Given the following data
`{:("Substance",DeltaH^(@)(KJ//mol),S^(@)(J//mol K),DeltaG^(@)(kJ//mol),),(FeO(s),-266.3,57.49,-245.12,),(C("Graphite"),0,5.74,0,),(Fe(s),0,5.74,0,),(CO(g),-110.5,197.6,-137.15,):}`
Determine at what temperature the following reaction is spontaneous?
`FeO(s)+C("Graphite")rarrFe(s)+CO(g)`

Options

A. `298 K`
B. `668 K`
C. `966 K`
D. `Delta G^(@)` is `+ve` hence the reaction will never be spontaneous

Explanation

To determine the temperature at which the reaction

\[ \text{FeO(s)} + \text{C(Graphite)} \rightarrow \text{Fe(s)} + \text{CO(g)} \]

is spontaneous, we need to analyze the Gibbs free energy change (\( \Delta G \)) for the reaction. A reaction is spontaneous when \( \Delta G < 0 \).

### Step 1: Calculate \( \Delta G \) for the reaction

The Gibbs free energy change for the reaction can be calculated using the formula:

\[ \Delta G = \Delta G_{\text{products}} - \Delta G_{\text{reactants}} \]

From the given data:

- \( \Delta G_{\text{Fe}} = 0 \, \text{kJ/mol} \)
- \( \Delta G_{\text{CO}} = -137.15 \, \text{kJ/mol} \)
- \( \Delta G_{\text{FeO}} = -245.12 \, \text{kJ/mol} \)
- \( \Delta G_{\text{C}} = 0 \, \text{kJ/mol} \)

Substituting these values into the equation:

\[ \Delta G = (0 + (-137.15)) - (-245.12 + 0) \]
\[ \Delta G = -137.15 + 245.12 \]
\[ \Delta G = 107.97 \, \text{kJ/mol} \]

### Step 2: Determine the change in enthalpy (\( \Delta H \)) and entropy (\( \Delta S \))

Using the same approach, we can calculate \( \Delta H \) and \( \Delta S \):

#### Calculate \( \Delta H \)

\[ \Delta H = \Delta H_{\text{products}} - \Delta H_{\text{reactants}} \]

From the given data:

- \( \Delta H_{\text{Fe}} = 0 \, \text{kJ/mol} \)
- \( \Delta H_{\text{CO}} = -110.5 \, \text{kJ/mol} \)
- \( \Delta H_{\text{FeO}} = -266.3 \, \text{kJ/mol} \)
- \( \Delta H_{\text{C}} = 0 \, \text{kJ/mol} \)

Substituting these values:

\[ \Delta H = (0 + (-110.5)) - (-266.3 + 0) \]
\[ \Delta H = -110.5 + 266.3 \]
\[ \Delta H = 155.8 \, \text{kJ/mol} \]

#### Calculate \( \Delta S \)

\[ \Delta S = \Delta S_{\text{products}} - \Delta S_{\text{reactants}} \]

From the given data:

- \( S_{\text{Fe}} = 5.74 \, \text{J/mol K} \)
- \( S_{\text{CO}} = 197.6 \, \text{J/mol K} \)
- \( S_{\text{FeO}} = 57.49 \, \text{J/mol K} \)
- \( S_{\text{C}} = 5.74 \, \text{J/mol K} \)

Substituting these values:

\[ \Delta S = (5.74 + 197.6) - (57.49 + 5.74) \]
\[ \Delta S = 203.34 - 63.23 \]
\[ \Delta S = 140.11 \, \text{J/mol K} \]

### Step 3: Find the temperature at which \( \Delta G = 0 \)

Using the Gibbs free energy equation:

\[ \Delta G = \Delta H - T \Delta S \]

Setting \( \Delta G = 0 \):

\[ 0 = \Delta H - T \Delta S \]

Rearranging gives:

\[ T = \frac{\Delta H}{\Delta S} \]

Substituting the values (note that \( \Delta H \) should be converted to J):

\[ \Delta H = 155.8 \, \text{kJ/mol} = 155800 \, \text{J/mol} \]
\[ \Delta S = 140.11 \, \text{J/mol K} \]

Calculating the temperature:

\[ T = \frac{155800 \, \text{J/mol}}{140.11 \, \text{J/mol K}} \approx 1112.4 \, \text{K} \]

### Conclusion

The reaction becomes spontaneous at temperatures higher than approximately \( 1112.4 \, \text{K} \).