The standard Gibbs free energy change of for the reaction, `2AB hArr A_2 +B_2` is 11.8 KJ at 230 K and 1 atm. Calculate degree of dissociation of AB at 230 K.

Options

A. 8.37 %
B. 0.837 %
C. 83.7 %
D. 0.0837 %

Explanation

To calculate the degree of dissociation of AB at 230 K for the reaction:

\[ 2AB \rightleftharpoons A_2 + B_2 \]

we will follow these steps:

### Step 1: Understanding the Reaction
Initially, we have 2 moles of AB, which dissociate into 1 mole of A2 and 1 mole of B2. Let the degree of dissociation be represented by \( \alpha \).

### Step 2: Setting Up Initial and Equilibrium Concentrations
- Initial moles of AB = 2
- At equilibrium:
- Moles of AB = \( 2(1 - \alpha) \)
- Moles of A2 = \( \alpha \)
- Moles of B2 = \( \alpha \)

### Step 3: Expression for Equilibrium Constant (K)
The equilibrium constant \( K \) for the reaction can be expressed as:

\[ K = \frac{[A_2][B_2]}{[AB]^2} \]

Substituting the equilibrium concentrations:

\[ K = \frac{\left(\frac{\alpha}{V}\right)\left(\frac{\alpha}{V}\right)}{\left(\frac{2(1 - \alpha)}{V}\right)^2} \]

This simplifies to:

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

### Step 4: Relating Gibbs Free Energy to K
The relationship between the Gibbs free energy change (\( \Delta G \)) and the equilibrium constant (\( K \)) is given by:

\[ \Delta G = -RT \ln K \]

Given that \( \Delta G = 11.8 \, \text{kJ} = 11800 \, \text{J} \) at \( T = 230 \, \text{K} \), we can rearrange this to find \( K \):

\[ K = e^{-\frac{\Delta G}{RT}} \]

### Step 5: Substituting Values
Using \( R = 8.314 \, \text{J/(mol K)} \):

\[ K = e^{-\frac{11800}{8.314 \times 230}} \]

Calculating the exponent:

\[ K = e^{-6.207} \approx 0.002 \]

### Step 6: Solving for Alpha
Now, substituting \( K \) back into the equilibrium expression:

\[ 0.002 = \frac{\alpha^2}{4(1 - \alpha)^2} \]

Cross-multiplying gives:

\[ 0.002 \cdot 4(1 - \alpha)^2 = \alpha^2 \]

This simplifies to:

\[ 0.008(1 - 2\alpha + \alpha^2) = \alpha^2 \]

Expanding and rearranging:

\[ 0.008 - 0.016\alpha + 0.008\alpha^2 = \alpha^2 \]

\[ 0.008\alpha^2 - \alpha^2 - 0.016\alpha + 0.008 = 0 \]

\[ -0.992\alpha^2 - 0.016\alpha + 0.008 = 0 \]

### Step 7: Solving the Quadratic Equation
Using the quadratic formula \( \alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = -0.992, b = -0.016, c = 0.008 \).

Calculating the discriminant:

\[ D = b^2 - 4ac = (-0.016)^2 - 4(-0.992)(0.008) \]

Calculating \( D \):

\[ D = 0.000256 + 0.031744 = 0.032 \]

Now substituting into the quadratic formula:

\[ \alpha = \frac{-(-0.016) \pm \sqrt{0.032}}{2 \cdot -0.992} \]

Calculating \( \alpha \):

\[ \alpha = \frac{0.016 \pm 0.179}{-1.984} \]

Taking the positive root:

\[ \alpha \approx 0.0837 \]

### Step 8: Calculating Percentage of Dissociation
To find the percentage of dissociation:

\[ \text{Percentage of dissociation} = \alpha \times 100 = 0.0837 \times 100 = 8.37\% \]

### Final Answer
The degree of dissociation of AB at 230 K is approximately **8.37%**.

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