Question

If equilibrium constant for the equation $ A_2 + B_2 \rightleftharpoons 2AB \quad \text{is} \, K_p, $ then find the equilibrium constant for the equation $ AB \rightleftharpoons \frac{1}{2} A_2 + \frac{1}{2} B_2. $

• A. \( \frac{1}{K_p} \)
• B. \( \frac{1}{\sqrt{K_p}} \)
• C. \( \sqrt{K_p} \)
• D. \( K_p^2 \)

Hint:

When adjusting the stoichiometry of a reaction, the equilibrium constant is affected by the power of the coefficient change.

Answer 1

The Correct Option is B

Step 1: Relate the two equilibrium constants.
For the first reaction, the equilibrium constant is given by:
\[ K_p = \frac{[AB]^2}{[A_2][B_2]}. \] Now, for the second reaction: \[ AB \rightleftharpoons \frac{1}{2} A_2 + \frac{1}{2} B_2, \] The equilibrium constant for this reaction, say \( K' \), can be written as: \[ K' = \frac{[A_2]^{1/2} [B_2]^{1/2}}{[AB]}. \] 
Step 2: Calculate the relationship between \( K_p \) and \( K' \).
Since the stoichiometry of the second reaction is half of that in the first, the equilibrium constant \( K' \) is related to \( K_p \) as: \[ K' = \frac{1}{\sqrt{K_p}}. \] Thus, the correct answer is \( \frac{1}{\sqrt{K_p}} \).