Consider the atomization of `Br_(2)(g)` according to the reaction :
`Br_(2)(g)hArr 2Br(g)" "{"Given antiloge "(-7.02)=9xx10-4}`
If the heat of atomization of bromine gas is 82 Kj/mol, standard molar entropies of `Br(g) and Br_(2)(g)` are 175 and `"245.4 JK"^(-1)` respectively, calculate degree of dissociation when the total pressure is 40 atm at 500 K.
(assume `alpha ltlt1` in your calculation)

Options

A. `2.37xx10^(-3)`
B. `3.20xx10^(-4)`
C. `4.30xx10^(-5)`
D. `3.60xx10^(-2)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Write the Reaction and Given Data
The reaction for the atomization of bromine gas is:
\[ \text{Br}_2(g) \rightleftharpoons 2 \text{Br}(g) \]

Given data:
- Heat of atomization, \( \Delta H^\circ = 82 \, \text{kJ/mol} = 82000 \, \text{J/mol} \)
- Standard molar entropy of \( \text{Br}(g) = 175 \, \text{J/K} \)
- Standard molar entropy of \( \text{Br}_2(g) = 245.4 \, \text{J/K} \)
- Total pressure \( P = 40 \, \text{atm} \)
- Temperature \( T = 500 \, \text{K} \)

### Step 2: Calculate the Change in Entropy (\( \Delta S^\circ \))
Using the formula for change in entropy:
\[ \Delta S^\circ = 2 S^\circ_{\text{Br}} - S^\circ_{\text{Br}_2} \]
Substituting the values:
\[ \Delta S^\circ = 2 \times 175 - 245.4 = 350 - 245.4 = 104.6 \, \text{J/K} \]

### Step 3: Calculate the Change in Gibbs Free Energy (\( \Delta G^\circ \))
Using the Gibbs free energy equation:
\[ \Delta G^\circ = \Delta H^\circ - T \Delta S^\circ \]
Substituting the values:
\[ \Delta G^\circ = 82000 - 500 \times 104.6 = 82000 - 52300 = 29700 \, \text{J} \]

### Step 4: Relate \( \Delta G^\circ \) to \( K_p \)
Using the relation:
\[ \Delta G^\circ = -R T \ln K_p \]
Where \( R = 8.314 \, \text{J/(K·mol)} \). Thus,
\[ 29700 = -8.314 \times 500 \ln K_p \]
Calculating:
\[ \ln K_p = -\frac{29700}{8.314 \times 500} \approx -7.13 \]
Taking the antilog:
\[ K_p = e^{-7.13} \approx 9.0 \times 10^{-4} \]

### Step 5: Express \( K_p \) in Terms of \( \alpha \)
For the dissociation of \( \text{Br}_2 \):
- Initial moles of \( \text{Br}_2 = a \)
- At equilibrium:
- Moles of \( \text{Br}_2 = a(1 - \alpha) \)
- Moles of \( \text{Br} = 2a\alpha \)

Total moles at equilibrium:
\[ n_{\text{total}} = a(1 - \alpha) + 2a\alpha = a(1 + \alpha) \]

Using the ideal gas law, the partial pressures can be expressed as:
\[ K_p = \frac{(2a\alpha)^2}{a(1 - \alpha)} \cdot \frac{P^2}{P} = \frac{4a^2\alpha^2}{a(1 - \alpha)} \cdot \frac{1}{40} \]
Thus,
\[ K_p = \frac{4\alpha^2}{40(1 - \alpha)} = \frac{\alpha^2}{10(1 - \alpha)} \]

### Step 6: Substitute \( K_p \) and Solve for \( \alpha \)
Substituting \( K_p \):
\[ 9.0 \times 10^{-4} = \frac{\alpha^2}{10(1 - \alpha)} \]
Assuming \( \alpha \ll 1 \), \( 1 - \alpha \approx 1 \):
\[ 9.0 \times 10^{-4} \approx \frac{\alpha^2}{10} \]
Thus,
\[ \alpha^2 \approx 9.0 \times 10^{-3} \]
Taking the square root:
\[ \alpha \approx 0.0949 \approx 9.49 \times 10^{-2} \]

### Final Answer
The degree of dissociation \( \alpha \) is approximately \( 0.0949 \) or \( 9.49 \times 10^{-2} \).