JEE Main & Advanced AIEEE Solved Paper-2004

  • 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\[{{k}_{q}}\]) species in solid state are not written (i.e., their molar concentrations are taken as 1) \[{{P}_{4}}(s)+5{{O}_{2}}(g){{P}_{4}}{{O}_{10}}(s)\] Thus,       \[{{K}_{c}}=\frac{1}{{{[{{O}_{2}}]}^{5}}}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner