Given is a concentrated solution of a weak electrolyte $A_xB_y$ of concentration $c$ and dissociation constant $K$. The degree of dissociation is given by:

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:

Answer 1

Step 1: Understanding the Question:
We need to derive the relationship between the degree of dissociation ($\alpha$) and the equilibrium constant ($K$) for a general electrolyte $\text{A}_x\text{B}_y$.
Step 2: Key Formula or Approach:
\[ \text{A}_x\text{B}_y \rightleftharpoons x\text{A}^{y+} + y\text{B}^{x-} \]
Initial concentration: $c, 0, 0$.
At equilibrium: $c(1-\alpha), cx\alpha, cy\alpha$.
Step 3: Detailed Explanation:
1. Equilibrium constant expression:
\[ K = \frac{[A^{y+}]^x [B^{x-}]^y}{[A_x B_y]} = \frac{(cx\alpha)^x (cy\alpha)^y}{c(1-\alpha)} \]
2. For a weak electrolyte, $\alpha \ll 1$, so $1 - \alpha \approx 1$.
\[ K = c^x x^x \alpha^x \cdot c^y y^y \alpha^y / c = c^{x+y-1} x^x y^y \alpha^{x+y} \]
3. Solving for $\alpha$:
\[ \alpha^{x+y} = \frac{K}{c^{x+y-1} x^x y^y} \]
\[ \alpha = \left[ \frac{K}{c^{x+y-1} x^x y^y} \right]^{\frac{1}{x+y}} \]
Step 4: Final Answer:
The expression for $\alpha$ matches option (B).

Given is a concentrated solution of a weak electrolyte \(A_xB_y\) of concentration \(c\) and dissociation constant \(K\). The degree of dissociation is given by:

The Correct Option is A

Solution and Explanation

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