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

  • question_answer
    The equilibrium constant for the reaction \[{{N}_{2(g)}}+{{O}_{2(g)}}\]⇌ \[2N{{O}_{(g)}}\] at temperature T is \[4\times {{10}^{-4}}\]. The value of \[{{K}_{c}}\]for the reaction \[N{{O}_{(g)}}\]⇌\[\frac{1}{2}{{N}_{2(g)}}+\frac{1}{2}{{O}_{2(g)}}\] at the same temperature is                                                        [AIEEE 2004]

    A)                 \[4\times {{10}^{-4}}\]  

    B)                         \[50\]

    C)                 \[2.5\times {{10}^{2}}\]

    D)                 0.02

    Correct Answer: B

    Solution :

               \[{{N}_{2}}(g)+{{O}_{2}}(g)\] ⇌ \[2NO(g)\]                    \[Kc=\frac{{{[NO]}^{2}}}{[{{N}_{2}}][{{O}_{2}}]}=4\times {{10}^{-4}}\]                    \[N{{O}_{2}}\] ⇌ \[\frac{1}{2}{{N}_{2}}(g)+\frac{1}{2}{{O}_{2}}(g)\]                    \[{{{K}'}_{c}}=\frac{{{[{{N}_{2}}]}^{1/2}}{{[{{O}_{2}}]}^{1/2}}}{[NO]}=\frac{1}{\sqrt{Kc}}=\frac{1}{\sqrt{4\times {{10}^{-4}}}}\]                                 \[=\frac{1}{2\times {{10}^{-2}}}=\frac{100}{2}=50\]


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