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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2011

  • question_answer
    Four moles of\[PC{{l}_{5}}\]are heated in a closed 4 \[d{{m}^{3}}\]container to reach equilibrium at 400 K. At equilibrium 50% of\[PC{{l}_{5}}\] is dissociated. What is the value of\[{{K}_{c}}\]for the dissociation of \[PC{{l}_{5}}\]into\[PC{{l}_{3}}\]and\[C{{l}_{2}}\]at\[400K\]?

    A)  0.50                                      

    B)  1.00

    C)  1.25                      

    D)         0.05

    E)  0.25

    Correct Answer: A

    Solution :

    \[PC{{l}_{5}}PC{{l}_{3}}+C{{l}_{2}}\]
    4 0 0 :initial moles
    4-2=2 2 2 : Moles at equilibrium
    \[\frac{2}{4}\] \[\frac{2}{4}\] \[\frac{2}{4}\] : Molar concentrations
    \[{{K}_{c}}=\frac{[PC{{l}_{3}}][C{{l}_{2}}]}{[PC{{l}_{5}}]}=\frac{\frac{2}{4}\times \frac{2}{4}}{\frac{2}{4}}\] \[\therefore \]  \[{{K}_{c}}=\frac{1}{2}=0.5\]


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