NEET Sample Paper NEET Sample Test Paper-15

  • 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)  + 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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