathrm{~kJ}+(-39\mathrm{~kJ})+(-54\mathrm{~kJ})=-70\mathrm{~kJ} $$ Therefore, the heat of reaction for the target reaction is \(\Delta H^{\circ}=-70\mathrm{~kJ}\).