A) 1
B) \[\frac{1}{4}\]
C) 4
D) 16
Correct Answer: C
Solution :
\[\begin{matrix} {} & A & + & 2B & & 2C+ & D \\ Initial & a & {} & 1.5a & {} & 0 & 0 \\ At.eq. & a-x & {} & 1.5a-2x & {} & 2x & x \\ \end{matrix}\] As given, at equilibrium, \[a-x=1.5a-2x\Rightarrow x=0.5a\] \[[C]=2x=0.5a\times 2=a,[D]=x=0.5a\] \[[B]=1.5a-2\times 0.5a=0.5a\] \[[A]=a-x=a-0.5a=0.5a\] \[k=\frac{{{[C]}^{2}}[D]}{[A]{{[B]}^{2}}}=\frac{{{a}^{2}}\times 0.5a}{{{(0.5a)}^{2}}\times 0.5a}=\frac{{{a}^{2}}}{0.25{{a}^{2}}}=4\]You need to login to perform this action.
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