Example Question - 1

Question: 1.00 mol of ethanoic acid and 1.00 mol of ethanol were mixed with a little acid catalyst and left to reach equilibrium at a constant temperature. At equilibrium the mixture contained 0.667 mol of ethyl ethanoate. Calculate the equilibrium constant K_c for the reaction:CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂OStep 1: find the equilibrium amounts. The molar ratio is 1:1:1:1, so 0.667 mol of ethanoic acid and 0.667 mol of ethanol reacted, and 0.667 mol of water formed. Ethanoic acid remaining = 1.00 - 0.667 = 0.333 mol, and ethanol remaining = 0.333 mol.Step 2: write the expression. K_c = [CH₃COOC₂H₅][H₂O] ÷ ([CH₃COOH][C₂H₅OH]).Step 3: each concentration is moles ÷ V. There are equal numbers of moles on each side, so the V terms cancel and the amounts in moles can be used directly: K_c = (0.667 × 0.667) ÷ (0.333 × 0.333) = 4.0.K_c has no units here, because the concentration units cancel. The most common error is to use the initial amounts instead of the equilibrium amounts.