Use the data in Appendix 4 for the following. a. Calculate \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for the reaction $$3 \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{CO}(g) \longrightarrow 2 \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{CO}_{2}(g)$$ that occurs in a blast furnace. b. Assume that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are independent of temperature. Calculate \(\Delta G^{\circ}\) at \(800 .{ }^{\circ} \mathrm{C}\) for this reaction.

To find the standard enthalpy change (\(\Delta H^{\circ}\)) for the reaction, we need to subtract the standard enthalpies of formation for the reactants from the standard enthalpies of formation for the products. Using the data from Appendix 4, we can write the equation for \(\Delta H^{\circ}\) as follows: $$\Delta H^{\circ} = \big(2 \times \Delta H^{\circ}_{3Fe_{2}O_{3}}\big) - \big(3 \times \Delta H^{\circ}_{Fe_{3}O_{4}} + \Delta H^{\circ}_{CO}\big)$$ Now, plug in the values for the standard enthalpies of formation for each compound: $$\Delta H^{\circ} = \big(2 \times (-1120.29\,\text{kJ/mol})\big) - \big(3 \times (-1118.44\,\text{kJ/mol}) + (-110.53\,\text{kJ/mol})\big)$$ Calculate the value of \(\Delta H^{\circ}\): $$\Delta H^{\circ} = (-2240.58\,\text{kJ/mol}) - (-3355.32\,\text{kJ/mol}) = 1114.74\,\text{kJ/mol}$$

To find the standard entropy change (\(\Delta S^{\circ}\)) for the reaction, we need to subtract the standard entropies for the reactants from the standard entropies for the products. Using the data from Appendix 4, we can write the equation for \(\Delta S^{\circ}\) as follows: $$\Delta S^{\circ} = \big(2 \times S^{\circ}_{3Fe_{2}O_{3}} + S^{\circ}_{CO_{2}}\big) - \big(3 \times S^{\circ}_{Fe_{3}O_{4}} + S^{\circ}_{CO}\big)$$ Now, plug in the values for the standard entropies for each compound: $$\Delta S^{\circ} = \big(2 \times (146.4\,\text{J/mol K}) + (213.8\,\text{J/mol K})\big) - \big(3 \times (200.0\,\text{J/mol K}) + (197.9\,\text{J/mol K})\big)$$ Calculate the value of \(\Delta S^{\circ}\): $$\Delta S^{\circ} = (506.6\,\text{J/mol K}) - (797.9\,\text{J/mol K}) = -291.3\,\text{J/mol K}$$

Now that we have the values for \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\), we can calculate the standard Gibbs free energy change (\(\Delta G^{\circ}\)) at 800°C using the following equation: $$\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}$$ Convert the temperature from Celsius to Kelvin: $$T = 800 + 273.15 = 1073.15\,\text{K}$$ Plug in the values for \(\Delta H^{\circ}\), \(\Delta S^{\circ}\), and T: $$\Delta G^{\circ} = 1114.74\,\text{kJ/mol} - (1073.15\,\text{K} \times -0.2913\,\text{kJ/mol K}) = 1114.74\,\text{kJ/mol} + 312.7\,\text{kJ/mol}$$ Calculate the value of \(\Delta G^{\circ}\): $$\Delta G^{\circ} = 1427.44\,\text{kJ/mol}$$ The standard Gibbs free energy change, \(\Delta G^{\circ}\), for the reaction at 800°C is 1427.44 kJ/mol.