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 $$ \mathrm{FeO}(s)+\mathrm{CO}(g) \longrightarrow \mathrm{Fe}(s)+\mathrm{CO}_{2}(g) $$

We need to reverse and multiply the given reactions to match the target reaction. Observing the given reactions, we notice that reaction 1 and reaction 3 need to be reversed based on products and reactants in the target reaction. Also, we need to manipulate them so that they can be combined in such a way that some reactants/products get canceled, and only the components of the target reaction are left. For reaction 1: Reverse the reaction and multiply it by \(\frac{1}{3}\). The revised reaction becomes: $$ \mathrm{Fe}(s) + \mathrm{CO}_{2}(g) \longrightarrow \mathrm{Fe}_{2}\mathrm{O}_{3}(s) + \frac{1}{3}\mathrm{CO}(g) \hspace{1cm} \Delta H^{\circ} = 23 \frac{\mathrm{kJ}}{3} $$ For reaction 3: Reverse the reaction. The revised reaction becomes: $$ 3\mathrm{FeO}(s) + \mathrm{CO}_{2}(g) \longrightarrow \mathrm{Fe}_{3}\mathrm{O}_{4}(s) + \mathrm{CO}(g) \hspace{1cm} \Delta H^{\circ} = -18 \mathrm{kJ} $$ #Step 3: Combine revised reactions#

To find the \(\Delta H^{\circ}\) for the target reaction, we need to sum the enthalpy changes of the revised reactions: $$ \Delta H^{\circ}_{\text{target}} = 23 \frac{\mathrm{kJ}}{3} - 18 \mathrm{kJ} - 39 \mathrm{kJ} $$ $$ \Delta H^{\circ}_{\text{target}} = 7.67 \mathrm{kJ} - 18 \mathrm{kJ} - 39 \mathrm{kJ} $$ $$ \Delta H^{\circ}_{\text{target}} = -49.33 \mathrm{kJ} $$ Therefore, the enthalpy change, \(\Delta H^{\circ}\), for the given reaction is approximately -49.33 kJ.