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CET Karnataka Medical CET - Karnataka Medical Solved Paper-2001

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