In a study of equilibrium,
$H_2\left(g\right)+I_2\left(g\right)\rightleftharpoons 2HI\left(g\right)$
1 mole of $H_2$ and 3 mole of $I_2$ gave rise at equilibrium to $x$ mole of $HI$.
Addition of a further 2 mole of $H_2$ gave an additional $x$ mole of $HI$. What is $x$?

The correct option is A 0.5

$H_2+I_2\rightleftharpoons 2HI$

$1$ $3$ $0$

$1-\frac{x}{2}$ $3-\frac{x}{2}$ $x$

$K_c=\frac{x^2}{(1-\frac{x}{2})(3-\frac{x}{2})}$

$H_2+I_2\rightleftharpoons 2HI$

$3$ $3$ $0$ $\Rightarrow$ $K_c=\frac{{4x}^2}{{(3-x)}^2}$

$3-x$ $3-x$ $2x$

$\Rightarrow \frac{{4x}^2}{(2-x)(6-x)}=\frac{{4x}^2}{{(3-x)}^2}\Rightarrow x=0.5$

Ans :- A