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JEE Main & Advanced JEE Main Paper (Held On 15 April 2018) Slot-II

  • question_answer
    At a certain temperature in a 5L vessel, 2moles of carbon monoxide and 3 moles of chlorine were allowed to reach equilibrium according to the reaction, \[CO+C{{l}_{2}}\rightleftharpoons COC{{l}_{2}}\] At equilibrium, if one mole of CO is present then equilibrium constant \[({{K}_{C}})\] for the reaction is? [JEE Online 15-04-2018 (II)]

    A)                     \[2.5\]                                  

    B) \[4\]     

    C) \[2\]                     

    D)          \[3\]

    Correct Answer: A

    Solution :

    Initially, 2 moles of CO are present. At equilibrium, 1 mole of CO is present. Hence, 2-1=1 mole of CO has reacted. 1 mole of CO will react with 1 mole of \[C{{l}_{2}}\] to form 1 mole of\[COC{{l}_{2}}\]. \[3-1=2\] moles of \[C{{l}_{2}}\] remains at equilibrium. The equilibrium constant \[{{K}_{C}}=\frac{[COC{{l}_{2}}]}{[CO][C{{l}_{2}}]}\] \[{{K}_{C}}=\frac{\frac{1mol}{5L}}{\frac{1mol}{5L}\times \frac{2mol}{5L}}\] \[{{K}_{C}}=2.5\]


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