The signs of Delta_r H^{circ} and Delta_r S^{ | vedclass

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(D) For a reaction to be spontaneous,the Gibbs free energy change $\Delta_r G^{\circ}$ must be negative. The relationship is given by the equation: $\

Vedclass · Accepted Answer

negative,positive

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(D) For a reaction to be spontaneous,the Gibbs free energy change $\Delta_r G^{\circ}$ must be negative. The relationship is given by the equation: $\Delta_r G^{\circ} = \Delta_r H^{\circ} - T \Delta_r S^{\circ}$. For the reaction to be spontaneous at all temperatures $(T)$,the term $\Delta_r G^{\circ}$ must remain negative regardless of the value of $T$. This occurs when $\Delta_r H^{\circ}$ is negative (exothermic) and $\Delta_r S^{\circ}$ is positive (increase in entropy). Thus,the correct signs are negative and positive.

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

The signs of $\Delta_r H^{\circ}$ and $\Delta_r S^{\circ}$ for a reaction to be spontaneous at all temperatures respectively are

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