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) : \(20\)

Answer 2

We are given:

• Reaction 1: 2A ⇌ B + C ; K₁ = 16
• Reaction 2: 2B + C ⇌ 2X ; K₂ = 25

We are asked to find the equilibrium constant K for the reaction:

A + ½B ⇌ X

Let’s manipulate the given reactions to obtain the target reaction. 

Step 1: Reverse Reaction 1 B + C ⇌ 2A ⇒ New K = 1/K₁ = 1/16

 Step 2: Take half of Reaction 2 B + 0.5C ⇌ X ⇒ New K = √K₂ = √25 = 5 

 Step 3: Add the two adjusted reactions:

• From Step 1: B + C ⇌ 2A
• From Step 2: B + 0.5C ⇌ X

Now subtract Step 1 from Step 2: → (B + 0.5C ⇌ X) − (B + C ⇌ 2A) = (−0.5C − 2A ⇌ X) which gives: A + 0.5B ⇌ X (multiply both sides by −1) So, total equilibrium constant:

K = (√K₂)/(K₁) = 5 / 16 = 5/16

But our desired reaction is A + ½B ⇌ X ⇒ Final K = 5/16 However, this value is not among the given options. Let’s double-check with a different route. 

 Alternate Method: Combine reactions directly to form A + ½B ⇌ X Start with:

• 2A ⇌ B + C ; K₁ = 16
• 2B + C ⇌ 2X ; K₂ = 25

Add the two:

• 2A + 2B + C ⇌ B + C + 2X
• Simplify: 2A + B ⇌ 2X

Divide entire equation by 2:

A + ½B ⇌ X

Then the new K = √(K₁ × K₂) = √(16 × 25) = √400 = 20 --- Correct Answer: 20