Question

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.

Short answer

The calculated values are \(\Delta H^{\circ} = 1114.74\,\text{kJ/mol}\), \(\Delta S^{\circ} = -291.3\,\text{J/mol K}\), and \(\Delta G^{\circ} = 1427.44\,\text{kJ/mol}\) at 800°C.

Step-by-step solution

Calculate \(\Delta H^{\circ}\)

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 \(\Delta H^{\circ}\): $$\Delta H^{\circ} = (-2240.58\,\text{kJ/mol}) - (-3355.32\,\text{kJ/mol}) = 1114.74\,\text{kJ/mol}$$

Calculate \(\Delta S^{\circ}\)

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 \(\Delta S^{\circ}\): $$\Delta S^{\circ} = (506.6\,\text{J/mol K}) - (797.9\,\text{J/mol K}) = -291.3\,\text{J/mol K}$$

Calculate \(\Delta G^{\circ}\) at 800°C

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.

Key concepts

Enthalpy Change

Enthalpy change, symbolized as \(\Delta H\), is a fundamental concept in thermodynamics in chemistry that quantifies the heat change in a reaction occurring at constant pressure. It's derived from the enthalpies of formation of the reactants and the products. If a reaction absorbs heat, it's called endothermic, and if it releases heat, it's exothermic.
In our exercise, the process is endothermic as the calculated \(\Delta H^{\circ}\) for the reaction is 1114.74 kJ/mol. This indicates the system needs to absorb 1114.74 kJ of energy per mole for the reaction to occur at standard conditions. To determine this, you sum up the enthalpies of the products and subtract the sum of the enthalpies of the reactants. This gives us insight into the energy required or released during the reaction.

Entropy Change

Entropy change, denoted \(\Delta S\), measures the degree of disorder or randomness within a system. In chemical reactions, an increase in entropy suggests a more dispersed energy state, while a decrease indicates more ordered energy.
In the given blast furnace reaction, \(\Delta S^{\circ}\) is calculated as -291.3 J/mol K. This negative value indicates that the reaction results in a decrease in randomness or disorder. Generally, reactions producing fewer gas molecules from more gas molecules tend to decrease in entropy, leading to a more ordered state. The calculation involves subtracting the sum of the entropies of the reactants from the sum of the entropies of the products, much like with enthalpy.

Gibbs Free Energy

The Gibbs Free Energy, symbolized \(\Delta G\), is an essential thermodynamic function combining enthalpy and entropy changes to determine a reaction's spontaneity. A negative \(\Delta G\) implies a reaction is spontaneous under constant temperature and pressure, while a positive \(\Delta G\) indicates non-spontaneity.
For our reaction at 800°C (1073.15 K), \(\Delta G^{\circ}\) is calculated to be 1427.44 kJ/mol. This positive value shows the reaction is not spontaneous under the given conditions. The free energy change is obtained using the equation \(\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\). The temperature-dependence of \(\Delta G\) highlights an important aspect: higher temperatures can shift the balance of spontaneity, as they increase the effect of \(-T\Delta S\). Thus, a reaction non-spontaneous at certain conditions might become spontaneous when conditions change.