chemistry

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)

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asked by Claire
  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

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    posted by DrBob222

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