At 320 K, a gas `A_(2)` is 20% dissociated to A(g). The standard free energy change at 320 K and 1 atm in `J mol^(-1)` is approximately : `(R =8.314 Jk^(-1)"mol"^(-1), "In" 2=0.693, "In" 3 = 1.098)`

Options

A. 4763
B. 2068
C. 1844
D. 4281

Explanation

To solve the problem, we need to determine the standard free energy change (ΔG°) for the dissociation of gas A₂ at 320 K, given that it is 20% dissociated. We'll follow these steps:

### Step 1: Determine Initial Concentrations
Assume we start with 1 mole of A₂ in a 1 L container. Therefore, the initial concentration of A₂ is:
\[ [A_2]_{initial} = 1 \, \text{M} \]
And the initial concentration of A is:
\[ [A]_{initial} = 0 \, \text{M} \]

### Step 2: Calculate Concentrations After Dissociation
Since 20% of A₂ is dissociated, the amount of A₂ that dissociates is:
\[ \text{Dissociated A}_2 = 1 \, \text{mol} \times 0.20 = 0.2 \, \text{mol} \]

This means that:
- Remaining A₂ concentration:
\[ [A_2] = 1 - 0.2 = 0.8 \, \text{M} \]
- The amount of A produced from the dissociation of A₂:
\[ \text{Produced A} = 2 \times 0.2 = 0.4 \, \text{mol} \]
Thus, the concentration of A is:
\[ [A] = 0 + 0.4 = 0.4 \, \text{M} \]

### Step 3: Calculate the Equilibrium Constant (K)
The equilibrium constant K for the reaction \( A_2 \rightleftharpoons 2A \) is given by:
\[ K = \frac{[A]^2}{[A_2]} \]
Substituting the concentrations we found:
\[ K = \frac{(0.4)^2}{0.8} = \frac{0.16}{0.8} = 0.2 \]

### Step 4: Calculate Standard Free Energy Change (ΔG°)
The formula for the standard free energy change is:
\[ \Delta G° = -RT \ln K \]
Where:
- R = 8.314 J/(K·mol)
- T = 320 K
- K = 0.2

Now, we need to calculate \( \ln(0.2) \). Using the relationship \( \ln(0.2) = \ln\left(\frac{2}{10}\right) = \ln(2) - \ln(10) \) and knowing that \( \ln(10) \approx 2.303 \):
\[ \ln(0.2) = \ln(2) - \ln(10) = 0.693 - 2.303 \approx -1.610 \]

Now substituting into the ΔG° formula:
\[ \Delta G° = - (8.314 \, \text{J/(K·mol)}) \times (320 \, \text{K}) \times (-1.610) \]
\[ \Delta G° \approx 8.314 \times 320 \times 1.610 \]
\[ \Delta G° \approx 4281 \, \text{J/mol} \]

### Conclusion
The standard free energy change at 320 K and 1 atm is approximately:
\[ \Delta G° \approx -4281 \, \text{J/mol} \]