A) \[0.08\]
B) \[0.8\]
C) \[8\]
D) \[80\]
Correct Answer: C
Solution :
\[\begin{align} & \underset{(1-x)}{\mathop{\underset{{}}{\mathop{\underset{1}{\mathop{{{A}_{(g)}}}}\,}}\,}}\,\,\,\,+\,\,\,\underset{(1-3X)}{\mathop{\underset{{}}{\mathop{\underset{1}{\mathop{3{{B}_{(g)}}}}\,}}\,}}\,\,\,\,\,\,\,\underset{4x}{\mathop{\underset{{}}{\mathop{\underset{0}{\mathop{4{{C}_{(g)}}}}\,}}\,}}\,\text{intial}\,\text{conce}\text{.} \\ & \text{final}\,\text{conc}\text{.} \\ \end{align}\] At equilibrium the concetrations of A and C are equal \[\therefore \] \[1-x=4x\] \[x=\frac{1}{5}\] For above reaction \[{{K}_{C}}=\frac{{{[C]}^{4}}}{[A]\,{{[B]}^{3}}}=\frac{{{(4x)}^{4}}}{(1-x)\,{{(1-3x)}^{3}}}\] \[{{K}_{C}}=\frac{{{\left( 4\times \frac{1}{5} \right)}^{4}}}{\left( 1-\frac{1}{5} \right){{\left( 1-3\times \frac{1}{5} \right)}^{3}}}=8.0\]
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