A) 1
B) 4
C) \[\frac{1}{4}\]
D) \[\frac{1}{2}\]
Correct Answer: A
Solution :
Key Idea:\[{{K}_{c}}=\frac{[C][D]}{[A][B]}\] where\[{{K}_{c}}=\]equilibrium constant. \[\begin{align} & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,A\,\,\,+\,\,\,B\,\,C+D \\ & \text{Initial}\,\text{moles}\,\,\,\,\,\,\,\,\,\,\,\,\,4\,\,\,\,\,\,\,\,\,\,\,4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,0 \\ & \text{Moles}\,\text{after}\,\,\,\,\,\,\,\,\,\,\,\,\,(4-2)\,(4-2)\,\,\,\,\,2\,\,\,\,\,\,2\,\, \\ & \text{reaction}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=2\,\,\,\,\,\,\,\,\,\,\,=2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\ \end{align}\] Let volume of container \[=V\,L\] \[\therefore \] \[{{K}_{c}}=\frac{(2/V)\times (2/V)}{(2/V)\times (2/V)}\] \[=1\]
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