Question
If an endothermic reaction is non-spontaneous at freezing point of water and becomes feasible at its boiling point, then
A.
$$\Delta H\,is - ve,\,\Delta S\,is\, + ve$$
B.
$$\Delta H\,and\,\,\Delta S\,\,both\,are\, + ve$$
C.
$$\Delta H\,and\,\Delta S\,\,both\,are\, - ve$$
D.
$$\Delta H\,is\, + ve,\,\Delta S\,is\, - ve$$
Answer :
$$\Delta H\,and\,\,\Delta S\,\,both\,are\, + ve$$
Solution :
TIPS/Formulae:
$$\Delta G = \Delta H - T\Delta S$$
Since $$\Delta G = \Delta H - T\Delta S$$ for an endothermic reaction,
$$\Delta H = + ve$$ and at low temperature $$\Delta S = + ve$$
Hence $$\Delta G = \left( + \right)\Delta H - T\left( + \right)\Delta S$$
and if $$T\Delta S < \Delta H$$ (at low temp)
$$\Delta G = + ve$$ (non spontaneous)
But at high temperature, reaction becomes spontaneous i.e. $$\Delta G = - ve.$$
because at higher temperature $$T\Delta S > \Delta H.$$