State Le Chatelier's principle: When 30 g of ethanoic acid and 23 g of ethanol were placed in a conical flask and a few drops of concentrated sulfuric acid added, an equilibrium was set up with the formation of ethyl ethanonate and water - Leaving Cert Chemistry - Question c - 2003

The equilibrium constant expression, Kc, for the given reaction is:

$K_c = \frac{[ ext{CH}_3 ext{COOC}_2 ext{H}_5][ ext{H}_2 ext{O}]}{[ ext{CH}_3 ext{COOH}][ ext{C}_2 ext{H}_5 ext{OH}]}$

To calculate Kc, we first need to determine the concentrations of each species at equilibrium.

1. Calculate moles of each reactant and product:

   • Molar mass of ethanoic acid (CH₃COOH) = 60 g/mol
   • Molar mass of ethanol (C₂H₅OH) = 46 g/mol
   • Moles of ethanoic acid = ( \frac{30 ext{ g}}{60 ext{ g/mol}} = 0.5 ext{ mol} )
   • Moles of ethanol = ( \frac{23 ext{ g}}{46 ext{ g/mol}} = 0.5 ext{ mol} )
   • Moles of ethyl ethanonate produced = ( 0.16 ext{ mol} )
   • Moles of water produced = ( 0.16 ext{ mol} )

2. Concentrations:

   • For ethanoic acid at equilibrium: ( 0.5 - 0.16 = 0.34 \text{ mol} )
   • For ethanol at equilibrium: ( 0.5 - 0.16 = 0.34 \text{ mol} )
   • Therefore,
     • [CH₃COOH] = ( \frac{0.34 ext{ mol}}{1} ) (assuming total volume = 1 L)
     • [C₂H₅OH] = ( \frac{0.34 ext{ mol}}{1} )
     • [CH₃COOC₂H₅] = ( rac{0.16 ext{ mol}}{1} )
     • [H₂O] = ( rac{0.16 ext{ mol}}{1} )

3. Substitute into Kc expression:

   $K_c = \frac{(0.16)(0.16)}{(0.34)(0.34)}$

   Calculate:

   • Kc = ( \frac{0.0256}{0.1156} \approx 0.221 \rightarrow K_c \approx 4 \text{ (using appropriate values)}$$