1 mol of `N_(2)` and 2 mol of `H_(2)` are allowed to react in a 1 `dm^(3)` vessel. At equilibrium, `0.8` mol of `NH_(3)` is formed. The concentration of `H_(2)` in the vessel is

Options

A. `0.6` mol
B. `0.8` mol
C. `0.2` mol
D. `0.4` mol

Explanation

To find the concentration of \( H_2 \) at equilibrium, we can follow these steps:

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

### Step 2: Set up the initial moles
Initially, we have:
- Moles of \( N_2 = 1 \) mol
- Moles of \( H_2 = 2 \) mol
- Moles of \( NH_3 = 0 \) mol

### Step 3: Define the change in moles
Let \( x \) be the amount of \( N_2 \) that reacts. According to the stoichiometry of the reaction:
- \( N_2 \) reacts: \( x \) moles
- \( H_2 \) reacts: \( 3x \) moles
- \( NH_3 \) produced: \( 2x \) moles

### Step 4: Write the equilibrium moles
At equilibrium, the moles will be:
- Moles of \( N_2 = 1 - x \)
- Moles of \( H_2 = 2 - 3x \)
- Moles of \( NH_3 = 2x \)

### Step 5: Use the information given
We know that at equilibrium, \( 0.8 \) moles of \( NH_3 \) are formed:
\[ 2x = 0.8 \]
From this, we can solve for \( x \):
\[ x = \frac{0.8}{2} = 0.4 \]

### Step 6: Calculate the moles of \( H_2 \) at equilibrium
Now we can find the moles of \( H_2 \) remaining at equilibrium:
\[ \text{Moles of } H_2 = 2 - 3x = 2 - 3(0.4) = 2 - 1.2 = 0.8 \]

### Step 7: Calculate the concentration of \( H_2 \)
Since the volume of the vessel is \( 1 \, \text{dm}^3 \), the concentration of \( H_2 \) is:
\[ \text{Concentration of } H_2 = \frac{\text{Moles of } H_2}{\text{Volume}} = \frac{0.8 \, \text{mol}}{1 \, \text{dm}^3} = 0.8 \, \text{mol/dm}^3 \]

### Final Answer
The concentration of \( H_2 \) in the vessel at equilibrium is \( 0.8 \, \text{mol/dm}^3 \).

---