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

  • question_answer
    For the reaction equilibrium \[{{N}_{2}}{{O}_{4}}\]⇌\[2N{{O}_{2(g)}}\], the concentrations of \[{{N}_{2}}{{O}_{4}}\] and \[N{{O}_{2}}\] at equilibrium are \[4.8\times {{10}^{-2}}\]and \[1.2\times {{10}^{-2}}\,mol\,litr{{e}^{-1}}\] respectively. The value of \[{{K}_{c}}\] for the reaction is [AIEEE 2003]

    A)                 \[3.3\times {{10}^{2}}\]\[mol\,\,litr{{e}^{-1}}\] 

    B)                         \[3\times {{10}^{-1}}\]\[mol\,\,litr{{e}^{-1}}\]

    C)                 \[3\times {{10}^{-3}}\]\[mol\,\,litr{{e}^{-1}}\]   

    D)                 \[3\times {{10}^{3}}\]\[mol\,\,litr{{e}^{-1}}\]

    Correct Answer: C

    Solution :

              \[K=\frac{{{[N{{O}_{2}}]}^{2}}}{[{{N}_{2}}{{O}_{4}}]}\]\[=\frac{{{[1.2\times {{10}^{-2}}]}^{2}}}{[4.8\times {{10}^{-2}}]}\]\[=0.3\times {{10}^{-2}}\]\[=3\times {{10}^{-3}}\]


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