• # question_answer 1 mol of$C{{H}_{3}}COOH$and 1 mol of ${{C}_{2}}{{H}_{5}}OH$reacts to produce $\frac{2}{3}$mol of $C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}.$The equilibrium constant is A)  2                     B)  + 2C)  - 4                   D)  + 4

$\underset{\text{At}\,\text{equilibrium}}{\mathop{\underset{\text{Initially}}{\mathop{C{{H}_{3}}COOH}}\,}}\,\,\,+\,\underset{=\frac{1}{3}}{\mathop{\,\underset{1-\frac{2}{3}}{\mathop{\underset{1}{\mathop{{{C}_{2}}{{H}_{5}}OH}}\,}}\,}}\,\to \underset{=\frac{1}{3}}{\mathop{\underset{1-\frac{2}{3}}{\mathop{\underset{1}{\mathop{C{{H}_{3}}}}\,}}\,}}\,\underset{\frac{2}{3}}{\mathop{\underset{0}{\mathop{COO{{C}_{2}}{{H}_{5}}}}\,}}\,\,\,+\,\,\underset{\frac{2}{3}}{\mathop{\underset{0}{\mathop{{{H}_{2}}O}}\,}}\,$ ${{K}_{C}}=\frac{[C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}][{{H}_{2}}O]}{[C{{H}_{3}}COOH][{{C}_{2}}{{H}_{5}}OH]}$ If Volume =V litre, then ${{C}_{2}}{{H}_{5}}OH=\frac{1}{3}\text{V}\,\text{mol}\,\text{per}\,\text{litre}$ $C{{H}_{3}}COO{{C}_{5}}=\frac{2}{3}\text{V}\,\text{mol}\,\text{per}\,\text{litre}$ ${{H}_{2}}O=\frac{2}{3}\text{V}\,\text{mol}\,\text{per}\,\text{litre}$ By putting the values, we obtain ${{K}_{C}}=\frac{2/3V\times 2/3\,V}{1/3\,V\times 1/3\,V}$ $=\frac{6V\times 6V}{3V\times 3V}=\frac{36V}{9V}=4.$ Hence, the correct option is [d].