A) \[{{10}^{-4}}\]
B) \[2\times {{10}^{-4}}\]
C) \[4\times {{10}^{-4}}\]
D) \[8\times {{10}^{-4}}\]
Correct Answer: B
Solution :
\[A(g)+2B(g)C(g)+D(g)\] At \[\underset{\approx y}{\mathop{e{{q}^{n}}0.5-x}}\,\underset{\approx 2y}{\mathop{1-x}}\,\underset{\approx 1}{\mathop{0.5+x}}\,\underset{\approx 4}{\mathop{3.5+x}}\,\] (due to very high value of equilibrium constant) \[{{10}^{12}}=\frac{1\times 4}{y\times {{(2y)}^{2}}};y={{10}^{-4}}\] \[[B(g)]=2y\to 2\times {{10}^{-4}}\]
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