(C) For a reaction to be spontaneous according to the $2^{nd}$ law of thermodynamics: $\Delta G < 0$. Since $\Delta G = \Delta H - T \Delta S$,we have

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(C) For a reaction to be spontaneous according to the $2^{nd}$ law of thermodynamics: $\Delta G Since $\Delta G = \Delta H - T \Delta S$,we have $\Delta H - T \Delta S At equilibrium,$\Delta G = 0$,so $\Delta H = T_e \Delta S$,which implies $T_e = \frac{\Delta H}{\Delta S}$. For the reaction to be spontaneous,$\Delta H - T \Delta S \Delta H$. Since $\Delta S$ is positive,$T > \frac{\Delta H}{\Delta S}$. Therefore,$T > T_e$.

Class 11 Chemistry Thermodynamics 2 nd Law of thermodynamics and Entropy

Medium

Let us consider a reversible reaction at temperature,$T$. In this reaction,both $\Delta H$ and $\Delta S$ were observed to have positive values. If the equilibrium temperature is $T_e$,then the reaction becomes spontaneous at:

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A
$T = T_e$
B
$T_e > T$
C
$T > T_e$
D
$T_e = 5 \ T$

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Thermodynamics

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2 nd Law of thermodynamics and Entropy

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Let us consider a reversible reaction at temperature,$T$. In this reaction,both $\Delta H$ and $\Delta S$ were observed to have positive values. If the equilibrium temperature is $T_e$,then the reaction becomes spontaneous at:

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