Answer:
(i)\[{{Q}_{c}}=\frac{[C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}][{{H}_{2}}O]}{[C{{H}_{3}}COOH][{{C}_{2}}{{H}_{5}}OH]}\]
(ii)\[\underset{\begin{smallmatrix}
t=0 \\
{{t}_{eq}}
\end{smallmatrix}}{\overset{{}}{\mathop{{}}}}\,\underset{\begin{align}
& 1 \\
& (1-x)
\\
&
(1-0.171) \\
& 0.829
\\
\end{align}}{\mathop{C{{H}_{3}}COOH(l)}}\,\,+\underset{\begin{align}
& 0.18
\\
&
(0.18-x) \\
&
(0.18-0.171) \\
& 0.009
\\
\end{align}}{\mathop{{{C}_{2}}{{H}_{5}}OH(l)}}\,\rightleftharpoons
\underset{\begin{align}
& 0 \\
& x \\
& 0.171
\\
& 0.171
\\
\end{align}}{\mathop{C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}(l)}}\,+\underset{\begin{align}
& 0 \\
& x \\
& 0.171
\\
& 0.171
\\
\end{align}}{\mathop{{{H}_{2}}O(l)}}\,\] [
\[{{K}_{c}}=\frac{[C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}][{{H}_{2}}O]}{[C{{H}_{3}}COOH][{{C}_{2}}{{H}_{5}}OH]}=\frac{0.17\times
0.171}{0.829\times 0.009}=3.92\](iii)\[\underset{\begin{smallmatrix}
t=0 \\
{{t}_{aq}}
\end{smallmatrix}}{\overset{{}}{\mathop{{}}}}\,\underset{\begin{align}
& 1 \\
& (1-x)
\\
&
(1-0.214) \\
& 0.786
\\
\end{align}}{\mathop{C{{H}_{3}}COOH(l)}}\,\,+\underset{\begin{align}
& 0.5
\\
&
(0.5-x) \\
& (0.5-0.214)
\\
& 0.286
\\
\end{align}}{\mathop{{{C}_{2}}{{H}_{5}}OH(l)}}\,\rightleftharpoons
\underset{\begin{align}
& 0 \\
& x \\
& 0.214
\\
& 0.214
\\
\end{align}}{\mathop{C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}(l)}}\,+\underset{\begin{align}
& 0 \\
& x \\
& 0.214
\\
& 0.214
\\
& \\
\end{align}}{\mathop{{{H}_{2}}O(l)}}\,\]
\[{{Q}_{c}}=\frac{[C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}][{{H}_{2}}O]}{[C{{H}_{3}}COOH][{{C}_{2}}{{H}_{5}}OH]}=\frac{0.214\times
0.214}{0.786\times 0.286}\]\[=0.204\]
\[{{Q}_{c}}\ne
{{K}_{c}}\]
\[\therefore \]Reaction
is not at equilibrium.
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