Esterification: equilibrium yield, catalysis and the limits of pushing a reaction
Ethyl ethanoate is prepared by refluxing 0.500 mol of ethanoic acid with 0.500 mol of ethanol and a few drops of concentrated sulfuric acid. After one hour the mixture is distilled and 32.0 g of ester (0.363 mol) is collected. A second student repeats the experiment exactly but omits the sulfuric acid; after three hours only about 5 g of ester has formed. At 298 K the equilibrium constant for
CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O
is Kc = 4.0 (concentrations in mol dm⁻³). The enthalpy change for the forward reaction is approximately −4 kJ mol⁻¹. A third student writes: 'Refluxing for longer, and raising the temperature to 120 °C, will push the yield up towards 100 %.' M(ethyl ethanoate) = 88.1 g mol⁻¹.
1. Calculate the equilibrium amount of ester obtainable at 298 K from 0.500 mol of each reactant, and compare it with the 0.363 mol actually collected.
Read the worked answer
Let x = mol of ester at equilibrium. Then water = x, and acid = alcohol = 0.500 − x. All four species are in the same volume, so the volumes cancel:
Kc = x²/(0.500 − x)² = 4.0
Taking the square root: x/(0.500 − x) = 2.0
x = 1.000 − 2x → 3x = 1.000 → x = 0.333 mol
Maximum equilibrium yield in a closed flask = 0.333/0.500 = 66.7 %
The first student collected 32.0/88.1 = 0.363 mol, i.e. 72.6 % - more than the closed-vessel equilibrium allows. That is not an error: the ester was removed by distillation as it formed. Removing a product makes the reaction quotient Q smaller than K, so the system is no longer at equilibrium and must respond by making more ester. The comparison therefore demonstrates Le Chatelier's principle in action, and it also means the simple Kc calculation gives only a lower bound for this apparatus.
2. Explain why the second student obtained only about 5 g in three hours, and why adding the acid catalyst could not have increased the equilibrium yield.
Read the worked answer
Why so little: the uncatalysed reaction has a much higher activation energy, so at reflux temperature only a small fraction of collisions is successful and the system is still far from equilibrium after three hours. Check this: 5 g = 0.057 mol, so Q = (0.057)²/((0.443)(0.443)) = 0.0166, which is far below Kc = 4.0. The mixture is simply crawling towards the same equilibrium position - given enough time it would reach the same 0.333 mol as the catalysed run.
Why the catalyst cannot change the yield: the acid protonates the carbonyl oxygen, making the carbonyl carbon more electrophilic and opening a lower-energy route (protonation, nucleophilic attack by ethanol, proton transfer, loss of water, deprotonation). This lowers the activation energy of both the forward and the reverse reaction by the same amount in terms of the pathway; it does not alter ΔG° or K. A catalyst changes the rate at which equilibrium is reached, not where it lies.
The student's 5 g is therefore a kinetic result, not a thermodynamic one - an important distinction when reporting a 'yield'.
3. Assess the third student's claim, and calculate how much ethanol would be needed to convert 90 % of the ethanoic acid at 298 K.
Read the worked answer
Longer reflux: cannot exceed the equilibrium limit in a closed system. Once Q = K, no further ester forms, however long the mixture is heated.
Raising the temperature to 120 °C: the reaction is weakly exothermic (ΔH ≈ −4 kJ mol⁻¹), so K falls slightly on heating. Using the van't Hoff relation:
ln(K₂/K₁) = −(ΔH/R)(1/T₂ − 1/T₁) = (4000/8.314)(1/393 − 1/298) ≈ −0.39
K₂ ≈ 4.0 × e⁻⁰·³⁹ ≈ 2.7, so the equilibrium yield would decrease, not increase. Heating mainly speeds up attainment of equilibrium. Also, 120 °C is above the boiling point of the mixture, which would need a sealed vessel.
The real levers are removing a product (distilling off the ester, or removing water with a drying agent or molecular sieve) or using an excess of one reactant.
Excess ethanol for 90 % conversion: x = 0.450 mol; acid left = 0.050 mol; let b = initial ethanol.
4.0 = (0.450)²/[0.050(b − 0.450)]
0.2025 = 0.200(b − 0.450) → b − 0.450 = 1.0125 → b ≈ 1.46 mol
So roughly 1.46 mol of ethanol (about three times the stoichiometric amount, i.e. ~0.96 mol excess) is needed. The claim of approaching 100 % is false; ~90 % is achievable, but only by manipulating the equilibrium, not by heating longer or harder.