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

Question

At a given temperature $T$, a compound $AB_{2(g)}$ dissociates according to the reaction \ $2AB_{2(g)} ightleftharpoons 2AB_{(g)} + B_{2(g)}$.\ The degree of dissociation $x$ is small compared with unity. Which of the following statements correctly describe the relationship between the degree of dissociation $x$, total pressure $P$, and volume $V$?

Professor PC Thomas Classes · Answer

Let the initial moles of $AB_2$ be $1$. At equilibrium, moles are $AB_2 = 1-x$, $AB = x$, $B_2 = x/2$. Total moles $\approx 1$ since $x$ is small. The equilibrium constant is $K_p = \frac{(xP)^2 (xP/2)}{(1-x)^2 P^2} \approx \frac{x^3 P}{2}$. Rearranging gives $x = (\frac{2K_p}{P})^{1/3}$. Option A is correct because $x \propto P^{-1/3}$. Option B is correct because at constant temperature, $P \propto 1/V$, so $x \propto V^{1/3}$. If $V$ is doubled, $x$ becomes $x \cdot 2^{1/3}$. Option C is correct because adding an inert gas at constant pressure increases the total volume, lowering the partial pressures of the reacting gases, which shifts the equilibrium towards the side with more gaseous moles (forward direction), thereby increasing $x$. Option D is incorrect; $K_p$ is a temperature-dependent constant and does not change with pressure.