At a certain temperature, the following equilibrium is established, $CO\left(g\right)+N{O}_2\left(g\right)\leftrightharpoons C{O}_2\left(g\right)+NO\left(g\right)$. One mole of each of four gases in mixed in a one-litre container and the reaction is allowed to reach an equilibrium state. When an excess of baryta water is added to the equilibrium mixture, the weight of white precipitate obtained is 236.49 g. The equilibrium constant $K_c$ of the reaction is

The correct option is D 2.25

$CO+N{O}_2\leftrightharpoons C{O}_2+NO$

Initial: 1 1 1 1

Equilbm: $1-x$ $1-x$ $1+x$ $1+x$

Again,

$C{O}_2+Ba(OH{)}_2\to BaC{O}_3+H_{2}O$

236.4 g of barium carbonate corresponds to $\frac{236.4}{197}=1.2$ moles of barium carbonate. The number of moles of carbon dioxide will also be 1.2.

Hence, $1+x=1.2$ or $x=0.2$.

The expression for the equilibrium constant is $K_c=(\frac{1+x}{1-x}{)}^2=(\frac{1.2}{0.8}{)}^2=2.25$.