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 : (Assume that `P_(NH_(3)) lt lt P_(total)` at equilibrium)

Options

A. `3^(3//2)K_(p)^(1//2)P^(2)/16`
B. `3^(3//2)K_(p)^(1//2)P^(2)/4`
C. `K_(p)^(1//2)P^(2)/16`
D. `K_(p)^(1//2)P^(2)/4`

Explanation

To find the partial pressure of ammonia at equilibrium after it dissociates, we can follow these steps:

### Step 1: Write the Balanced Reaction
The balanced reaction is:
\[ N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \]

### Step 2: Define the Equilibrium Constant Expression
The equilibrium constant \( K_p \) for the reaction is defined as:
\[ K_p = \frac{(P_{NH_3})^2}{(P_{N_2})(P_{H_2})^3} \]

### Step 3: Consider the Dissociation of Ammonia
When pure ammonia dissociates, we can represent this reaction as:
\[ 2NH_3(g) \rightleftharpoons N_2(g) + 3H_2(g) \]

### Step 4: Set Up the Initial Conditions
Assume we start with a pressure \( P_0 \) of ammonia. At the beginning (time = 0):
- \( P_{NH_3} = P_0 \)
- \( P_{N_2} = 0 \)
- \( P_{H_2} = 0 \)

### Step 5: Define Changes at Equilibrium
Let \( x \) be the amount of ammonia that dissociates at equilibrium. Thus, at equilibrium:
- \( P_{NH_3} = P_0 - 2x \)
- \( P_{N_2} = x \)
- \( P_{H_2} = 3x \)

### Step 6: Substitute into the Equilibrium Expression
Substituting these values into the \( K_p \) expression gives:
\[ K_p = \frac{(P_0 - 2x)^2}{(x)(3x)^3} \]
This simplifies to:
\[ K_p = \frac{(P_0 - 2x)^2}{27x^4} \]

### Step 7: Assume \( P_{NH_3} \ll P_{total} \)
Since we are told that \( P_{NH_3} \) is much less than \( P_{total} \), we can assume \( P_0 - 2x \approx P_0 \). This simplifies our equation to:
\[ K_p \approx \frac{P_0^2}{27x^4} \]

### Step 8: Rearrange to Solve for \( x \)
Rearranging gives:
\[ x^4 = \frac{P_0^2}{27K_p} \]
Taking the fourth root:
\[ x = \left(\frac{P_0^2}{27K_p}\right)^{1/4} \]

### Step 9: Find the Partial Pressure of Ammonia at Equilibrium
Now substituting \( x \) back into the expression for \( P_{NH_3} \):
\[ P_{NH_3} = P_0 - 2x \approx P_0 - 2\left(\frac{P_0^2}{27K_p}\right)^{1/4} \]

### Final Expression
Thus, the partial pressure of ammonia at equilibrium after dissociation is approximately:
\[ P_{NH_3} \approx P_0 - 2\left(\frac{P_0^2}{27K_p}\right)^{1/4} \]