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. \( \frac{2 K_p}{p} \)
• 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

For the dissociation equilibrium \( AB_2 \rightleftharpoons A + \frac{1}{2} B_2 \), the degree of dissociation \( x \) can be expressed as: \[ K_p = \frac{[A][B_2]^{1/2}}{[AB_2]} \] By applying the equilibrium expressions and assuming \( x \) is small, we get the correct expression for \( x \) as \( \frac{2 K_p}{p} \). 
Thus, the correct expression is \( \frac{2 K_p}{p} \).