Calculate `Delta_(f)H^(@)` (in kJ/mol) for `Cr_(2)O_(3)` from the `Delta_(r)G^(@)` and the `S^(@)` values provided at `27^(@)`
`4Cr(s)+3O_(2)(g)rarr2Cr_(2)O_(3)(s)," "Delta_(r)G^(@)=-2093.4kJ//mol`
`S^(@)("J//K mol") : S^(@)(Cr,s)=24," "S^(@)(O_(2),g)=205," "S^(@)(Cr_(2)O_(3),s)=81`

Options

A. `-2258.1kJ//mol`
B. `-1129.05kJ//mol`
C. `-964.35kJ//mol`
D. None of these

Explanation

To calculate the standard enthalpy of formation (Δ_fH^(@)) for Cr₂O₃ from the given Δ_rG^(@) and S^(@) values, we will follow these steps:

### Step 1: Write the Reaction
The reaction for the formation of chromium(III) oxide is:
\[ 4 \text{Cr}(s) + 3 \text{O}_2(g) \rightarrow 2 \text{Cr}_2\text{O}_3(s) \]

### Step 2: Calculate the Standard Entropy Change (ΔS^(@))
The standard entropy change for the reaction can be calculated using the formula:
\[ \Delta S^(@) = \sum \nu_i S^(@) \text{(products)} - \sum \nu_i S^(@) \text{(reactants)} \]

Where:
- \( S^(@)(\text{Cr}, s) = 24 \, \text{J/K mol} \)
- \( S^(@)(\text{O}_2, g) = 205 \, \text{J/K mol} \)
- \( S^(@)(\text{Cr}_2\text{O}_3, s) = 81 \, \text{J/K mol} \)

Calculating the entropy change:
\[ \Delta S^(@) = \left(2 \times S^(@)(\text{Cr}_2\text{O}_3)\right) - \left(4 \times S^(@)(\text{Cr}) + 3 \times S^(@)(\text{O}_2)\right) \]
\[ = \left(2 \times 81\right) - \left(4 \times 24 + 3 \times 205\right) \]
\[ = 162 - (96 + 615) \]
\[ = 162 - 711 = -549 \, \text{J/K mol} \]

### Step 3: Convert ΔS^(@) to kJ
To convert ΔS^(@) to kJ:
\[ \Delta S^(@) = -549 \, \text{J/K mol} \times \frac{1 \, \text{kJ}}{1000 \, \text{J}} = -0.549 \, \text{kJ/K mol} \]

### Step 4: Use the Gibbs Free Energy Equation
The relationship between Gibbs free energy, enthalpy, and entropy is given by:
\[ \Delta G^(@) = \Delta H^(@) - T \Delta S^(@) \]
Rearranging this gives:
\[ \Delta H^(@) = \Delta G^(@) + T \Delta S^(@) \]

### Step 5: Substitute the Values
Given:
- \( \Delta G^(@) = -2093.4 \, \text{kJ/mol} \)
- Temperature \( T = 27^\circ C = 300 \, \text{K} \)

Substituting the values:
\[ \Delta H^(@) = -2093.4 \, \text{kJ/mol} + (300 \, \text{K} \times -0.549 \, \text{kJ/K mol}) \]
\[ = -2093.4 \, \text{kJ/mol} - 164.7 \, \text{kJ/mol} \]
\[ = -2258.1 \, \text{kJ/mol} \]

### Step 6: Calculate the Enthalpy of Formation for 1 mole of Cr₂O₃
Since the reaction produces 2 moles of Cr₂O₃, the enthalpy of formation for 1 mole is:
\[ \Delta_f H^(@) = \frac{-2258.1 \, \text{kJ/mol}}{2} = -1129.05 \, \text{kJ/mol} \]

### Final Answer
The standard enthalpy of formation (Δ_fH^(@)) for Cr₂O₃ is:
\[ \Delta_f H^(@) = -1129.05 \, \text{kJ/mol} \]