In a chemical reaction, $A+2B\stackrel{K}{\rightleftarrows }2C+D,$the initial concentration of B was 1.5 times of the concentration of A.  but the equilibrium concentrations of A and B were found to be equal.  The equilibrium constant $\left(K\right)$ for the aforesaid chemical reaction is:

$A+2B\stackrel{K}{\rightleftarrows }2C+D$

$t=0 a_0 1.5a_0 0 0$

$t=t_{eq}$$(a_0-X)$ $(1.5a_0-2x)$  $2x$$x$

Given at equilibrium $\left[A\right]=\left[B\right]$

$\Rightarrow a_0-x-1.5a_0-2x$

$\Rightarrow x-0.5a_{0}or=2x$

Equilibrium constant $k=\frac{{\left[C\right]}^2\left[D\right]}{\left[A\right]{\left[B\right]}^2}$

$\Rightarrow k=\frac{{\left(2x\right)}^2\left(x\right)}{(a_0-x)(1.5a_0-2x)}$

$k=\frac{{\left(2x\right)}^2\left(0.5a_0\right)}{(a_0-0.5a_0){(1.5a_0-a_0)}^2}$

$k=\frac{0.5a_0^2}{0.5a_0{\left(0.5a_0\right)}^2}=4$