`K_(p)` for equilibrium `N_(2)O_(4) hArr 2NO_(2)` is `0.25` at `15^(@)C`. If the system is allowed to expand `&` `N_(2)` is added at a constant pressure of 1 atm. What will be the degree when partial of `N_(2)` is 0.6 atm.

Options

A. `0.38`
B. `0.23`
C. `0.61`
D. `0.55`

Explanation

To solve the problem step by step, we will follow the equilibrium principles and use the given data effectively.

### Step 1: Write the equilibrium expression
The equilibrium reaction is:
\[ N_2O_4 \rightleftharpoons 2 NO_2 \]

The equilibrium constant \( K_p \) is given by:
\[ K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} \]
Given \( K_p = 0.25 \) at \( 15^\circ C \).

### Step 2: Define initial conditions
Let the initial pressure of \( N_2O_4 \) be \( P_0 \) and the initial pressure of \( NO_2 \) be \( 0 \) at time \( t = 0 \).

At equilibrium, let \( x \) be the change in pressure of \( N_2O_4 \) that dissociates. Thus, the pressures at equilibrium will be:
- \( P_{N_2O_4} = P_0 - x \)
- \( P_{NO_2} = 2x \)

### Step 3: Substitute into the equilibrium expression
Substituting these into the \( K_p \) expression:
\[ K_p = \frac{(2x)^2}{P_0 - x} = 0.25 \]
This simplifies to:
\[ \frac{4x^2}{P_0 - x} = 0.25 \]
Multiplying both sides by \( P_0 - x \):
\[ 4x^2 = 0.25(P_0 - x) \]

### Step 4: Rearranging the equation
Rearranging gives:
\[ 4x^2 + 0.25x - 0.25P_0 = 0 \]
This is a quadratic equation in \( x \).

### Step 5: Total pressure after adding \( N_2 \)
When \( N_2 \) is added, the total pressure becomes \( 1 \, atm \). The total pressure can be expressed as:
\[ P_{total} = P_{N_2O_4} + P_{NO_2} + P_{N_2} \]
Where \( P_{N_2} = 0.6 \, atm \) (given).

Thus:
\[ P_{total} = (P_0 - x) + 2x + 0.6 = 1 \]
This simplifies to:
\[ P_0 + x + 0.6 = 1 \]
So:
\[ P_0 + x = 0.4 \]
Thus:
\[ P_0 = 0.4 - x \]

### Step 6: Substitute \( P_0 \) into the quadratic equation
Now substitute \( P_0 \) back into the quadratic equation:
\[ 4x^2 + 0.25x - 0.25(0.4 - x) = 0 \]
This simplifies to:
\[ 4x^2 + 0.25x - 0.1 + 0.25x = 0 \]
\[ 4x^2 + 0.5x - 0.1 = 0 \]

### Step 7: Solve the quadratic equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- \( a = 4, b = 0.5, c = -0.1 \)

Calculating the discriminant:
\[ b^2 - 4ac = (0.5)^2 - 4(4)(-0.1) = 0.25 + 1.6 = 1.85 \]

Now substituting into the formula:
\[ x = \frac{-0.5 \pm \sqrt{1.85}}{8} \]
Calculating \( \sqrt{1.85} \approx 1.36 \):
\[ x = \frac{-0.5 \pm 1.36}{8} \]

Calculating the two possible values for \( x \):
1. \( x = \frac{0.86}{8} \approx 0.1075 \)
2. \( x = \frac{-1.86}{8} \) (not valid since pressure cannot be negative)

### Step 8: Calculate \( P \)
Now substitute \( x \) back to find \( P \):
Using \( P = 4x^2 \):
\[ P = 4(0.1075)^2 \approx 0.046 \]

### Step 9: Calculate degree of dissociation \( \alpha \)
The degree of dissociation \( \alpha \) can be calculated using:
\[ \alpha = \frac{x}{P_0} \]
Where \( P_0 = 0.4 - x \):
\[ \alpha = \frac{0.1075}{0.4 - 0.1075} = \frac{0.1075}{0.2925} \approx 0.367 \]

### Final Result
Thus, the degree of dissociation is approximately \( 0.38 \).