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 \(\Delta_f H^\circ\) for \(Cr_2O_3\), we will use the provided values of \(\Delta_r G^\circ\) and the standard entropy \(S^\circ\) values at \(27^\circ C\) (or \(300 K\)).

### Step 1: Write the Reaction and Given Data
The reaction for the formation of chromium oxide is:
\[ 4Cr(s) + 3O_2(g) \rightarrow 2Cr_2O_3(s) \]
Given:
- \(\Delta_r G^\circ = -2093.4 \, \text{kJ/mol}\)
- \(S^\circ(Cr, s) = 24 \, \text{J/K mol}\)
- \(S^\circ(O_2, g) = 205 \, \text{J/K mol}\)
- \(S^\circ(Cr_2O_3, s) = 81 \, \text{J/K mol}\)

### Step 2: Calculate \(\Delta S^\circ\)
The change in entropy \(\Delta S^\circ\) for the reaction can be calculated using the formula:
\[ \Delta S^\circ = S^\circ_{products} - S^\circ_{reactants} \]
Substituting the values:
\[ \Delta S^\circ = 2 \times S^\circ(Cr_2O_3) - (4 \times S^\circ(Cr) + 3 \times S^\circ(O_2)) \]
\[ \Delta S^\circ = 2 \times 81 - (4 \times 24 + 3 \times 205) \]
Calculating each term:
\[ = 162 - (96 + 615) \]
\[ = 162 - 711 = -549 \, \text{J/K mol} \]

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

### Step 4: Use the Gibbs Free Energy Equation
We can relate \(\Delta G^\circ\), \(\Delta H^\circ\), and \(\Delta S^\circ\) using the equation:
\[ \Delta G^\circ = \Delta H^\circ - T \Delta S^\circ \]
Rearranging to find \(\Delta H^\circ\):
\[ \Delta H^\circ = \Delta G^\circ + T \Delta S^\circ \]
Substituting the values:
\[ \Delta H^\circ = -2093.4 \, \text{kJ/mol} + (300 \, \text{K}) \times (-0.549 \, \text{kJ/K mol}) \]
Calculating the second term:
\[ = -2093.4 \, \text{kJ/mol} - 164.7 \, \text{kJ/mol} \]
\[ = -2258.1 \, \text{kJ/mol} \]

### Step 5: Calculate \(\Delta_f H^\circ\) for \(Cr_2O_3\)
Since the reaction produces 2 moles of \(Cr_2O_3\), the enthalpy of formation for 1 mole of \(Cr_2O_3\) is:
\[ \Delta_f H^\circ(Cr_2O_3) = \frac{1}{2} \Delta H^\circ \]
\[ = \frac{1}{2} \times (-2258.1 \, \text{kJ/mol}) = -1129.05 \, \text{kJ/mol} \]

### Final Answer
\[ \Delta_f H^\circ(Cr_2O_3) = -1129.05 \, \text{kJ/mol} \]