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RAJASTHAN ­ PET Rajasthan PET Solved Paper-2008

  • question_answer
    The dissociation of\[X{{Y}_{2}}\]is as \[X{{Y}_{2}}(g)XY(g)+Y(g)\] The initial pressure of\[X{{Y}_{2}}\]is 600 mm of\[Hg\]. On establishing the equilibrium the total pressure becomes 800 mm of Hg. What is the value of K for the reaction, when the volume of system remains unchanged?

    A)  50              

    B)  100

    C)  166.6            

    D)  150

    Correct Answer: B

    Solution :

     The dissociation of\[X{{Y}_{2}}\]is as \[\underset{\begin{smallmatrix}  \,\,\,\,\,600\,mm \\  600-x\,mm \end{smallmatrix}}{\mathop{XY2(g)}}\,\underset{\begin{smallmatrix}  0 \\  x \end{smallmatrix}}{\mathop{XY(g)}}\,+\underset{\begin{smallmatrix}  0 \\  x \end{smallmatrix}}{\mathop{Y(g)}}\,\underset{\begin{smallmatrix}  in\text{ }initial\text{ }at \\  equilibrium \end{smallmatrix}}{\mathop{{}}}\,\] If \[600-x+x+x=800mm\] \[x=200\text{ }mm.\] \[K=\frac{{{p}_{(xy)}}{{p}_{(y)}}}{{{p}_{(x{{y}_{2}})}}}=\frac{200\times 200}{400}\] \[=100\]


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