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 question 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 the initial moles of reactants
Initially, we have:
- Moles of \(\text{NH}_3\) = 0.6 moles
- Moles of \(\text{N}_2\) = 0 moles
- Moles of \(\text{H}_2\) = 0 moles

### Step 3: Set up the change in moles at equilibrium
Let \(x\) be the change in moles of \(\text{N}_2\) formed at equilibrium. 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.
- Therefore, if \(x\) moles of \(\text{N}_2\) are formed, then \(2x\) moles of \(\text{NH}_3\) will dissociate, and \(3x\) moles of \(\text{H}_2\) will be formed.

### Step 4: Express the moles at equilibrium
At equilibrium, the moles of each species will be:
- Moles of \(\text{NH}_3\) = \(0.6 - 2x\)
- Moles of \(\text{N}_2\) = \(x\)
- Moles of \(\text{H}_2\) = \(3x\)

### Step 5: Use the information given in the problem
We know that at equilibrium, the vessel contains 0.15 moles of \(\text{H}_2\):
\[ 3x = 0.15 \]
From this, we can solve for \(x\):
\[ x = \frac{0.15}{3} = 0.05 \]

### Step 6: Calculate the moles of \(\text{NH}_3\) left at equilibrium
Now, substituting \(x\) back into the expression for \(\text{NH}_3\):
\[ \text{Moles of } \text{NH}_3 = 0.6 - 2(0.05) = 0.6 - 0.1 = 0.5 \text{ moles} \]

### Step 7: Calculate the moles of \(\text{N}_2\) at equilibrium
Using \(x\):
\[ \text{Moles of } \text{N}_2 = x = 0.05 \text{ moles} \]

### Step 8: Determine the concentration of \(\text{NH}_3\) at equilibrium
The concentration of \(\text{NH}_3\) can be calculated using the formula:
\[ \text{Concentration} = \frac{\text{Moles}}{\text{Volume}} \]
Given the volume of the reaction vessel is \(2 \, \text{dm}^3\):
\[ \text{Concentration of } \text{NH}_3 = \frac{0.5}{2} = 0.25 \, \text{mol/dm}^3 \]

### Conclusion
Now we can analyze the statements:
1. **0.15 mole of original NH3 had dissociated at equilibrium.** (False, only 0.1 moles dissociated)
2. **0.55 mole of ammonia is left in the vessel.** (False, 0.5 moles are left)
3. **At equilibrium, the vessel contains 0.45 moles of N2.** (False, only 0.05 moles of N2 are present)
4. **The concentration of NH3 at equilibrium is 0.25 mole per dm³.** (True)

Thus, the correct statement is that the concentration of \(\text{NH}_3\) at equilibrium is \(0.25 \, \text{mol/dm}^3\).