Question

The dissociation equilibrium of a gas ${\mathrm{AB}}_2$ can be represented as:

$2{\mathrm{AB}}_2\left(\mathrm{g}\right)\rightleftharpoons 2\mathrm{AB}\left(\mathrm{g}\right)+{\mathrm{B}}_2\left(\mathrm{g}\right)$
The degree of dissociation is $\mathrm{x}$ and is small compared to $1$. The expression relating the degree of dissociation $\mathrm{x}$ with equilibrium constant ${\mathrm{K}}_{\mathrm{p}}$ and total pressure $\mathrm{p}$ is?

Answer 1

Step 1:Given data:

The dissociation equilibrium of gas ${\mathrm{AB}}_2$ has been given as:

$2{\mathrm{AB}}_2\left(\mathrm{g}\right)\rightleftharpoons 2\mathrm{AB}\left(\mathrm{g}\right)+{\mathrm{B}}_2\left(\mathrm{g}\right)$

Step 2: expression for Kp:

Let $\mathrm{P}$ be the initial pressure of ${\mathrm{AB}}_2$
Whereas the equilibrium pressures of ${\mathrm{AB}}_2,\mathrm{AB}\mathrm{and}{\mathrm{B}}_2$​ are $\mathrm{P}(1-\mathrm{x}),\mathrm{xP}\mathrm{and}0.5\mathrm{xP}$ respectively.
Hence, the equilibrium constant can be expressed as :
$$\begin{gathered} {\mathrm{K}}_{\mathrm{p}}=\frac{{{\mathrm{P}}_{\mathrm{AB}}}^2{{\mathrm{P}}_{\mathrm{B}}}^2}{{({{\mathrm{P}}_{\mathrm{AB}}}_2)}^2} \\ \Rightarrow {\mathrm{K}}_{\mathrm{p}}=\frac{{\left[\mathrm{xP}\right]}^2\times 0.5\mathrm{xP}}{{\left[\mathrm{P}\right(1-\mathrm{x}\left)\right]}^2} \\ \Rightarrow {\mathrm{K}}_{\mathrm{p}}=\mathrm{P}\frac{0.5{\mathrm{x}}^3}{{\left[\right(1-\mathrm{x}\left)\right]}^2} \end{gathered}$$
(Since the degree of dissociation is small compared to 1, hence $1-\mathrm{x}\approx 1$)

${\mathrm{K}}_{\mathrm{p}}=\mathrm{P}\times 0.5{\mathrm{x}}^3....\left(1\right)$

Step 3: expression for total pressure:
The total pressure can be calculated as:
$$\begin{gathered} \mathrm{P}(1-\mathrm{x})+\mathrm{Px}+0.5\mathrm{Px}=\mathrm{P}(1+0.5\mathrm{x})=\mathrm{p} \\ \Rightarrow \mathrm{P}=\frac{\mathrm{p}}{1+0.5\mathrm{x}}.....\left(2\right) \end{gathered}$$
Step 4:Finding relation for Kp with $\mathrm{x}$ and total pressure $\mathrm{p}$ :

By Substituting (2) in (1), we can get :
$\mathrm{Kp}=\frac{\mathrm{p}}{1+0.5\mathrm{x}}\times 0.5{\mathrm{x}}^3$
(Since the degree of dissociation is small compared to $1,1+0.5\mathrm{x}\approx 1$)
$$\begin{gathered} {\mathrm{K}}_{\mathrm{p}}=\mathrm{p}\times 0.5{\mathrm{x}}^3 \\ \Rightarrow \mathrm{x}={\left(\frac{2{\mathrm{K}}_{\mathrm{p}}}{\mathrm{p}}\right)}^{1/3} \end{gathered}$$
Hence, the expression relating the degree of dissociation $\mathrm{x}$ with equilibrium constant ${\mathrm{K}}_{\mathrm{p}}$ and total pressure $\mathrm{p}$ is $\mathrm{x}={\left(\frac{2{\mathrm{K}}_{\mathrm{p}}}{\mathrm{p}}\right)}^{1/3}$