BCECE Medical BCECE Medical Solved Papers-2009

  • question_answer
    For the reaction, \[2B+AC\], the equilibrium constant is

    A)  \[\frac{[A]\,{{[B]}^{3}}}{[C]}\]

    B)  \[\frac{[C]}{[A]\,[2B]}\]

    C)  \[\frac{[C]}{[A]\,{{[B]}^{2}}}\]

    D)  \[\frac{[A]\,[B]}{[C]}\]

    Correct Answer: C

    Solution :

    For the reaction, \[2B+AC\] Rate of forward reaction \[={{k}_{f}}[A]\,{{[B]}^{2}}\] Rate of backward reaction \[={{k}_{b}}[C]\] At equilibrium, Rate of forward reaction = Rate of backward reaction                 \[{{k}_{f}}[A]\,{{[B]}^{2}}={{k}_{b}}[C]\] \[\frac{{{k}_{f}}}{{{k}_{b}}}=\frac{[C]}{[A]\,{{[B]}^{2}}}={{K}_{c}}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner