Question

The dissociation equilibrium of a gas $ AB_2 $ can be represented as $2 A B_2(g) \rightleftharpoons 2 A B (g) + B_2 (g) $ The degree of dissociation is $x$ and is small compared to $1$. The expression relating the degree of dissociation (x) with equilibrium constant $K_p$ and total pressure $p$ is

• A. $ (2K_p/p) $
• B. $ (2K_p/p)^{1/3} $
• C. $ (2K_p/p)^{1/2} $
• D. $ (K_p/p) $

Answer 1

The Correct Option is B

To solve the problem of finding the degree of dissociation x for the equilibrium reaction: 2AB_2(g) \rightleftharpoons 2AB(g) + B_2(g), we will follow these steps.

1. Initial and Equilibrium Moles:
   • Suppose we start with 2 moles of AB_2.
   • Let the degree of dissociation be x, which means at equilibrium:
     • 2(1 - x) moles of AB_2 remain.
     • 2x moles of AB are formed.
     • x moles of B_2 are formed.
2. Total Moles at Equilibrium:
   • Total = (2(1-x) + 2x + x = 2 + x)
3. Partial Pressures:
   • The partial pressure of each gas can be found using Dalton's Law, P_i = \frac{\text{moles of } i}{\text{Total moles}} \times \text{Total pressure} (p).
   • Partial pressure of AB_2: \frac{2(1-x)}{2+x} \cdot p
   • Partial pressure of AB: \frac{2x}{2+x} \cdot p
   • Partial pressure of B_2: \frac{x}{2+x} \cdot p
4. Expression for Equilibrium Constant:
   • For the given reaction, the equilibrium constant K_p is given by: \[ K_p = \frac{(P_{AB})^2 \times P_{B_2}}{(P_{AB_2})^2} \]
   • Substitute the partial pressures: \[ K_p = \frac{\left( \frac{2x}{2+x} \cdot p \right)^2 \times \left( \frac{x}{2+x} \cdot p \right)}{\left( \frac{2(1-x)}{2+x} \cdot p \right)^2} \]
5. Simplifying the Expression:
   • After simplification, taking the assumption \( x \) is very small and 2+x \approx 2: \[ K_p \approx \frac{(2x^2 \cdot p^3)}{2^3 \cdot (1-x)^2 \cdot p^2} \] \[ K_p \approx \frac{8x^3 \cdot p}{8 \cdot p^2} \] \[ K_p \approx \frac{x^3 \cdot p}{p^2} \] \[ x^3 \approx \frac{K_p}{p} \]
   • Solve for x: \[ x = \left( \frac{2 K_p}{p} \right)^{1/3} \]

Thus, the correct expression relating the degree of dissociation with equilibrium constant K_p is \(\left( \frac{2K_p}{p} \right)^{1/3}\), which matches with the second option.