$\mathrm{\Delta H}$ and $\mathrm{\Delta S}$ for a reaction are + 30.558 kJ $\mathrm{mol}$ and $0.066 \mathrm{kJ} {\mathrm{K}}^{-} {\mathrm{mol}}^{-}$ at 1 atm pressure. The temperature at which, free energy change is equal to zero and the nature of the reaction below this temperature are:

$\begin{array}{l}\mathrm{\Delta H}=+30.558\mathrm{kJ}{\mathrm{mol}}^{-}\\ \mathrm{\Delta S}=0.066{\mathrm{kJK}}^{-}{\mathrm{mol}}^{-};\end{array}$

We know: $\begin{array}{r}\mathrm{\Delta G}=\mathrm{\Delta H}-\mathrm{T}∆\mathrm{S},0=+30.558-\mathrm{T}\times 0.066\\ \mathrm{T}=\frac{30.558}{0.066}=463\mathrm{K}\end{array}$

$(\mathrm{dG}{)}_{\text{T.P }}=0$

As temperature decreases, the TΔS term decreases and as ΔH is positive, the ΔG terms becomes greater than zero which means that reaction becomes non-spontaneous.