Question

At temperature $T$, the compound $A{B}_2\left(g\right)$ dissociates according to the reaction,
$2A{B}_2\left(g\right)\rightleftharpoons 2AB\left(g\right)+B_2\left(g\right)$
with a degree of dissociation, $x$. which is small compared with unity. Deduce the expression for $x$ in terms of the equilibrium constant, $K_p$ and the total pressure, $P$.

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

Answer 1

The correct option is B ${\left(\frac{2K_P}{P}\right)}^{1/3}$

$\begin{array}{cccccc}& 2A{B}_2\left(g\right)& \rightleftharpoons & 2AB\left(g\right)& +& B_2\left(g\right)\\ \text{At equilibrium}& (1-x)& & x& & x/2\\ \text{Total moles at equilibrium}& =1-x+x+x/2& & & & \\ & =\frac{2+x}{2}& & & & \end{array}$

$P_{A{B}_2}=\frac{2(1-x)}{(2+x)}\cdot P;P_{AB}=\frac{2x}{(2+x)}\cdot P;p_{B_2}=\frac{x}{(2+x)}\cdot P;$
$K_p=\frac{\left(p_{AB}{)}^2\right(p_{B_2})}{(p_{A{B}_2}{)}^2}=\frac{{\left[\frac{2x}{(2+x)}P\right]}^2\left(\frac{x}{2+x}\right)P}{{\left(\frac{2(1-x)}{(2+x)}P\right)}^2}$
$=\frac{x^{3}P}{(2+x)(1-x{)}^2}$
As $x$ is very small, $(2+x)\to 2$ and $(1-x)\to 1$
So, $K_P=\frac{x^{3}P}{2}$
$\Rightarrow x^3=\frac{2K_P}{P}$
$\Rightarrow x={\left(\frac{2K_P}{P}\right)}^{1/3}$