\[ \Delta G^\circ = -RT \ln K_p \]
\[ \Delta G^\circ = -8.314 \times 10^{-3} \times 298 \times \ln(2.47 \times 10^{-29}) \]
\[ = -8.314 \times 10^{-3} \times 298 \times (-65.87) \]
\[ = 163.19 \, \text{kJ} \]
If \[ \frac{dy}{dx} + 2y \sec^2 x = 2 \sec^2 x + 3 \tan x \cdot \sec^2 x \] and
and \( f(0) = \frac{5}{4} \), then the value of \[ 12 \left( y \left( \frac{\pi}{4} \right) - \frac{1}{e^2} \right) \] equals to: