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. 1844
B. 4763
C. 2068
D. 4281

Answer

To find the standard free energy change (ΔG) for the dissociation of gas A₂ at 320 K and 1 atm, we can follow these steps:

### Step 1: Determine the initial and final amounts of A₂ and A
Initially, we have 1 mole of A₂. Since A₂ is 20% dissociated, the amount of A₂ that dissociates is:
\[ \text{Amount dissociated} = 20\% \text{ of } 1 \text{ mole} = 0.2 \text{ moles} \]
Thus, the amount of A formed is:
\[ \text{Amount of A} = 2 \times 0.2 = 0.4 \text{ moles} \]
After dissociation, the amounts are:
- A₂ remaining: \(1 - 0.2 = 0.8 \text{ moles}\)
- A formed: \(0.4 \text{ moles}\)

### Step 2: Calculate the equilibrium constant (Kp)
The equilibrium constant for the reaction \(A_2(g) \rightleftharpoons 2A(g)\) can be expressed as:
\[ K_p = \frac{(P_A)^2}{(P_{A_2})} \]
Where:
- \(P_A\) is the partial pressure of A
- \(P_{A_2}\) is the partial pressure of A₂

Assuming ideal gas behavior and that the total pressure is 1 atm, we can express the partial pressures:
\[ P_{A_2} = \frac{0.8}{1} \text{ atm} = 0.8 \text{ atm} \]
\[ P_A = \frac{0.4}{1} \text{ atm} = 0.4 \text{ atm} \]
Thus:
\[ K_p = \frac{(0.4)^2}{0.8} = \frac{0.16}{0.8} = 0.2 \]

### Step 3: Calculate ΔG using the formula
The standard free energy change (ΔG) is calculated using the formula:
\[ \Delta G = -RT \ln K_p \]
Where:
- \(R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}\)
- \(T = 320 \text{ K}\)
- \(K_p = 0.2\)

Substituting the values:
\[ \Delta G = -8.314 \times 320 \times \ln(0.2) \]

### Step 4: Calculate ln(0.2)
Using the properties of logarithms:
\[ \ln(0.2) = \ln\left(\frac{2}{10}\right) = \ln(2) - \ln(10) \]
Given that \(\ln(2) = 0.693\) and \(\ln(10) \approx 2.303\):
\[ \ln(0.2) = 0.693 - 2.303 = -1.610 \]

### Step 5: Substitute ln(0.2) back into the ΔG equation
Now substituting back into the ΔG equation:
\[ \Delta G = -8.314 \times 320 \times (-1.610) \]
Calculating this gives:
\[ \Delta G = 8.314 \times 320 \times 1.610 \approx 4281 \text{ J mol}^{-1} \]

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