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Chemistry Question on Thermodynamics

Consider the following reaction at 298 K.
32​O2(g)​⇌O3(g)​,Kp​=2.47×10−29\frac{3}{2}​O_{2(g)}​⇌O_{3(g)}​,K_p​=2.47×10^{−29}.
Δr​GθΔ_r​G^{\theta} for the reaction is ______ kJ. (Given R=8.314  JK−1mol−18.314\;J K^{−1}mol^{−1})

Answer

ΔG∘=−RTln⁡Kp\Delta G^\circ = -RT \ln K_p
ΔG∘=−8.314×10−3×298×ln⁡(2.47×10−29)\Delta G^\circ = -8.314 \times 10^{-3} \times 298 \times \ln(2.47 \times 10^{-29})
=−8.314×10−3×298×(−65.87)= -8.314 \times 10^{-3} \times 298 \times (-65.87)
=163.19 kJ= 163.19 \, \text{kJ}