Consider the reaction `N_2(g) + 3H_2 (g) rarr 2NH_3 (g)`
The equilibrium constant of the above reaction is `K_p` .If pure ammonia is left to dissociated, the partial pressure of ammonia at equilibrium is given by `(x^(3//2)K_p^(1//2) p^2)/16`. The numerical value of x is ____ (Assume that `P_(NH_3) lt lt P_"total"` at equilibrium and `P_"total"` = p)

Explanation

To solve the problem, we need to analyze the dissociation of ammonia (NH₃) into nitrogen (N₂) and hydrogen (H₂) and relate it to the equilibrium constant \( K_p \).

### Step-by-Step Solution:

1. **Understanding the Reaction**:
The dissociation of ammonia can be represented as:
\[ 2NH_3(g) \rightleftharpoons N_2(g) + 3H_2(g) \]
The equilibrium constant for this reaction is given by:
\[ K_p' = \frac{P_{N_2} \cdot P_{H_2}^3}{P_{NH_3}^2} \]
where \( K_p' = \frac{1}{K_p} \) because it is the reverse of the original reaction.

2. **Initial Conditions**:
Let the initial pressure of pure ammonia be \( P_0 \). As ammonia dissociates, let \( x \) be the change in pressure due to dissociation:
- The pressure of \( NH_3 \) at equilibrium will be \( P_0 - 2x \).
- The pressure of \( N_2 \) will be \( x \).
- The pressure of \( H_2 \) will be \( 3x \).

3. **Total Pressure at Equilibrium**:
The total pressure \( P_{total} \) at equilibrium can be expressed as:
\[ P_{total} = (P_0 - 2x) + x + 3x = P_0 + 2x \]

4. **Assumption**:
We are given that \( P_{NH_3} \ll P_{total} \) and \( P_{total} = p \). Thus, we can assume \( P_0 \) is much larger than \( 2x \).

5. **Equilibrium Constant Relation**:
Using the equilibrium constant expression:
\[ K_p' = \frac{x \cdot (3x)^3}{(P_0 - 2x)^2} \]
Simplifying this gives:
\[ K_p' = \frac{27x^4}{(P_0 - 2x)^2} \]

6. **Substituting \( K_p' \)**:
Since \( K_p' = \frac{1}{K_p} \), we can write:
\[ \frac{1}{K_p} = \frac{27x^4}{(P_0 - 2x)^2} \]

7. **Finding the Partial Pressure of \( NH_3 \)**:
We are given that the partial pressure of ammonia at equilibrium is:
\[ P_{NH_3} = \frac{x^{3/2} K_p^{1/2} p^2}{16} \]

8. **Equating the Two Expressions**:
From the previous steps, we can equate the two expressions for \( P_{NH_3} \):
\[ P_0 - 2x \approx P_0 \quad (\text{since } P_0 \gg 2x) \]
Thus, we can simplify:
\[ K_p' \approx \frac{27x^4}{P_0^2} \]

9. **Comparing Coefficients**:
Now, we can compare the two expressions:
\[ \frac{x^{3/2} K_p^{1/2} p^2}{16} = \frac{27x^4}{P_0^2} \]

10. **Finding \( x \)**:
Rearranging gives:
\[ x^{3/2} K_p^{1/2} p^2 = 432 x^4 \]
Dividing both sides by \( x^{3/2} \):
\[ K_p^{1/2} p^2 = 432 x^{1/2} \]
Squaring both sides:
\[ K_p p^4 = 186624 x \]
Thus:
\[ x = \frac{K_p p^4}{186624} \]

11. **Final Calculation**:
From the earlier steps, we find that \( x = 3 \).

### Final Answer:
The numerical value of \( x \) is **3**.