`N_(2(g)) + 3H_(2(g)) hArr 2NH_(3(g))` for the reaction initially 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 partial 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 need to analyze the given chemical reaction and the conditions at equilibrium. Let's break it down step by step.

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

### Step 2: Determine initial moles
According to the problem, the initial mole ratio of \(N_2\) to \(H_2\) is 1:3. Therefore, we can assume:
- Initial moles of \(N_2 = 1\)
- Initial moles of \(H_2 = 3\)
- Initial moles of \(NH_3 = 0\)

### Step 3: Determine moles at equilibrium
It is given that 50% of each reactant has reacted at equilibrium. This means:
- \(N_2\) reacts: \(1 \times 0.5 = 0.5\) moles
- \(H_2\) reacts: \(3 \times 0.5 = 1.5\) moles

Now we can calculate the moles at equilibrium:
- Moles of \(N_2\) at equilibrium = \(1 - 0.5 = 0.5\)
- Moles of \(H_2\) at equilibrium = \(3 - 1.5 = 1.5\)
- Moles of \(NH_3\) formed = \(2 \times 0.5 = 1\) (from the stoichiometry of the reaction)

### Step 4: Calculate total moles at equilibrium
Now, we can calculate the total moles at equilibrium:
\[ \text{Total moles} = \text{Moles of } N_2 + \text{Moles of } H_2 + \text{Moles of } NH_3 = 0.5 + 1.5 + 1 = 3 \]

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

### 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 \(P\):
\[ \text{Partial pressure of } NH_3 = \frac{1}{3} \times P = \frac{P}{3} \]

### Conclusion
Thus, the partial pressure of \(NH_3\) at equilibrium is:
\[ \text{Partial pressure of } NH_3 = \frac{P}{3} \]