At `207^@C`, for the reaction :
`N_2(g) + 3H_2(g) hArr 2NH_3(g)`
it is found that at equilibrium, the mixture contains 52% of ammonia. If the pressure of the mixture is 10 atm, calculate `K_p` and `K_c` for the reaction.

Explanation

To solve the problem of calculating \( K_p \) and \( K_c \) for the reaction

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

given that at equilibrium, the mixture contains 52% ammonia and the total pressure is 10 atm, we can follow these steps:

### Step 1: Determine the partial pressures of the gases

Since the mixture contains 52% ammonia, we can find the partial pressure of ammonia (\( P_{NH_3} \)) and the partial pressures of nitrogen (\( P_{N_2} \)) and hydrogen (\( P_{H_2} \)).

\[ P_{NH_3} = 0.52 \times 10 \, \text{atm} = 5.2 \, \text{atm} \]

The remaining percentage of the gases is 48%, which is composed of nitrogen and hydrogen. The reaction shows that for every 2 moles of ammonia produced, 1 mole of nitrogen and 3 moles of hydrogen are consumed.

Let \( x \) be the total pressure of nitrogen and hydrogen combined:

\[ P_{N_2} + P_{H_2} = 0.48 \times 10 \, \text{atm} = 4.8 \, \text{atm} \]

### Step 2: Establish the relationship between the partial pressures

From the stoichiometry of the reaction, we can express the partial pressures of nitrogen and hydrogen in terms of a variable \( y \):

Let \( P_{N_2} = y \) and \( P_{H_2} = 3y \) (since for every 1 mole of \( N_2 \), 3 moles of \( H_2 \) are required).

Thus, we can write:

\[ y + 3y = 4.8 \, \text{atm} \]

\[ 4y = 4.8 \, \text{atm} \]

\[ y = 1.2 \, \text{atm} \]

So,

\[ P_{N_2} = 1.2 \, \text{atm} \quad \text{and} \quad P_{H_2} = 3 \times 1.2 \, \text{atm} = 3.6 \, \text{atm} \]

### Step 3: Calculate \( K_p \)

The equilibrium constant \( K_p \) for the reaction is given by the expression:

\[ K_p = \frac{(P_{NH_3})^2}{(P_{N_2})(P_{H_2})^3} \]

Substituting the values we found:

\[ K_p = \frac{(5.2)^2}{(1.2)(3.6)^3} \]

Calculating the values:

\[ K_p = \frac{27.04}{1.2 \times 46.656} = \frac{27.04}{55.9872} \approx 0.483 \]

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

To convert \( K_p \) to \( K_c \), we use the relationship:

\[ K_p = K_c(RT)^{\Delta n} \]

Where:
- \( R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \)
- \( T = 207 + 273 = 480 \, \text{K} \)
- \( \Delta n = n_{products} - n_{reactants} = 2 - (1 + 3) = -2 \)

Thus,

\[ K_p = K_c (0.0821 \times 480)^{-2} \]

Calculating \( RT \):

\[ RT = 0.0821 \times 480 \approx 39.408 \]

Now substituting back:

\[ 0.483 = K_c (39.408)^{-2} \]

Calculating \( (39.408)^{-2} \):

\[ (39.408)^{-2} \approx 0.000645 \]

Now, substituting to find \( K_c \):

\[ K_c = \frac{0.483}{0.000645} \approx 748.84 \]

### Final Answers:

\[ K_p \approx 0.483 \quad \text{and} \quad K_c \approx 748.84 \]