`K_(c)` for the reaction `N_(2)+3H_(2) hArr 2NH_(3)` is `0.5 mol^(-2) L^(2)` at `400 K`. Find `K_(p)`.
Given `R=0.082 L-"atm K"^(-1) mol^(-1)`

Explanation

To find \( K_p \) for the reaction \( N_2 + 3H_2 \rightleftharpoons 2NH_3 \) given \( K_c = 0.5 \, \text{mol}^{-2} \, \text{L}^2 \) at \( 400 \, K \), we can follow these steps:

### Step 1: Write the reaction and identify the components
The reaction is:
\[ N_2 + 3H_2 \rightleftharpoons 2NH_3 \]

### Step 2: Calculate \( \Delta N_g \)
Calculate the change in the number of moles of gas (\( \Delta N_g \)):
- Products: \( 2 \, \text{moles of } NH_3 \)
- Reactants: \( 1 \, \text{mole of } N_2 + 3 \, \text{moles of } H_2 = 4 \, \text{moles} \)

Thus,
\[ \Delta N_g = \text{moles of products} - \text{moles of reactants} = 2 - 4 = -2 \]

### Step 3: Use the formula to find \( K_p \)
The relationship between \( K_p \) and \( K_c \) is given by the formula:
\[ K_p = K_c \cdot R^T \cdot (RT)^{\Delta N_g} \]

### Step 4: Substitute the values
We know:
- \( K_c = 0.5 \, \text{mol}^{-2} \, \text{L}^2 \)
- \( R = 0.082 \, \text{L} \cdot \text{atm} \cdot \text{K}^{-1} \cdot \text{mol}^{-1} \)
- \( T = 400 \, K \)
- \( \Delta N_g = -2 \)

Substituting these values into the equation:
\[ K_p = K_c \cdot R^T \cdot (RT)^{-2} \]

Calculating \( R \cdot T \):
\[ R \cdot T = 0.082 \, \text{L} \cdot \text{atm} \cdot \text{K}^{-1} \cdot \text{mol}^{-1} \cdot 400 \, K = 32.8 \, \text{L} \cdot \text{atm} \cdot \text{mol}^{-1} \]

Now, substituting into the equation:
\[ K_p = 0.5 \cdot (32.8)^{-2} \]

Calculating \( (32.8)^{-2} \):
\[ (32.8)^{-2} = \frac{1}{32.8^2} = \frac{1}{1075.84} \approx 0.000928 \]

Thus,
\[ K_p \approx 0.5 \cdot 0.000928 \approx 0.000464 \]

### Step 5: Final result
Therefore,
\[ K_p \approx 4.648 \times 10^{-4} \, \text{atm}^{-2} \]