For the reversible reaction
`N_(2)(g)+3H_(2)(g)rarrNH_(3)(g)`
at `500^(@)C` the value of `K_(p)` is `1.44xx10^(-5)` when partial pressure is measured in atmosphere. The corresponding value of `K_(e)` with concentration in mol/L is

Options

A. `(1.44 xx 10^(-7))/((0.082 xx 773)^(-2))`
B. `(1.44 xx 10^(-5))/((0.082 xx 773)^(-2))`
C. `(1.44 xx 10^(-5))/((8.314 xx 500)^(-2))`
D. `(1.44 xx 10^(-5))/((0.082 xx 500)^(-2))`

Answer

To find the value of \( K_c \) for the reaction

\[ N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \]

at \( 500^\circ C \) given that \( K_p = 1.44 \times 10^{-5} \), we will use the relationship between \( K_p \) and \( K_c \).

### Step-by-Step Solution:

1. **Identify the Reaction and Calculate \(\Delta n_g\)**:
- The balanced reaction is:
\[ N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \]
- Calculate the change in the number of moles of gas (\(\Delta n_g\)):
\[ \Delta n_g = \text{moles of products} - \text{moles of reactants} \]
- Moles of products = 2 (from \( 2NH_3 \))
- Moles of reactants = 1 (from \( N_2 \)) + 3 (from \( 3H_2 \)) = 4
- Thus,
\[ \Delta n_g = 2 - 4 = -2 \]

2. **Use the Relationship Between \( K_p \) and \( K_c \)**:
- The relationship is given by:
\[ K_p = K_c (RT)^{\Delta n_g} \]
- Rearranging gives:
\[ K_c = \frac{K_p}{(RT)^{\Delta n_g}} \]

3. **Substitute the Known Values**:
- Given \( K_p = 1.44 \times 10^{-5} \)
- The gas constant \( R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \)
- Convert the temperature to Kelvin:
\[ T = 500^\circ C + 273 = 773 \, \text{K} \]
- Now substitute into the equation:
\[ K_c = \frac{1.44 \times 10^{-5}}{(0.0821 \times 773)^{-2}} \]

4. **Calculate \( RT \)**:
- Calculate \( RT \):
\[ RT = 0.0821 \times 773 \approx 63.5 \]
- Now calculate \( (RT)^{-2} \):
\[ (63.5)^{-2} \approx 0.000248 \]

5. **Final Calculation of \( K_c \)**:
- Substitute back to find \( K_c \):
\[ K_c = \frac{1.44 \times 10^{-5}}{0.000248} \approx 5.81 \times 10^{-2} \]

### Final Answer:
\[ K_c \approx 5.81 \times 10^{-2} \, \text{mol/L} \]