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 determine the initial concentrations of \(N_2\) and \(O_2\) for the reaction:

\[ N_2 + O_2 \rightleftharpoons 2NO \]

Given the equilibrium concentrations:
- \([N_2] = 0.25 \, \text{moles/L}\)
- \([O_2] = 0.05 \, \text{moles/L}\)
- \([NO] = 1.0 \, \text{moles/L}\)

### Step 1: Set up the initial concentrations
Let the initial concentration of \(N_2\) be \(A\) and the initial concentration of \(O_2\) be \(B\). Initially, the concentration of \(NO\) is \(0\).

### Step 2: Define the change in concentrations
Let \(x\) be the amount of \(N_2\) and \(O_2\) that reacts at equilibrium. According to the stoichiometry of the reaction:
- The change in concentration of \(N_2\) will be \(-x\)
- The change in concentration of \(O_2\) will be \(-x\)
- The change in concentration of \(NO\) will be \(+2x\)

### Step 3: Write the equilibrium concentrations
At equilibrium, we can express the concentrations as follows:
- \([N_2] = A - x\)
- \([O_2] = B - x\)
- \([NO] = 2x\)

### Step 4: Substitute the equilibrium values
From the problem, we know:
1. \(A - x = 0.25\)
2. \(B - x = 0.05\)
3. \(2x = 1.0\)

### Step 5: Solve for \(x\)
From equation (3):
\[ 2x = 1 \implies x = 0.5 \]

### Step 6: Substitute \(x\) back to find \(A\) and \(B\)
Now, substitute \(x\) into equations (1) and (2):

For \(N_2\):
\[ A - 0.5 = 0.25 \implies A = 0.25 + 0.5 = 0.75 \]

For \(O_2\):
\[ B - 0.5 = 0.05 \implies B = 0.05 + 0.5 = 0.55 \]

### Conclusion
The initial concentrations are:
- \([N_2] = 0.75 \, \text{moles/L}\)
- \([O_2] = 0.55 \, \text{moles/L}\)