Answer: 463 K, non-spontaneous

$\Delta G = \Delta H - T \Delta S = 0$

$\therefore T = \frac{\Delta H}{\Delta S} = \frac{+30.558 \text{ kJ } mol^{-1}}{0.066 \text{ kJ }K^{-1} mol^{-1}}$ =** 463 K**

$\Delta G = \Delta H - T \Delta S = (30.558 - T \times 0.066)$ kJ $mol^{-1}$

At T<463 K,

0.066 T < 463 $\times$ 0.066

or, 0.066 T < 30.558

or, 0 < 30.558 - 0.066 T

or, 0 < $\Delta G$

Thus, at T < 463 K, $\Delta G$ > 0 i.e. , process is** non-spontaneous.**