The equilibrium `N_(2) (g) + O_(2) (g)hArr 2NO (g)` is established in a reaction vessel of `2.5` L capacity. The amounts of `N_(2)` and `O_(2)` taken at the start were respectively 2 moles and 4 moles. Half a mole of nitrogen has been used up at equilibrium. The molar concentration of nitric oxide is:

Options

A. (A) `0.2`
B. (B) `0.4`
C. (C) `0.6`
D. (D) `0.1`

Explanation

To solve the problem step by step, we will analyze the given chemical equilibrium reaction and the amounts of reactants and products involved.

### Step 1: Write the balanced chemical equation
The balanced chemical equation is:
\[ N_2(g) + O_2(g) \rightleftharpoons 2NO(g) \]

### Step 2: Identify initial amounts of reactants
From the question, we know that:
- Initial moles of \( N_2 = 2 \) moles
- Initial moles of \( O_2 = 4 \) moles
- Initial moles of \( NO = 0 \) moles

### Step 3: Determine the change in moles at equilibrium
We are told that half a mole of nitrogen (\( N_2 \)) has been used up at equilibrium. Therefore:
- Change in moles of \( N_2 = -0.5 \)
- Change in moles of \( O_2 = -0.5 \) (since the stoichiometry of the reaction shows that 1 mole of \( N_2 \) reacts with 1 mole of \( O_2 \))
- Change in moles of \( NO = +1 \) (since 2 moles of \( NO \) are produced for every mole of \( N_2 \) reacted)

### Step 4: Calculate the equilibrium amounts of each substance
Using the initial amounts and the changes:
- Moles of \( N_2 \) at equilibrium:
\[ 2 - 0.5 = 1.5 \, \text{moles} \]
- Moles of \( O_2 \) at equilibrium:
\[ 4 - 0.5 = 3.5 \, \text{moles} \]
- Moles of \( NO \) at equilibrium:
\[ 0 + 1 = 1 \, \text{mole} \]

### Step 5: Calculate the molar concentration of \( NO \)
The molar concentration (\( C \)) is calculated using the formula:
\[ C = \frac{\text{Number of moles}}{\text{Volume in liters}} \]
Given that the volume of the reaction vessel is \( 2.5 \) L, we can calculate the concentration of \( NO \):
\[ C_{NO} = \frac{1 \, \text{mole}}{2.5 \, \text{L}} = 0.4 \, \text{mol/L} \]

### Final Answer
The molar concentration of nitric oxide (\( NO \)) at equilibrium is:
\[ \boxed{0.4 \, \text{mol/L}} \]

---