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

  • question_answer
    The decomposition of \[{{N}_{2}}{{O}_{4}}\] to \[N{{O}_{2}}\] is carried out at \[280K\] in chloroform. When equilibrium has been established, 0.2 mol of \[{{N}_{2}}{{O}_{4}}\] and \[2\times {{10}^{-3}}\] mol of \[N{{O}_{2}}\]  are present in 2 litre solution. The equilibrium constant for reaction \[{{N}_{2}}{{O}_{4}}\] ⇌ \[2N{{O}_{2}}\] is                                  [AIIMS 1984]

    A)                 \[1\times {{10}^{-2}}\]  

    B)                 \[2\times {{10}^{-3}}\]

    C)                 \[1\times {{10}^{-5}}\]  

    D)                 \[2\times {{10}^{-5}}\]

    Correct Answer: C

    Solution :

              \[K=\frac{{{[N{{O}_{2}}]}^{2}}}{[{{N}_{2}}{{O}_{4}}]}=\frac{{{\left[ 2\times {{\frac{10}{2}}^{-3}} \right]}^{2}}}{\left[ \frac{.2}{2} \right]}=\frac{{{10}^{-6}}}{{{10}^{-1}}}={{10}^{-5}}\].


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