Answer to Question #236913 in Physical Chemistry for shyla

Question #236913

For the reaction shown below

O3(g) → 3/2 O2(g)

at 298 K, ΔG= -163 kJ/mol. The value of the equilibrium constant, Kp, for this reaction is?


1
Expert's answer
2021-09-15T02:28:01-0400

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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