For the reaction, `N_2(g) +O_2(g) hArr 2NO(g)`
Equilibrium constant `k_c=2`
Degree of association is

Options

A. `1/(1-sqrt2)`
B. `1/(1+sqrt2)`
C. `2/(1+sqrt2)`
D. `2/(1-sqrt2)`

Answer

To find the degree of association for the reaction \( N_2(g) + O_2(g) \rightleftharpoons 2NO(g) \) with an equilibrium constant \( K_c = 2 \), we can follow these steps:

### Step 1: Set up the initial conditions
Assume we start with 1 mole of \( N_2 \) and 1 mole of \( O_2 \) at the beginning (time \( t = 0 \)). Initially, there are 0 moles of \( NO \).

- Initial concentrations:
- \( [N_2] = 1 \) M
- \( [O_2] = 1 \) M
- \( [NO] = 0 \) M

### Step 2: Define the change in concentration
Let \( x \) be the amount of \( N_2 \) and \( O_2 \) that reacts to form \( NO \). According to the stoichiometry of the reaction:

- At equilibrium:
- \( [N_2] = 1 - x \)
- \( [O_2] = 1 - x \)
- \( [NO] = 2x \) (since 2 moles of \( NO \) are produced for every mole of \( N_2 \) and \( O_2 \) that reacts)

### Step 3: Write the expression for the equilibrium constant
The equilibrium constant \( K_c \) for the reaction is given by:

\[ K_c = \frac{[NO]^2}{[N_2][O_2]} = \frac{(2x)^2}{(1-x)(1-x)} \]

### Step 4: Substitute the values into the equilibrium expression
Given \( K_c = 2 \):

\[ 2 = \frac{(2x)^2}{(1-x)^2} \]

### Step 5: Simplify the equation
This simplifies to:

\[ 2 = \frac{4x^2}{(1-x)^2} \]

Cross-multiplying gives:

\[ 2(1-x)^2 = 4x^2 \]

Expanding the left side:

\[ 2(1 - 2x + x^2) = 4x^2 \]

This leads to:

\[ 2 - 4x + 2x^2 = 4x^2 \]

Rearranging gives:

\[ 2 - 4x - 2x^2 = 0 \]

### Step 6: Rearrange into standard quadratic form
Rearranging leads to:

\[ 2x^2 + 4x - 2 = 0 \]

Dividing through by 2 simplifies to:

\[ x^2 + 2x - 1 = 0 \]

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

Here, \( a = 1, b = 2, c = -1 \):

\[ x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} \]

Calculating the discriminant:

\[ x = \frac{-2 \pm \sqrt{4 + 4}}{2} = \frac{-2 \pm \sqrt{8}}{2} = \frac{-2 \pm 2\sqrt{2}}{2} = -1 \pm \sqrt{2} \]

### Step 8: Choose the valid solution
Since \( x \) must be a positive value, we take:

\[ x = -1 + \sqrt{2} \]

### Step 9: Calculate the degree of association
The degree of association \( \alpha \) can be defined as the fraction of the original reactants that have reacted:

\[ \alpha = \frac{x}{1} = -1 + \sqrt{2} \]

### Final Result
Thus, the degree of association is:

\[ \alpha = -1 + \sqrt{2} \]