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RAJASTHAN PMT Rajasthan - PMT Solved Paper-1999

  • question_answer
    \[PC{{l}_{5}}PC{{l}_{3}}+C{{l}_{2}}\]in this reversible reaction the moles of \[PC{{l}_{5}},\] \[PC{{l}_{3}},\] and \[C{{l}_{2}}\] are a, b and c respectively and total pressure is P then value of \[{{K}_{p}}\] is:

    A)  \[\frac{bc}{a}.RT\]       

    B)                         \[\frac{b}{(a+b+c)}.P\]

    C)                         \[\frac{bc.P}{a(a+b+c)}\]

    D)                         \[\frac{c}{(a+b+c)}.P\]

    Correct Answer: C

    Solution :

    \[{{K}_{P}}=\frac{P'PC{{l}_{3}}\times P'C{{l}_{2}}}{P'PC{{l}_{5}}}\] \[=\frac{\frac{b}{(a+b+c)}.P\times \frac{c}{(a+b+c)}\times P}{\frac{a}{(a+b+c)}P}\] \[{{K}_{p}}=\frac{bc.\,P}{a(a+b+c)}\] \[\underset{Acetylene}{\mathop{CH\equiv CH}}\,\xrightarrow[Ni]{{{H}_{2}}}C{{H}_{2}}=C{{H}_{2}}\]


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