The ester, ethyl acetate is formed by the reaction between ethanol and acetic acid and equilibrium is represented as:
`CH_(3)COOH_((l))+C_(2)H_(5)OH_((l))hArrCH_(3)COOC_(2)H_(5_((aq)))+H_(2)O_((l))`
(`a`) Write the concentration ratio (reaction quotient), `Q_(e)`, for this reaction. Note that water is not in excess and is not a solvent in this reaction.
(`b`) At `293 K`, if one starts with `1.00 "mole"` of acetic acid and `0.180` of ethanol, there is `0.171 "mole"` of ehtyl acetate in the final equilibrium mixture. Calculate the equilibrium constant.
(`c`) Starting with `0.500 "mole"` of ethanol and `1.000 "mole"` of acetic acid and maintaining it at `293 K`, `0.214 "mole"` of ethyl acetate is found after some time. Has equilibrium been reached?

Answer

(i) `Q_(C)=([CH_(3)COOC_(2)H_(5)][H_(2)O])/([CH_(3)COOH][C_(2)H_(5)OH])` (ii) `K_(c)3.5` (iii) `"AS"Q_(c)=K_(c)`, equilibrium has not been attained.

Explanation

(i) `Q_(c)=([CH_(3)COOC_(2)H_(5)][H_(2)O])/([CH_(3)COOH][C_(2)H_(5)OH])`
(ii) `CH_(3)COOH+C_(2)H_(5)OHhArrCH_(3)COOC_(2)+H_(2)O`
(ii) `{:(Initial,1.00mol,0.180mol,,,),("At eqm".,1-0.17,0.180-0.17,0.17,0.17,):}`
`K_(c)=([CH_(3)COOC_(2)H_(5)][H_(2)O])/([CH_(3)COOH][C_(2)H_(5)OH])=(0.17xx0.17)/(0.83xx0.01)=(1)/(6)`
(iii) `CH_(3)COOH+C_(2)H_(5)OHhArrCH_(3)COOc_(2)H_(5)+H_(2)O`
`{:(Initial,1.000mol,0.500mol,,,),(After,1-0.214,0.500-0.214,0.214mol,0.214mol,):}`
Reaction quotient `(Q_(c))=((0.214//V)(0.214//V))/((0.786//V//)(0.286//V))=0.204`
As `Q_(c)=K_(c)`,equilibrium has not been attained.