In a chemical reaction, $\text{A}+\text{2B}\stackrel{K}{\rightleftharpoons }\text{2C}+\text{D,}$  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 (K) for the aforesaid chemical reaction is.

$A+2B\to 2C+D$

Intial                   a              1.5a          0           0
At equation       a-x           1.5a-x       2x         x
At eq. (A) = (B)
(a-x) = 1.5a-2x
X = 0.5a
(A) = a- 0.5a = 0.5a
(B) = 1.5a-2 x 0.5a = 0.5a
(C) = 2 x 0.5a = a
(D) = 0.5a
$K_C=\frac{\left(D\right){\left(C\right)}^2}{\left(A\right){\left(B\right)}^2}=\frac{0.5a\times a^2}{0.5a\times 0.5a\times 0.5a}$

$K_C=4$