1. Reaction and degree of dissociation

The dissociation is:

$A_2(g) \rightleftharpoons 2A(g)$

Given degree of dissociation:

$\alpha = 20\% = 0.2$

2. Equilibrium composition starting with 1 mole of $A_2$

Initially:

• $A_2 = 1$
• $A = 0$

At equilibrium:

• $A_2 = 1-\alpha = 0.8$
• $A = 2\alpha = 0.4$

Total moles at equilibrium:

$n_{\text{total}} = 0.8 + 0.4 = 1.2$

3. Partial pressures at total pressure 1 atm

Since total pressure is $1\,\text{atm}$,

$p_{A_2} = \frac{0.8}{1.2} = \frac{2}{3}\,\text{atm}$

$p_A = \frac{0.4}{1.2} = \frac{1}{3}\,\text{atm}$

4. Equilibrium constant $K_p$

For

$A_2(g) \rightleftharpoons 2A(g)$

$K_p = \frac{(p_A)^2}{p_{A_2}}$

Substitute:

$K_p = \frac{\left(\frac{1}{3}\right)^2}{\frac{2}{3}} = \frac{1}{9}\cdot\frac{3}{2} = \frac{1}{6}$

So,

$K_p = \frac{1}{6}$

5. Standard free energy change

Use:

$\Delta G^\circ = -RT \ln K_p$

Since $K_p = 1/6$,

$\Delta G^\circ = -RT\ln\left(\frac{1}{6}\right)=RT\ln 6$

Now,

$\ln 6 = \ln 2 + \ln 3 = 0.693 + 1.098 = 1.791$

Therefore,

$\Delta G^\circ = 8.314 \times 320 \times 1.791$

First,

$8.314 \times 320 = 2660.48$

Then,

$\Delta G^\circ = 2660.48 \times 1.791 \approx 4764\,\text{J mol}^{-1}$

So approximately,

$\Delta G^\circ \approx 4763\,\text{J mol}^{-1}$

6. Match with options

This corresponds to:

Option A: 4763

7. Comparison with stored answer

Stored correct answer is D (4281), but the calculated answer is A (4763).

The stored answer appears inconsistent with the equilibrium calculation for 20% dissociation at 1 atm.