Fischer esterification: equilibrium, yield and scale-up
In a pre-university synthesis project, a student refluxes 23.0 g of ethanol (0.500 mol) with 30.0 g of glacial acetic acid (0.500 mol) and 1.0 mL of concentrated sulfuric acid. After 1 hour the mixture is cooled, diluted, and titrated: 0.200 mol of acetic acid remains. The student isolates 0.280 mol of ethyl acetate by distillation and reports a yield of 56% based on acetic acid. The teacher points out that the yield should be judged against the equilibrium, not against a hypothetical 100% reaction. The student wants to use the same apparatus to make a larger batch and asks whether adding more ethanol or removing water is better, and why the mixture sometimes turns brown at higher temperature.
1. Calculate the equilibrium constant and the equilibrium conversion. How should the isolated 0.280 mol be judged?
Read the worked answer
Acid reacted = 0.500 - 0.200 = 0.300 mol. Because ethanol and acid react 1:1, the equilibrium amounts are: acetic acid 0.200 mol, ethanol 0.200 mol, ester 0.300 mol, water 0.300 mol (assuming negligible initial water).
K = [ester][water]/[acid][ethanol] = (0.300 × 0.300)/(0.200 × 0.200) = 2.25.
The equilibrium conversion is 0.300/0.500 = 60%. The isolated 0.280 mol is 93% of the equilibrium amount, so the 56% figure understates the chemical performance; the loss is mainly mechanical and in work-up, not a failure to reach equilibrium.
2. Compare using 1.00 mol ethanol with 0.50 mol acetic acid against removing water. Calculate the new equilibrium conversion for the excess-alcohol case.
Read the worked answer
With 1.00 mol ethanol and 0.50 mol acid, let x react: K = x²/[(0.50 - x)(1.00 - x)] = 2.25.
This gives 1.25x² - 3.375x + 1.125 = 0, so x = 0.390 mol. Conversion = 0.390/0.50 = 78%, higher than 60%.
Removing water has a similar Le Chatelier effect and can be combined with excess ethanol; a Dean–Stark trap or distilling the ester as it forms shifts the equilibrium further. Excess ethanol is simpler but must be separated and recovered; water removal is more atom-efficient and avoids a large recycle.
3. Explain the brown colour at higher temperature and state practical precautions for the larger batch.
Read the worked answer
At higher temperature, concentrated H2SO4 can dehydrate ethanol to ethene and diethyl ether, oxidize some organic material, and char impurities; the brown colour is an early sign of these side reactions.
Keep the reflux moderate (about 70–80 °C), use a heating mantle rather than an open flame, add boiling chips, and do not let the acid become concentrated by losing ethanol or water. For a larger batch, scale by volume with adequate stirring and keep the same acid concentration; monitor the colour and stop heating if it darkens.