`N_(3) +3H_(2) hArr 2NH_(3)` Starting with one mole of nitrogen and 3 moles of hydrogen, at equiliibrium `50%` of each had reacted. If the equilibrium pressure is `P`, the partial pressure of hydrogen at equilibrium would be

Options

A. `P//2`
B. `P//3`
C. `P//4`
D. `P//6`

Explanation

To solve the problem step by step, we will analyze the reaction and the information given:

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

### Step 2: Determine initial moles of reactants
According to the question, we start with:
- 1 mole of \( N_2 \)
- 3 moles of \( H_2 \)

### Step 3: Determine the change in moles at equilibrium
It is given that at equilibrium, 50% of each reactant has reacted. Therefore:
- Moles of \( N_2 \) reacted = \( 0.5 \) moles
- Moles of \( H_2 \) reacted = \( 1.5 \) moles (since it reacts in a 1:3 ratio with \( N_2 \))

### Step 4: Calculate moles at equilibrium
Now we can calculate the moles of each substance at equilibrium:
- Moles of \( N_2 \) at equilibrium = \( 1 - 0.5 = 0.5 \) moles
- Moles of \( H_2 \) at equilibrium = \( 3 - 1.5 = 1.5 \) moles
- Moles of \( NH_3 \) formed = \( 2 \times 0.5 = 1 \) mole (since 2 moles of \( NH_3 \) are produced for every mole of \( N_2 \) reacted)

### Step 5: Total moles at equilibrium
Total moles at equilibrium = Moles of \( N_2 \) + Moles of \( H_2 \) + Moles of \( NH_3 \)
\[ = 0.5 + 1.5 + 1 = 3 \text{ moles} \]

### Step 6: Calculate the mole fraction of hydrogen
The mole fraction of \( H_2 \) at equilibrium is given by:
\[ \text{Mole fraction of } H_2 = \frac{\text{Moles of } H_2}{\text{Total moles}} = \frac{1.5}{3} = 0.5 \]

### Step 7: Calculate the partial pressure of hydrogen
The partial pressure of a gas can be calculated using the formula:
\[ P_{H_2} = \text{Mole fraction of } H_2 \times P \]
Substituting the values we have:
\[ P_{H_2} = 0.5 \times P = \frac{P}{2} \]

### Final Answer
The partial pressure of hydrogen at equilibrium is:
\[ \frac{P}{2} \]

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