= Solution
\b[Excess <ethanol> drives a reversible <ester> exchange mainly through the actual mixture composition, not through a large new bond-energy advantage.] The net reaction exchanges one <ester> and one <alcohol> for a closely related <ester> and <alcohol>. It preserves the main bond types and has no obvious large exothermicity. Solvation and <ion> pairing can contribute to its <enthalpy>, but there is no basis here for asserting a large negative standard reaction <enthalpy>.
There are two <molecules> on each side, so an increase in particle count is not the <entropy> explanation. Instead use
$$
\Delta_rG=\Delta_rG^\circ+RT\ln Q,\qquad
Q=\frac{a_{\rm ethyl\ acetate}a_{\rm methanol}}{a_{\rm methyl\ acetate}a_{\rm ethanol}}.
$$
With <ethanol> as the solvent and little <methanol> initially, the <reaction quotient> is small; product formation and mixing lower the <Gibbs free energy> until the exchange equilibrium is reached. <ethanol> can be treated approximately as having fixed activity near its pure-solvent value, rather than assigned an arbitrarily enormous activity, while <methanol> remains dilute. This composition-dependent chemical-potential contribution is the appropriate entropic/mixing argument. It explains forward conversion under these conditions without claiming that $\Delta_rS^\circ$ must be positive, or that complete conversion follows without an <equilibrium constant> or removal of <methanol>.
Back to article page
