Consider the atomization of `Br_(2)(g)` according to the reaction :
`Br_(2)(g)hArr 2Br(g)" "{"Given antilog "(-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 of calculating the degree of dissociation of bromine gas (Br₂) at a given temperature and pressure, we will follow these steps:

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

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

### Step 2: Calculate \( \Delta S \)
The change in entropy \( \Delta S \) for the reaction can be calculated using the formula:
\[ \Delta S = \sum S_{\text{products}} - \sum S_{\text{reactants}} = 2 \cdot S_{\text{Br}} - S_{\text{Br}_2} \]
Substituting the values:
\[ \Delta S = 2 \cdot 175 \, \text{J/K} - 245.4 \, \text{J/K} = 350 - 245.4 = 104.6 \, \text{J/K} \]

### Step 3: Calculate \( \Delta G \)
Using the Gibbs free energy equation:
\[ \Delta G = \Delta H - T \Delta S \]
Substituting the values:
\[ \Delta G = 82000 \, \text{J/mol} - 500 \cdot 104.6 \, \text{J/mol} \]
Calculating \( T \Delta S \):
\[ T \Delta S = 500 \cdot 104.6 = 52300 \, \text{J/mol} \]
Now substituting back:
\[ \Delta G = 82000 - 52300 = 29700 \, \text{J/mol} \]

### Step 4: Relate \( \Delta G \) to \( K_p \)
The relationship between \( \Delta G \) and the equilibrium constant \( K_p \) is given by:
\[ \Delta G = -RT \ln K_p \]
Rearranging gives:
\[ \ln K_p = -\frac{\Delta G}{RT} \]
Substituting \( R = 8.314 \, \text{J/(K} \cdot \text{mol)} \):
\[ \ln K_p = -\frac{29700}{8.314 \cdot 500} \]
Calculating:
\[ \ln K_p = -\frac{29700}{4157} \approx -7.15 \]
Thus,
\[ K_p = e^{-7.15} \approx 9.03 \times 10^{-4} \]

### Step 5: Set Up the Expression for \( K_p \)
Let the initial moles of \( \text{Br}_2 \) be \( A \) and the degree of dissociation be \( \alpha \). At equilibrium:
- Moles of \( \text{Br}_2 = A(1 - \alpha) \)
- Moles of \( \text{Br} = 2A\alpha \)

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

### Step 6: Calculate Mole Fractions and Partial Pressures
The mole fraction of \( \text{Br}_2 \) and \( \text{Br} \):
\[ \text{Mole fraction of } \text{Br}_2 = \frac{A(1 - \alpha)}{A(1 + \alpha)} = \frac{1 - \alpha}{1 + \alpha} \]
\[ \text{Mole fraction of } \text{Br} = \frac{2A\alpha}{A(1 + \alpha)} = \frac{2\alpha}{1 + \alpha} \]

The partial pressures:
\[ P_{\text{Br}_2} = \frac{(1 - \alpha)}{(1 + \alpha)} \cdot 40 \]
\[ P_{\text{Br}} = \frac{2\alpha}{(1 + \alpha)} \cdot 40 \]

### Step 7: Substitute into the \( K_p \) Expression
The expression for \( K_p \) is:
\[ K_p = \frac{(P_{\text{Br}})^2}{P_{\text{Br}_2}} = \frac{\left(\frac{2\alpha}{(1 + \alpha)} \cdot 40\right)^2}{\frac{(1 - \alpha)}{(1 + \alpha)} \cdot 40} \]
This simplifies to:
\[ K_p = \frac{(2\alpha)^2 \cdot 40}{(1 - \alpha)(1 + \alpha)} \cdot 40 \]
Substituting \( K_p \):
\[ 9.03 \times 10^{-4} = \frac{4\alpha^2 \cdot 1600}{(1 - \alpha)(1 + \alpha)} \]

### Step 8: Solve for \( \alpha \)
Assuming \( \alpha \) is small, we can approximate \( (1 - \alpha) \approx 1 \) and \( (1 + \alpha) \approx 1 \):
\[ 9.03 \times 10^{-4} \approx 6400\alpha^2 \]
Thus,
\[ \alpha^2 \approx \frac{9.03 \times 10^{-4}}{6400} \approx 1.41 \times 10^{-7} \]
Taking the square root gives:
\[ \alpha \approx \sqrt{1.41 \times 10^{-7}} \approx 0.000375 \]

### Final Result
The degree of dissociation \( \alpha \) is approximately \( 0.000375 \) or \( 3.75 \times 10^{-4} \).