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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 29

  • question_answer
    For the reaction, \[X2Y\] and \[ZP+Q\] occurring at two different pressure \[{{P}_{1}}\] and \[{{P}_{2}},\] respectively. The ratio of the two pressure is\[1:3\]. What will be the ratio of equilibrium constant, if the degree of dissociation of X and Z are equal?

    A) \[1:36\]                

    B)        \[1:12\]                 

    C) \[1:9\]                   

    D)        \[4:3\]

    Correct Answer: D

    Solution :

    [d] \[\frac{K{{p}_{1}}}{K{{p}_{2}}}=\frac{x}{y}=\frac{{{(2)}^{2}}{{P}_{1}}}{{{P}_{2}}}=\frac{4\times 1}{3}=4:3\]


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