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

Compare the given reactions with the desired reaction. We can see that all the substances in the target reaction are present in the given reactions. We will use the given reactions to extract the necessary information to find the enthalpy change for the target reaction.

First, we have to manipulate the given reactions to match the target reaction. We can find the desired equation by adding a combination of these reactions. We notice that to cancel out the \(\mathrm{Fe}_{2}\mathrm{O}_{3}\), we need to reverse the direction of the first two reactions. $$-\Delta H^{\circ} = -[\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{CO}(g) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{CO}_{2}(g)]$$ $$-\Delta H^{\circ} = -[3 \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{CO}(g) \longrightarrow 2 \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{CO}_{2}(g)]$$ Before we add up these equations, we need to ensure the substances we don't need to cancel each other out. Thus, we can rewrite the equations as: $$\frac{1}{3}\times(-\Delta H^{\circ} = -[\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{CO}(g) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{CO}_{2}(g)])$$ $$\frac{1}{2}\times(-\Delta H^{\circ} = -[3 \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{CO}(g) \longrightarrow 2 \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{CO}_{2}(g)])$$ $$\Delta H^{\circ} = [\mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{CO}(g) \longrightarrow 3 \mathrm{FeO}(s)+\mathrm{CO}_{2}(g)]$$ Now we can add up these equations getting: $$\mathrm{FeO}(s)+\mathrm{CO}(g) \longrightarrow \mathrm{Fe}(s)+\mathrm{CO}_{2}(g)$$

Now that we have obtained the linear combination of the given reactions to get the desired reaction, we can find the enthalpy change for this reaction by summing the enthalpy change of the given reactions after manipulating them. We have: $$\Delta H^{\circ}_{target} = \frac{-1}{3}\times(-23 \,\mathrm{kJ}) + \frac{-1}{2}\times(-39\, \mathrm{kJ}) + 18\, \mathrm{kJ}$$ $$\Delta H^{\circ}_{target} = \frac{23}{3} \,\mathrm{kJ} + \frac{39}{2}\, \mathrm{kJ} - 18\, \mathrm{kJ}$$ Calculating the enthalpy change for the target reaction: $$\Delta H^{\circ}_{target} = 7.67 \,\mathrm{kJ} + 19.5\, \mathrm{kJ} - 18\, \mathrm{kJ} =\boldsymbol{9.17\, \mathrm{kJ}}$$ The calculated enthalpy change for the desired reaction is \(\Delta H^{\circ}_{\mathrm{FeO}(s)+\mathrm{CO}(g) \longrightarrow \mathrm{Fe}(s)+\mathrm{CO}_{2}(g)} = 9.17 \,\mathrm{kJ}\).