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 $\alpha$/2, 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. $P\frac{{\alpha }^3}{4}$
• C. $P\frac{{\alpha }^3}{16}$
• D. None of the above

Answer 1

The correct option is C $P\frac{{\alpha }^3}{16}$

The equilibrium reaction is as shown below:

$2A{B}_2\left(g\right)\rightleftharpoons 2AB\left(g\right)+B_2\left(g\right)$

$P(1-\frac{\alpha }{2})P\frac{\alpha }{2}P\frac{\alpha }{4}$

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.

$Kp=\frac{(P\alpha /2{)}^2\cdot (\frac{P\alpha }{4})}{\left(P\right(1-\frac{\alpha }{2}){)}^2}$. But $\alpha <<1$

Hence, $Kp=\frac{(P\alpha {)}^2\cdot (\frac{P\alpha }{4})}{4(P{)}^2}$ .

$\therefore Kp=\frac{P{\alpha }^3}{16}$.