A) 2
B) + 2
C) - 4
D) + 4
Correct Answer: D
Solution :
\[\underset{\text{At}\,\text{equilibrium}}{\mathop{\underset{\text{Initially}}{\mathop{C{{H}_{3}}COOH}}\,}}\,\,\,+\,\underset{=\frac{1}{3}}{\mathop{\,\underset{1-\frac{2}{3}}{\mathop{\underset{1}{\mathop{{{C}_{2}}{{H}_{5}}OH}}\,}}\,}}\,\to \underset{=\frac{1}{3}}{\mathop{\underset{1-\frac{2}{3}}{\mathop{\underset{1}{\mathop{C{{H}_{3}}}}\,}}\,}}\,\underset{\frac{2}{3}}{\mathop{\underset{0}{\mathop{COO{{C}_{2}}{{H}_{5}}}}\,}}\,\,\,+\,\,\underset{\frac{2}{3}}{\mathop{\underset{0}{\mathop{{{H}_{2}}O}}\,}}\,\] \[{{K}_{C}}=\frac{[C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}][{{H}_{2}}O]}{[C{{H}_{3}}COOH][{{C}_{2}}{{H}_{5}}OH]}\] 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].
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