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JEE Main & Advanced AIEEE Solved Paper-2004

  • question_answer
    The equilibrium constant for the reaction \[{{N}_{2}}(g)+{{O}_{2}}(g)2NO(g)\] at temperature T is \[4\times {{10}^{-4}}\]. The value of\[{{K}_{c}}\] for the reaction \[NO(g)\frac{1}{2}{{N}_{2}}(g)+\frac{1}{2}{{O}_{2}}(g)\]at the same temperature is

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

    B) 50                          

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

    D)        0.02

    Correct Answer: B

    Solution :

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


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