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  3. CEE Kerala Engineering

CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2001

  • question_answer
    When 3 moles of the reactant A and 1 mole of the reactant B are mixed in a vessel of volume 1 L, the   following   reaction   takes   place, \[A(g)+B(g)2C(g)\]. If 1.5 moles of C is formed at equilibrium, the equilibrium constant\[({{K}_{c}})\]of the reaction is:

    A) 0.12                                       

    B) 0.50

    C) 0.25                       

    D)        2.25

    E) 4.00

    Correct Answer: E

    Solution :

    \[A(g)+B(g)2C(g)\]
    a b 0
    3 mol 1 mol 0 Initially
    \[(a-x)\] \[(b-x)\] \[2x\]
    \[\left( 3-\frac{1.5}{2} \right)\] \[\left( 1-\frac{1.5}{2} \right)\] \[=1.5\] At equilibrium
    \[=2.25\] \[=0.25\]
    \[2x=1.5\] \[x=\frac{1.5}{2}=0.75\] \[{{K}_{c}}=\frac{{{[C]}^{2}}}{[A][B]}\]                   ?..(i) Now, substituting the value of [A], [B] and [C] in the Eq. (i) \[{{K}_{c}}=\frac{{{[1.5]}^{2}}}{[2.25][0.25]}=\frac{1.5\times 1.5}{2.25\times 0.25}\] \[=4.00\]


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