2. Solved Papers
  3. BHU PMT

BHU PMT BHU PMT (Screening) Solved Paper-2005

  • question_answer
    \[A+BC+D\]. If initially the concentration of A and B are both equal but at equilibrium, concentration of D will be twice of that of A, then what will be the equilibrium constant of reaction?

    A)  \[\frac{4}{9}\]                                  

    B)  \[\frac{9}{4}\]

    C)   \[\frac{1}{9}\]                                 

    D)  4

    Correct Answer: D

    Solution :

                     Key Idea: We will use the expression for equilibrium constant\[({{K}_{c}}),\]substitute the values and solve the question.
    \[\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,A\,\,\,+\,\,\,B\,\,\,\,\,\,\,\,\,\,\,\,C\,\,\,\,+\,\,\,\,D\]
    Initial conc. \[x\] \[x\] 0 0
    At. Equ. \[x-y\] \[x-y\] \[y\] \[y\]
    \[{{K}_{c}}=\frac{[C][D]}{[A][B]}\] Since, concentration ofD is twice of A \[\therefore \] \[y=2(x-y)\]                 \[{{K}_{c}}=\frac{[y][y]}{[x-y][x-y]}\]                 \[=\frac{[2(x-y)[2(x-y)]}{[x-y][x-y]}\]                 \[{{K}_{c}}=4\]


You need to login to perform this action.
You will be redirected in 3 sec spinner