Given:

Question

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)

Answer 1

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