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

  • question_answer
    In the reaction \[A+2B\]⇌\[2C\], if 2 moles of \[A,\,\,3.0\] moles of \[B\] and 2.0 moles of \[C\] are placed in a \[2.0\,\,l\] flask and the equilibrium concentration of \[C\] is 0.5 mole/\[l\]. The equilibrium constant \[({{K}_{c}})\] for the reaction is                [KCET 1996]

    A)                 0.073    

    B)                 0.147

    C)                 0.05       

    D)                 0.026

    Correct Answer: C

    Solution :

                               A     +      2B    ⇌    2C            Initial conc.  2                 3              2            at eqm.        2.5              4              1            Molar          \[\frac{2.5}{2}=1.25\]\[\frac{4}{2}=2\]   \[\frac{1}{2}=0.5\]                                 \[K=\frac{{{[0.5]}^{2}}}{[1.25]\times {{[2]}^{2}}}=0.05\]


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