At 800 K, the equilibrium constant for the reaction,
`N_2(g) + 3H_2(g) hArr 2NH_3(g)` is `6.05xx10^(-2)L^2 "mol"^(-2)`. Calculate `K_p` for the reaction at the same temperature.

Explanation

To calculate \( K_p \) for the reaction \( N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \) at 800 K, given that \( K_c = 6.05 \times 10^{-2} \, L^2 \, mol^{-2} \), we can use the relationship between \( K_p \) and \( K_c \):

### Step 1: Write the relationship between \( K_p \) and \( K_c \)

The relationship is given by the equation:
\[ K_p = K_c \cdot (R \cdot T)^{\Delta n_g} \]
where:
- \( R \) is the universal gas constant (0.0821 L·atm/(K·mol)),
- \( T \) is the temperature in Kelvin,
- \( \Delta n_g \) is the change in the number of moles of gas.

### Step 2: Calculate \( \Delta n_g \)

For the reaction:
- Reactants: \( N_2(g) + 3H_2(g) \) → Total moles = 1 + 3 = 4 moles
- Products: \( 2NH_3(g) \) → Total moles = 2 moles

Now, calculate \( \Delta n_g \):
\[ \Delta n_g = \text{moles of products} - \text{moles of reactants} = 2 - 4 = -2 \]

### Step 3: Substitute the values into the equation

Now we can substitute \( K_c \), \( R \), \( T \), and \( \Delta n_g \) into the equation:
\[ K_p = K_c \cdot (R \cdot T)^{\Delta n_g} \]
Substituting the values:
- \( K_c = 6.05 \times 10^{-2} \, L^2 \, mol^{-2} \)
- \( R = 0.0821 \, L \cdot atm/(K \cdot mol) \)
- \( T = 800 \, K \)
- \( \Delta n_g = -2 \)

Calculating \( R \cdot T \):
\[ R \cdot T = 0.0821 \cdot 800 = 65.68 \, L \cdot atm/mol \]

Now, substitute into the equation:
\[ K_p = 6.05 \times 10^{-2} \cdot (65.68)^{-2} \]

### Step 4: Calculate \( K_p \)

Calculating \( (65.68)^{-2} \):
\[ (65.68)^{-2} = \frac{1}{65.68^2} \approx 0.000230 \]

Now substituting back:
\[ K_p = 6.05 \times 10^{-2} \cdot 0.000230 \approx 1.39 \times 10^{-5} \, atm^{-2} \]

### Final Answer:
\[ K_p \approx 1.39 \times 10^{-5} \, atm^{-2} \]