`N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g)`
For the reaction intially the mole ratio was `1:3 ` of `N_(2):H_(2)`.At equilibrium 50% of each has reacted .If the equilibrium pressure is P, the parial pressure of `NH_(3)` at equilibrium is :

Options

A. `(p)/(3)`
B. `(P)/(4)`
C. `(P)/(6)`
D. `(p)/(8)`

Answer

To solve the problem, we will follow these steps:

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

### Step 2: Define initial moles
Given the initial mole ratio of \( N_2 \) to \( H_2 \) is \( 1:3 \), we can assume:
- Let the initial moles of \( N_2 \) be \( a \).
- Therefore, the initial moles of \( H_2 \) will be \( 3a \).
- The initial moles of \( NH_3 \) is \( 0 \).

### Step 3: Determine moles at equilibrium
We are told that 50% of each reactant has reacted at equilibrium. This means:
- Moles of \( N_2 \) that reacted = \( 0.5a \)
- Moles of \( H_2 \) that reacted = \( 0.5 \times 3a = 1.5a \)

Now, we can calculate the moles at equilibrium:
- Moles of \( N_2 \) at equilibrium = \( a - 0.5a = 0.5a \)
- Moles of \( H_2 \) at equilibrium = \( 3a - 1.5a = 1.5a \)
- Moles of \( NH_3 \) formed = \( 2 \times 0.5a = a \)

### Step 4: Calculate total moles at equilibrium
Now, we can find the total number of moles at equilibrium:
\[ \text{Total moles} = \text{Moles of } N_2 + \text{Moles of } H_2 + \text{Moles of } NH_3 \]
\[ = 0.5a + 1.5a + a = 3a \]

### Step 5: Calculate the mole fraction of \( NH_3 \)
The mole fraction of \( NH_3 \) at equilibrium is given by:
\[ \text{Mole fraction of } NH_3 = \frac{\text{Moles of } NH_3}{\text{Total moles}} = \frac{a}{3a} = \frac{1}{3} \]

### Step 6: Calculate the partial pressure of \( NH_3 \)
The partial pressure of a gas can be calculated using the formula:
\[ \text{Partial pressure of } NH_3 = \text{Mole fraction of } NH_3 \times \text{Total pressure} \]
Given that the total pressure at equilibrium is \( P \):
\[ \text{Partial pressure of } NH_3 = \frac{1}{3} \times P = \frac{P}{3} \]

### Final Answer
The partial pressure of \( NH_3 \) at equilibrium is:
\[ \frac{P}{3} \]

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