JEE Main & Advanced Chemistry Equilibrium / साम्यावस्था Question Bank Mock Test - Chemical Equilibrium

  • question_answer
    For the reaction equilibrium;\[{{N}_{2}}{{O}_{4}}(g)\rightleftarrows 2N{{O}_{2}}(g)\]; the concentration of \[{{N}_{2}}{{O}_{4}}\]and \[N{{O}_{2}}\]at equilibrium are \[4.8\times {{10}^{-2}}\]and \[1.2\times {{10}^{-2}}\]mol/L respectively. The value of \[{{K}_{C}}\] for the reaction is:

    A) \[3\times {{10}^{-3}}M\]         

    B) \[3\times {{10}^{3}}M\]

    C) \[3.3\times {{10}^{2}}M\]                    

    D) \[3\times {{10}^{-1}}M\]

    Correct Answer: A

    Solution :

    [a] \[{{K}_{C}}=\frac{{{[N{{O}_{2}}]}^{2}}}{[{{N}_{2}}{{O}_{4}}]}=\frac{{{(1.2\times {{10}^{-2}})}^{2}}}{4.8\times {{10}^{-2}}}=3.0\times {{10}^{-3}}M\]  


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