JEE Chemistry 2025 Question with Solution

Given: The reaction is

$$X_2Y(\text{g}) \rightleftharpoons X_2(\text{g}) + \frac{1}{2}Y_2(\text{g})$$

Initial pressure of $X_2Y(\text{g})$ is $p$ and degree of dissociation is $x$.

Find: The relation between $x$ and $K_p$.

At equilibrium, the partial pressures are:

$$p_{X_2Y} = p(1-x), \quad p_{X_2} = px, \quad p_{Y_2} = \frac{px}{2}$$

The equilibrium constant used in the solution is

$$K_p = \frac{(p_{X_2})(p_{\frac{1}{2}Y_2})}{p_{X_2Y}}$$

Substituting the equilibrium pressures,

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

For small dissociation, assume

$$1-x \approx 1$$

Then,

$$K_p = \frac{px^2}{2}$$

So,

$$x = \sqrt{\frac{2K_p}{p}}$$

Therefore, the correct option is B.

Given:

$$X_2Y(\text{g}) \rightleftharpoons X_2(\text{g}) + \frac{1}{2}Y_2(\text{g})$$

Initial pressure of $X_2Y(\text{g})$ is $p$.

Find: The correct relation between degree of dissociation $x$ and equilibrium constant $K_p$.

1. Let the degree of dissociation be $x$.
2. Then the equilibrium pressures are:

$$p_{X_2Y} = p(1-x)$$ $$p_{X_2} = px$$ $$p_{Y_2} = \frac{px}{2}$$

3. Total pressure at equilibrium is:

$$P_T = p(1-x) + px + \frac{px}{2} = p\left(1 + \frac{x}{2}\right)$$

4. Using the expression shown in the solution,

$$K_p = \frac{(P_{X_2})\left(P_{\frac{1}{2}Y_2}\right)}{P_{X_2Y}}$$

5. Substitute the equilibrium pressures:

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

6. For small $x$,

$$1-x \approx 1$$

Hence,

$$K_p = \frac{px^2}{2}$$

7. Solving for $x$,

$$x = \sqrt{\frac{2K_p}{p}}$$

Therefore, the correct relationship is $x = \sqrt{\frac{2K_p}{p}}$, so the correct option is B.