If the equilibrium constant for the reaction `2AB Leftrightarrow A_(2)+B_(2)` is 36. What is the equilibrium constant for `AB Leftrightarrow 1/2 A_(2)+1/2 B_(2)`

Options

A. 49
B. 24
C. 6
D. 2

Answer

To find the equilibrium constant for the reaction \( AB \leftrightarrow \frac{1}{2} A_2 + \frac{1}{2} B_2 \) given that the equilibrium constant for the reaction \( 2AB \leftrightarrow A_2 + B_2 \) is 36, we can follow these steps:

### Step 1: Write the given equilibrium reaction and its constant
The given reaction is:
\[ 2AB \leftrightarrow A_2 + B_2 \]
The equilibrium constant for this reaction is:
\[ K_1 = 36 \]

### Step 2: Relate the new reaction to the given reaction
We need to find the equilibrium constant for the reaction:
\[ AB \leftrightarrow \frac{1}{2} A_2 + \frac{1}{2} B_2 \]
Notice that this new reaction can be derived from the original reaction by dividing all coefficients by 2.

### Step 3: Apply the property of equilibrium constants
According to the properties of equilibrium constants, if we multiply the entire reaction by a factor \( n \), the equilibrium constant \( K \) is raised to the power of \( n \). In our case, since we are dividing the coefficients by 2, we can express this mathematically as:
\[ K_2 = K_1^{\frac{1}{2}} \]

### Step 4: Substitute the value of \( K_1 \)
Now substitute the value of \( K_1 \):
\[ K_2 = 36^{\frac{1}{2}} \]

### Step 5: Calculate \( K_2 \)
Calculating \( K_2 \):
\[ K_2 = \sqrt{36} = 6 \]

### Conclusion
Thus, the equilibrium constant for the reaction \( AB \leftrightarrow \frac{1}{2} A_2 + \frac{1}{2} B_2 \) is:
\[ K_2 = 6 \]