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) \[{{K}_{1}}=\frac{1}{{{K}_{2}}}=\frac{1}{{{({{K}_{3}})}^{2}}}\]You need to login to perform this action.
You will be redirected in
3 sec