AMU Medical AMU Solved Paper-2002

  • question_answer
    2 moles of \[PC{{l}_{5}}\] is heated in one litre vessel. At equilibrium 0.4 mole \[C{{l}_{2}}\] is formed. The value of equilibrium constant will be

    A)  \[1\times {{10}^{-3}}\]                                 

    B)  \[1\times {{10}^{-2}}\]

    C)  \[2\times {{10}^{-1}}\]                                 

    D)  \[1\times {{10}^{-1}}\]

    Correct Answer: D

    Solution :

                     Key Idea For the reaction                 \[{{n}_{1}}A+{{n}_{2}}B{{m}_{1}}C+{{m}_{2}}D\] Equilibrium constant                 \[({{K}_{c}})=\frac{{{[C]}^{{{m}_{1}}}}{{[D]}^{{{m}_{2}}}}}{{{[A]}^{{{n}_{1}}}}{{[B]}^{{{n}_{2}}}}}\] When \[PC{{l}_{5}}\] is heated, then following reaction takes place                                 \[PC{{l}_{5}}PC{{l}_{3}}+C{{l}_{2}}\] Initial     2              0              0 At equi.                1.6          0.4          0.4                 \[{{K}_{c}}=\frac{[PC{{l}_{3}}]\,\,[C{{l}_{2}}]}{[PC{{l}_{5}}]}\]                 \[=\frac{(0.4)\,\,(0.4)}{(1.6)}=\frac{0.16}{1.6}\]                 \[=1\times {{10}^{-1}}\]


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