WB JEE Medical WB JEE Medical Solved Paper-2007

  • question_answer
    In the reaction, \[3A+2B\xrightarrow{{}}2C,\]the equilibrium constant \[{{K}_{c}}\]is given by

    A)  \[\frac{[3A]\times [2B]}{[C]}\]                 

    B)  \[\frac{{{[A]}^{3}}\times [B]}{[C]}\]

    C)  \[\frac{{{[C]}^{2}}}{{{[A]}^{3}}\times {{[B]}^{2}}}\]                         

    D)  \[\frac{[C]}{[3A]\times [2B]}\]

    Correct Answer: C

    Solution :

                     \[3A+2B\xrightarrow{{}}2C\] \[{{\text{K}}_{\text{C}}}\text{=}\frac{\text{concentration}\,\text{of}\,\text{products}}{\text{conentration}\,\text{of}\,\text{reactants}}\] \[=\frac{{{[C]}^{2}}}{{{[A]}^{3}}\times {{[B]}^{2}}}\]


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