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 start with the given reaction:

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

### Step 1: Write the expression for the equilibrium constant \( K_p \)

The equilibrium constant \( K_p \) for the reaction can be expressed as:

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

### Step 2: Define the initial conditions

If we start with pure ammonia, the initial partial pressure of ammonia \( P_{NH_3} \) is equal to \( p \) (the total pressure), and the initial pressures of \( N_2 \) and \( H_2 \) are both 0.

### Step 3: Set up the changes in pressures at equilibrium

Let \( P_{NH_3} \) dissociate to an extent \( y \). At equilibrium, the pressures will be:

- \( P_{NH_3} = p - 2y \)
- \( P_{N_2} = y \)
- \( P_{H_2} = \frac{3}{2}y \)

### Step 4: Substitute the equilibrium pressures into the \( K_p \) expression

Substituting these values into the expression for \( K_p \):

\[ K_p = \frac{(p - 2y)^2}{(y)\left(\frac{3}{2}y\right)^3} \]

### Step 5: Simplify the expression

This simplifies to:

\[ K_p = \frac{(p - 2y)^2}{y \cdot \frac{27}{8}y^3} = \frac{8(p - 2y)^2}{27y^4} \]

### Step 6: Assume \( P_{NH_3} \ll P_{total} \)

Given that \( P_{NH_3} \ll P_{total} \), we can assume \( p - 2y \approx p \). Thus, we can simplify further:

\[ K_p \approx \frac{8p^2}{27y^4} \]

### Step 7: Rearranging for \( y \)

Rearranging gives:

\[ y^4 = \frac{8p^2}{27K_p} \]

Taking the fourth root:

\[ y = \left(\frac{8p^2}{27K_p}\right)^{1/4} \]

### Step 8: Relate \( y \) to \( x \)

From the problem statement, we have:

\[ P_{NH_3} = \frac{x^{3/2} K_p^{1/2} p^2}{16} \]

Setting the two expressions for \( P_{NH_3} \) equal gives:

\[ \frac{x^{3/2} K_p^{1/2} p^2}{16} = p - 2y \approx p \]

### Step 9: Solve for \( x \)

Substituting \( y \) into the equation and simplifying gives:

\[ \frac{x^{3/2} K_p^{1/2} p^2}{16} = p \]

This leads to:

\[ x^{3/2} K_p^{1/2} = 16 \]

### Step 10: Isolate \( x \)

Now we can isolate \( x \):

\[ x^{3/2} = 16 K_p^{-1/2} \]

Taking both sides to the power of \( \frac{2}{3} \):

\[ x = \left(16 K_p^{-1/2}\right)^{\frac{2}{3}} = 16^{\frac{2}{3}} K_p^{-1/3} \]

Calculating \( 16^{\frac{2}{3}} \):

\[ 16^{\frac{2}{3}} = 4^{2} = 16 \]

Thus, we find:

\[ x = 16 K_p^{-1/3} \]

### Final Answer

The numerical value of \( x \) is \( 16 \).

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