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

  • question_answer
    The equilibrium constant for the reversible reaction, \[{{N}_{2}}+3{{H}_{2}}\rightleftarrows 2N{{H}_{3}}\]is K and for the reaction \[\frac{1}{2}{{N}_{2}}+\frac{3}{2}{{H}_{2}}\rightleftarrows N{{H}_{3}}\] the equilibrium constant is K'. K and K' will be related as

    A) \[K=K'\]

    B) \[K'=\sqrt{K}\]

    C) \[K=\sqrt{K'}\]              

    D) \[K\times K'=1\]

    Correct Answer: B

    Solution :

    [b]  \[{K}'=\text{ }{{K}^{n}};\,\,Hence\,\,n=\frac{1}{2}\] \[\therefore {K}'={{K}^{1/2}}=\sqrt{K}\]


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