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

  • question_answer
    In the reaction \[PC{{l}_{5(g)}}\]⇌ \[PC{{l}_{3(g)}}\]\[+C{{l}_{2(g)}}.\] The equilibrium concentrations of \[PC{{l}_{5}}\] and \[PC{{l}_{3}}\] are 0.4 and 0.2 mole/litre respectively. If the value of \[{{K}_{c}}\]is 0.5 what is the concentration of \[C{{l}_{2}}\] in moles/litre             [EAMCET 2003]

    A)                 2.0         

    B)                 1.5

    C)                 1.0         

    D)                 0.5

    Correct Answer: C

    Solution :

              \[{{K}_{c}}=\frac{[PC{{l}_{3}}]\,\,[C{{l}_{2}}]}{[PC{{l}_{5}}]}=\frac{0.2\times x}{0.4}=0.5\], \[x=1\]


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