$$2O_3(g) \rightleftharpoons 3O_2(g)$$
At $$300$$ K, ozone is fifty percent dissociated. The standard free energy change at this temperature and $$1$$ atm pressure is $$(-)$$ ______ J mol$$^{-1}$$. (Nearest integer)
[Given: $$\ln 1.35 = 0.3$$ and $$R = 8.3$$ J K$$^{-1}$$ mol$$^{-1}$$]

We need to find the standard free energy change for the reaction: $$2O_3(g) \rightleftharpoons 3O_2(g)$$ at 300 K where ozone is 50% dissociated.

Let us start with 2 moles of $$O_3$$ to match the stoichiometric coefficient. At 50% dissociation, the degree of dissociation $$\alpha = 0.50$$. Starting with 2 moles of $$O_3$$, 50% dissociation means 1 mole of $$O_3$$ reacts (since 50% of 2 = 1 mole reacts according to the forward reaction with coefficient 2, i.e., the reaction proceeds by $$x = 0.5$$ times). Using extent of reaction $$\xi$$: if $$\alpha = 0.5$$, then from 2 mol of $$O_3$$, 50% dissociates = 1 mol $$O_3$$ reacts. The initial amounts are $$O_3 = 2$$ mol and $$O_2 = 0$$ mol; the change is $$O_3: -1$$ mol and $$O_2: +1.5$$ mol (by stoichiometry: for every 2 mol $$O_3$$ that react, 3 mol $$O_2$$ form, so 1 mol of $$O_3$$ reacting gives 1.5 mol of $$O_2$$); at equilibrium $$O_3 = 1$$ mol and $$O_2 = 1.5$$ mol.

From the equilibrium amounts the total moles are $$n_{total} = 1 + 1.5 = 2.5$$ mol, giving mole fractions $$x_{O_3} = \frac{1}{2.5} = 0.4$$ and $$x_{O_2} = \frac{1.5}{2.5} = 0.6$$.

At total pressure P = 1 atm the partial pressures are $$p_{O_3} = 0.4 \times 1 = 0.4$$ atm and $$p_{O_2} = 0.6 \times 1 = 0.6$$ atm.

The equilibrium constant in terms of partial pressures is $$K_p = \frac{(p_{O_2})^3}{(p_{O_3})^2} = \frac{(0.6)^3}{(0.4)^2}$$ and $$K_p = \frac{0.216}{0.16} = 1.35$$.

Using the formula for standard free energy change gives $$\Delta G° = -RT \ln K_p$$, then $$\Delta G° = -(8.3)(300) \ln(1.35)$$, followed by $$\Delta G° = -(8.3)(300)(0.3)$$, and thus $$\Delta G° = -747 \text{ J mol}^{-1}$$.

The magnitude of the standard free energy change is 747 J mol$$^{-1}$$.