Consider the reaction $$X_{2}Y(g)=X_{2}(g)+\frac{1}{2}Y_{2}(g)$$ The equation representing correct relationship between the degree of dissociation (x) of $$X_{2}Y(g)$$ with its equilibrium constant Kp is ______ . Assume x to be very very small.

For the reaction $$X_2Y(g) \rightleftharpoons X_2(g) + \frac{1}{2}Y_2(g)$$, we want to find the relationship between the degree of dissociation $$x$$ and $$K_p$$.

We begin by setting up the equilibrium composition. Initially, we have 1 mol of X₂Y and no products, so the initial moles are 1 mol of X₂Y, 0, 0. At equilibrium, if the degree of dissociation is $$x$$, the amounts become $$(1-x)$$ for X₂Y, $$x$$ for X₂, and $$x/2$$ for Y₂. The total number of moles is therefore

$$1 - x + x + \tfrac{x}{2} = 1 + \tfrac{x}{2}.$$

Since $$x$$ is very small, we approximate the total moles as 1.

Next, with a total pressure of $$P$$, the partial pressures are given by

$$P_{X_2Y} = (1-x)P \approx P,$$

$$P_{X_2} = xP,$$

$$P_{Y_2} = \tfrac{x}{2}P.$$

The equilibrium constant $$K_p$$ is written in terms of these partial pressures as:

$$K_p = \frac{P_{X_2} \cdot P_{Y_2}^{1/2}}{P_{X_2Y}} = \frac{xP \cdot \left(\frac{x}{2}P\right)^{1/2}}{P}$$

$$= xP \cdot \frac{x^{1/2}}{2^{1/2}} \cdot \frac{P^{1/2}}{P}$$

$$= \frac{x^{3/2} P^{1/2}}{2^{1/2}}.$$

To solve for $$x$$, we rewrite the expression as:

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

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

$$x^3 = \frac{2K_p^2}{P}$$

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

The correct answer is Option 2: \boxed{x = \sqrt[3]{\frac{2K_p^2}{P}}}.