Two moles of `N_2` and two moles of `H_2` are taken in a closed vessel of 5 litres capacity and suitable conditions are provided for the reaction. When the equilibrium is reached ,it is found that a half mole of `N_2` is used up. The equilibrium concentration of `NH_3` is

Options

A. 0.3
B. 0.4
C. 0.2
D. 0.1

Answer

To find the equilibrium concentration of ammonia (NH₃) when 2 moles of nitrogen (N₂) and 2 moles of hydrogen (H₂) are reacted in a closed vessel of 5 liters, we can follow these steps:

### Step 1: Write the balanced chemical equation
The balanced equation for the synthesis of ammonia is:
\[ N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \]

### Step 2: Determine initial moles
Initially, we have:
- Moles of N₂ = 2
- Moles of H₂ = 2
- Moles of NH₃ = 0

### Step 3: Determine moles at equilibrium
It is given that half a mole of N₂ is used up at equilibrium. Therefore:
- Moles of N₂ used = 0.5
- Remaining moles of N₂ = 2 - 0.5 = 1.5

Now, according to the stoichiometry of the reaction, for every 1 mole of N₂ that reacts, 3 moles of H₂ are required. Therefore, for 0.5 moles of N₂ used:
- Moles of H₂ used = 3 × 0.5 = 1.5
- Remaining moles of H₂ = 2 - 1.5 = 0.5

Since 0.5 moles of N₂ produce 1 mole of NH₃ (from the balanced equation), we have:
- Moles of NH₃ produced = 2 × 0.5 = 1

### Step 4: Summary of equilibrium moles
At equilibrium, we have:
- Moles of N₂ = 1.5
- Moles of H₂ = 0.5
- Moles of NH₃ = 1

### Step 5: Calculate the concentration of NH₃
Concentration is calculated using the formula:
\[ \text{Concentration} = \frac{\text{Number of moles}}{\text{Volume in liters}} \]
For NH₃:
- Number of moles of NH₃ = 1
- Volume of the vessel = 5 L

Thus, the concentration of NH₃ is:
\[ \text{Concentration of NH₃} = \frac{1 \text{ mole}}{5 \text{ L}} = 0.2 \text{ M} \]

### Final Answer
The equilibrium concentration of NH₃ is **0.2 M**.

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