Consider the following reaction at 298K . (3)/(2) O _ 2(g) ⇌ O _...

Given $R = 8 . 314{JK}^{– 1}{mol}^{– 1}$

$\dfrac{3}{2}{O}_{2\left(g\right)} ⇌ {O}_{3\left(g\right)} . {K}_{p} = 2 . 47 \times {10}^{- 29}$

Standard Gibbs free energy of the formation of a compound is basically the change of Gibbs free energy that is followed by the formation of one mole of that substance from its component element available at their standard states or the most stable form of the element which is at $25 ° C$  and $100$ kPa. Its symbol is ${\Delta}_{f}G ˚$.

${∆}_{r}{G}^{0} = - RTln{K}_{p}$

$= - 8 . 314 \times {10}^{– 3} \times 298 \times \ln ( 2 . 47 \times {10}^{– 29} )$

$= – 8 . 314 \times {10}^{– 3} \times 298 \times ( – 65 . 87 )$

$= 163 . 19kJ$