Question

The equilibrium constants for the following two reactions at 298k are given below:
2A ⇋ B+C; K₁ =16
2B + C ⇋ 2X; K₂ = 25
What is the value of k for the reaction, \(A+\frac{1}{2}B \rightleftharpoons X\) at 298K?

• A. \(\frac{1}{5}\)
• B. \(\frac{1}{40}\)
• C. \(\frac{5}{4}\)
• D. \(\frac{4}{5}\)
• E. 20

Answer 1

Answer: (E)

We are given the following two reactions with their equilibrium constants:
1. \( 2A \rightleftharpoons B + C \), with \( K_1 = 16 \) 2. \( 2B + C \rightleftharpoons 2X \), with \( K_2 = 25 \)
We need to determine the equilibrium constant \( K \) for the reaction: \[ A + \frac{1}{2} B \rightleftharpoons X \]Step 1: Manipulate the given reactions
The first reaction is \( 2A \rightleftharpoons B + C \) with \( K_1 = 16 \). - Dividing this reaction by 2 gives: \[ A \rightleftharpoons \frac{1}{2} B + \frac{1}{2} C \] The equilibrium constant for this modified reaction will be \( \sqrt{K_1} \), so: \[ K_1' = \sqrt{K_1} = \sqrt{16} = 4 \]
The second reaction is \( 2B + C \rightleftharpoons 2X \) with \( K_2 = 25 \).
Dividing this reaction by 2 gives: \[ B + \frac{1}{2} C \rightleftharpoons X \] The equilibrium constant for this modified reaction will be \( \sqrt{K_2} \), so: \[ K_2' = \sqrt{K_2} = \sqrt{25} = 5 \]Step 2: Combine the reactions Now, to get the desired reaction \( A + \frac{1}{2} B \rightleftharpoons X \), we combine the reactions: - The first modified reaction: \( A \rightleftharpoons \frac{1}{2} B + \frac{1}{2} C \) with \( K_1' = 4 \)
The second modified reaction: \( B + \frac{1}{2} C \rightleftharpoons X \) with \( K_2' = 5 \) The equilibrium constant for the overall reaction is the product of the individual equilibrium constants: \[ K = K_1' \times K_2' = 4 \times 5 = 20 \]

The correct option is (E) : \(20\)