At `60^(@) N_(2)O_(4)` is 50% dissociated at this temperature and one atmosphere pressure the standard free energy change is

Options

A. `-863.8Jmol^(-1)`
B. `-963.8Jmol^(-1)`
C. `-973.8Jmol^(-1)`
D. `-796.9Jmol^(-1)`

Explanation

To find the standard free energy change (ΔG°) for the dissociation of N₂O₄ at 60°C and 1 atm, we can follow these steps:

### Step 1: Write the dissociation reaction
The dissociation of dinitrogen tetroxide (N₂O₄) can be represented as:
\[ \text{N}_2\text{O}_4 (g) \rightleftharpoons 2 \text{NO}_2 (g) \]

### Step 2: Determine the degree of dissociation
Given that N₂O₄ is 50% dissociated, we can denote the degree of dissociation (α) as:
\[ \alpha = 0.5 \]

### Step 3: Calculate the moles at equilibrium
Assuming we start with 1 mole of N₂O₄:
- Initial moles of N₂O₄ = 1
- Moles of N₂O₄ at equilibrium = \( 1 - \alpha = 1 - 0.5 = 0.5 \)
- Moles of NO₂ produced = \( 2\alpha = 2 \times 0.5 = 1 \)

Thus, at equilibrium:
- Moles of N₂O₄ = 0.5
- Moles of NO₂ = 1
- Total moles at equilibrium = \( 0.5 + 1 = 1.5 \)

### Step 4: Calculate the partial pressures at equilibrium
Using the total pressure of 1 atm:
- Partial pressure of N₂O₄:
\[ P_{\text{N}_2\text{O}_4} = \frac{0.5}{1.5} \times 1 \text{ atm} = \frac{1}{3} \text{ atm} \]

- Partial pressure of NO₂:
\[ P_{\text{NO}_2} = \frac{1}{1.5} \times 1 \text{ atm} = \frac{2}{3} \text{ atm} \]

### Step 5: Calculate Kp
The equilibrium constant \( K_p \) for the reaction is given by:
\[ K_p = \frac{(P_{\text{NO}_2})^2}{P_{\text{N}_2\text{O}_4}} \]
Substituting the values:
\[ K_p = \frac{\left(\frac{2}{3}\right)^2}{\frac{1}{3}} = \frac{\frac{4}{9}}{\frac{1}{3}} = \frac{4}{3} \approx 1.33 \]

### Step 6: Calculate ΔG°
Using the formula:
\[ \Delta G° = -2.303 \cdot R \cdot T \cdot \log K_p \]
Where:
- \( R = 8.314 \, \text{J/mol·K} \)
- \( T = 60°C = 333 \, \text{K} \)

Substituting the values:
\[ \Delta G° = -2.303 \cdot 8.314 \cdot 333 \cdot \log(1.33) \]

Calculating \( \log(1.33) \):
\[ \log(1.33) \approx 0.1238 \]

Now substituting this into the equation:
\[ \Delta G° = -2.303 \cdot 8.314 \cdot 333 \cdot 0.1238 \]
\[ \Delta G° \approx -789.67 \, \text{J/mol} \]

### Conclusion
The standard free energy change (ΔG°) at 60°C for the dissociation of N₂O₄ is approximately:
\[ \Delta G° \approx -789.67 \, \text{J/mol} \]