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

  • question_answer
    A mixture of 0.3 mole of \[{{H}_{2}}\] and 0.3 mole of \[{{I}_{2}}\] is allowed to react in a 10 litre  evacuated flask at \[{{500}^{o}}C\]. The reaction is \[{{H}_{2}}+{{I}_{2}}\]⇌ \[2HI\], the \[K\] is found to be 64. The amount of unreacted \[{{I}_{2}}\] at equilibrium is       [KCET 1990]

    A)                 0.15 mole           

    B)                 0.06 mole

    C)                 0.03 mole           

    D)                 0.2 mole

    Correct Answer: B

    Solution :

           \[{{K}_{c}}=\frac{{{[HI]}^{2}}}{[{{H}_{2}}]\,\,[{{I}_{2}}]}\];  \[64=\frac{{{x}^{2}}}{0.03\times 0.03}\]                    \[{{x}^{2}}=64\times 9\times {{10}^{-4}}\]                    \[x=8\times 3\times {{10}^{-2}}=0.24\]                    x is the amount of HI at equilibrium amount of \[{{I}_{2}}\]at equilibrium will be                                 \[0.30-0.24=0.06\]


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