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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2001

  • question_answer
    A quantity of \[\text{PC}{{\text{l}}_{\text{5}}}\]was heated in a \[\text{10 d}{{\text{m}}^{\text{3}}}\]vessel at \[\text{250}{{\,}^{\text{o}}}\text{C}\]\[\text{PC}{{\text{l}}_{5}}(g)\rightleftharpoons PC{{l}_{3}}(g)C{{l}_{2}}(g)\] At equilibrium the vessel contains\[0.1\]mole of \[\text{PC}{{\text{l}}_{\text{5}}}\]and\[0.2\]mole of \[\text{C}{{\text{l}}_{\text{2}}}\text{.}\] The equilibrium constant of the reaction is:

    A)  \[0.05\]                              

    B) \[0.02\]

    C)  \[0.025\]                            

    D)  \[0.04\]

    Correct Answer: D

    Solution :

    The volume of vessel \[=10\text{ }d{{m}^{3}}\] \[PC{{l}_{5}}(g)\rightleftharpoons PC{{l}_{3}}(g)+C{{l}_{2}}(g)\] At equilibrium: \[PC{{l}_{5}}=\frac{0.1}{10}=0.01\] \[C{{l}_{2}}=\frac{0.2}{10}=0.02\] The equilibrium constant, \[{{K}_{c}}=\frac{[PC{{l}_{3}}][C{{l}_{2}}]}{[PC{{l}_{5}}]}\] \[=\frac{0.02\times 0.02}{0.01}=0.04\]


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