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

To solve this problem, we need to find the degree of dissociation, $x$, in terms of the equilibrium constant $K_p$ and the total pressure $p$ for the given reaction.

1. Understanding the Equilibrium:
The reaction is: $AB_{2(g)} \rightleftharpoons AB_{(g)} + \frac{1}{2} B_{2(g)}$. We will use an ICE table to relate the initial pressure, the change in pressure due to dissociation, and the equilibrium pressures.

2. Setting Up the ICE Table:
Assume initial pressure of $AB_2$ is $p_0$.

| | $AB_{2(g)}$ | $AB_{(g)}$ | $B_{2(g)}$ | | :---------- | :---------- | :-------- | :---------------- | | 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 Pressure at Equilibrium:
The total pressure $p$ is the sum of the partial pressures at equilibrium: $p = p_0(1-x) + xp_0 + \frac{1}{2}xp_0 = p_0(1 + \frac{1}{2}x)$
Since $x \ll 1$, $p \approx p_0$.

4. Expressing $K_p$ in Terms of Partial Pressures:
$K_p = \frac{P_{AB} \cdot (P_{B_2})^{1/2}}{P_{AB_2}} = \frac{(xp_0) \cdot (\frac{1}{2}xp_0)^{1/2}}{p_0(1-x)}$

5. Simplifying the Expression:
Since $x \ll 1$, we can approximate $(1-x) \approx 1$: $K_p \approx \frac{(xp_0) \cdot (\frac{1}{\sqrt{2}}\sqrt{x}\sqrt{p_0})}{p_0} = \frac{x^{3/2}p_0^{3/2}}{\sqrt{2}p_0} = \frac{x^{3/2}\sqrt{p_0}}{\sqrt{2}}$
Since $p \approx p_0$, $K_p \approx \frac{x^{3/2}\sqrt{p}}{\sqrt{2}}$.

6. Solving for $x$:
$x^{3/2} = \frac{\sqrt{2}K_p}{\sqrt{p}}$
$x = \left(\frac{\sqrt{2}K_p}{\sqrt{p}}\right)^{2/3} = \left(\frac{2K_p^2}{p}\right)^{1/3} = \sqrt[3]{\frac{2K_p^2}{p}}$

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

Answer 2

Step 1: Write the given equilibrium reaction.
The given dissociation equilibrium is:
\[ AB_2 \rightleftharpoons A + \frac{1}{2}B_2 \]
Let the initial pressure of \( AB_2 \) be \( p \), and the degree of dissociation be \( x \) (where \( x \) is small compared to unity).

Step 2: Express partial pressures at equilibrium.
At equilibrium:
- Pressure of \( AB_2 = p(1 - x) \)
- Pressure of \( A = px \)
- Pressure of \( B_2 = \frac{1}{2}px \)
Total pressure at equilibrium:
\[ p_{total} = p(1 - x) + px + \frac{1}{2}px = p\left(1 + \frac{x}{2}\right) \]

Step 3: Write the expression for \( K_p \).
The equilibrium constant in terms of partial pressures is:
\[ K_p = \frac{(p_A)(p_{B_2})^{1/2}}{p_{AB_2}} \] Substitute the values:
\[ K_p = \frac{(px) \left(\frac{1}{2}px\right)^{1/2}}{p(1 - x)} \] Simplify assuming \( x \) is small, so \( (1 - x) \approx 1 \):
\[ K_p = p^{1/2}x^{3/2}\frac{1}{\sqrt{2}} \] \[ K_p = \frac{p^{1/2}x^{3/2}}{\sqrt{2}} \]

Step 4: Solve for \( x \).
Rearranging the above equation:
\[ x^{3/2} = K_p \sqrt{\frac{2}{p}} \] Raise both sides to the power of \( \frac{2}{3} \):
\[ x = \left(\frac{2K_p^2}{p}\right)^{1/3} \]

Final Answer:
\[ \boxed{x = \sqrt[3]{\frac{2K_p^2}{p}}} \]