A) 0.02
B) 50
C) \[4\times {{10}^{-4}}\]
D) \[2.5\times {{10}^{-2}}\]
Correct Answer: B
Solution :
[b] \[{{N}_{2}}+{{O}_{2}}\rightleftharpoons 2NO\] \[{{K}_{C1}}=\frac{{{[NO]}^{2}}}{[{{N}_{2}}][{{O}_{2}}]}=4\times {{10}^{-4}}\] \[NO\rightleftharpoons 1/2{{N}_{2}}+1/2{{O}_{2}}\] \[{{K}_{{{C}_{2}}}}=\frac{{{[{{N}_{2}}]}^{1/2}}{{[{{O}_{2}}]}^{1/2}}}{[NO]}\] \[{{K}_{{{C}_{2}}}}=\sqrt{\frac{1}{{{K}_{{{c}_{1}}}}}}=\sqrt{\frac{1}{4\times {{10}^{-4}}}}=50\]You need to login to perform this action.
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