$\text { Initial moles } \;\; 1 \quad \;\;\;\;\quad 0 \quad \quad \quad \quad 0$

At equil. $\quad 2(1-x)\quad \quad 2 x\quad \quad \quad x$

where, $x=$ degree of dissociation

Total moles at equilibrium

$=2-2 x+2 x+x=(2+x)$

$\text { So, } p_{A B 2}= \frac{2(1-x) p}{(2+x)}, p_{A B}=\frac{2 x p}{(2+x)}$

$p_{B 2}= \frac{x p}{(2+x)}$

$K_{p}=\frac{p_{A} s}{\left(p_{A}\right)^{2}\left(P_{B 2}\right)}$

$=\frac{\left(\frac{2 r p}{2+p}\right)^{2}\left[\left(\frac{x}{2+x}\right) p\right]}{\left[\left(\frac{2 a-x}{(2+x)}\right) p\right]}$

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

$=\frac{x^{3} p}{2}[\cdots x < < < 1 \text { and } 2]$

$x =\left(\frac{2 K p}{p}\right)^{1 / 3}$

So, $(1-x) \approx ; 1(2+x) \approx 2$