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, we will follow these steps:

### Step 1: Determine the initial moles of ammonia
We start with 2 moles of ammonia (NH₃) before the reaction begins.

### Step 2: Calculate the extent of dissociation
Given that ammonia dissociates to an extent of 96%, we can express this as:
\[ \text{Extent of dissociation} (\alpha) = 0.96 \]

### Step 3: Calculate the moles at equilibrium
- Moles of NH₃ that dissociated = \(2 \times 0.96 = 1.92\) moles
- Remaining moles of NH₃ = \(2 - 1.92 = 0.08\) moles
- According to the stoichiometry of the reaction:
- For every 2 moles of NH₃ that dissociate, 1 mole of N₂ and 3 moles of H₂ are produced.
- Moles of N₂ produced = \(1.92 / 2 = 0.96\) moles
- Moles of H₂ produced = \(1.92 \times \frac{3}{2} = 2.88\) moles

### Step 4: Total moles at equilibrium
Total moles at equilibrium = Moles of NH₃ + Moles of N₂ + Moles of H₂
\[ = 0.08 + 0.96 + 2.88 = 3.92 \text{ moles} \]

### Step 5: Calculate the partial pressures
Using the formula for partial pressure:
\[ P_i = \left(\frac{n_i}{n_{total}}\right) \times P_{total} \]
where \(P_{total} = 10 \text{ atm}\).

- Partial pressure of NH₃:
\[ P_{NH₃} = \left(\frac{0.08}{3.92}\right) \times 10 = 0.2041 \text{ atm} \]

- Partial pressure of N₂:
\[ P_{N₂} = \left(\frac{0.96}{3.92}\right) \times 10 = 2.4480 \text{ atm} \]

- Partial pressure of H₂:
\[ P_{H₂} = \left(\frac{2.88}{3.92}\right) \times 10 = 7.3469 \text{ atm} \]

### Step 6: Write the expression for Kp
The expression for \(K_p\) for the reaction:
\[ 2NH₃(g) \rightleftharpoons N₂(g) + 3H₂(g) \]
is given by:
\[ K_p = \frac{(P_{N₂})(P_{H₂})^3}{(P_{NH₃})^2} \]

### Step 7: Substitute the values into the Kp expression
Substituting the calculated partial pressures:
\[ K_p = \frac{(2.4480)(7.3469)^3}{(0.2041)^2} \]

### Step 8: Calculate Kp
Calculating the values:
\[ K_p = \frac{(2.4480)(395.096)}{0.0417} \approx 23318.28 \]

### Final Answer
Thus, the value of \(K_p\) for the reaction is approximately:
\[ K_p \approx 23318.28 \text{ atm}^2 \]

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