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

(i)`Q_c=([CH_3COOC_2H_5][H_2O])/([CH_3COOH][C_2H_5OH])`
(ii)`{:(,CH_3COOH+, C_2H_5OH hArr , CH_3COOC_2H_5+,H_2O),("Initial conc.",1.00,0.18,0,0),("At equi.",1-0.171, 0.18-0.171,0.171,0.171),(,=0.829,0.009,0.171,0.171),(,"mol","mol","mol","mol"),("Molar conc.","0.829/V","0.009/V","0.717V/V","0.171/V"):}`
`K_c=((0.171//V)xx(0.171//V))/((0.829//V)xx(0.009//V))`=3.92
(iii)`{:(,CH_3COOH +,C_2H_5OH hArr, CH_3COOC_2H_5+,H_2O),("Initial conc.","1.00 mol","0.500 mol",,),("At equi",1-0.214,0.500-0.214,,),(,=0.786,=0.286,0.214,0.214),(,"mol","mol","mol","mol") :}`
`Q_c=((0.214//V)xx(0.214//V))/((0.786//V)xx(0.286//V))`=0.204
Since , `Q_c ne K_c`, equilibrium has not been reached.