The equilibrium constant for the following reaction is $K_{p} = 3.44 \times 10^{24}$ at $25 ^{\circ}C$. Calculate the value of $\Delta_{f}G^{o}(SO_2)$. Given that the value of $\Delta_{f}G^{o}(SO_3)$ is $-88.52 \ kcal/mol$. The reaction is $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$.

The standard Gibbs free energy change for the reaction is given by $\Delta_{r}G^{o} = -RT \ln K_{p}$.
Given $R = 1.987 \times 10^{-3} \ kcal/(mol \cdot K)$,$T = 298 \ K$,and $K_{p} = 3.44 \times 10^{24}$.
$\Delta_{r}G^{o} = -(1.987 \times 10^{-3}) \times 298 \times \ln(3.44 \times 10^{24})$.
$\Delta_{r}G^{o} = -0.592 \times (\ln(3.44) + 24 \ln(10)) \approx -0.592 \times (1.235 + 55.288) \approx -34.5 \ kcal/mol$.
Also,$\Delta_{r}G^{o} = 2 \Delta_{f}G^{o}(SO_3) - 2 \Delta_{f}G^{o}(SO_2) - \Delta_{f}G^{o}(O_2)$.
Since $\Delta_{f}G^{o}(O_2) = 0$,we have $-34.5 = 2(-88.52) - 2 \Delta_{f}G^{o}(SO_2)$.
$-34.5 = -177.04 - 2 \Delta_{f}G^{o}(SO_2)$.
$2 \Delta_{f}G^{o}(SO_2) = -177.04 + 34.5 = -142.54$.
$\Delta_{f}G^{o}(SO_2) = -71.27 \ kcal/mol$.