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JCECE Medical JCECE Medical Solved Paper-2002

  • question_answer
    The rate constant of forward reaction is \[2.38\times {{10}^{-4}}\]and the rate constant of backward reaction is \[4.76\times {{10}^{-5}}.\] The equilibrium constant for the reaction will be:

    A)  5            

    B) \[5\times {{10}^{-2}}\]

    C) \[2\times {{10}^{-4}}\]      

    D)  none of these

    Correct Answer: A

    Solution :

     Key Idea: Equilibrium constant\[{{K}_{c}}=\frac{{{k}_{f}}}{{{k}_{b}}}\] Given \[{{k}_{f}}=2.38\times {{10}^{-4}}\] \[{{k}_{b}}=4.76\times {{10}^{-5}}\] \[\therefore \] \[{{K}_{c}}=\frac{2.38\times {{10}^{-4}}}{4.76\times {{10}^{-5}}}=5\]


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