Pure ammonia is placed in a vessel at a temperature where its dissociation constant `(alpha)` is appreciable. At equilibrium,

Options

A. `K_(rho)` does not change significantly with pressure
B. Concentration of `N_(2)` does not change with pressure
C. Concentration of `NH_(3)` does not change with pressure
D. `K_(rho)` changes with pressure but `alpha` does not change

Answer

To solve the problem regarding the dissociation of pure ammonia in a vessel at equilibrium, we will analyze the equilibrium reaction and the behavior of the equilibrium constant \( K_p \) with respect to pressure changes.

### Step-by-Step Solution:

1. **Identify the Dissociation Reaction of Ammonia**:
The dissociation of ammonia (\( NH_3 \)) can be represented as:
\[ 2NH_3(g) \rightleftharpoons N_2(g) + 3H_2(g) \]

2. **Write the Expression for the Equilibrium Constant \( K_p \)**:
The equilibrium constant \( K_p \) for the reaction at a given temperature is expressed as:
\[ K_p = \frac{P_{N_2} \cdot (P_{H_2})^3}{(P_{NH_3})^2} \]
where \( P_{N_2} \), \( P_{H_2} \), and \( P_{NH_3} \) are the partial pressures of nitrogen, hydrogen, and ammonia, respectively.

3. **Understand the Effect of Pressure on \( K_p \)**:
The equilibrium constant \( K_p \) is only affected by changes in temperature. It does not change with changes in pressure, volume, or concentration. Therefore, we conclude that:
\[ K_p \text{ does not change significantly with pressure.} \]

4. **Analyze the Concentration Changes with Pressure**:
According to Le Chatelier's principle, if the pressure of a gaseous system at equilibrium is increased, the equilibrium will shift towards the side with fewer moles of gas. In this case:
- The left side (reactants) has 2 moles of \( NH_3 \).
- The right side (products) has \( 1 + 3 = 4 \) moles of gas (1 mole of \( N_2 \) and 3 moles of \( H_2 \)).
Thus, if the pressure is increased, the equilibrium will shift to the left (towards the formation of ammonia), resulting in a decrease in the concentration of \( N_2 \) and \( H_2 \).

5. **Conclusion on the Options**:
- **Option A**: \( K_p \) does not change significantly with pressure. **(Correct)**
- **Option B**: Concentration of nitrogen does not change with pressure. **(Incorrect)**
- **Option C**: Concentration of ammonia does not change with pressure. **(Incorrect)**
- **Option D**: \( K_p \) changes with pressure but \( \alpha \) does not change. **(Incorrect)**

Thus, the correct answer is **Option A**.