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 $P$.

Answer 1

$2{AB}_2\left(g\right)\rightleftharpoons 2AB\left(g\right)+B_2\left(g\right)$ total pressure $P$

$a$ $0$ $0$

$a(1-x)$ $\frac{2ax}{2}$ $ax/2$
Total no. of moles at eqm. $=a-ax+\frac{2ax}{2}+\frac{ax}{2}=a(1+\frac{x}{2})$
Partial pressure = mole fraction $\times$ total pressure.

$K_P=\frac{P_{AB}^2{P}_{B_2}}{{P^2}_{{AB}_2}}\Rightarrow \frac{\frac{ax/2}{a(1+x/2)}P\times {\left(\frac{ax}{a(1+x/2)}P\right)}^2}{{\left(\frac{a(1-x)}{a(1+x/2)}P\right)}^2}$

$\Rightarrow \frac{\frac{x}{2(1+x/2)}\times \frac{x^2}{{(1+x/2)}^2}\times P^3}{\frac{{(1-x)}^2}{{(1+x/2)}^2}{P}^2}\Rightarrow P\frac{x^3}{2{(1-x)}^2(1+x/2)}=K_P$
According to the Qs, the degree of dissociation x is small compared with unity.

$\frac{{Px}^3}{\left(1\right)\left(1\right)\left(2\right)}=K_P$

$x={\left(\frac{2K_P}{P}\right)}^{1/3}$