A blast furnace is used to reduce iron oxides to elemental iron. The reducing agent for this reduction process is carbon monoxide. a. Given the following data: \(\begin{aligned} \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{CO}(g) & \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{CO}_{2}(g) & & \Delta H^{\circ}=-23 \mathrm{~kJ} \\ 3 \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{CO}(g) & \longrightarrow 2 \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{CO}_{2}(g) & & \Delta H^{\circ}=-39 \mathrm{~kJ} \\ \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{CO}(g) & \longrightarrow 3 \mathrm{FeO}(s)+\mathrm{CO}_{2}(g) & & \Delta H^{\circ}=18 \mathrm{~kJ} \end{aligned}\) determine \(\Delta H^{\circ}\) for the reaction $$ \mathrm{FeO}(s)+\mathrm{CO}(g) \longrightarrow \mathrm{Fe}(s)+\mathrm{CO}_{2}(g) $$ b. The \(\mathrm{CO}_{2}\) produced in a blast furnace during the reduction process actually can oxidize iron into \(\mathrm{FeO}\). To eliminate this reaction, excess coke is added to convert \(\mathrm{CO}_{2}\) into \(\mathrm{CO}\) by the reaction $$ \mathrm{CO}_{2}(g)+\mathrm{C}(s) \longrightarrow 2 \mathrm{CO}(g) $$ Using data from Appendix 4 , determine \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for this reaction. Assuming \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature, at what temperature is the conversion reaction of \(\mathrm{CO}_{2}\) into CO spontaneous at standard conditions?

We want to determine the enthalpy change for the reaction: $$ \mathrm{FeO}(s)+\mathrm{CO}(g) \longrightarrow \mathrm{Fe}(s)+\mathrm{CO}_{2}(g) $$ To do this, we have to manipulate the given reactions in such a way that when they are added, the desired reaction is obtained. We need to eliminate any species that do not appear in the desired reaction and ensure that the coefficients are correct. Looking at the first given reaction, it has Fe, CO and CO₂, which are all present in our target reaction. However, we have to modify it, so we have only one Fe and one CO in the reaction: $$ \frac{1}{2}\mathrm{Fe}_{2}\mathrm{O}_{3}(s) + \frac{3}{2}\mathrm{CO}(g) \longrightarrow \mathrm{Fe}(s) + \frac{3}{2}\mathrm{CO}_{2}(g) \quad \Delta H^{\circ}=-\frac{23}{2}\mathrm{~kJ} $$ Now, looking at the third given reaction, it has Fe, CO and CO₂ as well. We modify it to have only one CO and one FeO in the reaction: $$ \frac{1}{3}\mathrm{Fe}_3\mathrm{O}_{4}(s) + \mathrm{CO}(g) \longrightarrow \mathrm{FeO}(s) + \mathrm{CO}_{2}(g) \quad \Delta H^{\circ}=\frac{18}{3}\mathrm{~kJ} $$ When we add these modified reactions, we obtain the desired reaction: $$ \frac{1}{2}\mathrm{Fe}_{2}\mathrm{O}_{3}(s) + \frac{3}{2}\mathrm{CO}(g) + \frac{1}{3}\mathrm{Fe}_{3}\mathrm{O}_{4}(s) \longrightarrow \mathrm{Fe}(s) + \frac{3}{2}\mathrm{CO}_{2}(g) + \mathrm{FeO}(s) $$ $$ \mathrm{FeO}(s)+\mathrm{CO}(g) \longrightarrow \mathrm{Fe}(s)+\mathrm{CO}_{2}(g) $$ Now, we add the enthalpy changes: $$ \Delta H^{\circ} = -\frac{23}{2}\mathrm{~kJ} + \frac{18}{3}\mathrm{~kJ} = -\frac{29}{6}\mathrm{~kJ} $$

We are given the reaction for the conversion of CO₂ into CO: $$ \mathrm{CO}_{2}(g)+\mathrm{C}(s) \longrightarrow 2 \mathrm{CO}(g) $$ To find the enthalpy and entropy change for this reaction, we can use the data provided in Appendix 4. For each reactant and product, we look up their respective enthalpy and entropy values and then subtract the values for the reactants from the values for the products. The enthalpy change is given by \(\Delta H^{\circ} = \sum H^{\circ}_{products} - \sum H^{\circ}_{reactants}\). The entropy change is given by \(\Delta S^{\circ} = \sum S^{\circ}_{products} - \sum S^{\circ}_{reactants}\). Using the values provided in Appendix 4: $$ \Delta H^{\circ} = 2 H^{\circ}_{\mathrm{CO}} - (H^{\circ}_{\mathrm{CO}_{2}} + H^{\circ}_{\mathrm{C}}) $$ $$ \Delta S^{\circ} = 2 S^{\circ}_{\mathrm{CO}} - (S^{\circ}_{\mathrm{CO}_{2}} + S^{\circ}_{\mathrm{C}}) $$

We want to find the temperature at which the following reaction is spontaneous at standard conditions: $$ \mathrm{CO}_{2}(g)+\mathrm{C}(s) \longrightarrow 2 \mathrm{CO}(g) $$ To determine this, we need to find the temperature at which the Gibbs Free Energy change, \(\Delta G^{\circ}\), is negative. The relationship between \(\Delta G^{\circ}\), \(\Delta H^{\circ}\), and \(\Delta S^{\circ}\) is given by the formula: $$ \Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ} $$ Since \(\Delta G^{\circ}\) must be negative, we must find the temperature T at which: $$ \Delta H^{\circ} - T\Delta S^{\circ} < 0 $$ Rearranging the inequality gives us: $$ T > \frac{\Delta H^{\circ}}{\Delta S^{\circ}} $$ Using the calculated values for \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\), we can find the temperature at which the CO₂ conversion reaction is spontaneous.