2. Solved Papers
  3. BHU PMT

BHU PMT BHU PMT (Screening) Solved Paper-2010

  • question_answer
    Two moles of HI were heated in a sealed tube at\[440{}^\circ C\]till the equilibrium was reached.\[HI\] was found to be 22% decomposed. The equilibrium constant for dissociation is

    A)  0.282                                   

    B)  0.0796

    C)  0.0199                                 

    D)  1.99

    Correct Answer: C

    Solution :

                     
    \[2HI{{H}_{2}}+{{I}_{2}}\]
    Initially 2 0 0
    at equilibrium \[2-\frac{2\times 22}{100}\] \[\frac{2\times 22}{100}\] \[\frac{2\times 22}{100}\]
    = 2 - 0.44 = 0.22 = 0.22
    = 1.56


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