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?

Explanation

a. `Q_(C )=([CH_(3)COOC_(2)H_(5)][H_(2)O])/([CH_(3)COOH][C_(2)H_(5)OH])`
b. `CH_(3)COOH+C_(2)H_(5)OH hArr CH_(3)COOC_(2)H_(5)+H_(2)O`
`{:(1.00 mol,0.180 mol,,,"Initial"),(1-0.171,0.180-0.171,,0.171 mol,0.171 mol),(=0.829,=0.009,,,"At equilibrium"),(mol,mol,,,),(0.829//V,0.009//V,,0.171//V,0.171//V),(,,,,"Molar concentration"):}`
`K_(c )=([CH_(3)COOC_(2)H_(5)][H_(2)O])/([CH_(3)COOH][C_(2)H_(5)OH])`
`=((0.171//V)(0.171//V))/((0.829//V)(0.009//V))=3.92`
c. `CH_(3)COOH+C_(2)H_(5)OH hArr CH_(3)COOC_(2)H_(5)+H_(2)O`
`{:(1.000 mol,0.500 mol,,,Initial),(1-0.214,0.500-0.214,,0.214,0.214),(=0.786,=0.286,,mol,mol),(mol,mol,,,"After time t"):}`
Reaction quotient `(Q_(c ))=((0.214//V)(0.214//V))/((0.786//V)(0.286//V))`
`=0.204`
As `Q_(c )ne K_(c )`, equilibrium has not been attained.