Urgent chem help needed!! first correct answer gets 10 points!?

Use tabulated thermodynamic data to calculate K at 298 K for the following reaction.

H2(g) + O2(g) <=> H2O2(l)

heres the data:
for H2- delta h=0, delta g=0, delta s=130.6
for O2- delta h=0, delta g=0, delta s=205
for H2O2- delta h= -187.8, delta g=-120.4, delta s=110

Answer:
-120400 = -(8.3145) (298) ln K

K = 1.27 x 10^21
First we need to calculate the Gibbs Free energy from

Gibbs free energy of reaction = Gibbs free energy of products - Gibbs free energy of reactants

Gibbs Free Energy of reaction = Gibbs free energy of hydrogen peroxide - Gibbs free energy of oxygen - Gibbs free energy of oxygen

Gibbs free energy of reaction = -120. 4- 0 -0 = -120.4 kJ/mol

GIbbs free energy of reaction= -gas constant *Temperature*lnK

K= exp(- GIbbs free energy of reaction/gas constant /Temperature) = exp( 120400/8.314/298)

K= 1.273383 x 10^21.

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