For the reaction shown below
O3(g) → 3/2 O2(g)
at 298 K, ΔGo = -163 kJ/mol. The value of the equilibrium constant, Kp, for this reaction is?
The free energy is calculated as
ΔrG0 =−RTlnKp
Where T=298K
R=8.314
∆rG0 = -163kJ/mol
-163kJ/mol = 8.314×298×ln(Kp)
Kp = 2.47×10-29
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