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RAJASTHAN PMT Rajasthan - PMT Solved Paper-2002

  • question_answer
    For  \[{{N}_{2}}+3{{H}_{2}}2N{{H}_{3}}\] .equilibrium constant is K then equilibrium constant for \[2{{N}_{2}}+6{{H}_{2}}4N{{H}_{3}}\]:

    A)  \[\sqrt{K}\]                       

    B)         \[{{K}^{2}}\]                     

    C)  \[\frac{K}{2}\]                 

    D)         \[\sqrt{\frac{1}{K}}+1\]

    Correct Answer: B

    Solution :

    \[{{N}_{2}}+3{{H}_{2}}2N{{H}_{3}},\] equilibrium constant = K                 \[K=\frac{{{[N{{H}_{3}}]}^{2}}}{[{{N}_{2}}]\,{{[{{H}_{2}}]}^{3}}}\]              ?..(1)                 \[2{{N}_{2}}+6{{H}_{2}}4N{{H}_{3}},\] equilibrium constant = K                 \[K=\frac{{{[N{{H}_{3}}]}^{2}}}{{{[{{N}_{2}}]}^{2}}\,{{[{{H}_{2}}]}^{6}}}\] \[={{K}^{2}}\] [from equation (1)]


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