Question

At temperature $T$, a compound $A{B}_2(g)$ dissociates according to the reaction, $2A{B}_2(g)\rightleftharpoons 2AB(g)+B_2(g)$ 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. $\frac{P{x}^2}{2}$

Answer 1

Answer: (A)

$\therefore K_p=\frac{P_{B_2}\times P_{AB}^2}{P_{A{B}_2}^2}$
$=\frac{\left(\frac{\frac{x}{2}}{1+\frac{x}{2}}\right)P\times \frac{x^2}{{\left(1+\frac{x}{2}\right)}^2}{P}^2}{\frac{{\left(1-x\right)}^2}{{\left(1+\frac{x}{2}\right)}^2}}=\frac{P{x}^3}{2\left(1+\frac{x}{2}\right){\left(1-x\right)}^2}$
$\therefore x<<1$
$\Rightarrow K_p=\frac{P{x}^3}{2\times 1\times 1}=\frac{P{x}^3}{2}$
Also,
Degree of dissociation $=\frac{\text{ Number of moles disgociated }}{\text{ Number of moles taken }}$