I am currently reading chapter 26 of Clayden's Organic Chemistry, and a diagram for the elementary steps of the Claisen condensation with ethyl acetate in ethoxide is provided. It is obvious why the net reaction is favorable. The 1,3-dicarbonyl compound has a pKa of ~10 while the pKa of ethanol is 16 (so the more acidic 1,3-dicarbonyl compound will be the product in its deprotonated form). Even though the first few elementary steps may not favor the product, the last step is essentially irreversible.
I am having trouble understanding why the second equilibrium arrow claims to favor the formation of the reactants and not the products.
The first equilibrium arrow is obvious, the pKa of ethanol is 16, and the pKa of the ester enolate is ~25 (when protonated), so the less acidic ester will be favored in equilibrium. However, in the second equilibrium arrow, the pKa of the starting enolate is ~25 (when protonated) while the pKa (when protonated) of the product alkoxide is likely somewhere around 16. Wouldn't this mean the product for the second equilibrium arrow would be favored thermodynamically? I have considered that this step does result in a decrease in entropy, but I wouldn't expect this to have a drastic impact on the direction of the equillibrium.