1. Set up the reaction and mole balance

The reaction is $X_2(g) \rightleftharpoons 2X(g)$

Initially:

• moles of $X_2 = 1$
• moles of $X = 0$

Let $\beta$ be the number of moles of $X$ formed at equilibrium.

Since 1 mole of $X_2$ produces 2 moles of $X$, the moles of $X_2$ decomposed are: $\frac{\beta}{2}$

So at equilibrium: $n_{X_2}=1-\frac{\beta}{2}, \qquad n_X=\beta$

Total moles at equilibrium: $n_{\text{tot}}=1-\frac{\beta}{2}+\beta=1+\frac{\beta}{2}$

2. Write mole fractions

$y_{X_2}=\frac{1-\beta/2}{1+\beta/2}$

$y_X=\frac{\beta}{1+\beta/2}$

Given total pressure, $P=2\,\text{bar}$

Hence partial pressures are: $p_{X_2}=y_{X_2}P=\frac{1-\beta/2}{1+\beta/2}\cdot 2$

$p_X=y_XP=\frac{\beta}{1+\beta/2}\cdot 2$

3. Expression for $K_p$

For the reaction $X_2(g) \rightleftharpoons 2X(g)$

$K_p=\frac{(p_X)^2}{p_{X_2}}$

Substitute the above expressions: $K_p=\frac{\left(\dfrac{2\beta}{1+\beta/2}\right)^2}{\left(\dfrac{2(1-\beta/2)}{1+\beta/2}\right)}$

Now simplify: $K_p=\frac{\dfrac{4\beta^2}{(1+\beta/2)^2}}{\dfrac{2(1-\beta/2)}{1+\beta/2}}$

$K_p=\frac{4\beta^2}{(1+\beta/2)^2}\cdot \frac{1+\beta/2}{2(1-\beta/2)}$

$K_p=\frac{2\beta^2}{(1+\beta/2)(1-\beta/2)}$

Use $(1+\beta/2)(1-\beta/2)=1-\frac{\beta^2}{4}$

So, $K_p=\frac{2\beta^2}{1-\beta^2/4}$

Multiply numerator and denominator by 4: $K_p=\frac{8\beta^2}{4-\beta^2}$

At equilibrium, $\beta=\beta_{\text{equilibrium}}$, hence $K_p=\frac{8\beta_{\text{equilibrium}}^2}{4-\beta_{\text{equilibrium}}^2}$

4. Match with options

This corresponds to: $\boxed{\text{B}}$

5. Comparison with stored correct answer

Stored correct answer: B

Derived answer: B

So, the derived answer agrees with the stored correct answer.