A sample of pure `NO_2` gas heated to 1000 K decomposes :
`2NO_2(g) hArr 2NO(g)+O_(2)(g)`
The equilibrium constant `K_P` is 100 atm. Analysis shows that the partial pressure of `O_2` is 0.25 atm. At equilibrium.The partial pressure of `NO_2` at equilibrium is:

Options

A. 0.03
B. 0.25
C. 0.025
D. 0.04

Answer

To solve the problem, we will follow these steps:

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

### Step 2: Define the initial conditions and changes
Let’s assume we start with an initial pressure of \(NO_2\) as \(P_{NO_2}^0\). At equilibrium, the changes in pressure can be defined as follows:
- The change in pressure of \(NO_2\) is \(-2x\) (since 2 moles of \(NO_2\) decompose).
- The change in pressure of \(NO\) is \(+2x\).
- The change in pressure of \(O_2\) is \(+x\).

### Step 3: Express equilibrium pressures
At equilibrium, the pressures can be expressed as:
- \(P_{NO_2} = P_{NO_2}^0 - 2x\)
- \(P_{NO} = 2x\)
- \(P_{O_2} = x\)

### Step 4: Use the given information
From the problem, we know:
- The equilibrium constant \(K_P = 100 \, \text{atm}\)
- The partial pressure of \(O_2\) at equilibrium is \(0.25 \, \text{atm}\).

Thus, we can set \(x = 0.25 \, \text{atm}\).

### Step 5: Calculate the pressure of \(NO\)
Using the relationship for \(NO\):
\[ P_{NO} = 2x = 2(0.25) = 0.5 \, \text{atm} \]

### Step 6: Substitute into the equilibrium expression
The equilibrium expression for \(K_P\) is:
\[ K_P = \frac{(P_{NO})^2 \cdot (P_{O_2})}{(P_{NO_2})^2} \]
Substituting the known values:
\[ 100 = \frac{(0.5)^2 \cdot (0.25)}{(P_{NO_2})^2} \]

### Step 7: Simplify the equation
Calculating the numerator:
\[ (0.5)^2 = 0.25 \]
Thus:
\[ 100 = \frac{0.25 \cdot 0.25}{(P_{NO_2})^2} \]
\[ 100 = \frac{0.0625}{(P_{NO_2})^2} \]

### Step 8: Rearranging to find \(P_{NO_2}\)
Now, rearranging gives:
\[ (P_{NO_2})^2 = \frac{0.0625}{100} \]
\[ (P_{NO_2})^2 = 0.000625 \]

### Step 9: Solve for \(P_{NO_2}\)
Taking the square root:
\[ P_{NO_2} = \sqrt{0.000625} = 0.025 \, \text{atm} \]

### Final Answer
The partial pressure of \(NO_2\) at equilibrium is:
\[ P_{NO_2} = 0.025 \, \text{atm} \]

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