Question

Esterification is the reaction of a carboxylic acid (RCOOH) with an alcohol (R'OH) to form an ester (RCOOR') with a loss of water. Equation [1] is an example of an intermolecular esterification reaction. Equation [2] is an example of an intramolecular esterification reaction; that is, the carboxylic acid and alcohol are contained in the same starting material, forming a cyclic ester as the product. The equilibrium constants for both reactions are given. Explain why ${\mathbf{K}}_{\mathbf{e}\mathbf{q}}$is different for these two apparently similar reactions.

[1]

[2]

Short answer

Answer

Entropy does not play a role in reaction[1] as the number of molecules present in the starting material and the products are similar. The entropy increases in reaction[2] as the small molecule of starting material lead to product formation.

Step-by-step solution

Step-by-Step SolutionStep 1: Esterification

Esterification reactions are reactions where an ester is created when alcohol and carboxylic acid are combined. Acid anhydrides can also combine with alcohol to generate an ester.

Equilibrium constant Keq

The equilibrium constant connects the quantity of starting material with the product at equilibrium.The equilibrium acts to the right if the equilibrium constant ${\mathrm{K}}_{\mathrm{eq}}$ is greater than one. It acts to the left if the equilibrium constant is less than one.

Difference in the equilibrium constant for the given reactions

Entropy does not have a role in reaction[1] as the number of molecules of reactants and products is similar. In reaction [2], the single-molecule of starting material leads to form two products, so the entropy increases.

This leads $∆\mathrm{G}°$ to becoming more favorable, thereby leading to an increase in ${\mathrm{K}}_{\mathrm{eq}}$.