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MGIMS WARDHA MGIMS WARDHA Solved Paper-2003

  • question_answer
    \[PC{{l}_{5}}\]was heated in a\[10\text{ }d{{m}^{3}}\]container at \[250{}^\circ C\]\[PC{{l}_{5}}(g)PC{{l}_{3}}(g)+C{{l}_{2}}(g)\]0.1 mol\[PC{{l}_{5}}\] and 0.2 mol\[C{{l}_{2}}\]was found at equilibrium. The equilibrium constant for the reaction is:

    A)  0.025                                   

    B)  0.04

    C)  0.02                                      

    D)  0.05

    Correct Answer: B

    Solution :

                                               \[\underset{a}{\mathop{PC{{l}_{5}}}}\,\underset{0}{\mathop{PC{{l}_{3}}}}\,+\underset{0}{\mathop{C{{l}_{2}}}}\,\] In equil, \[(a-x)=0.1\text{ }mol.\,\,\,\,\,\,\text{ }0.2\text{ }mol.\text{ }\,\,\,\,\,\,\,\,0.2\text{ }mol.\] Hence,  \[{{K}_{c}}=\frac{[PC{{l}_{3}}][C{{l}_{2}}]}{[PC{{l}_{5}}]}=\frac{\left( \frac{0.2}{10} \right)\times \left( \frac{0.2}{10} \right)}{\left( \frac{0.1}{10} \right)}\]\[=0.04\]


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