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:
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:
- \( K_1^2 = 2K_2 \)
- \( K_2 = \frac{K_1}{2} \)
- \( K_1 = \frac{1}{\sqrt{K_2}} \)
- \( K_2 = \frac{1}{\sqrt{K_1}} \)
The Correct Option is D
Solution and Explanation
The second equilibrium reaction is the inverse of the first reaction, raised to the power of one-half.
Thus, the equilibrium constants are related as follows:
\[K_{2}=\frac{1}{\sqrt{K_{1}}}\]
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