`N_2 and H_2` are taken in 1:3 molar ratio in a closed vessel to attained the following equilibrium , `N_2(g)+3H_2(g) hArr 2NH_3(g)`
Find `K_p` for reaction at total pressure of 2P if `P_(N_2)` at equilibrium is `P/3`:

Options

A. `1/(3P^2)`
B. `4/(3P^2)`
C. `(4P^2)/3`
D. None of these

Answer

To solve the question step by step, we will follow these instructions:

### Step 1: Identify the initial conditions
Given that nitrogen (N₂) and hydrogen (H₂) are taken in a 1:3 molar ratio, we can denote the initial moles of N₂ as \( n \) and the initial moles of H₂ as \( 3n \).

### Step 2: Write the equilibrium expression
The equilibrium reaction is:
\[ N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \]

### Step 3: Define the pressures at equilibrium
Let the equilibrium pressure of \( N_2 \) be \( P_{N_2} = \frac{P}{3} \) (given in the question).

Since the total pressure at equilibrium is given as \( 2P \), we can denote the equilibrium pressures of the gases as follows:
- \( P_{N_2} = \frac{P}{3} \)
- Let \( P_{H_2} = P_H \)
- Let \( P_{NH_3} = P_N \)

### Step 4: Use the total pressure equation
The total pressure at equilibrium can be expressed as:
\[ P_{N_2} + P_{H_2} + P_{NH_3} = 2P \]
Substituting the known values:
\[ \frac{P}{3} + P_H + P_N = 2P \]

### Step 5: Express \( P_H \) and \( P_N \) in terms of \( P \)
From the equation:
\[ P_H + P_N = 2P - \frac{P}{3} \]
\[ P_H + P_N = \frac{6P}{3} - \frac{P}{3} \]
\[ P_H + P_N = \frac{5P}{3} \]

### Step 6: Relate \( P_H \) to \( P_{N_2} \)
Since the initial moles of \( H_2 \) are three times that of \( N_2 \), and knowing that the volume is constant, the pressure of \( H_2 \) at equilibrium will also be three times that of \( N_2 \):
\[ P_H = 3 \times P_{N_2} = 3 \times \frac{P}{3} = P \]

### Step 7: Substitute \( P_H \) back into the total pressure equation
Now substituting \( P_H \) into the equation:
\[ P + P_N = \frac{5P}{3} \]
\[ P_N = \frac{5P}{3} - P \]
\[ P_N = \frac{5P}{3} - \frac{3P}{3} = \frac{2P}{3} \]

### Step 8: Calculate \( K_p \)
Now we can calculate \( K_p \) using the equilibrium pressures:
\[ K_p = \frac{(P_{NH_3})^2}{(P_{H_2})^3 \cdot (P_{N_2})} \]
Substituting the values:
\[ K_p = \frac{(P_N)^2}{(P_H)^3 \cdot (P_{N_2})} = \frac{\left(\frac{2P}{3}\right)^2}{(P)^3 \cdot \left(\frac{P}{3}\right)} \]

### Step 9: Simplify the expression
\[ K_p = \frac{\frac{4P^2}{9}}{P^3 \cdot \frac{P}{3}} = \frac{\frac{4P^2}{9}}{\frac{P^4}{3}} = \frac{4P^2}{9} \cdot \frac{3}{P^4} = \frac{12}{9P^2} = \frac{4}{3P^2} \]

### Final Answer
Thus, the value of \( K_p \) is:
\[ K_p = \frac{4}{3} P^2 \]