Question

At certain temperature 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 the degree of dissociation $\alpha$, which is small compared with unity. The expression of $K_p$, in terms of $\alpha$ and initial pressure P is:

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

Answer 1

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

The equilibrium reaction is as shown below:

$2A{B}_2\left(g\right)\rightleftharpoons 2AB\left(g\right)+B_2\left(g\right)$
$P(1-\alpha )P\alpha P\frac{\alpha }{2}$

Here, P is the initial pressure and $\alpha$ is the degree of dissociation.

The expression for the equilibrium constant is

$K_p=\frac{P_{AB}^2\times P_{B_2}}{P_{A{B}_2}^2}$

Substitute values in the above expression.

$Kp=\frac{(P\alpha {)}^2\cdot (\frac{P\alpha }{2})}{\left(P\right(1-\alpha ){)}^2}$. But $\alpha <<1$

Hence, $Kp=\frac{(P\alpha {)}^2\cdot (\frac{P\alpha }{2})}{(P{)}^2}$ .

$\therefore Kp=\frac{P{\alpha }^3}{2}$.