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

Answer: (D)

$K_2 = \frac{1}{\sqrt{K_1}}$

Answer 2

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

Therefore, the relationship between the equilibrium constants is:

$K_{2}=\frac{1}{\sqrt{K_{1}}}$