If an endothermic reaction is non-spontaneous | vedclass

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(B) For a reaction to be spontaneous,the Gibbs free energy change $\Delta G = \Delta H - T\Delta S$ must be negative $(\Delta G < 0)$.Given that the r

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$\Delta H$ and $\Delta S$ both are $+ve$

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(B) For a reaction to be spontaneous,the Gibbs free energy change $\Delta G = \Delta H - T\Delta S$ must be negative $(\Delta G Given that the reaction is endothermic,$\Delta H > 0$ (positive).At the freezing point of water $(T = 273 \ K)$,the reaction is non-spontaneous,meaning $\Delta G > 0$,so $\Delta H > T\Delta S$.At the boiling point of water $(T = 373 \ K)$,the reaction becomes feasible (spontaneous),meaning $\Delta G Since $\Delta H$ is positive,for $\Delta G$ to become negative as $T$ increases,$\Delta S$ must also be positive. Thus,both $\Delta H$ and $\Delta S$ are positive.

$\Delta_{\text{r}}H^0$	$\Delta_{\text{r}}S^0$	$\Delta_{\text{r}}G^0$	Description
$(a) (+)$	$(-)$	$(+)$	$(i) \text{ Non-spontaneous at all temperatures}$
$(b) (-)$	$(-)$	$(-)$	$(ii) \text{ Spontaneous at low temperatures}$
$(c) (-)$	$(+)$	$(-)$	$(iii) \text{ Spontaneous at all temperatures}$

If an endothermic reaction is non-spontaneous at the freezing point of water and becomes feasible at its boiling point,then:

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