1. Reaction and given data

   The reaction is $2O_3(g) \rightleftharpoons 3O_2(g)$

   At $300\,\text{K}$, ozone is $50\%$ dissociated.

2. Assume initial moles

   Let initial moles of $O_3 = 1$ and initially no $O_2$.

   Since dissociation is $50\%$, moles of $O_3$ dissociated $=0.5$.

   From $2O_3 \rightarrow 3O_2$

   if $0.5$ mol of $O_3$ is consumed, then moles of $O_2$ formed are $0.5\times \frac{3}{2}=0.75$

   So at equilibrium:

   • $O_3 = 1-0.5=0.5$
   • $O_2 = 0.75$
   • Total moles $=0.5+0.75=1.25$

3. Equilibrium partial pressures at total pressure $1\,\text{atm}$

   $p_{O_3}=\frac{0.5}{1.25}=0.4\,\text{atm}$ $p_{O_2}=\frac{0.75}{1.25}=0.6\,\text{atm}$

4. Equilibrium constant in terms of pressure

   $K_p=\frac{(p_{O_2})^3}{(p_{O_3})^2}$

   Substituting: $K_p=\frac{(0.6)^3}{(0.4)^2}$ $K_p=\frac{0.216}{0.16}=1.35$

5. Relation between standard free energy and equilibrium constant

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

   Given: $R=8.3\,\text{J K}^{-1}\text{mol}^{-1},\quad T=300\,\text{K},\quad \ln 1.35=0.3$

   Therefore, $\Delta G^\circ=-(8.3)(300)(0.3)$ $\Delta G^\circ=-747\,\text{J mol}^{-1}$

6. Final integer asked

   Since the question is of the form "The standard free energy change is $(-)\,\underline{\hspace{1cm}}\,\text{J mol}^{-1}$", the blank should be filled by $747$

Comparison with stored answer: Stored correct answer is $747$, which matches the derived answer.