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(B) At equilibrium,the Gibbs free energy change $\Delta G = 0$. Using the relation $\Delta G = \Delta H - T \Delta S$,at equilibrium we have $\Delta H

Vedclass · Accepted Answer

$T > T_e$

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(B) At equilibrium,the Gibbs free energy change $\Delta G = 0$. Using the relation $\Delta G = \Delta H - T \Delta S$,at equilibrium we have $\Delta H - T_e \Delta S = 0$,which gives $T_e = \frac{\Delta H}{\Delta S}$. For a reaction to be spontaneous,the condition is $\Delta G Substituting the expression for $\Delta G$,we get $\Delta H - T \Delta S Dividing by $\Delta S$ (since $\Delta S > 0$),we get $T > \frac{\Delta H}{\Delta S}$. Since $T_e = \frac{\Delta H}{\Delta S}$,the condition for spontaneity is $T > T_e$.

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

For a particular reversible reaction at temperature $T$,$\Delta H$ and $\Delta S$ were found to be both positive. If $T_e$ is the temperature at equilibrium,the reaction would be spontaneous when

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