2. Solved Papers
  3. JEE Main & Advanced

JEE Main & Advanced JEE Main Paper (Held On 10 April 2016)

  • question_answer
    A solid XY kept in an evacuated sealed container undergoes decomposition to form a mixture of gases X and Y at temperature T. The equilibrium pressure is 10 bar in this vessel.                                                                                                 \[{{K}_{P}}\] for this reaction is :   JEE Main Online Paper (Held On 10 April 2016)

    A) 25                                          

    B) 5

    C) 0                                             

    D) 100

    Correct Answer: A

    Solution :

                    \[XY(s)X\underset{P}{\mathop{(g)}}\,+\underset{P}{\mathop{Y(g)}}\,\] At eq. Total pressure = 2P = 10 bar\[\Rightarrow \]\[P=5\] Now, \[{{K}_{P}}=({{P}_{X}})({{P}_{Y}})={{P}^{2}}=25\]


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