At 1000 K , a sample of pure `NO_(2)` gases decomposes as :
`2NO_(2)(g)hArr2NO(g)+O_(2)(g)`
The equilibrium constant `K_(P)` is 156.25 atm .Analysis showns that the partial pressure of `O_(2)` is 0.25 atm at equilibrium .The parital pressure o f`NO_(2)` at equilibrium is :

Options

A. `0.03`
B. `0.02`
C. `0.025`
D. `0.04`

Answer

To find the partial pressure of \( NO_2 \) at equilibrium, we will follow these steps:

### Step 1: Write the balanced chemical equation
The decomposition of \( NO_2 \) is given by the equation:
\[ 2NO_2(g) \rightleftharpoons 2NO(g) + O_2(g) \]

### Step 2: Define the changes in pressure
Let the initial pressure of \( NO_2 \) be \( P_{NO_2}^0 \). At equilibrium, let \( x \) be the change in pressure of \( NO_2 \) that decomposes. The changes in pressure can be represented as follows:
- Initial: \( P_{NO_2}^0 \), \( 0 \), \( 0 \) (for \( NO \) and \( O_2 \))
- Change: \( -2x \), \( +2x \), \( +x \) (for \( NO_2 \), \( NO \), and \( O_2 \))
- Equilibrium: \( P_{NO_2}^0 - 2x \), \( 2x \), \( x \)

### Step 3: Use the given equilibrium information
We know that at equilibrium, the partial pressure of \( O_2 \) is given as:
\[ P_{O_2} = x = 0.25 \, \text{atm} \]

### Step 4: Calculate the partial pressure of \( NO \)
Since the stoichiometry of the reaction shows that 2 moles of \( NO \) are produced for every mole of \( O_2 \), we can find the partial pressure of \( NO \):
\[ P_{NO} = 2x = 2 \times 0.25 \, \text{atm} = 0.5 \, \text{atm} \]

### Step 5: Write the expression for the equilibrium constant \( K_p \)
The expression for the equilibrium constant \( K_p \) is given by:
\[ K_p = \frac{(P_{NO})^2 \cdot (P_{O_2})}{(P_{NO_2})^2} \]
Substituting the known values:
\[ 156.25 = \frac{(0.5)^2 \cdot (0.25)}{(P_{NO_2})^2} \]

### Step 6: Rearrange to find \( P_{NO_2} \)
Rearranging the equation gives:
\[ (P_{NO_2})^2 = \frac{(0.5)^2 \cdot (0.25)}{156.25} \]

### Step 7: Calculate \( P_{NO_2} \)
Calculating the numerator:
\[ (0.5)^2 = 0.25 \quad \text{and} \quad 0.25 \cdot 0.25 = 0.0625 \]
Now substituting back:
\[ (P_{NO_2})^2 = \frac{0.0625}{156.25} = 0.0004 \]
Taking the square root gives:
\[ P_{NO_2} = \sqrt{0.0004} = 0.02 \, \text{atm} \]

### Final Answer
The partial pressure of \( NO_2 \) at equilibrium is:
\[ \boxed{0.02 \, \text{atm}} \]