Question

At temperature T, compound AB2 dissociates as \( AB_2 \rightleftharpoons A + \frac{1}{2} B_2 \), having degree of dissociation \( x \) (small compared to unity). The correct expression for \( x \) in terms of \( K_p \) and \( p \) is:

• A. \( \frac{p}{K_p p} \)
• B. \( \frac{2 K_p}{p} \)
• C. $\sqrt[3]{\frac{2 K_p^2}{p}}$
• D. \( \frac{2 K_p^2}{p} \)

Hint:

For dissociation reactions, the degree of dissociation can be found using the equilibrium constant and the initial concentration or pressure of the species.

Answer 1

The Correct Option is C

The objective is to determine the degree of dissociation, denoted by $x$, as a function of the equilibrium constant $K_p$ and the total pressure $p$ for the specified reaction.

1. Equilibrium Analysis:
The reaction in question is $AB_{2(g)} \rightleftharpoons AB_{(g)} + \frac{1}{2} B_{2(g)}$. An ICE (Initial, Change, Equilibrium) table will be employed to establish relationships between initial pressures, pressure changes due to dissociation, and equilibrium partial pressures.

2. ICE Table Construction:
Let the initial pressure of $AB_2$ be $p_0$.

| | $AB_{2(g)}$ | $AB_{(g)}$ | $B_{2(g)}$ | | :---------- | :---------- | :-------- | :---------------- | | Initial | $p_0$ | $0$ | $0$ | | Change | $-xp_0$ | $+xp_0$ | $+\frac{1}{2}xp_0$ | | Equilibrium | $p_0(1-x)$ | $xp_0$ | $\frac{1}{2}xp_0$ |

3. Total Equilibrium Pressure:
The total pressure $p$ at equilibrium is the sum of all partial pressures: $p = p_0(1-x) + xp_0 + \frac{1}{2}xp_0 = p_0(1 + \frac{1}{2}x)$
Given that $x \ll 1$, we can approximate $p \approx p_0$.

4. $K_p$ Expression in Partial Pressures:
The equilibrium constant $K_p$ is defined as: $K_p = \frac{P_{AB} \cdot (P_{B_2})^{1/2}}{P_{AB_2}} = \frac{(xp_0) \cdot (\frac{1}{2}xp_0)^{1/2}}{p_0(1-x)}$

5. Expression Simplification:
Applying the approximation $(1-x) \approx 1$ for $x \ll 1$: $K_p \approx \frac{(xp_0) \cdot (\frac{1}{\sqrt{2}}\sqrt{x}\sqrt{p_0})}{p_0} = \frac{x^{3/2}p_0^{3/2}}{\sqrt{2}p_0} = \frac{x^{3/2}\sqrt{p_0}}{\sqrt{2}}$
Using the approximation $p \approx p_0$, we get $K_p \approx \frac{x^{3/2}\sqrt{p}}{\sqrt{2}}$.

6. Solving for $x$:
Rearranging the simplified $K_p$ expression to solve for $x$: $x^{3/2} = \frac{\sqrt{2}K_p}{\sqrt{p}}$
Raising both sides to the power of $2/3$: $x = \left(\frac{\sqrt{2}K_p}{\sqrt{p}}\right)^{2/3} = \left(\frac{2K_p^2}{p}\right)^{1/3} = \sqrt[3]{\frac{2K_p^2}{p}}$

Final Result:
The degree of dissociation $x$ is expressed as $\sqrt[3]{\frac{2 K_p^2}{p}}$.