`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. (A) `(1)/(3P^(2)`
B. (B) `(4)/(3P^(2))`
C. (C) `(4P^(2))/(3)`
D. (D) none

Explanation

To solve the problem, we need to find the equilibrium constant \( K_p \) for the reaction:

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

given the conditions of the problem.

### Step 1: Determine Initial Moles and Pressures

Given that \( N_2 \) and \( H_2 \) are taken in a 1:3 molar ratio, we can assume:

- Let the number of moles of \( N_2 \) be \( n \).
- Then, the number of moles of \( H_2 \) will be \( 3n \).

### Step 2: Calculate Total Initial Moles

The total initial moles before the reaction starts will be:

\[ n + 3n = 4n \]

### Step 3: Determine Partial Pressure of \( N_2 \)

It is given that at equilibrium, the partial pressure of \( N_2 \) is \( P_{N_2} = \frac{P}{3} \).

### Step 4: Calculate Total Pressure

The total pressure at equilibrium is given as \( P_{total} = 2P \).

### Step 5: Determine Partial Pressure of \( H_2 \)

Since the initial pressure of \( N_2 \) is \( \frac{P}{3} \), we can find the partial pressure of \( H_2 \):

Let \( P_{H_2} \) be the partial pressure of \( H_2 \) at equilibrium.

Using the relationship of total pressure:

\[ P_{total} = P_{N_2} + P_{H_2} + P_{NH_3} \]

We know:

\[ 2P = \frac{P}{3} + P_{H_2} + P_{NH_3} \]

### Step 6: Set Up the Equilibrium Expression

Let \( P_{NH_3} = x \) (the partial pressure of ammonia).

Thus, we can rewrite the equation:

\[ 2P = \frac{P}{3} + P_{H_2} + x \]

### Step 7: Solve for \( P_{H_2} \) and \( x \)

We know that:

\[ P_{H_2} = 3 \times P_{N_2} = 3 \times \frac{P}{3} = P \]

Substituting \( P_{H_2} \) into the total pressure equation:

\[ 2P = \frac{P}{3} + P + x \]

This simplifies to:

\[ 2P = \frac{P}{3} + \frac{3P}{3} + x \]

\[ 2P = \frac{4P}{3} + x \]

Now, rearranging gives:

\[ x = 2P - \frac{4P}{3} = \frac{6P}{3} - \frac{4P}{3} = \frac{2P}{3} \]

Thus, the partial pressure of \( NH_3 \) is:

\[ P_{NH_3} = \frac{2P}{3} \]

### Step 8: Calculate \( K_p \)

The expression for \( K_p \) is given by:

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

Substituting the values we found:

\[ K_p = \frac{\left(\frac{2P}{3}\right)^2}{\left(\frac{P}{3}\right) \left(P\right)^3} \]

Calculating this gives:

\[ K_p = \frac{\frac{4P^2}{9}}{\frac{P}{3} \cdot P^3} = \frac{\frac{4P^2}{9}}{\frac{P^4}{3}} = \frac{4P^2}{9} \cdot \frac{3}{P^4} = \frac{12}{9P^2} = \frac{4}{3P^2} \]

### Final Answer

Thus, the equilibrium constant \( K_p \) is:

\[ K_p = \frac{4}{3P^2} \]

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