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In reaction $\mathrm{A}+2 \mathrm{~B} \rightleftharpoons 2 \mathrm{C}+\mathrm{D}$, initial concentration of $\mathrm{B}$ was $1.5$ times of $\mathrm{[A]}$, but at equilibrium the concentrations of $\mathrm{A}$ and $\mathrm{B}$ became equal. The equilibrium constant for the reaction is :
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The correct answer is:
$4$
$4$
Hence $\mathrm{K}_{\mathrm{C}}=\frac{(2 x)^2 \times x}{(a-x)(1.5 a-2 x)^2}$
Given, at equilibrium
$\begin{array}{ll}\therefore & (a-x)=(1.5 a-2 x) \\ \therefore & a=2 x\end{array}$
On solving $K_C=4$
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