A) \[4\times {{10}^{-4}}\]
B) \[50\]
C) \[2.5\times {{10}^{2}}\]
D) 0.02
Correct Answer: B
Solution :
\[{{N}_{2}}(g)+{{O}_{2}}(g)\] ⇌ \[2NO(g)\] \[Kc=\frac{{{[NO]}^{2}}}{[{{N}_{2}}][{{O}_{2}}]}=4\times {{10}^{-4}}\] \[N{{O}_{2}}\] ⇌ \[\frac{1}{2}{{N}_{2}}(g)+\frac{1}{2}{{O}_{2}}(g)\] \[{{{K}'}_{c}}=\frac{{{[{{N}_{2}}]}^{1/2}}{{[{{O}_{2}}]}^{1/2}}}{[NO]}=\frac{1}{\sqrt{Kc}}=\frac{1}{\sqrt{4\times {{10}^{-4}}}}\] \[=\frac{1}{2\times {{10}^{-2}}}=\frac{100}{2}=50\]You need to login to perform this action.
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