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

  • question_answer
    2 moles of \[PC{{l}_{5}}\] were heated in a closed vessel of 2 litre capacity. At equilibrium, 40% of \[PC{{l}_{5}}\] is dissociated into \[PC{{l}_{3}}\] and \[C{{l}_{2}}\]. The value of equilibrium constant is [MP PMT 1989; RPMT 2000; UPSEAT 2004; Kerala CET 2005]

    A)                 0.266    

    B)                 0.53

    C)                 2.66       

    D)                 5.3

    Correct Answer: A

    Solution :

           \[\underset{2}{\mathop{PC{{l}_{5}}}}\,\]⇌ \[\underset{0}{\mathop{PC{{l}_{3}}}}\,+\underset{0}{\mathop{C{{l}_{2}}}}\,\]                    \[\frac{2\times 60}{100}\] \[\frac{2\times 40}{100}\] \[\frac{2\times 40}{100}\]                    Volume of container = 2 litre.                                 \[{{K}_{c}}=\frac{\frac{2\times 40}{100\times 2}\times \frac{2\times 40}{100\times 2}}{\frac{2\times 60}{100\times 2}}=0.266\].


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