JEE Main & Advanced Chemistry Equilibrium / साम्यावस्था Question Bank Topic Test - Chemical Equilibrium (6-5-21)

  • question_answer
    Equilibrium constants \[{{K}_{1}}\]and \[{{K}_{2}}\]for the following equilibria       \[NO(g)+\frac{1}{2}{{O}_{2}}\]            \[N{{O}_{2}}(g)\]             and \[2N{{O}_{2}}(g)\]           \[2NO(g)+{{O}_{2}}(g)\] are related as 

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

    B) \[{{K}_{2}}=K_{1}^{2}\]

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

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

    Correct Answer: D

    Solution :

    [d] \[{{K}_{1}}=\frac{[N{{O}_{2}}]}{[NO]{{[{{O}_{2}}]}^{1/2}}}\];  \[{{K}_{2}}=\frac{{{[NO]}^{2}}[{{O}_{2}}]}{{{[N{{O}_{2}}]}^{2}}}\]
             \[\Rightarrow \] \[\frac{{{[N{{O}_{2}}]}^{2}}}{{{[NO]}^{2}}[{{O}_{2}}]}=\frac{1}{{{K}_{2}}}\]\[\Rightarrow \] \[\frac{[N{{O}_{2}}]}{[NO]\ {{[{{O}_{2}}]}^{1/2}}}=\frac{1}{\sqrt{{{K}_{2}}}}\]   
                \[\Rightarrow \] \[{{K}_{1}}=\frac{1}{\sqrt{{{K}_{2}}}}\]; \[{{K}_{2}}=\frac{1}{K_{1}^{2}}\].


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