JEE Main & Advanced Chemistry Equilibrium / साम्यावस्था Question Bank Kp and Kc Relationship and Characteristics of K

  • question_answer
    If equilibrium constants of reaction, \[{{N}_{2}}+{{O}_{2}}\]⇌ \[2NO\] is \[{{K}_{1}}\]and \[\]\[\frac{1}{2}{{N}_{2}}+\frac{1}{2}{{O}_{2}}\]⇌ \[NO\] is \[{{K}_{2}}\], then               [BHU 2004]

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

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

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

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

    Correct Answer: B

    Solution :

               \[{{N}_{2}}+{{O}_{2}}\]   ⇌  \[2NO\]                                            ?..(i)                    \[\frac{1}{2}{{N}_{2}}+\frac{1}{2}{{O}_{2}}\]⇌\[NO\]                               ??(ii)                    For equation number (i)                    \[{{K}_{1}}=\frac{{{[NO]}^{2}}}{[{{N}_{2}}]\,[{{O}_{2}}]}\]                         ?.. (iii)                    For equation number (ii)                    \[{{K}_{2}}=\frac{[NO]}{{{[{{N}_{2}}]}^{1/2}}{{[{{O}_{2}}]}^{1/2}}}\]                     ?... (iv)                    From equation (iii) & (iv) it is clear that                                 \[{{K}_{2}}={{({{K}_{1}})}^{1/2}}=\sqrt{{{K}_{1}}}\];         Hence, \[{{K}_{2}}=\sqrt{{{K}_{1}}}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner