
Equation to calculate concentration from number of moles and volume
Ethanoic acid and ethanol react according to the following equation:
CH3COOH (I) + C2H5OH (I)?? CH3COOC2H5?(I) + H2O (I)
At equilibrium, 500 cm3?of the reaction mixture contained 0.235 mol of ethanoic acid and 0.035 mol of ethanol together with 0.182 mol of ethyl ethanoate and 0.182 mol of water.
Calculate the value of?Kc for this reaction.
Answer
Step 1:?Calculate the concentrations of the reactants and products

Step 2:?Write out the balanced chemical equation with the concentrations of beneath each substance

Step 3:?Write the equilibrium constant for this reaction in terms of concentration

Step 4:?Substitute the equilibrium concentrations into the expression

Step 5:?Deduce the correct units for?Kc

All units cancel out
Therefore,?Kc?=?4.03
Note that the smallest number of significant figures used in the question is 3, so the final answer should also be given to 3 significant figures
Calculating Kc?of ethyl ethanoate
Ethyl ethanoate is hydrolysed by water:
CH3COOC2H5(I) +?H2O(I)?? CH3COOH(I) + C2H5OH(I)
0.1000 mol of ethyl ethanoate are added to 0.1000 mol of water. A little acid catalyst is added and the mixture made up to 1dm3. At equilibrium 0.0654 mol of water are present. Use this data to calculate a value of Kc?for this reaction.
Answer
Step 1:?Write out the balanced chemical equation with the concentrations of beneath each substance using an initial, change and equilibrium table

Step 2:?Calculate the concentrations of the reactants and products

Step 3:?Write the equilibrium constant for this reaction in terms of concentration

Step 4:?Substitute the equilibrium concentrations into the expression

= 0.28
Step 5:?Deduce the correct units for?Kc

All units cancel out
Therefore,?Kc=?0.28
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