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JEE Main & Advanced JEE Main Paper (Held On 12 May 2012)

  • question_answer
    8 mol of\[A{{B}_{3}}(g)\]are introduced into a 1.0 dm3 vessel. If it dissociates as  \[2A{{B}_{3}}(g){{A}_{2}}(g)+3{{B}_{2}}(g).\]At equilibrium, 2 mol of \[{{A}_{2}}\] are found to be present. The equilibrium constant of this reaction is     JEE Main Online Paper (Held On 12 May 2012)

    A) 2                                             

    B)                        3

    C)                        27     

    D)                        36

    Correct Answer: C

    Solution :

     
      \[\underset{8}{\mathop{2A{{B}_{3}}}}\,(g)\] \[{{A}_{2}}\] \[\underset{0}{\mathop{(g)}}\,+3\] \[{{B}_{2}}\underset{0}{\mathop{(g)}}\,\]
    at  \[t=0\] \[8\]   \[0\] \[0\]
    at eq. \[(8-2\times 2)\]   \[2\] \[3\times 2\]
      \[=4\]   \[2\] \[6\]
    now\[{{K}_{C}}=\frac{[{{A}_{2}}]{{[{{B}_{2}}]}^{3}}}{{{[A{{B}_{3}}]}^{2}}}=\frac{2/1\times {{[6/1]}^{3}}}{{{[4/1]}^{2}}}=27\]                                


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