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EAMCET Medical EAMCET Medical Solved Paper-2005

  • question_answer
    At \[\text{600}{{\,}^{\text{o}}}\text{C,}{{\text{K}}_{\text{p}}}\] for the following reaction is 1 atm. \[\text{X(g)}Y(g)+Z(g)\] At equilibrium, 50% of \[\text{X(g)}\] is dissociated. The total pressure of the equilibrium system is P atm. What is the partial pressure (in atm) ofA1[g) at equilibrium?

    A)  1                                            

    B)  4

    C)  2                                            

    D)  0.5

    Correct Answer: A

    Solution :

                     \[X(g)Y(g)+Z(g)\]
    1 0 0 Intial
    \[\begin{align}   & \underline{\begin{align}   & 0.5 \\  & 0.5 \\ \end{align}}\,P \\  & 1.5 \\ \end{align}\] \[\begin{align}   & \underline{\begin{align}   & 0.5 \\  & 0.5 \\ \end{align}}\,P \\  & 1.5 \\ \end{align}\] \[\begin{align}   & \underline{\begin{align}   & 0.5 \\  & 0.5 \\ \end{align}}\,P \\  & 1.5 \\ \end{align}\] At equilibrium Partial pressure
    \[\therefore \]  \[{{K}_{p}}=\frac{{{P}_{Y}}.{{P}_{Z}}}{{{P}_{X}}}\] \[1=\frac{\frac{P}{3}\times \frac{P}{3}}{\frac{P}{3}}\] \[\therefore \]  \[P=3\,atm.\] Partial pressure of \[X=\frac{P}{3}=\frac{3}{3}=1\,atm.\]


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