For a reaction at $298 \ K$,the equilibrium constant is ${K_p} = 0.17 \times {10^{12}}$. Find the standard Gibbs free energy change $\Delta {G^\Theta }$. (Given: $R = 8.314 \ J \ mol^{-1} \ K^{-1}$)

The relationship between standard Gibbs free energy change and the equilibrium constant is given by the formula: $\Delta {G^\Theta } = -RT \ln {K_p}$
Substituting the given values:
$\Delta {G^\Theta } = -(8.314 \ J \ mol^{-1} \ K^{-1}) \times (298 \ K) \times \ln(0.17 \times 10^{12})$
$\Delta {G^\Theta } = -2477.572 \times \ln(1.7 \times 10^{11})$
Using the property $\ln(a \times b) = \ln(a) + \ln(b)$:
$\Delta {G^\Theta } = -2477.572 \times [\ln(1.7) + 11 \times \ln(10)]$
$\Delta {G^\Theta } = -2477.572 \times [0.5306 + 11 \times 2.303]$
$\Delta {G^\Theta } = -2477.572 \times [0.5306 + 25.333]$
$\Delta {G^\Theta } = -2477.572 \times 25.8636 \approx -64078 \ J \ mol^{-1} \approx -64.08 \ kJ \ mol^{-1}$