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

  • question_answer
    The rate constant for forward and backward reactions of hydrolysis of ester are \[1.1\times {{10}^{-2}}\] and \[1.5\times {{10}^{-3}}\] per minute respectively. Equilibrium constant for the reaction is \[C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}+{{H}_{2}}O\]⇌\[C{{H}_{3}}COOH\]\[+{{C}_{2}}{{H}_{5}}OH\] [AIIMS 1999]

    A)                 4.33       

    B)                 5.33

    C)                 6.33       

    D)                 7.33

    Correct Answer: D

    Solution :

           \[{{K}_{f}}=1.1\times {{10}^{-2}}\];  \[{{K}_{b}}=1.5\times {{10}^{-3}}\]                                 \[{{K}_{c}}=\frac{{{K}_{f}}}{{{K}_{b}}}=\frac{1.1\times {{10}^{-2}}}{1.5\times {{10}^{-3}}}=7.33\].


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