In which of the following equilibria is K_{eq | vedclass

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(NONE) For an equilibrium to have $K_{eq} > 1$,the reaction must proceed in the forward direction,meaning the products must be more stable than the re

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(NONE) For an equilibrium to have $K_{eq} > 1$,the reaction must proceed in the forward direction,meaning the products must be more stable than the reactants or the acid-base reaction must favor the formation of a weaker acid.$A$: The $pK_a$ of propyne $(HC \equiv C-CH_3)$ is $\approx 25$,while the $pK_a$ of water is $\approx 15.7$. Since the reactant acid is much weaker than the product acid (water),the equilibrium lies far to the left $(K_{eq} $B$: In the first conformer,$-OH$ is equatorial and $-F$ is axial. In the second,$-OH$ is axial and $-F$ is equatorial. The equatorial position is generally more stable. Since $-OH$ is bulkier than $-F$,the conformer with equatorial $-OH$ is more stable. Thus,the equilibrium favors the left side $(K_{eq} $C$: In the first conformer,$-Et$ is equatorial and $-Me$ is axial. In the second,$-Et$ is axial and $-Me$ is equatorial. Since the ethyl group $(-Et)$ is bulkier than the methyl group $(-Me)$,the conformer with equatorial $-Et$ is significantly more stable. Thus,the equilibrium favors the left side $(K_{eq} $D$: Phenol $(C_6H_5OH)$ has a $pK_a$ of $\approx 10$,while carbonic acid $(H_2CO_3)$ has a $pK_a$ of $\approx 6.4$. Since phenol is a weaker acid than carbonic acid,it cannot react with $NaHCO_3$ to form $CO_2$. The equilibrium lies far to the left $(K_{eq} Note: Based on standard chemical principles,none of the provided options have $K_{eq} > 1$. However,if this is a multiple-choice question where one must be selected,there may be an error in the question statement or options.

In which of the following equilibria is $K_{eq} > 1$?

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