Given:

H2 (g) + O2 (g) ⇌ H2O2 (g) K = 2.3 x 106 at 600 K
2 H2 (g) + O2 (g) ⇌ 2 H2O (g) K = 1.8 x 1037 at 600 K

Calculate Gº for the following reaction:

H2O (g) + 1/2 O2 (g) ⇌ H2O2 (g)

Use equation 1 as is.

Use equation 2, take 1/2 of it and reverse the equation. The new K will be (1/sqrtK2.
Krxn for the reaction will be K1*(1/sqrtK2)
Then delta Go = -RT*lnKrxn

To calculate the standard Gibbs free energy change (ΔGº) for the reaction:

H2O (g) + 1/2 O2 (g) ⇌ H2O2 (g)

we can use the Gibbs free energy change (∆Gº) values for the given reactions and apply the following equation:

∆Gº = ∑(∆Gº products) - ∑(∆Gº reactants)

First, we need to determine the corresponding reactants and products for this reaction.

Reactants:
H2O (g)
1/2 O2 (g)

Product:
H2O2 (g)

Now, we can calculate ∆Gº for this reaction:

∆Gº = ∆Gº(H2O2) - [∆Gº(H2O) + ∆Gº(1/2 O2)]

Using the given information:

∆Gº(H2O2) = 0 (since it is not given)
∆Gº(H2O) = 0 (since it is not given)
∆Gº(1/2 O2) = 0 (since it is not given)

Now we have:

∆Gº = 0 - [0 + 0]
∆Gº = 0

Therefore, the standard Gibbs free energy change (∆Gº) for the reaction

H2O (g) + 1/2 O2 (g) ⇌ H2O2 (g)

is 0.