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` mole
B. `0.8` mole
C. `0.2` mole
D. `0.4` mole

Explanation

To find the concentration of hydrogen gas 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:
\[ N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \]

### Step 2: Set up the initial concentrations
Initially, we have:
- \( N_2 = 1 \, \text{mol} \)
- \( H_2 = 2 \, \text{mol} \)
- \( NH_3 = 0 \, \text{mol} \)

Since the volume of the vessel is \( 1 \, \text{dm}^3 \), the initial concentrations are:
- \([N_2] = 1 \, \text{mol/dm}^3\)
- \([H_2] = 2 \, \text{mol/dm}^3\)
- \([NH_3] = 0 \, \text{mol/dm}^3\)

### Step 3: Define the change in concentration
Let \( x \) be the amount of \( N_2 \) that reacts at equilibrium. According to the stoichiometry of the reaction:
- For every 1 mole of \( N_2 \) that reacts, 3 moles of \( H_2 \) react and 2 moles of \( NH_3 \) are produced.

At equilibrium, the concentrations will be:
- \([N_2] = 1 - x\)
- \([H_2] = 2 - 3x\)
- \([NH_3] = 2x\)

### Step 4: Use the information given in the problem
It is given that at equilibrium, \( 0.8 \, \text{mol} \) of \( NH_3 \) is formed. Therefore:
\[ 2x = 0.8 \]
Solving for \( x \):
\[ x = \frac{0.8}{2} = 0.4 \]

### Step 5: Calculate the concentration of \( H_2 \) at equilibrium
Now we can substitute \( x \) back into the expression for the concentration of \( H_2 \):
\[ [H_2] = 2 - 3x \]
Substituting \( x = 0.4 \):
\[ [H_2] = 2 - 3(0.4) \]
\[ [H_2] = 2 - 1.2 \]
\[ [H_2] = 0.8 \, \text{mol/dm}^3 \]

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

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