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Haryana PMT Haryana PMT Solved Paper-2011

  • question_answer
    \[2moles\]of \[PC{{l}_{5}}\] were heated in a closed vessel of 2L capacity. At equilibrium, 40% of \[PC{{l}_{5}}\] is dissociated into \[PC{{l}_{3}}\] and\[C{{l}_{2}}\]. The value of equilibrium constant is

    A)  \[0.266\]                            

    B)  \[0.366\]

    C)  \[2.66\]                              

    D)  \[3.66\]

    Correct Answer: A

    Solution :

    \[\underset{0}{\mathop{PC{{l}_{5}}}}\,\underset{0}{\mathop{PC{{l}_{3}}}}\,+\underset{0}{\mathop{C{{l}_{2}}}}\,\underset{initilly}{\mathop{{}}}\,\] \[\frac{2\times 60}{100}\] \[\frac{2\times 40}{100}\]  \[\frac{2\times 40}{100}\] at equilibrium Volume of container \[=2L\] \[{{K}_{c}}=\frac{\left( \frac{2\times 40}{100\times 2} \right)\left( \frac{2\times 40}{100\times 2} \right)}{\frac{2\times 60}{100\times 2}}\] \[=0.266\]


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