In the reaction `N_(2(g))+O_(2(g))rArr2NO_((g))`at equilibrium the concentrations of `N_(2),O_(2)andNO` are 0.25 , 0.05 and 1 M respectively . Calculate intial concentrations of `N_(2)andO_(2)`.

Options

A. 0.25 M , 0.35 M
B. 0.05 M , 0.35 M
C. 0.75 M , 0.55 M
D. 0.75 M , 0.45 M

Answer

To solve the problem, we need to find the initial concentrations of \( N_2 \) and \( O_2 \) in the reaction:

\[ N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)} \]

Given the equilibrium concentrations:
- \([N_2] = 0.25 \, M\)
- \([O_2] = 0.05 \, M\)
- \([NO] = 1 \, M\)

### Step 1: Set up the equilibrium expression
Let the initial concentrations of \( N_2 \) and \( O_2 \) be \( A \) and \( B \) respectively.

At equilibrium, we can express the changes in concentration as follows:
- Change in \( N_2 \): \( -x \)
- Change in \( O_2 \): \( -x \)
- Change in \( NO \): \( +2x \)

Thus, we can write the equilibrium concentrations as:
- \([N_2] = A - x\)
- \([O_2] = B - x\)
- \([NO] = 2x\)

### Step 2: Relate \( x \) to the concentration of \( NO \)
From the equilibrium concentration of \( NO \):
\[ 2x = 1 \, M \implies x = 0.5 \, M \]

### Step 3: Substitute \( x \) to find \( A \)
Using the equilibrium concentration of \( N_2 \):
\[ A - x = 0.25 \, M \]
Substituting \( x = 0.5 \):
\[ A - 0.5 = 0.25 \implies A = 0.25 + 0.5 = 0.75 \, M \]

### Step 4: Substitute \( x \) to find \( B \)
Using the equilibrium concentration of \( O_2 \):
\[ B - x = 0.05 \, M \]
Substituting \( x = 0.5 \):
\[ B - 0.5 = 0.05 \implies B = 0.05 + 0.5 = 0.55 \, M \]

### Final Answer
The initial concentrations are:
- \([N_2] = 0.75 \, M\)
- \([O_2] = 0.55 \, M\)