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

  • question_answer
    For the equilibrium \[{{N}_{2}}+3{{H}_{2}}\]⇌\[2N{{H}_{3}},{{K}_{c}}\] at 1000K is \[2.37\times {{10}^{-3}}\]. If at equilibrium \[[{{N}_{2}}]=2M,\,[{{H}_{2}}]=3M\], the concentration of \[N{{H}_{3}}\] is                 [JIPMER 2000]

    A)                 0.00358 M          

    B)                 0.0358 M

    C)                 0.358 M               

    D)                 3.58 M

    Correct Answer: C

    Solution :

             \[{{K}_{c}}=\frac{{{[N{{H}_{3}}]}^{2}}}{[{{N}_{2}}]\,\,{{[{{H}_{2}}]}^{3}}}\]            \[2.37\times {{10}^{-3}}=\frac{{{x}^{2}}}{[2]\,\,{{[3]}^{3}}}={{x}^{2}}=0.12798\]                                 x = 0.358 M.


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