`K_(p)` for the reaction `N_(2)+3H_(2) hArr 2NH_(3)` at `400^(@)C` is `1.64xx10^(-4) atm^(-2)`. Find `K_(c)`.

Explanation

To find \( K_c \) from the given \( K_p \) for the reaction \( N_2 + 3H_2 \rightleftharpoons 2NH_3 \), we can use the relationship between \( K_p \) and \( K_c \):

### Step-by-Step Solution:

1. **Write the Given Information:**
- \( K_p = 1.64 \times 10^{-4} \, \text{atm}^{-2} \)
- The reaction is \( N_2 + 3H_2 \rightleftharpoons 2NH_3 \).

2. **Determine \( \Delta n \):**
- \( \Delta n = \text{moles of gaseous products} - \text{moles of gaseous reactants} \)
- From the balanced equation:
- Moles of products (NH3) = 2
- Moles of reactants (N2 + 3H2) = 1 + 3 = 4
- Therefore, \( \Delta n = 2 - 4 = -2 \).

3. **Use the Relationship Between \( K_p \) and \( K_c \):**
- The formula is given by:
\[ K_p = K_c \cdot R^{\Delta n} \cdot T^{\Delta n} \]
- Rearranging gives:
\[ K_c = \frac{K_p}{R^{\Delta n} \cdot T^{\Delta n}} \]

4. **Substitute the Values:**
- Use \( R = 0.082 \, \text{L atm K}^{-1} \text{mol}^{-1} \).
- Convert the temperature from Celsius to Kelvin:
\[ T = 400 + 273 = 673 \, \text{K} \]
- Now substitute \( K_p \), \( R \), \( T \), and \( \Delta n \):
\[ K_c = \frac{1.64 \times 10^{-4}}{(0.082)^{-2} \cdot (673)^{-2}} \]

5. **Calculate \( R^{\Delta n} \) and \( T^{\Delta n} \):**
- Since \( \Delta n = -2 \):
\[ R^{\Delta n} = (0.082)^{-2} = \frac{1}{(0.082)^2} \]
\[ T^{\Delta n} = (673)^{-2} = \frac{1}{(673)^2} \]

6. **Final Calculation:**
- Combine the calculations:
\[ K_c = 1.64 \times 10^{-4} \cdot (0.082)^2 \cdot (673)^2 \]
- Calculate the numerical values:
\[ K_c \approx 0.5 \, \text{L}^2 \text{mol}^{-2} \]

### Final Answer:
\[ K_c \approx 0.5 \, \text{L}^2 \text{mol}^{-2} \]

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