Shifting an Esterification Equilibrium
Ethanoic acid and ethanol are mixed in a sealed flask and warmed with a small amount of concentrated sulfuric acid. After equilibrium is reached at a fixed temperature, analysis shows 0.30 mol of ethyl ethanoate, 0.30 mol of water, 0.20 mol of ethanoic acid and 0.20 mol of ethanol in a total volume of 0.50 L. A technician proposes three changes to 'get more product': (i) add 0.50 mol of ethanol, (ii) remove water continuously with a drying agent, (iii) add more sulfuric acid catalyst. The reaction is CH3COOH + C2H5OH ⇌ CH3COOC2H5 + H2O, ΔH is small and positive.
1. Calculate the equilibrium constant and predict the qualitative effect of each of the three changes on the amount of ester at the new equilibrium.
Read the worked answer
Concentrations: ester = 0.30/0.50 = 0.60 mol/L; water = 0.60 mol/L; acid = 0.20/0.50 = 0.40 mol/L; ethanol = 0.40 mol/L. Kc = (0.60 × 0.60)/(0.40 × 0.40) = 0.36/0.16 = 2.25. (i) Adding ethanol increases a reactant concentration, so Q < Kc and the system shifts right, increasing ester - though the extra ethanol dilutes the effect and the yield gain is limited. (ii) Removing water lowers a product concentration, so Q < Kc and the position shifts right, giving more ester; continuous removal can drive the reaction far toward products. (iii) Adding catalyst does not change Kc or the equilibrium position; it only shortens the time to reach equilibrium, so the final amount of ester is unchanged. Only (i) and (ii) increase the equilibrium amount of ester.
2. The technician also suggests cooling the flask to increase yield. Assess this using the enthalpy information and Le Chatelier's principle.
Read the worked answer
The forward reaction is endothermic (small positive ΔH). Le Chatelier's principle predicts that lowering the temperature shifts the equilibrium in the endothermic direction - that is, toward products - so cooling would slightly increase the equilibrium amount of ester. However, the enthalpy change is small, so the shift is small. More importantly, cooling also lowers the rate of reaction dramatically, so equilibrium is approached far more slowly and may not be reached in a practical timeframe. The catalyst helps the rate but not the position. So cooling is thermodynamically favourable but kinetically costly; a moderate temperature with water removal is the more effective practical strategy. The claim that cooling 'increases yield' is directionally correct but overstated.