The reaction,
`MgO(s) + C(s) to Mg(s) + CO(g),` for which `Delta_(r)H^(@)= +491.1 " " kJ " " mol^(-1)` and `Delta_(r)S^(@)= 198.0 JK^(-1)mol^(-1)`, is not feasible at 289 K . Temperature above which reaction will be feasible is :

Options

A. `2480.3 K `
B. `1890.0K `
C. `2380.5K`
D. `2040.5K `

Explanation

To determine the temperature above which the reaction

\[ \text{MgO(s) + C(s) } \rightarrow \text{ Mg(s) + CO(g)} \]

is feasible, we will use the Gibbs free energy equation:

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

### Step 1: Understand the conditions for feasibility
For the reaction to be feasible, the Gibbs free energy change (\(\Delta G^\circ\)) must be less than zero:

\[ \Delta G^\circ < 0 \]

### Step 2: Substitute the known values
We know from the problem statement:

- \(\Delta H^\circ = +491.1 \, \text{kJ/mol} = 491100 \, \text{J/mol}\) (convert kJ to J)
- \(\Delta S^\circ = 198.0 \, \text{J/K·mol}\)

Substituting these values into the Gibbs free energy equation gives:

\[ 491100 - T \cdot 198 < 0 \]

### Step 3: Rearrange the equation
Rearranging the inequality to solve for \(T\):

\[ 491100 < T \cdot 198 \]

\[ T > \frac{491100}{198} \]

### Step 4: Calculate the temperature
Now, we perform the division:

\[ T > \frac{491100}{198} \approx 2475.76 \, \text{K} \]

### Step 5: Conclusion
Thus, the temperature above which the reaction will be feasible is approximately:

\[ T > 2475.76 \, \text{K} \]

### Final Answer
The temperature above which the reaction will be feasible is **approximately 2476 K**.

---