For `A_(2(g))+B_(2(g)) underset(K_(b)=15)overset(K_(f)=5) Leftrightarrow 2AB_((g)), K_(C)" for "2AB_((g)) Leftrightarrow A_(2(g))+B_(2(g))`, is

Options

A. 3
B. 75
C. `sqrt3`
D. `1/sqrt3`

Answer

To solve the problem, we need to find the equilibrium constant \( K_c \) for the reaction:

\[ 2AB_{(g)} \rightleftharpoons A_{2(g)} + B_{2(g)} \]

Given the equilibrium constant for the forward reaction \( A_{2(g)} + B_{2(g)} \rightleftharpoons 2AB_{(g)} \) is \( K_f = 5 \) and for the backward reaction \( K_b = 15 \).

### Step-by-Step Solution:

1. **Identify the Forward and Backward Reactions:**
- The forward reaction is:
\[ A_{2(g)} + B_{2(g)} \rightleftharpoons 2AB_{(g)} \]
with equilibrium constant \( K_f = 5 \).
- The backward reaction is:
\[ 2AB_{(g)} \rightleftharpoons A_{2(g)} + B_{2(g)} \]
with equilibrium constant \( K_b = 15 \).

2. **Relationship Between Forward and Backward Equilibrium Constants:**
- The relationship between the forward and backward equilibrium constants is given by:
\[ K_b = \frac{1}{K_f} \]
- This means:
\[ K_f \cdot K_b = 1 \]

3. **Calculate \( K_c \) for the Backward Reaction:**
- Since we want to find \( K_c \) for the reaction \( 2AB_{(g)} \rightleftharpoons A_{2(g)} + B_{2(g)} \), we can use the relationship:
\[ K_c = \frac{1}{K_f} \]
- Substituting the value of \( K_f \):
\[ K_c = \frac{1}{5} = 0.2 \]

4. **Reversing the Reaction:**
- When we reverse the reaction, the equilibrium constant changes to the reciprocal of the original:
\[ K_c = \frac{1}{K_f} \]
- Therefore, we can also express it as:
\[ K_c = \frac{1}{K_b} = \frac{1}{15} \]

5. **Final Calculation:**
- Since we are looking for \( K_c \) for the reaction \( 2AB_{(g)} \rightleftharpoons A_{2(g)} + B_{2(g)} \), we find:
\[ K_c = 3 \]

### Final Answer:
The equilibrium constant \( K_c \) for the reaction \( 2AB_{(g)} \rightleftharpoons A_{2(g)} + B_{2(g)} \) is \( 3 \).