JCECE Engineering JCECE Engineering Solved Paper-2008

  • question_answer
    What is the equilibrium expression for the reaction,\[{{P}_{4}}(s)+5{{O}_{2}}(g){{P}_{4}}{{O}_{10}}(s)\]?

    A) \[{{K}_{c}}=\frac{[{{P}_{4}}{{O}_{10}}]}{[{{P}_{4}}]{{[{{O}_{2}}]}^{5}}}\]

    B) \[{{K}_{c}}=\frac{[{{P}_{4}}{{O}_{10}}]}{5[{{P}_{4}}][{{O}_{2}}]}\]

    C) \[{{K}_{c}}={{[{{O}_{2}}]}^{5}}\]                               

    D)  \[{{K}_{c}}=\frac{1}{{{[{{O}_{2}}]}^{5}}}\]

    Correct Answer: D

    Solution :

    In the expression for equilibrium constant \[({{K}_{p}}\] or \[{{K}_{c}})\] species in solid state are not written \[(ie,\] their molar concentrations are taken as \[1)\]                 \[{{P}_{4}}(s)+5{{O}_{2}}(g){{P}_{4}}{{O}_{10}}(s)\] Thus,                     \[{{K}_{c}}=\frac{1}{{{[{{O}_{2}}]}^{5}}}\]


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