1 mol of `N_2` and 4 mol of `H_2` are allowed to react in a vessel and after reaction, `H_2O` is added. Aqueous solution required 1 mol of HCI for neutralization. Mol fraction of `H_2` in the mixture after reaction is :

Options

A. `1/2`
B. `5/6`
C. `1/3`
D. `3/4`

Answer

To solve the problem step by step, we will follow the chemical reaction and the calculations required to find the mole fraction of \( H_2 \) in the mixture after the reaction.

### Step 1: Write the balanced chemical equation
The reaction between nitrogen (\( N_2 \)) and hydrogen (\( H_2 \)) to form ammonia (\( NH_3 \)) is given by:
\[ N_2 + 3H_2 \rightarrow 2NH_3 \]

### Step 2: Determine the initial moles of reactants
From the question, we have:
- Moles of \( N_2 = 1 \) mol
- Moles of \( H_2 = 4 \) mol

### Step 3: Calculate moles of \( NH_3 \) produced
According to the stoichiometry of the reaction, 1 mole of \( N_2 \) reacts with 3 moles of \( H_2 \) to produce 2 moles of \( NH_3 \).

Since we have 1 mole of \( N_2 \) and 4 moles of \( H_2 \), we can determine how much \( H_2 \) is consumed:
- To produce 2 moles of \( NH_3 \), we need 3 moles of \( H_2 \).

Thus, \( H_2 \) consumed = 3 moles, and \( NH_3 \) produced = 2 moles.

### Step 4: Calculate remaining moles of \( H_2 \)
Initial moles of \( H_2 \) = 4 moles
Moles of \( H_2 \) consumed = 3 moles
Remaining moles of \( H_2 \) = \( 4 - 3 = 1 \) mole

### Step 5: Calculate remaining moles of \( N_2 \)
Initial moles of \( N_2 \) = 1 mole
Moles of \( N_2 \) reacted = 0.5 moles (since 1 mole of \( N_2 \) produces 2 moles of \( NH_3 \), and we used 3 moles of \( H_2 \) which corresponds to 0.5 moles of \( N_2 \))
Remaining moles of \( N_2 \) = \( 1 - 0.5 = 0.5 \) moles

### Step 6: Total moles after the reaction
Total moles in the mixture after the reaction:
- Remaining \( H_2 = 1 \) mole
- Remaining \( N_2 = 0.5 \) moles
- Moles of \( NH_3 = 2 \) moles (produced)

Total moles = \( 1 + 0.5 + 2 = 3.5 \) moles

### Step 7: Calculate the mole fraction of \( H_2 \)
The mole fraction of a component is given by the formula:
\[ \text{Mole fraction of } H_2 = \frac{\text{Moles of } H_2}{\text{Total moles}} \]
Substituting the values:
\[ \text{Mole fraction of } H_2 = \frac{1}{3.5} = \frac{2}{7} \approx 0.2857 \]

### Final Answer
The mole fraction of \( H_2 \) in the mixture after the reaction is approximately \( 0.2857 \) or \( \frac{2}{7} \).

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