Consider the reaction $$\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{H}_{2} \mathrm{O}(g)$$ a. Use \(\Delta G_{f}^{\circ}\) values in Appendix 4 to calculate $\Delta G^{\circ}$ for this reaction. b. Is this reaction spontaneous under standard conditions at 298 $\mathrm{K} ?$ c. The value of \(\Delta H^{\circ}\) for this reaction is 100 . kJ. At what temperatures is this reaction spontaneous at standard conditions? Assume that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature.

To calculate ΔG° for this reaction, use the formula: ΔG° = Σ ΔGf°(products) - Σ ΔGf°(reactants) From Appendix 4, we get the following ΔGf° values: ΔGf°(Fe₂O₃) = -740 kJ/mol ΔGf°(H₂) = 0 kJ/mol (since it's a diatomic gas) ΔGf°(Fe) = 0 kJ/mol (since it's in its standard state) ΔGf°(H₂O) = -228.58 kJ/mol Now we can calculate ΔG°: ΔG° = (2 x 0 + 3 x (-228.58)) - (1 x (-740) + 3 x 0) ΔG° = -685.74 kJ/mol

To check if the reaction is spontaneous, compare the calculated ΔG° value with zero: If ΔG° < 0, then the reaction is spontaneous. If ΔG° > 0, then the reaction is non-spontaneous. In our case, since ΔG° = -685.74 kJ/mol, which is less than 0, the reaction is spontaneous at 298 K.

To determine the temperature range for which the reaction is spontaneous, we can use the Gibbs-Helmholtz equation: ΔG° = ΔH° - TΔS° We need to find ΔS° using the given ΔH° value: ΔH° = 100 kJ/mol Rearrange the equation to find ΔS°: ΔS° = (ΔH° - ΔG°) / T Using the given temperature of 298 K: ΔS° = (100 - (-685.74)) / 298 ΔS° = 2.637 kJ/mol K Now, we need to find the temperature at which ΔG° = 0: 0 = ΔH° - TΔS° Rearrange to find T: T = ΔH° / ΔS° T = 100 / 2.637 T = 37.92 K Since the problem asks for the temperature range where the reaction is spontaneous, we can conclude that this reaction is spontaneous for all temperatures above 37.92 K under standard conditions, given that ΔH° and ΔS° do not depend on temperature.