Question

At temperature $T$, a compound ${AB}_2\left(g\right)$ 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 as compared to unity. The expression for $K_p$ in term 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

The correct option is A $\frac{P{x}^3}{2}$

The given reaction is :-

${2AB}_2\rightleftharpoons 2AB+B_2$

Initial pressure : $Y$ $0$ $0$

At eqm : $Y(1-x)$ $Yx$ $Yx/2$ (given degree of dissociation = x)

Now, total pressure at equilibrium $=Y(1-x)+Yx+Yx/2$

$=Y+Yx/2$

$\Rightarrow P=Y+\frac{Yx}{2}\Rightarrow 2P=2Y+Yx$

$\Rightarrow \frac{2P}{(2+x)}=Y$

Now, $K_P=\frac{{\left(p_{AB}\right)}^2.p_{B_2}}{{\left(p_{{AB}_2}\right)}^2}$

$=\frac{\left(\frac{Yx}{2}\right).{\left(Yx\right)}^2}{Y^2{(1-x)}^2}$

$K_P=\frac{{Yx}^3}{2}(\because x<<1)$

$\Rightarrow K_P=\frac{2P}{(2+x)}.\frac{x^3}{2}=\frac{x^{3}P}{2+x}$

$\Rightarrow K_P=\frac{x^{3}P}{2}(\because x<<1)$