In an experiment, at a total of 10 atmospheres and `400^(@)C`, in the equilibrium mixture `2NH_(3)(g) hArr N_(2)(g)+3H_(2)(g)` the ammonia was found to have dissociated to an extent of 96%. The `K_(p)` for the reaction will be

Options

A. 11880
B. 71432
C. 23320
D. 32456

Explanation

To solve the problem of finding the equilibrium constant \( K_p \) for the reaction

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

at a total pressure of 10 atmospheres and a temperature of \( 400^\circ C \) with ammonia dissociating to an extent of 96%, follow these steps:

### Step 1: Define Initial Conditions
Assume we start with 2 moles of \( NH_3 \) (since the stoichiometry of the reaction shows that 2 moles of ammonia dissociate). The initial amounts of \( N_2 \) and \( H_2 \) are both 0.

- Initial moles:
- \( NH_3 = 2 \)
- \( N_2 = 0 \)
- \( H_2 = 0 \)

### Step 2: Calculate Change in Moles
Given that ammonia dissociates to an extent of 96%, we can calculate the change in moles:

- \( \alpha = 0.96 \) (extent of dissociation)
- Moles of \( NH_3 \) that dissociate = \( 2 \times 0.96 = 1.92 \)
- Moles of \( N_2 \) formed = \( \frac{1.92}{2} = 0.96 \)
- Moles of \( H_2 \) formed = \( 3 \times 0.96 = 2.88 \)

### Step 3: Calculate Equilibrium Moles
Now, we can find the equilibrium moles:

- Equilibrium moles:
- \( NH_3 = 2 - 1.92 = 0.08 \)
- \( N_2 = 0 + 0.96 = 0.96 \)
- \( H_2 = 0 + 2.88 = 2.88 \)

### Step 4: Calculate Total Moles at Equilibrium
Total moles at equilibrium can be calculated as:

\[ \text{Total moles} = 0.08 + 0.96 + 2.88 = 3.92 \]

### Step 5: Calculate Partial Pressures
To find the partial pressures, we need the mole fractions of each component multiplied by the total pressure (10 atm):

- Partial pressure of \( N_2 \):
\[ P_{N_2} = \left( \frac{0.96}{3.92} \right) \times 10 = 2.45 \text{ atm} \]

- Partial pressure of \( H_2 \):
\[ P_{H_2} = \left( \frac{2.88}{3.92} \right) \times 10 = 7.36 \text{ atm} \]

- Partial pressure of \( NH_3 \):
\[ P_{NH_3} = \left( \frac{0.08}{3.92} \right) \times 10 = 0.20 \text{ atm} \]

### Step 6: Calculate \( K_p \)
Using the expression for \( K_p \):

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

Substituting the values:

\[ K_p = \frac{(2.45) \cdot (7.36)^3}{(0.20)^2} \]

Calculating \( K_p \):

1. Calculate \( (7.36)^3 = 396.14 \)
2. Substitute back into the equation:
\[ K_p = \frac{(2.45) \cdot (396.14)}{(0.04)} = \frac{970.78}{0.04} = 24269.5 \text{ atm}^2 \]

### Final Result
Thus, the value of \( K_p \) is approximately:

\[ K_p \approx 24269.5 \text{ atm}^2 \]