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. 1890.0 K
C. 2380.5 K
D. 2040.5 K

Explanation

To determine the temperature above which the reaction `MgO(s) + C(s) → Mg(s) + CO(g)` becomes feasible, we can use the Gibbs free energy equation:

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

For the reaction to be feasible (spontaneous), we need:

\[ \Delta G < 0 \]

This leads us to the condition:

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

Rearranging this gives:

\[ T \Delta S > \Delta H \]

Thus, we can express the temperature \( T \) as:

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

### Step 1: Convert the values of ΔH and ΔS

Given:
- \(\Delta H = +491.1 \, \text{kJ/mol}\)
- \(\Delta S = 198.0 \, \text{J/K/mol}\)

First, we need to convert \(\Delta H\) from kJ to J:

\[ \Delta H = 491.1 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = 491100 \, \text{J/mol} \]

### Step 2: Substitute the values into the equation

Now we can substitute the values into the temperature equation:

\[ T > \frac{491100 \, \text{J/mol}}{198.0 \, \text{J/K/mol}} \]

### Step 3: Calculate the temperature

Now, we perform the division:

\[ T > \frac{491100}{198.0} \approx 2480.3 \, \text{K} \]

### Conclusion

Thus, the temperature above which the reaction becomes feasible is approximately:

\[ T > 2480.3 \, \text{K} \]