A) 4
B) 8
C) 12
D) 16
Correct Answer: A
Solution :
In the given equation \[\begin{matrix} \text{A+2B} & & \text{2C+D} \\ \text{1}\text{.01}\text{.5} & {} & \text{ 00}\,\text{intial}\,\text{conc}\text{.} \\ \text{(1-x)}\,\text{(1}\text{.5-2x)} & {} & \text{2x}\,\text{x}\,\text{final}\,\text{conc}\text{.} \\ \end{matrix}\] At equilibrium \[[A]=[B]\] \[1-x=1.5-2x\] \[\Rightarrow \] \[x\overset{\bullet }{\mathop{=}}\,0.5\] \[{{K}_{c}}\frac{{{[C]}^{2}}[D]}{[A]{{[B]}^{2}}}=\frac{{{(2x)}^{2}}(x)}{(1-x){{(1.5-2x)}^{2}}}\] \[=\frac{1\times 1\times 0.5}{0.5\times 0.5\times 0.5}=4\]You need to login to perform this action.
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