A) 0.02
B) \[2.5\times {{10}^{2}}\]
C) \[4\times {{10}^{-4}}\]
D) 50.0
Correct Answer: D
Solution :
\[{{N}_{2}}+{{O}_{2}}\xrightarrow{{}}2NO\] \[K=4\times {{10}^{-4}}\] \[NO\xrightarrow{{}}\frac{1}{2}{{N}_{2}}+\frac{1}{2}{{O}_{2}}\] \[K'=\frac{1}{\sqrt{K}}=\frac{1}{\sqrt{4\times {{10}^{-4}}}}=50\]You need to login to perform this action.
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