JEE Main & Advanced Chemistry Equilibrium / साम्यावस्था Question Bank Topic Test - Chemical Equilibrium (6-5-21)

  • question_answer
    In a chemical equilibrium, the rate constant of the backward reaction is \[7.5\times {{10}^{-4}}\] and the equilibrium constant is 1.5. So the rate constant of the forward reaction is          

    A) \[5\times {{10}^{-4}}\]

    B) \[2\times {{10}^{-3}}\]

    C) \[1.125\times {{10}^{-3}}\]

    D) \[9.0\times {{10}^{-4}}\]

    Correct Answer: C

    Solution :

    [c] \[{{K}_{c}}=\frac{{{K}_{f}}}{{{K}_{b}}}\]
     \[{{K}_{f}}={{K}_{c}}\times {{K}_{b}}=1.5\times 7.5\times {{10}^{-4}}\]\[=1.125\times {{10}^{-3}}\]

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