Question

At a given temperature and pressure, the equilibrium constant values for the equilibria are given below:
$ 3A_2 + B_2 \rightleftharpoons 2A_3B, \, K_1 $
$ A_3B \rightleftharpoons \frac{3}{2}A_2 + \frac{1}{2}B_2, \, K_2 $
The relation between $ K_1 $ and $ K_2 $ is:

• A. \( K_1^2 = 2K_2 \)
• B. \( K_2 = \frac{K_1}{2} \)
• C. \( K_1 = \frac{1}{\sqrt{K_2}} \)
• D. \( K_2 = \frac{1}{\sqrt{K_1}} \)

Answer 1

The Correct Option is D

The second equilibrium reaction is the reverse of the first reaction divided by 2.

Therefore, the relationship between the equilibrium constants is:

\[K_{2}=\frac{1}{\sqrt{K_{1}}}\]