Calculate the standard free energy change for the reaction `: Fe_(2)O_(3)(s) + 4H_(2)(g) rarr2Fe(s) + 3H_(2)O(l)` Given that the standard free energies of formation of`Fe_(2)O_(3)` and `H_(2)O` are - 741.0 and -237.2kJ `mol^(-1)` respectively.

Answer

To calculate the standard free energy change (ΔG°) for the reaction:

\[ \text{Fe}_2\text{O}_3(s) + 4\text{H}_2(g) \rightarrow 2\text{Fe}(s) + 3\text{H}_2\text{O}(l) \]

we will use the standard free energies of formation for the reactants and products.

### Step 1: Write the formula for ΔG°
The standard free energy change for a reaction can be calculated using the following formula:

\[ \Delta G^\circ = \Delta G^\circ_{\text{products}} - \Delta G^\circ_{\text{reactants}} \]

### Step 2: Identify the standard free energies of formation
From the problem statement, we have:
- Standard free energy of formation of \(\text{Fe}_2\text{O}_3(s)\) = \(-741.0 \, \text{kJ/mol}\)
- Standard free energy of formation of \(\text{H}_2\text{O}(l)\) = \(-237.2 \, \text{kJ/mol}\)

The standard free energies of formation for elements in their standard states (like \(\text{Fe}(s)\) and \(\text{H}_2(g)\)) are \(0 \, \text{kJ/mol}\).

### Step 3: Calculate ΔG° for the products
The products of the reaction are \(2\text{Fe}(s)\) and \(3\text{H}_2\text{O}(l)\). The ΔG° for the products is calculated as follows:

\[ \Delta G^\circ_{\text{products}} = 2 \times \Delta G^\circ_{\text{Fe}} + 3 \times \Delta G^\circ_{\text{H}_2\text{O}} \]
\[ \Delta G^\circ_{\text{products}} = 2 \times 0 + 3 \times (-237.2) = 0 - 711.6 = -711.6 \, \text{kJ} \]

### Step 4: Calculate ΔG° for the reactants
The reactants of the reaction are \(\text{Fe}_2\text{O}_3(s)\) and \(4\text{H}_2(g)\). The ΔG° for the reactants is calculated as follows:

\[ \Delta G^\circ_{\text{reactants}} = \Delta G^\circ_{\text{Fe}_2\text{O}_3} + 4 \times \Delta G^\circ_{\text{H}_2} \]
\[ \Delta G^\circ_{\text{reactants}} = -741.0 + 4 \times 0 = -741.0 \, \text{kJ} \]

### Step 5: Substitute values into the ΔG° formula
Now, we can substitute the values into the ΔG° formula:

\[ \Delta G^\circ = \Delta G^\circ_{\text{products}} - \Delta G^\circ_{\text{reactants}} \]
\[ \Delta G^\circ = (-711.6) - (-741.0) \]
\[ \Delta G^\circ = -711.6 + 741.0 = 29.4 \, \text{kJ} \]

### Conclusion
The standard free energy change for the reaction is:

\[ \Delta G^\circ = 29.4 \, \text{kJ} \]

This indicates that the reaction is non-spontaneous at standard conditions since ΔG° is greater than 0.