The ester , ethyl acetate is formed by the reaction of ethanol and acetic acid and the equilibrium is represented as :
`CH_(3) COOH(l) +C_(2)H_(5)OH(l)hArr CH_(3)COOC_(2)H_(5)(l) +H_(2)O(l)`
(i) Write the concentration ratio (concentration quotient) Q for this reaction. Note that water is not in excess and is not a solvent in this reaction.
(ii) At 293 K, if one starts with 1.000 mol of acetic acid 0.180 mol of ethanol, there is 0.171 mol of ethyl acetate in the final equilibrium mixture . Calculate the equilibrium constant.
(iii) Starting with 0.50 mol of ethanol and 1.000 mol of acetic acid and maintaining it at 293 K, 0.214 mol of ethyl acetate is found after some time. Has equilibrium been reached?

Explanation

(`a`) `Q_(c)=([CH_(3)COOC_(2)H_(5)][H_(2)O])/([CH_(3)COOH][C_(2)H_(5)OH])`
[`:' ` Water is not a solvent here.]
(`b`) `K_(c)=((0.171)xx(0.171))/((1-0.171)xx(0.180-0.171))=3.92`
(`c`) `Q_(c)=((0.214)xx(0.214))/((0.786)xx(0.286))=0.204 ne K_(c)`
`:.` Equilibrium has not been reached.