The gas $A_2$ in the left gass bulb allowed to react with gas $B_2$ present in right glass bulb as shown in figure : 

$A_2\left(g\right)+B_2\left(g\right)\underset{ }{\rightleftharpoons }2A{B}_{\left(g\right)}$, $K_C = 4 at {27}^{\circ }C$ what is the concentration of AB when equilibrium is established.

$$\begin{gathered} {\mathrm{A}}_{\mathrm{2}}\hspace{0.17em}+\hspace{0.17em}{\mathrm{B}}_{\mathrm{2}}\mathrm{\hspace{0.17em}}\underset{ }{\overset{ }{\rightleftharpoons }}\hspace{0.17em}\hspace{0.17em}2\mathrm{AB} \\ \mathrm{At} \mathrm{t}=0, 2 4 0 \\ \mathrm{at} \mathrm{equi}. 2-\mathrm{x} 4-\mathrm{x} 2\mathrm{x} \end{gathered}$$

where, x=degree of dissociation

$\text{con\hspace{0.17em}\hspace{0.17em}=\hspace{0.17em}}\frac{\text{n}}{\text{V}}$               V = 1 + 3 = 4

$\left[{\mathrm{A}}_{\mathrm{2}}\right]\mathrm{\hspace{0.17em}=\hspace{0.17em}}\frac{\mathrm{2-}\mathrm{x}}{\mathrm{4}}\mathrm{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[{\mathrm{B}}_{\mathrm{2}}\right]\mathrm{\hspace{0.17em}=}\frac{\mathrm{4-}\mathrm{x}}{\mathrm{4}}\mathrm{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{AB}\right]\mathrm{\hspace{0.17em}=\hspace{0.17em}}\frac{2\mathrm{x}}{\mathrm{4}}\mathrm{=}\frac{\mathrm{x}}{2}$

${\text{K}}_{\text{C}}\text{=\hspace{0.17em}}\frac{{\left(\frac{\text{x}}{\text{2}}\right)}^{\text{2}}}{\left(\frac{\text{2-x}}{\text{4}}\right)\left(\frac{\text{4-x}}{\text{4}}\right)}\text{=}\frac{{\text{x}}^{\text{2}}\text{\hspace{0.17em}×\hspace{0.17em}4\hspace{0.17em}×\hspace{0.17em}4}}{\text{4\hspace{0.17em}×\hspace{0.17em}}\left(\text{2-x}\right)\left(\text{4-x}\right)}$

${\text{K}}_{\text{C}}\text{=\hspace{0.17em}\hspace{0.17em}}\frac{{\text{x}}^{\text{2}}\text{\hspace{0.17em}×\hspace{0.17em}\hspace{0.17em}4}}{{\text{8-6x\hspace{0.17em}\hspace{0.17em}+\hspace{0.17em}x}}^{\text{2}}}\Rightarrow K_C = 4$

$\text{4\hspace{0.17em}=\hspace{0.17em}}\frac{{\text{x}}^{\text{2}}\text{\hspace{0.17em}×\hspace{0.17em}4}}{{\text{8-6x\hspace{0.17em}+\hspace{0.17em}x}}^{\text{2}}}$

$x^{2 }= 8 – 6x + x^2$

$\text{6x\hspace{0.17em}=\hspace{0.17em}8\hspace{0.17em}\hspace{0.17em}}\Rightarrow \text{\hspace{0.17em}\hspace{0.17em}x\hspace{0.17em}= }\frac{\text{4}}{\text{3}}$

$\left[\text{AB}\right]\text{\hspace{0.17em}=\hspace{0.17em}}\frac{\text{x}}{\text{2}}\text{=}\frac{\text{4}}{\text{3\hspace{0.17em}×\hspace{0.17em}2}}\text{\hspace{0.17em}=\hspace{0.17em}0.66}$