Question

At temperature T, compound \( {AB}_2 \) 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. \( \sqrt{K_p p} \)
• B. \( \frac{2K_p}{p} \)
• C. \( \frac{2K_p}{p} \)
• D. \( \frac{2K_p^2}{p} \)

Hint:

For dissociation reactions with small \( x \), the change in concentration can often be approximated as \( p - x p \approx p \), which simplifies the equilibrium expression and makes the calculation manageable.

Answer 1

The Correct Option is C

Let the initial amount of \( {AB}_2 \) be \( p \). Upon dissociation, the amount of \( {A} \) formed will be \( x p \), and the amount of \( {B}_2 \) formed will be \( \frac{x p}{2} \). The equilibrium expression for \( K_p \) is given by: \[ K_p = \frac{[{A}][{B}_2]^{1/2}}{[{AB}_2]} = \frac{x p \left(\frac{x p}{2}\right)^{1/2}}{p - x p} \] Simplifying the above expression, we find that: \[ x = \frac{2K_p}{p} \] Thus, the correct expression for \( x \) is \( \frac{2K_p}{p} \).