JEE MAIN - Chemistry (2002 - No. 1)
If an endothermic reaction is non-spontaneous at freezing point of water and becomes feasible at its boiling point, then :
$$\Delta H$$ is -ve, $$\Delta S$$ is +ve
$$\Delta H$$ and $$\Delta S$$ are both +ve
$$\Delta H$$ and $$\Delta S$$ are both -ve
$$\Delta H$$ is +ve, $$\Delta S$$ is -ve
Explanation
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 $$T\Delta S > \Delta H.$$
$$\Delta G = \Delta H - T\Delta S$$
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 $$T\Delta S > \Delta H.$$
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