0.6 mole of `NH_(3)` in a reaction vessel of `2dm^(3)` capacity was brought to equilibrium. The vessel was then found to contain 0.15 mole of `H_(2)` formed by the reaction `2NH_(3)(g) hArr N_(2)(g)+3H_(2)(g)`. Which of the following statements is true?

Options

A. 0.15 mole of the original `NH_(3)` had dissociated at equilibrium
B. 0.55 mole of ammonia is left in the vessel
C. At equilibrium the vessel contained 0.45 mole of `N_(2)`
D. The concentration of `NH_(3)` at equilibrium is 0.25 mole per `dm^(3)`

Answer

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

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the reaction is:
\[ 2 \text{NH}_3(g) \rightleftharpoons \text{N}_2(g) + 3 \text{H}_2(g) \]

### Step 2: Determine initial moles and concentrations
We are given:
- Initial moles of \(\text{NH}_3 = 0.6 \, \text{moles}\)
- Volume of the vessel = \(2 \, \text{dm}^3\)

The initial concentration of \(\text{NH}_3\) can be calculated as:
\[ \text{Concentration of } \text{NH}_3 = \frac{\text{moles}}{\text{volume}} = \frac{0.6}{2} = 0.3 \, \text{mol/dm}^3 \]

### Step 3: Define the change in moles at equilibrium
Let \(X\) be the degree of dissociation of \(\text{NH}_3\). According to the stoichiometry of the reaction:
- For every \(2\) moles of \(\text{NH}_3\) that dissociate, \(1\) mole of \(\text{N}_2\) and \(3\) moles of \(\text{H}_2\) are formed.

At equilibrium:
- Moles of \(\text{NH}_3\) remaining = \(0.6 - 2X\)
- Moles of \(\text{N}_2\) formed = \(X\)
- Moles of \(\text{H}_2\) formed = \(3X\)

### Step 4: Use the information about \(\text{H}_2\) at equilibrium
We know that at equilibrium, the vessel contains \(0.15 \, \text{moles}\) of \(\text{H}_2\):
\[ 3X = 0.15 \implies X = \frac{0.15}{3} = 0.05 \]

### Step 5: Calculate remaining moles of \(\text{NH}_3\)
Now, we can find the remaining moles of \(\text{NH}_3\):
\[ \text{Moles of } \text{NH}_3 \text{ at equilibrium} = 0.6 - 2(0.05) = 0.6 - 0.1 = 0.5 \, \text{moles} \]

### Step 6: Calculate moles of \(\text{N}_2\) at equilibrium
Using the value of \(X\):
\[ \text{Moles of } \text{N}_2 \text{ at equilibrium} = X = 0.05 \, \text{moles} \]

### Step 7: Calculate concentration of \(\text{NH}_3\) at equilibrium
The concentration of \(\text{NH}_3\) at equilibrium can be calculated as:
\[ \text{Concentration of } \text{NH}_3 = \frac{0.5 \, \text{moles}}{2 \, \text{dm}^3} = 0.25 \, \text{mol/dm}^3 \]

### Conclusion: Evaluate the statements
1. **Statement 1**: Incorrect. \(0.15\) moles of \(\text{H}_2\) were formed, not dissociated from \(\text{NH}_3\).
2. **Statement 2**: Correct. \(0.5\) moles of \(\text{NH}_3\) remain.
3. **Statement 3**: Incorrect. Only \(0.05\) moles of \(\text{N}_2\) are present.
4. **Statement 4**: Correct. The concentration of \(\text{NH}_3\) is \(0.25 \, \text{mol/dm}^3\).