A) 8
B) 4
C) 12
D) 6
Correct Answer: D
Solution :
Let degree of dissociation \[=x\] \[A\,\,\,\,\,+\,\,\,\,2B\,\underset{{}}{\leftrightarrows}\,2C\,\,+\,\,\,D\] Initial cone. 1 1.5 0 0 At equilibrium \[(1-x)\] \[(1.5-2x)\] \[2x\] \[x\] Given, \[(1-x)=(1.5-2x)\] \[1=1.5-2x+x\] \[1=1.5-x\] \[x=1.5-1=0.5\] Equilibrium constant for the reaction \[{{K}_{C}}=\frac{{{[C]}^{2}}[D]}{[A]{{[B]}^{2}}}=\frac{{{(2x)}^{2}}\,(x)}{(1-x)\,(1.5-2x)}\] \[\because \] \[x=0.5\] \[{{K}_{C}}=\frac{{{(2\times 0.5)}^{2}}\,(0.5)}{(1-0.5)\,{{(1.5-2\times 0.5)}^{2}}}\] \[=\frac{(1)\times (0.5)}{(0.5)\,{{(0.5)}^{2}}}=\frac{0.5}{0.5\times 0.25}=\frac{0.5}{0.125}=4\]You need to login to perform this action.
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