We know, ΔG=ΔH−TΔS So, lets find the equilibrium temperature, i.e. at which $\begin{aligned}
& \Delta \mathrm{G}=0 \
& \Delta \mathrm{H}=\mathrm{T} \Delta \mathrm{S} \
& \mathrm{T}=\frac{179.1 \times 1000}{160.2} \
& =1118 \mathrm{~K}
\end{aligned}$ So, at temperature above this, the reaction will become spontaneous. Hence, (D) is correct answer.