A) \[11\]
B) \[\frac{1}{121}\]
C) \[\frac{1}{11}\]
D) \[121\]
Correct Answer: C
Solution :
\[{{N}_{2}}+{{O}_{2}}2NO\] \[{{K}_{c}}=\frac{{{[NO]}^{2}}}{[{{N}_{2}}][{{O}_{2}}]}=121\] \[NO\frac{1}{2}{{N}_{2}}+\frac{1}{2}{{O}_{2}}\] \[K_{c}^{'}=\frac{{{[{{N}_{2}}]}^{1/2}}{{[{{O}_{2}}]}^{1/2}}}{[NO]}\] On squarring \[{{(K_{c}^{'})}^{2}}=\frac{[{{N}_{2}}][{{O}_{2}}]}{{{[NO]}^{2}}}=\frac{1}{{{K}_{c}}}\] Hence, \[{{K}_{c}}=\frac{1}{{{(K_{c}^{'})}^{2}}}\] \[K_{c}^{'}=\frac{1}{\sqrt{{{K}_{c}}}}=\frac{1}{\sqrt{121}}=\frac{1}{11}\]
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