question_answer 1)

The ester, ethyl acetate, is formed by the reaction of ethanol and acetic acid and the equilibrium is represented as: \[C{{H}_{3}}COOH(l)+{{C}_{2}}{{H}_{5}}OH(l)\rightleftharpoons C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}(l)+{{H}_{2}}O(l)\](i) Write the concentration ratio, Qfor this reaction. Note that water is not in excess and is not a solvent in this reaction. (ii) At 293 K, if one starts with 1.000 mol of acetic acid and 0.180 mol of ethanol, there is 0.171 mol of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant. (iii) Starting with 0.50 mol of ethanol and 1.000 mol of acetic acid and maintaining it at 293 K, 0.214 mol of ethyl acetate is found after some time. Has equilibrium been reached?

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.