Question

For the reaction shown below

O_3(g) → 3/2 O_2(g)

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

Answer 1

The free energy is calculated as

ΔrG⁰ =−RTlnK_p

Where T=298K

R=8.314

∆_rG⁰ = -163kJ/mol

-163kJ/mol = 8.314×298×ln(K_p)

K_p = 2.47×10^-29