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JEE Main & Advanced Chemistry Equilibrium / साम्यावस्था Question Bank Law of equilibrium and Equilibrium constant

  • question_answer
    9.2 grams of \[{{N}_{2}}{{O}_{4(g)}}\] is taken in a closed one litre vessel and heated till the following equilibrium is reached \[{{N}_{2}}{{O}_{4(g)}}\]⇌ \[2N{{O}_{2(g)}}\]. At equilibrium, 50% \[{{N}_{2}}{{O}_{4(g)}}\] is dissociated. What is the equilibrium constant (in mol litre?1) (Molecular weight of \[{{N}_{2}}{{O}_{4}}=92)\]       [MP PET 2003]

    A)                 0.1         

    B)                         0.4

    C)                 0.2         

    D)                 2

    Correct Answer: C

    Solution :

              \[{{K}_{c}}=\frac{{{[N{{O}_{2}}]}^{2}}}{[{{N}_{2}}{{O}_{4}}]}\]\[=\frac{4\times {{(0.05)}^{2}}}{0.05}=4\times 0.05=0.2\]


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