2NOCl(g) ⇌ 2NO(g) + Cl _ 2 (g) In an experiment, 2 . 0 moles of...

$2NOCl\left(g\right) ⇌ 2NO\left(g\right) + {Cl}_{2}\left(g\right)$

$\begin{cases}t = 0 & 2M\end{cases}$

$\begin{cases}t = {t}_{eq}\left(2 - x\right)M & xM & \dfrac{x}{2}M\end{cases}$

$∵ x = 0 . 4M$

$∴ {\left[NOCl\right]}_{eq} = 1 . 6M$

${\left[NO\right]}_{eq} = 0 . 4M$

${\left[{Cl}_{2}\right]}_{eq} = 0 . 2M$

$⇒ {K}_{c} = \dfrac{{\left[NO\right]}^{2}\left[{Cl}_{2}\right]}{{\left[NOCl\right]}^{2}} = \dfrac{{\left[0 . 4\right]}^{2}\left[0 . 2\right]}{{\left[1 . 6\right]}^{2}}$

${K}_{c} = \dfrac{32}{2 . 56} \times {10}^{- 3}$

${K}_{c} = 12 . 5 \times {10}^{- 3}$

${K}_{c} = 125 \times {10}^{- 4}$

Integer answer is $125$