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