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JCECE Engineering JCECE Engineering 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    \[{{k}_{f}}=2.38\times {{10}^{-4}}\] \[\therefore \]  \[{{K}_{c}}=\frac{2.38\times {{10}^{-4}}}{4.76\times {{10}^{-5}}}=5\]


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