Question

At TK, a compound $AB_2(g)$ dissociates according to the reaction $2AB_2(g) \rightleftharpoons 2AB(g) + B_2(g)$, with a degree of dissociation 'x' which is small compared with unity. The expression for 'x' in terms of the equilibrium constant, $K_p$ and the total pressure P is

• A. $\frac{K_p}{P}$
• B. $(K_p)^{1/3}$
• C. $\left(\frac{2K_p}{P}\right)^{1/3}$
• D. $\left(\frac{K_p}{P}\right)^{1/3}$

Answer 1

Answer: (C)

$\left(\frac{2K_p}{P}\right)^{1/3}$

Answer 2

Starting with 1 mole of $AB_2$, at equilibrium we have $(1-x)$ moles of $AB_2$, $x$ moles of $AB$, and $x/2$ moles of $B_2$. The total moles are $1 - x + x + x/2 = 1 + x/2$. The partial pressures are: $P_{AB_2} = \frac{1-x}{1+x/2}P$ $P_{AB} = \frac{x}{1+x/2}P$ $P_{B_2} = \frac{x/2}{1+x/2}P$ The equilibrium constant $K_p$ is given by: $K_p = \frac{(P_{AB})^2 P_{B_2}}{(P_{AB_2})^2} = \frac{\left(\frac{x}{1+x/2}P\right)^2 \left(\frac{x/2}{1+x/2}P\right)}{\left(\frac{1-x}{1+x/2}P\right)^2} = \frac{x^3/2 \cdot P}{(1+x/2)(1-x)^2}$ Since $x$ is small, $1+x/2 \approx 1$ and $1-x \approx 1$. $K_p \approx \frac{x^3/2 \cdot P}{1 \cdot 1^2} = \frac{x^3 P}{2}$ Solving for $x$: $x^3 = \frac{2K_p}{P} \implies x = \left(\frac{2K_p}{P}\right)^{1/3}$