Esterification Equilibrium: Adding Water Does Not Push the Reaction Right
Ethyl ethanoate is formed by the reversible reaction:
CH3COOH(l) + C2H5OH(l) ⇌ CH3COOC2H5(l) + H2O(l)
A student mixes 0.50 mol of ethanoic acid with 0.50 mol of ethanol in a sealed 1.00 dm3 flask at 25 °C, using a little concentrated sulfuric acid as catalyst. At equilibrium, the flask contains 0.33 mol of ethyl ethanoate. The student then prepares a second mixture in a 1.00 dm3 flask at 25 °C containing 0.30 mol ethanoic acid, 0.10 mol ethanol, 0.50 mol ethyl ethanoate and 0.20 mol water, with the same catalyst. A classmate suggests adding 0.50 mol of water to the second mixture to 'push the equilibrium to the right and make more ester'. The student wants to know whether the second mixture is already at equilibrium and how the ester yield could be improved.
1. Calculate Kc for the esterification at 25 °C. Include units if appropriate.
Read the worked answer
Let the equilibrium amount of ethyl ethanoate be 0.33 mol. From the stoichiometry, 0.33 mol of water is also formed, and the amounts of acid and alcohol left are 0.50 - 0.33 = 0.17 mol each. Since the flask volume is 1.00 dm3, the equilibrium concentrations are [CH3COOH] = 0.17 mol dm-3, [C2H5OH] = 0.17 mol dm-3, [CH3COOC2H5] = 0.33 mol dm-3 and [H2O] = 0.33 mol dm-3. Therefore Kc = (0.33 × 0.33)/(0.17 × 0.17) = 0.1089/0.0289 = 3.77. There are two moles of reactant and two moles of product, so the concentration units cancel and Kc is dimensionless under these conditions.
2. For the second mixture, calculate the reaction quotient Q and predict the direction of net change.
Read the worked answer
The second mixture initially contains [CH3COOH] = 0.30 mol dm-3, [C2H5OH] = 0.10 mol dm-3, [CH3COOC2H5] = 0.50 mol dm-3 and [H2O] = 0.20 mol dm-3. The reaction quotient has the same form as Kc: Q = ([CH3COOC2H5][H2O])/([CH3COOH][C2H5OH]) = (0.50 × 0.20)/(0.30 × 0.10) = 0.100/0.030 = 3.33. Since Q is less than Kc (3.77), the reaction is not at equilibrium. The system will shift to the right, increasing the amounts of ester and water and decreasing the amounts of acid and alcohol, until Q reaches Kc.
3. Evaluate the classmate's suggestion to add 0.50 mol of water. Explain whether it would increase the ester yield, and suggest a better change.
Read the worked answer
Adding water increases the concentration of a product. In the expression Q = ([ester][water])/([acid][alcohol]), increasing [water] increases Q. The mixture would then have Q greater than Kc, so the equilibrium would shift to the left, consuming ester and water and producing more acid and alcohol. The ester yield would therefore decrease, not increase. Adding water does not 'push the reaction to the right'. A better change would be to remove ethyl ethanoate or water as it forms, which lowers Q and shifts the equilibrium to the right. Alternatively, using an excess of ethanoic acid or ethanol would increase the equilibrium yield of ester, although the catalyst only speeds up attainment of equilibrium and does not change Kc.