Equilibrium constant for the reaction , `N_(2) + 3 H_(2) hArr 2 NH_(3) ` is K, then equilibrium constant for the reaction, `NH_(3) hArr 1/2 N_(2) + 3/2 H_(2) ` will be ` ……….. .`

Answer

To find the equilibrium constant for the reaction \( \text{NH}_3 \rightleftharpoons \frac{1}{2} \text{N}_2 + \frac{3}{2} \text{H}_2 \) given that the equilibrium constant for the reaction \( \text{N}_2 + 3 \text{H}_2 \rightleftharpoons 2 \text{NH}_3 \) is \( K \), we can follow these steps:

### Step 1: Write the original reaction and its equilibrium constant
The original reaction is:
\[ \text{N}_2 + 3 \text{H}_2 \rightleftharpoons 2 \text{NH}_3 \]
The equilibrium constant for this reaction is given as \( K \).

### Step 2: Reverse the original reaction
When we reverse the reaction, we get:
\[ 2 \text{NH}_3 \rightleftharpoons \text{N}_2 + 3 \text{H}_2 \]
The equilibrium constant for the reversed reaction is the reciprocal of \( K \):
\[ K' = \frac{1}{K} \]

### Step 3: Adjust the coefficients
Now, we need to adjust the coefficients to match the desired reaction:
\[ \text{NH}_3 \rightleftharpoons \frac{1}{2} \text{N}_2 + \frac{3}{2} \text{H}_2 \]
To do this, we divide the entire equation by 2:
\[ \text{NH}_3 \rightleftharpoons \frac{1}{2} \text{N}_2 + \frac{3}{2} \text{H}_2 \]

### Step 4: Calculate the new equilibrium constant
When we divide the coefficients of a balanced equation by a factor, we raise the equilibrium constant to the power of that factor. Since we divided by 2, we take the square root of the equilibrium constant from the reversed reaction:
\[ K'' = (K')^{\frac{1}{2}} = \left(\frac{1}{K}\right)^{\frac{1}{2}} = \frac{1}{\sqrt{K}} \]

### Final Answer
Thus, the equilibrium constant for the reaction \( \text{NH}_3 \rightleftharpoons \frac{1}{2} \text{N}_2 + \frac{3}{2} \text{H}_2 \) is:
\[ \frac{1}{\sqrt{K}} \]