'a' moles of ethanoic acid and 'b' moles of ethanol were reacted together at 50°C until equilibrium was established. At equilibrium, x mol of the ester were found to be present. Write an expression for the equilibrium constant, Kc, for the reaction in terms of a, b and x.

Step 1: Write the balanced chemical equation and set up an ICE (Initial, Change, Equilibrium) table for moles. The reaction is between ethanoic acid and ethanol to form ethyl ethanoate (ester) and water. CH_3COOH(l) + CH_3CH_2OH(l) CH_3COOCH_2CH_3(l) + H_2O(l) Initial moles: Ethanoic acid = a mol, Ethanol = b mol, Ester = 0 mol, Water = 0 mol. At equilibrium, x mol of the ester are present. According to the stoichiometry (1:1:1:1), if x moles of ester are formed, then x moles of water are also formed, and x moles of ethanoic acid and x moles of ethanol must have reacted. | Species | Initial (mol) | Change (mol) | Equilibrium (mol) | | :-------------- | :------------ | :----------- | :---------------- | | CH_3COOH | a | -x | a - x | | CH_3CH_2OH | b | -x | b - x | | CH_3COOCH_2CH_3 | 0 | +x | x | | H_2O | 0 | +x | x | Step 2: Write the equilibrium constant expression in terms of concentrations. Let the volume of the reaction vessel be V dm^3. The equilibrium concentration of each species is its equilibrium moles divided by the volume V. [CH_3COOH] = (a-x)/(V) [CH_3CH_2OH] = (b-x)/(V) [CH_3COOCH_2CH_3] = (x)/(V) [H_2O] = (x)/(V) The equilibrium constant K_c is given by: K_c = [CH_3COOCH_2CH_3][H_2O][CH_3COOH][CH_3CH_2OH] Step 3: Substitute the equilibrium concentrations into the K_c expression. K_c = ((x)/(V))((x)/(V))((a-x)/(V))((b-x)/(V)) K_c = (x^2)/(V^2)((a-x)(b-x))/(V^2) The V^2 terms in the numerator and denominator cancel out. K_c = (x^2)/((a-x)(b-x)) The expression for the equilibrium constant K_c is K_c = (x^2)/((a-x)(b-x)). What's next?