The ester, ethyl acetate is formed by the reaction between ethanol and acetic
acid and equilibrium is represented as :
\(\mathrm{CH}_{3} \mathrm{COOH}_{(\mathrm{l})}+\mathrm{C}_{2} \mathrm{H}_{5}
\mathrm{OH}_{(\mathrm{l})} \rightleftharpoons \mathrm{CH}_{3}
\mathrm{COOC}_{2} \mathrm{H}_{5(\mathrm{aq})}+\mathrm{H}_{2}
\mathrm{O}_{(\mathrm{l})}\)
(a) Write the concentration ratio (reaction quotient), \(Q_{\mathrm{e}}\), for
this reaction. Note that water is not in excess and is not a solvent in this
reaction.
(b) At \(293 \mathrm{~K}\), if one starts with \(1.00\) mole of acetic acid and
\(0.180\) of ethanol, there is \(0.171\) mole of ethyl 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 \(293 \mathrm{~K}, 0.214\) mole of ethyl acetate is found
after some time. Has equilibrium been reached?