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JEE Main & Advanced Chemistry Equilibrium / साम्यावस्था Question Bank Law of equilibrium and Equilibrium constant

  • question_answer
    The equilibrium constant (Kc) for the reaction \[\text{HA}+\text{B}\]⇌\[\text{B}{{\text{H}}^{+}}+{{\text{A}}^{-}}\] is 100. If the rate constant for the forward reaction is 105, then rate constant for the backward reaction is              [CBSE PMT 2002]

    A)                 \[{{10}^{7}}\]    

    B)                 \[{{10}^{3}}\]

    C)                 \[{{10}^{-3}}\]  

    D)                 \[{{10}^{-5}}\]

    Correct Answer: B

    Solution :

                    \[{{K}_{c}}=\frac{{{K}_{f}}}{{{K}_{b}}}\,\,\therefore \,{{K}_{b}}=\frac{{{K}_{f}}}{{{K}_{c}}}=\frac{{{10}^{5}}}{100}={{10}^{3}}\]


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