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

  • question_answer
    A quantity of \[PC{{l}_{5}}\] was heated in a 10 litre vessel at \[{{250}^{o}}C\]; \[PC{{l}_{5}}(g)\]⇌ \[PC{{l}_{3}}(g)+C{{l}_{2}}(g)\]. At equilibrium the vessel contains 0.1 mole of \[PC{{l}_{5}}\,0.20\]  mole of \[PC{{l}_{3}}\] and 0.2 mole of \[C{{l}_{2}}\]. The equilibrium constant of the reaction is [KCET 1993, 2001; MP PMT 2003]

    A)                 0.02       

    B)                 0.05

    C)                 0.04       

    D)                 0.025

    Correct Answer: C

    Solution :

              \[{{K}_{c}}=\frac{[PC{{l}_{3}}]\,\,[C{{l}_{2}}]}{[PC{{l}_{5}}]}=\frac{\frac{0.2}{10}\times \frac{0.2}{10}}{\left[ {0.1}/{10}\; \right]}=0.04\].


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