1. Write the dissociation reaction

$AB_{2(g)} \rightleftharpoons AB_{(g)} + \frac{1}{2}B_{2(g)}$

Let the initial pressure of $AB_2$ be $p$.

If the degree of dissociation is $x$ and $x \ll 1$, then at equilibrium:

• $AB_2$: $p(1-x)$
• $AB$: $px$
• $B_2$: $\dfrac{px}{2}$

But these are proportional amounts. Since total moles change, we must first compute partial pressures properly.

2. Take 1 mole initially

Assume initially 1 mole of $AB_2$ at total pressure $p$.

If degree of dissociation is $x$, then at equilibrium:

$AB_2 = 1-x$ $AB = x$ $B_2 = \frac{x}{2}$

Total moles at equilibrium:

$n_{\text{tot}} = 1-x+x+\frac{x}{2} = 1+\frac{x}{2}$

Hence partial pressures are:

$p_{AB_2} = \frac{1-x}{1+x/2}p$ $p_{AB} = \frac{x}{1+x/2}p$ $p_{B_2} = \frac{x/2}{1+x/2}p$

3. Write the expression for $K_p$

For

$AB_2 \rightleftharpoons AB + \frac{1}{2}B_2$

$K_p = \frac{p_{AB}\,(p_{B_2})^{1/2}}{p_{AB_2}}$

Substitute the partial pressures:

$K_p = \frac{\left(\frac{xp}{1+x/2}\right)\left(\frac{(x/2)p}{1+x/2}\right)^{1/2}}{\left(\frac{(1-x)p}{1+x/2}\right)}$

Simplify:

$K_p = \frac{xp}{1+x/2}\cdot \frac{\sqrt{(x/2)p}}{\sqrt{1+x/2}}\cdot \frac{1+x/2}{(1-x)p}$

$K_p = \frac{x\sqrt{(x/2)p}}{(1-x)\sqrt{1+x/2}}$

4. Use the condition $x \ll 1$

Since $x$ is very small:

$1-x \approx 1, \qquad 1+\frac{x}{2} \approx 1$

Therefore,

$K_p \approx x\sqrt{\frac{xp}{2}}$

$K_p \approx \sqrt{\frac{p}{2}}\;x^{3/2}$

5. Solve for $x$

$x^{3/2} = K_p\sqrt{\frac{2}{p}}$

Raise both sides to the power $\frac{2}{3}$:

$x = \left(K_p\sqrt{\frac{2}{p}}\right)^{2/3}$

$x = \left(K_p^2\cdot \frac{2}{p}\right)^{1/3}$

So,

$\boxed{x = \sqrt[3]{\frac{2K_p^2}{p}}}$

6. Match with options

This corresponds to:

$\boxed{\text{Option B}}$

7. Compare with stored correct answer

Stored correct answer = B

Our derived answer = B

So they agree.