An alcohol, $\text{ROH}$, undergoes a reversible reaction with ethanoic acid to form an ester. $\text{ROH}(l) + \text{CH}_3\text{COOH}(l) \rightleftharpoons \text{CH}_3\text{COOR}(l) + \text{H}_2\text{O}(l)$ A mixture is made by combining $3.0\,\text{mol}$ of $\text{ROH}$, $2.0\,\text{mol}$ of ethanoic acid and $1.0\,\text{mol}$ of water. When equilibrium is reached, $1.5\,\text{mol}$ of $\text{CH}_3\text{COOR}$ is present. Calculate the equilibrium constant, $K_c$, for this reaction.
- A$0.20$
- B$0.25$
- C$2.00$
- D$5.00$