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

To find the relation between the equilibrium constants \( K_1 \) and \( K_2 \), let's analyze the given equilibrium reactions.

The first equilibrium given is:

\(3A_2 + B_2 \rightleftharpoons 2A_3B, \, K_1\) 

This means that for the above reaction, the equilibrium constant is \( K_1 \).

The second equilibrium is the reverse of this to some extent:

\(A_3B \rightleftharpoons \frac{3}{2}A_2 + \frac{1}{2}B_2, \, K_2\)

This reaction is essentially the reverse of the formation of \( A_3B \) with half of the quantities from the initial equilibrium.

To relate \( K_1 \) and \( K_2 \), let's write out their expressions:

• For the reaction \( 3A_2 + B_2 \rightleftharpoons 2A_3B \), the equilibrium constant \( K_1 \) is given by: \(K_1 = \frac{{[A_3B]^2}}{{[A_2]^3[B_2]}}\)
• For the reaction \( A_3B \rightleftharpoons \frac{3}{2}A_2 + \frac{1}{2}B_2 \), the equilibrium constant \( K_2 \) is given by: \(K_2 = \frac{{[A_2]^{\frac{3}{2}}[B_2]^{\frac{1}{2}}}}{{[A_3B]}}\)

Since the second reaction is the reverse half of the first, we calculate the relation using their inverses and powers:

The inverse of the first equilibrium (the reverse reaction) would have the constant \( \frac{1}{K_1} \). Since the coefficients are halved in the second reaction, the constants' relation involves a square root due to the balance of moles:

• \( K_2 \) is the equilibrium constant for a reaction which is half of the reverse of the \( K_1 \) reaction. It follows: \(K_2 = \sqrt{\frac{1}{K_1}}\)

The correct relationship can be expressed as: \(K_2 = \frac{1}{\sqrt{K_1}}\)

Therefore, the correct option is: \(K_2 = \frac{1}{\sqrt{K_1}}\)