Question

The equilibrium constants for the following two reactions are given below at T(K): \[ 2A(g) \leftrightarrow B(g) + C(g), \quad K = 16 \quad \text{and} \quad 2B(g) \leftrightarrow D(g), \quad K = 25. \] What is the value of the equilibrium constant (K) for the reaction given below at T(K)? \[ \frac{1}{2} A(g) \leftrightarrow B(g) \]

• A. 100
• B. 50
• C. 20
• D. 75

Hint:

Combining equilibrium constants involves adjusting the reactions and constants based on their stoichiometry. When halving a reaction, the equilibrium constant is the square root of the original constant.

Answer 1

The Correct Option is C

To solve this, we need to use the given equilibrium constants and manipulate them based on the stoichiometry of the reactions. We will combine the reactions and equilibrium constants as needed. 

Step 1: Write the given reactions and their equilibrium constants We are given the following two reactions: 1. \( 2A(g) \leftrightarrow B(g) + C(g), \quad K_1 = 16 \) 2. \( 2B(g) \leftrightarrow D(g), \quad K_2 = 25 \) Now, the reaction we need to analyze is: \[ \frac{1}{2} A(g) \leftrightarrow B(g) \] 

Step 2: Manipulate the given reactions to match the desired reaction We can manipulate the first reaction \( 2A(g) \leftrightarrow B(g) + C(g) \) to match the desired reaction. To do this, divide the entire equation by 2: \[ A(g) \leftrightarrow \frac{1}{2}B(g) + \frac{1}{2}C(g) \] Since dividing the entire reaction by 2 will square the equilibrium constant, we now have: \[ K_3 = \sqrt{K_1} = \sqrt{16} = 4 \] 

Step 3: Calculate the equilibrium constant for the desired reaction Now, the desired reaction is \( \frac{1}{2}A(g) \leftrightarrow B(g) \), and since this is just the first reaction without the extra factor of \( C(g) \), the equilibrium constant for the desired reaction is \( K_3 \times K_2 \). That is: \[ K = K_3 \times K_2 = 4 \times 5 = 20 \] Thus, the equilibrium constant for the reaction \( \frac{1}{2} A(g) \leftrightarrow B(g) \) is: \[ \boxed{20} \]