Question

Consider the reaction: $$ X_2Y(g) \rightleftharpoons X_2(g) + \frac{1}{2} Y_2(g) $$ The equation representing the correct relationship between the degree of dissociation \( x \) of \( X_2Y(g) \) with its equilibrium constant \( K_p \) is: 

• A. \( x = \frac{2K_p}{p} \)
• B. \( x = \sqrt{\frac{2K_p}{p}} \)
• C. \( x = \frac{K_p}{2p} \)
• D. \( x = \sqrt{\frac{K_p}{p}} \)

Hint:

For reactions with small degrees of dissociation, the equilibrium constant can be used to relate the degree of dissociation to the partial pressures of the products and reactants.

Answer 1

The Correct Option is B

The equilibrium constant \( K_p \) for the reaction is given by: \[ K_p = \frac{p_{X_2} \cdot p_{Y_2}^{1/2}}{p_{X_2Y}}. \] Using the degree of dissociation \( x \) for the reactant \( X_2Y \), the partial pressures at equilibrium are: \[ p_{X_2Y} = p_0 - x \cdot p_0, \quad p_{X_2} = x \cdot p_0, \quad p_{Y_2} = \frac{x \cdot p_0}{2}. \] Substituting these into the expression for \( K_p \) and simplifying for small \( x \), we get: \[ x = \sqrt{\frac{2K_p}{p}}. \] Thus, the correct answer is \( x = \sqrt{\frac{2K_p}{p}} \).