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

To establish the relationship between the degree of dissociation, denoted by \(x\), of \(X_2Y(g)\) and its equilibrium constant, \(K_p\), we must first examine the provided reaction: \(X_2Y(g) \rightleftharpoons X_2(g) + \frac{1}{2} Y_2(g)\).

1. Assume the initial pressure of \(X_2Y(g)\) is \(p\).
2. At equilibrium, if the degree of dissociation of \(X_2Y(g)\) is \(x\), the corresponding equilibrium pressures are:
   • For \(X_2Y(g)\): \(p(1-x)\)
   • For \(X_2(g)\): \(px\)
   • For \(\frac{1}{2} Y_2(g)\): \(\frac{px}{2}\)
3. The total equilibrium pressure, \(P_T\), can be calculated as:
   • \(P_T = p(1-x) + px + \frac{px}{2} = p(1 + \frac{x}{2})\)
4. The equilibrium constant, \(K_p\), is defined by the equation:
   • \(K_p = \frac{(P_{X_2})(P_{\frac{1}{2}Y_2})}{P_{X_2Y}}\)
   • Substituting the equilibrium pressures yields:
   • \(K_p = \frac{(px)(\frac{px}{2})}{p(1-x)}\)
   • \(K_p = \frac{p^2 x^2 }{2p(1-x)}\)
5. Rearranging the expression to solve for \(x\):
   • \(K_p = \frac{px^2}{2(1-x)}\)
   • Under the assumption that \(x\) is small, \(1 - x \approx 1\).
   • \(K_p = \frac{px^2}{2}\)
   • Solving for \(x\) produces:
   • \(x = \sqrt{\frac{2K_p}{p}}\)
6. Therefore, the established relationship between \(x\) and \(K_p\) is:
   • \(x = \sqrt{\frac{2K_p}{p}}\)

This derivation confirms that the correct answer is the equation \(x = \sqrt{\frac{2K_p}{p}}\).