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.

Reaction	$\Delta H \ (kJ \ mol^{-1}) / \Delta S \ (J \ K^{-1} \ mol^{-1})$
$I$	$+98.0, +14.8$
$II$	$-55.5, -84.6$
$III$	$+28.3, -17.0$
$IV$	$-40.5, +24.6$
$V$	$+34.7, 0.0$

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

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