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JEE Main & Advanced JEE Main Paper (Held on 7 May 2012)

  • question_answer
    \[{{K}_{1}},{{K}_{2}}\]and\[{{K}_{3}}\] are the equilibrium constants of the following reactions (I), (II) and (III)respectively: (i)\[{{N}_{2}}+2{{O}_{2}}2N{{O}_{2}}\](ii)\[2N{{O}_{2}}{{N}_{2}}+2{{O}_{2}}\]                 (iii)\[N{{O}_{2}}\frac{1}{2}{{N}_{2}}+{{O}_{2}}\] The correct relation from the following is   JEE Main Online Paper (Held On 07 May 2012)

    A) \[{{K}_{1}}=\frac{1}{{{K}_{2}}}=\frac{1}{{{K}_{3}}}\]        

    B)                        \[{{K}_{1}}=\frac{1}{{{K}_{2}}}=\frac{1}{{{({{K}_{3}})}^{2}}}\]

    C)                        \[{{K}_{1}}=\sqrt{{{K}_{2}}}={{K}_{3}}\]

    D)                        \[{{K}_{1}}=\frac{1}{{{K}_{2}}}={{K}_{3}}\]

    Correct Answer: B

    Solution :

                    (I)\[{{N}_{2}}+2{{O}_{2}}2N{{O}_{2}}\] \[{{K}_{1}}=\]                                    ?(i) (II)\[2N{{O}_{2}}{{N}_{2}}+2{{O}_{2}}\] \[{{K}_{2}}=\frac{[{{N}_{2}}]{{[{{O}_{2}}]}^{2}}}{{{[N{{O}_{2}}]}^{2}}}\]                                  ?(ii) (III)\[N{{O}_{2}}\frac{1}{2}{{N}_{2}}+{{O}_{2}}\] \[{{K}_{3}}=\frac{{{[{{N}_{2}}]}^{1/2}}[{{O}_{2}}]}{[N{{O}_{2}}]}\] \[\therefore \]\[{{\left( {{K}_{3}} \right)}^{2}}=\frac{[{{N}_{2}}]{{[{{O}_{2}}]}^{2}}}{{{[N{{O}_{2}}]}^{2}}}\]                                                ?(ii) \[\therefore \]from equation (i), (ii) and (iii) \[{{K}_{1}}=\frac{1}{{{K}_{2}}}=\frac{1}{{{({{K}_{3}})}^{2}}}\]


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