For the reaction `N_(2)+O_(2)hArr 2NO` at equilibrium number of moles of `N_(2),O_(2) and NO` pressent in the system per litre are 0.25 mole, 0.05 moles and 1.0 mole respectively What will be the initial concentration of `N_(2) and O_(2)` respectively

Options

A. 0.75, 0.55
B. 0.25, 0.75
C. 0.20, 0.80
D. 0.30, 0.70

Answer

To solve the problem, we need to find the initial concentrations of \( N_2 \) and \( O_2 \) before the reaction reached equilibrium. We know the equilibrium concentrations of all species involved in the reaction:

\[ N_2 + O_2 \rightleftharpoons 2NO \]

Given:
- Equilibrium concentration of \( N_2 \) = 0.25 moles/L
- Equilibrium concentration of \( O_2 \) = 0.05 moles/L
- Equilibrium concentration of \( NO \) = 1.0 moles/L

### Step 1: Determine the change in concentration
Let \( x \) be the amount of \( N_2 \) and \( O_2 \) that reacted to form \( NO \). According to the stoichiometry of the reaction, for every 1 mole of \( N_2 \) and \( O_2 \) that reacts, 2 moles of \( NO \) are produced.

From the equilibrium concentration of \( NO \):
\[ 2x = 1.0 \quad \Rightarrow \quad x = 0.5 \]

### Step 2: Write the expressions for the initial concentrations
Let \( A \) be the initial concentration of \( N_2 \) and \( B \) be the initial concentration of \( O_2 \).

At equilibrium, the concentrations can be expressed as:
- For \( N_2 \): \( A - x \)
- For \( O_2 \): \( B - x \)

### Step 3: Set up equations based on equilibrium concentrations
Using the equilibrium concentrations given:
1. For \( N_2 \):
\[ A - x = 0.25 \]
Substituting \( x = 0.5 \):
\[ A - 0.5 = 0.25 \quad \Rightarrow \quad A = 0.25 + 0.5 = 0.75 \, \text{moles/L} \]

2. For \( O_2 \):
\[ B - x = 0.05 \]
Substituting \( x = 0.5 \):
\[ B - 0.5 = 0.05 \quad \Rightarrow \quad B = 0.05 + 0.5 = 0.55 \, \text{moles/L} \]

### Final Answer:
- The initial concentration of \( N_2 \) is **0.75 moles/L**.
- The initial concentration of \( O_2 \) is **0.55 moles/L**.