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 find the degree of dissociation for \( A_2 \) or \( B_2 \) in the reaction \( A_2 + B_2 \rightleftharpoons 2AB \) with an equilibrium constant \( K_c = 2 \), we can follow these steps:

### Step 1: Set up the initial concentrations
Assume the initial concentrations of \( A_2 \) and \( B_2 \) are both 1 M. At the start of the reaction, no products are formed, so:
- \([A_2] = 1\)
- \([B_2] = 1\)
- \([AB] = 0\)

### Step 2: Define the change in concentration
Let \( x \) be the degree of dissociation of \( A_2 \) and \( B_2 \). When \( A_2 \) and \( B_2 \) dissociate, the changes in concentration will be:
- \([A_2] = 1 - x\)
- \([B_2] = 1 - x\)
- \([AB] = 2x\) (since 2 moles of \( AB \) are produced for every mole of \( A_2 \) or \( B_2 \) that dissociates)

### Step 3: Write the expression for the equilibrium constant
The equilibrium constant \( K_c \) is given by the formula:
\[ K_c = \frac{[\text{products}]^{\text{coefficients}}}{[\text{reactants}]^{\text{coefficients}}} \]
For our reaction, this becomes:
\[ K_c = \frac{(2x)^2}{(1-x)(1-x)} \]
Given \( K_c = 2 \), we can set up the equation:
\[ 2 = \frac{(2x)^2}{(1-x)^2} \]

### Step 4: Simplify the equation
Expanding the equation gives:
\[ 2 = \frac{4x^2}{(1-x)^2} \]
Cross-multiplying leads to:
\[ 2(1-x)^2 = 4x^2 \]
Expanding the left side:
\[ 2(1 - 2x + x^2) = 4x^2 \]
This simplifies to:
\[ 2 - 4x + 2x^2 = 4x^2 \]
Rearranging gives:
\[ 2 - 4x - 2x^2 = 0 \]

### Step 5: Solve the quadratic equation
Rearranging the equation:
\[ 2x^2 + 4x - 2 = 0 \]
Dividing through by 2:
\[ x^2 + 2x - 1 = 0 \]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1 \), \( b = 2 \), and \( c = -1 \):
\[ x = \frac{-2 \pm \sqrt{(2)^2 - 4(1)(-1)}}{2(1)} = \frac{-2 \pm \sqrt{4 + 4}}{2} = \frac{-2 \pm \sqrt{8}}{2} = \frac{-2 \pm 2\sqrt{2}}{2} \]
This simplifies to:
\[ x = -1 \pm \sqrt{2} \]
Since \( x \) must be a positive value, we take:
\[ x = -1 + \sqrt{2} \]

### Step 6: Find the degree of dissociation
The degree of dissociation \( \alpha \) is given by \( x \):
\[ \alpha = -1 + \sqrt{2} \]

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