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 concentrations
Assume we start with 1 mole of \( N_2 \) and 1 mole of \( O_2 \), and no \( NO \) is present initially. Therefore, the initial concentrations can be represented as:
- \([N_2] = 1\)
- \([O_2] = 1\)
- \([NO] = 0\)

### Step 2: Define the change in concentration
Let \( x \) be the degree of association of \( N_2 \) and \( O_2 \). As the reaction proceeds towards equilibrium, the changes in concentration will be:
- \([N_2] = 1 - x\)
- \([O_2] = 1 - x\)
- \([NO] = 2x\)

### Step 3: Write the expression for the equilibrium constant \( K_c \)
The equilibrium constant \( K_c \) for the reaction is given by:
\[ K_c = \frac{[NO]^2}{[N_2][O_2]} \]
Substituting the concentrations at equilibrium:
\[ K_c = \frac{(2x)^2}{(1 - x)(1 - x)} = \frac{4x^2}{(1 - x)^2} \]

### Step 4: Set up the equation using the given \( K_c \)
We know that \( K_c = 2 \), so we can set up the equation:
\[ \frac{4x^2}{(1 - x)^2} = 2 \]

### Step 5: Cross-multiply and simplify
Cross-multiplying gives:
\[ 4x^2 = 2(1 - x)^2 \]
Expanding the right side:
\[ 4x^2 = 2(1 - 2x + x^2) \]
\[ 4x^2 = 2 - 4x + 2x^2 \]
Bringing all terms to one side:
\[ 4x^2 - 2x^2 + 4x - 2 = 0 \]
This simplifies to:
\[ 2x^2 + 4x - 2 = 0 \]

### Step 6: Solve the quadratic equation
Dividing the entire equation by 2:
\[ x^2 + 2x - 1 = 0 \]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-1)}}{2(1)} = \frac{-2 \pm \sqrt{4 + 4}}{2} = \frac{-2 \pm \sqrt{8}}{2} = \frac{-2 \pm 2\sqrt{2}}{2} \]
This simplifies to:
\[ x = -1 \pm \sqrt{2} \]
Since \( x \) must be a positive value (as it represents the degree of association), we take:
\[ x = -1 + \sqrt{2} \]

### Step 7: Calculate the degree of association
The degree of association \( \alpha \) can be expressed as:
\[ \alpha = \frac{x}{\text{initial concentration}} = -1 + \sqrt{2} \]

### Final Answer
Thus, the degree of association is:
\[ \alpha = -1 + \sqrt{2} \]