Question

At temperature $T$, a compound $AB _{2}( g )$ dissociates according to the reaction, $2 AB _{2}( g ) \rightleftharpoons 2 AB ( g )+ B _{2}( g )$ with a degree of dissociation ' $x$ ' which is small as compared to unity. The expression for $K _{ p }$ in terms of ' $x$' and total pressure $P$ is

• A. $\frac{P x^{3}}{2}$
• B. $\frac{P x^{2}}{3}$
• C. $\frac{P x^{3}}{3}$
• D. $\frac{ Px ^{2}}{2}$

Answer 1

Answer: (A)

$\therefore K_{p}=\frac{P_{B_{2}} \times P_{A B}^{2}}{P_{A B_{2}}^{2}} $
$=\frac{\left(\frac{\frac{x}{2}}{1+\frac{x}{2}}\right)P \times\frac{x^{2}}{\left(1 + \frac{x}{2}\right)^{2}}P^{2}}{\frac{\left(1 - x\right)^{2}}{\left(1 +\frac{x}{2}\right)^{2}}} = \frac{Px^{3}}{2\left(1+\frac{x}{2}\right)\left(1-x\right)^{2}}$
$\therefore x < < 1 $
$\Rightarrow K_{p}=\frac{P x^{3}}{2 \times 1 \times 1}=\frac{P x^{3}}{2}$
Also,
Degree of dissociation $=\frac{\text { Number of moles disgociated }}{\text { Number of moles taken }}$