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

  • question_answer
    The equilibrium constant for the reaction; \[{{N}_{2(g)}}+{{O}_{2(g)}}\rightleftarrows 2N{{O}_{(g)}}\] at temperature T is \[4\times {{10}^{-4.}}\].The value of \[{{K}_{c}}\]for the reaction. \[N{{O}_{(g)}}\rightleftarrows 1/2{{N}_{2(g)}}+1/2{{O}_{2(g)}}\]at the temperature is

    A) 0.02     

    B) 50

    C)  \[4\times {{10}^{-4}}\]                       

    D) \[2.5\times {{10}^{-2}}\]

    Correct Answer: B

    Solution :

    [b] \[{{N}_{2}}+{{O}_{2}}\rightleftharpoons 2NO\] \[{{K}_{C1}}=\frac{{{[NO]}^{2}}}{[{{N}_{2}}][{{O}_{2}}]}=4\times {{10}^{-4}}\] \[NO\rightleftharpoons 1/2{{N}_{2}}+1/2{{O}_{2}}\] \[{{K}_{{{C}_{2}}}}=\frac{{{[{{N}_{2}}]}^{1/2}}{{[{{O}_{2}}]}^{1/2}}}{[NO]}\] \[{{K}_{{{C}_{2}}}}=\sqrt{\frac{1}{{{K}_{{{c}_{1}}}}}}=\sqrt{\frac{1}{4\times {{10}^{-4}}}}=50\]


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