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 solve this problem, we need to find the correct relationship between the degree of dissociation \(x\) of \(X_2Y(g)\) and its equilibrium constant \(K_p\). Let's analyze the given reaction:

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

1. Let the initial pressure of \(X_2Y(g)\) be \(p\).
2. At equilibrium, let the degree of dissociation of \(X_2Y(g)\) be \(x\). Thus, the pressures at equilibrium are:
   • For \(X_2Y(g)\): \(p(1-x)\)
   • For \(X_2(g)\): \(px\)
   • For \(\frac{1}{2} Y_2(g)\): \(\frac{px}{2}\)
3. Using these expressions, the total pressure at equilibrium \(P_T\) is:
   • \(P_T = p(1-x) + px + \frac{px}{2} = p(1 + \frac{x}{2})\)
4. The equilibrium constant, \(K_p\), is defined as:
   • \(K_p = \frac{(P_{X_2})(P_{\frac{1}{2}Y_2})}{P_{X_2Y}}\)
   • Substitute the equilibrium pressures:
   • \(K_p = \frac{(px)(\frac{px}{2})}{p(1-x)}\)
   • \(K_p = \frac{p^2 x^2 }{2p(1-x)}\)
5. Rearranging for \(x\) gives:
   • \(K_p = \frac{px^2}{2(1-x)}\)
   • Assuming \(x\) is small, \(1 - x \approx 1\).
   • \(K_p = \frac{px^2}{2}\)
   • Solving for \(x\), we get:
   • \(x = \sqrt{\frac{2K_p}{p}}\)
6. Hence, the correct relationship between \(x\) and \(K_p\) is:
   • \(x = \sqrt{\frac{2K_p}{p}}\)

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

Answer 2

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 \( x = \sqrt{\frac{2K_p}{p}} \).