Consider the graph of Gibbs free energy G vs. | vedclass

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(D) For a spontaneous process,the slope of the $G$ vs. extent of reaction graph is negative,i.e.,$dG/d\xi < 0$.For a process at equilibrium,the slope

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(D) For a spontaneous process,the slope of the $G$ vs. extent of reaction graph is negative,i.e.,$dG/d\xi For a process at equilibrium,the slope is zero,i.e.,$dG/d\xi = 0$.For a non-spontaneous process,the slope is positive,i.e.,$dG/d\xi > 0$.At point $(a)$,the slope is negative,so the reaction is spontaneous.At point $(b)$,the slope is zero,so the reaction is at equilibrium.At point $(c)$,the slope is positive,so the reaction is non-spontaneous.Evaluating the statements:$A$. Reaction is spontaneous at $(a)$ and $(b)$: False (at $(b)$ it is at equilibrium).$B$. Reaction is at equilibrium at point $(b)$ and non-spontaneous at point $(c)$: True.$C$. Reaction is spontaneous at $(a)$ and non-spontaneous at $(c)$: True.$D$. Reaction is non-spontaneous at $(a)$ and $(b)$: False.Therefore,there are $2$ true statements ($B$ and $C$).

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