Let the initial moles of $AB_2$ be 1. The dissociation reaction is: $2AB_2(g) \rightleftharpoons 2AB(g) + B_2(g)$ Initial moles: 1, 0, 0 Moles at equilibrium: $1-x$, $x$, $x/2$ Total number of moles at equilibrium = $1 - x + x + x/2 = 1 + x/2$ Partial pressures of the species at equilibrium: $p_{AB_2} = \frac{1-x}{1+x/2} \times p$ $p_{AB} = \frac{x}{1+x/2} \times p$ $p_{B_2} = \frac{x/2}{1+x/2} \times p$ The equilibrium constant $K_p$ is given by: $K_p = \frac{(p_{AB})^2 (p_{B_2})}{(p_{AB_2})^2}$ $K_p = \frac{\left(\frac{x}{1+x/2}p\right)^2 \left(\frac{x/2}{1+x/2}p\right)}{\left(\frac{1-x}{1+x/2}p\right)^2} = \frac{x^3 \cdot p}{2(1-x)^2(1+x/2)}$ Since $x$ is small compared to 1, we can approximate $1-x \approx 1$ and $1+x/2 \approx 1$. Therefore, $K_p = \frac{x^3 p}{2}$ $x^3 = \frac{2K_p}{p}$ $x = \left(\frac{2K_p}{p}\right)^{1/3}$