2 is the equilibrium constant for the reaction `A_(2)+B_(2)hArr 2AB` at a given temperature. What is the degree of dissociation for `A_(2)" or "B_(2)`

Options

A. 0.2
B. 0.5
C. `(1)/(1+sqrt2)`
D. `(sqrt2)/(1+sqrt2)`

Answer

To solve the problem, we need to determine the degree of dissociation (α) for the reaction:

\[ A_2 + B_2 \rightleftharpoons 2AB \]

Given that the equilibrium constant \( K_c \) for this reaction is 2 at a certain temperature, we can follow these steps:

### Step 1: Set up the initial concentrations
Assume we start with 1 mole of \( A_2 \) and 1 mole of \( B_2 \) in a 1-liter container. Therefore, the initial concentrations are:
- \([A_2] = 1 \, \text{mol/L}\)
- \([B_2] = 1 \, \text{mol/L}\)
- \([AB] = 0 \, \text{mol/L}\)

### Step 2: Define the change in concentration
Let \( x \) be the amount of \( A_2 \) and \( B_2 \) that dissociates. At equilibrium, the concentrations will be:
- \([A_2] = 1 - x\)
- \([B_2] = 1 - x\)
- \([AB] = 2x\) (since 2 moles of \( AB \) are produced for every mole of \( A_2 \) and \( B_2 \) that reacts)

### Step 3: Write the expression for the equilibrium constant
The equilibrium constant \( K_c \) is given by the expression:
\[ K_c = \frac{[AB]^2}{[A_2][B_2]} \]
Substituting the equilibrium concentrations:
\[ K_c = \frac{(2x)^2}{(1-x)(1-x)} = \frac{4x^2}{(1-x)^2} \]

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

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

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

### Step 7: Find the degree of dissociation
The degree of dissociation \( \alpha \) is defined as:
\[ \alpha = \frac{x}{\text{initial concentration}} = \frac{-1 + \sqrt{2}}{1} = -1 + \sqrt{2} \]

### Final Answer
Thus, the degree of dissociation for \( A_2 \) or \( B_2 \) is:
\[ \alpha = -1 + \sqrt{2} \]