Question: 1.00 mol of ethanoic acid and 1.00 mol of ethanol were mixed and left to reach equilibrium. At equilibrium, 0.67 mol of ethyl ethanoate was present. Calculate Kc.
CH3COOH + C2H5OH ⇌ CH3COOC2H5 + H2OStep 1: use the equation to find the equilibrium moles. The ratio is 1:1:1:1, so 0.67 mol of water also formed and 0.67 mol of each reactant was used up. This leaves 1.00 - 0.67 = 0.33 mol of ethanoic acid and 0.33 mol of ethanol.Step 2: write the Kc expression: Kc = [CH3COOC2H5][H2O] / [CH3COOH][C2H5OH]Step 3: each concentration is moles divided by volume V. There are two concentration terms on the top and two on the bottom, so V cancels and the moles can be used directly.Kc = (0.67 × 0.67) / (0.33 × 0.33) = 4.1. Kc has no units here because the units on the top and bottom cancel.
CH3COOH + C2H5OH ⇌ CH3COOC2H5 + H2OStep 1: use the equation to find the equilibrium moles. The ratio is 1:1:1:1, so 0.67 mol of water also formed and 0.67 mol of each reactant was used up. This leaves 1.00 - 0.67 = 0.33 mol of ethanoic acid and 0.33 mol of ethanol.Step 2: write the Kc expression: Kc = [CH3COOC2H5][H2O] / [CH3COOH][C2H5OH]Step 3: each concentration is moles divided by volume V. There are two concentration terms on the top and two on the bottom, so V cancels and the moles can be used directly.Kc = (0.67 × 0.67) / (0.33 × 0.33) = 4.1. Kc has no units here because the units on the top and bottom cancel.