In conversion of lime-stone to lime, $$CaCO_3(s) \longrightarrow CaO(s) + CO_2(g)$$ the values of $$\Delta H^\circ$$ and $$\Delta S^\circ$$ are $$+179.1$$ kJ $$mol^{-1}$$ and 160.2 J/K respectively at 298 K and 1 bar. Assuming that $$\Delta H^\circ$$ do not change with temperature, temperature above which conversion of limestone to lime will be spontaneous is

For spontaneity, the Gibbs free-energy change must be negative:

$$\Delta G^\circ \;=\; \Delta H^\circ - T\,\Delta S^\circ$$

The reaction becomes just spontaneous when $$\Delta G^\circ = 0$$. Setting the above expression to zero gives the threshold temperature $$T_c$$:

$$0 = \Delta H^\circ - T_c\,\Delta S^\circ \quad\Longrightarrow\quad T_c = \frac{\Delta H^\circ}{\Delta S^\circ}$$

Convert all units to joule so they are consistent:
$$\Delta H^\circ = 179.1\ \text{kJ mol}^{-1} = 179.1 \times 10^{3}\ \text{J mol}^{-1}$$
$$\Delta S^\circ = 160.2\ \text{J K}^{-1}\,\text{mol}^{-1}$$

Now compute $$T_c$$:

$$T_c \;=\; \frac{179.1 \times 10^{3}}{160.2} \;=\; 1118.4\ \text{K}$$

Above this temperature ($$T \gt 1118\ \text{K}$$) the reaction $$CaCO_3(s) \rightarrow CaO(s) + CO_2(g)$$ becomes spontaneous.

Therefore, the required temperature is approximately $$1118\ \text{K}$$.

Option D which is: 1118 K