At temperature $T$, a compound $AB_{2(g)}$ dissociates according to the reaction $2AB_{2(g)} \rightl

Question

At temperature $T$, a compound $AB_{2(g)}$ dissociates according to the reaction $2AB_{2(g)} \rightleftharpoons 2AB_{(g)} + B_{2(g)}$. The degree of dissociation $x$ is very small compared with unity. Which of the following expressions correctly relates $x$ to the equilibrium constant $K_p$ and the total pressure $P$?

Professor PC Thomas Classes · Answer

For the reaction $2AB_2 \rightleftharpoons 2AB + B_2$, let initial moles of $AB_2$ be 1. At equilibrium, moles of $AB_2 = 1-x$, $AB = x$, and $B_2 = x/2$. Total moles = $1 - x + x + x/2 = 1 + x/2 \approx 1$ (since $x \ll 1$). The partial pressures are $P_{AB_2} \approx 1 \cdot P = P$, $P_{AB} \approx x \cdot P$, and $P_{B_2} \approx (x/2) \cdot P$. The equilibrium constant $K_p = \frac{(P_{AB})^2 (P_{B_2})}{(P_{AB_2})^2} = \frac{(xP)^2 (xP/2)}{P^2} = \frac{x^3 P}{2}$. Solving for $x$ gives $x = (\frac{2K_p}{P})^{1/3}$.