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) in the given reaction, we can follow these steps:

### Step-by-Step Solution:

1. **Write the Balanced Chemical Equation:**
The reaction between nitrogen (N₂) and hydrogen (H₂) to form ammonia (NH₃) is:
\[ N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \]

2. **Identify Initial Moles:**
Initially, we have:
- Moles of N₂ = 2 moles
- Moles of H₂ = 2 moles

3. **Determine Moles Used at Equilibrium:**
According to the problem, half a mole of N₂ is used up at equilibrium:
- Moles of N₂ used = 0.5 moles

4. **Calculate Moles of NH₃ Produced:**
From the balanced equation, 1 mole of N₂ produces 2 moles of NH₃. Therefore, if 0.5 moles of N₂ are used:
\[ \text{Moles of NH₃ produced} = 0.5 \, \text{moles of N₂} \times 2 \, \text{moles of NH₃/mole of N₂} = 1 \, \text{mole of NH₃} \]

5. **Calculate the Total Moles at Equilibrium:**
At equilibrium, the moles of each component are:
- Moles of N₂ remaining = 2 - 0.5 = 1.5 moles
- Moles of H₂ remaining = 2 - (3/2) × 0.5 = 2 - 0.75 = 1.25 moles (since 3 moles of H₂ are required for 1 mole of N₂)
- Moles of NH₃ = 1 mole

6. **Calculate the Equilibrium Concentration of NH₃:**
The concentration is given by the formula:
\[ \text{Concentration} = \frac{\text{Number of moles}}{\text{Volume in liters}} \]
Given that the volume of the vessel is 5 liters:
\[ \text{Concentration of NH₃} = \frac{1 \, \text{mole}}{5 \, \text{liters}} = 0.2 \, \text{M} \]

### Final Answer:
The equilibrium concentration of NH₃ is **0.2 M** (molar).

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