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RAJASTHAN PMT Rajasthan - PMT Solved Paper-1999

  • question_answer
    In a reversible reaction if the concentration of reactants and products are doubled, the value of \[{{K}_{c}}\]will be:

    A)  half of the initial value

    B)                         double of initial value

    C)                         one fourth of the initial value

    D)                         same the initial value

    Correct Answer: B

    Solution :

    As    \[AB\] \[{{K}_{c}}=\frac{[B]}{[A]}\] Let concentration is double                 \[2A2B\]                 \[{{K}_{c}}=\frac{{{[B]}^{2}}}{{{[A]}^{2}}}\] Double of initial value.


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