The reaction $$MgO(s) + C(s) \to Mg(s) + CO(g)$$, for which $$\Delta H° = +491.1$$ kJ mol$$^{-1}$$ and $$\Delta S° = 198.0$$ JK$$^{-1}$$ mol$$^{-1}$$ is not feasible at 298 K. Temperature above which reaction will be feasible is

For deciding whether a reaction is spontaneous, we use the Gibbs free-energy change formula

$$\Delta G^{\circ}=\Delta H^{\circ}-T\Delta S^{\circ}$$

A process is feasible (spontaneous) only when $$\Delta G^{\circ}<0$$. Thus we want

$$\Delta H^{\circ}-T\Delta S^{\circ}<0$$

Rearranging this inequality, we obtain

$$-T\Delta S^{\circ}<-\Delta H^{\circ}\;,\qquad\text{ so }$$

$$T\Delta S^{\circ}>\Delta H^{\circ}$$

and finally

$$T>\dfrac{\Delta H^{\circ}}{\Delta S^{\circ}}$$

Now we substitute the numerical values. First, convert the enthalpy change from kilojoules to joules so that both $$\Delta H^{\circ}$$ and $$\Delta S^{\circ}$$ are expressed in the same units:

$$\Delta H^{\circ}=+491.1\text{ kJ mol}^{-1}=491.1\times10^{3}\text{ J mol}^{-1}$$

Given $$\Delta S^{\circ}=198.0\text{ J K}^{-1}\text{ mol}^{-1}$$, we write

$$T>\dfrac{491.1\times10^{3}}{198.0}$$

Carrying out the division step by step:

$$\dfrac{491.1\times10^{3}}{198.0}= \dfrac{491100}{198}$$

We compute the quotient:

$$198\times2000 = 396000$$

$$198\times400 = 79200 \quad\;\;(\text{cumulative } 396000+79200=475200)$$

$$198\times80 = 15840 \quad\;\;(\text{cumulative } 475200+15840=491040)$$

Adding these contributions, we obtain $$2000+400+80=2480$$ with a small remainder:

$$491100-491040=60$$

Dividing the remaining 60 by 198 gives approximately $$0.3$$.

So,

$$T>2480.3\text{ K}$$

Therefore the reaction becomes thermodynamically feasible only when the temperature exceeds about $$2480.3\text{ K}$$.

Hence, the correct answer is Option C.