A reaction has both Delta H and Delta S negat | vedclass

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(B) For a reaction where $\Delta H < 0$ and $\Delta S < 0$,the spontaneity is governed by the Gibbs free energy equation: $\Delta G = \Delta H - T\Del

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increases with decrease in temperature

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(B) For a reaction where $\Delta H For the reaction to be spontaneous,$\Delta G$ must be negative. Since $\Delta S$ is negative,the term $-T\Delta S$ is positive. As temperature $(T)$ decreases,the magnitude of the positive term $-T\Delta S$ decreases,making $\Delta G$ more negative. Thus,the reaction becomes more thermodynamically favorable at lower temperatures. In the context of this question,the term "rate" refers to the thermodynamic favorability or the extent of the reaction.

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

$A$ reaction has both $\Delta H$ and $\Delta S$ negative. The rate of reaction

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