Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as:
$CH_3COOH_{(l)} + C_2H_5OH_{(l)} \longleftrightarrow CH_3COOC_2H_{5(l)} + H_2O_{(l)}$
$(i)$ Write the concentration ratio (reaction quotient),$Q_c$,for this reaction (note: water is not in excess and is not a solvent in this reaction).
$(ii)$ At $293 \ K$,if one starts with $1.00 \ mol$ of acetic acid and $0.18 \ mol$ of ethanol,there is $0.171 \ mol$ of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant.
$(iii)$ Starting with $0.5 \ mol$ of ethanol and $1.0 \ mol$ of acetic acid and maintaining it at $293 \ K$,$0.214 \ mol$ of ethyl acetate is found after sometime. Has equilibrium been reached?

$(i)$ The reaction quotient $Q_c$ is given by the ratio of the product of concentrations of products to the product of concentrations of reactants: $Q_c = \frac{[CH_3COOC_2H_5][H_2O]}{[CH_3COOH][C_2H_5OH]}$.
$(ii)$ Initial moles: $CH_3COOH = 1.00$,$C_2H_5OH = 0.18$,$CH_3COOC_2H_5 = 0$,$H_2O = 0$.
At equilibrium,moles of ethyl acetate = $0.171 \ mol$.
Therefore,moles of $CH_3COOH = 1.00 - 0.171 = 0.829$,$C_2H_5OH = 0.18 - 0.171 = 0.009$,$H_2O = 0.171$.
Since the volume $V$ cancels out in the expression for $K_c$,we use moles: $K_c = \frac{0.171 \times 0.171}{0.829 \times 0.009} \approx 3.92$.
$(iii)$ Initial moles: $CH_3COOH = 1.0$,$C_2H_5OH = 0.5$,$CH_3COOC_2H_5 = 0.214$,$H_2O = 0.214$.
Equilibrium moles: $CH_3COOH = 1.0 - 0.214 = 0.786$,$C_2H_5OH = 0.5 - 0.214 = 0.286$.
$Q_c = \frac{0.214 \times 0.214}{0.786 \times 0.286} \approx \frac{0.0458}{0.2248} \approx 0.204$.
Since $Q_c (0.204) \neq K_c (3.92)$,equilibrium has not been reached.