Ethanoic acid is combined with ethanol. The ethanol contains a small quantity of methanol as an impurity. The equilibria below are set up. $\text{CH}_3\text{CO}_2\text{H}(l) + \text{CH}_3\text{CH}_2\text{OH}(l) \rightleftharpoons \text{CH}_3\text{CO}_2\text{CH}_2\text{CH}_3(l) + \text{H}_2\text{O}(l)$, $K_c = K_1$ $\text{CH}_3\text{CO}_2\text{H}(l) + \text{CH}_3\text{OH}(l) \rightleftharpoons \text{CH}_3\text{CO}_2\text{CH}_3(l) + \text{H}_2\text{O}(l)$, $K_c = K_2$ Which statement about the equilibrium mixture is correct?
- AOnly ethyl ethanoate will be formed because there is much more ethanol present than methanol.
- BIn this mixture $\frac{[\text{CH}_3\text{CO}_2\text{CH}_2\text{CH}_3]}{[\text{CH}_3\text{CO}_2\text{CH}_3]} = \frac{K_1}{K_2}$.
- CAdding water to the mixture will alter the mole ratio of the two esters.
- DAdding methyl ethanoate to the mixture will increase the number of moles of ethyl ethanoate.