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The overall reaction for the corrosion(rusting) of iron by oxygen isUsing the following data, calcu

The overall reaction for the corrosion(rusting) of iron by oxygen is Using the following data, calculate the equilibrium constant for this reaction at 25C 4Fe(s) + 3O2(g) → 2 Fe2O3(s)


Solution: ΔG° = -RTlnK, where R =8.31 J/molK = 0.00831Kj/mol K, T is temperature in Kelvin and K is the equibrium constant. First we need to calculate ΔG°, ΔG°=ΔH°-TΔS°

To calculate ΔH°, we do ΔH°f of products -ΔH°f reactants including their coefficients.

ΔH° = 2(-286)-0 = -1652 kJ/mol

We do the same thing for ΔS° using S°values

ΔS° = 2(90) - (3*205+4*27) = -.543Kj/mol (notice we converted from J to Kj to make units match)

Now, let's calculate ΔG°


ΔG°=ΔH°-TΔS°

ΔG°= -1652 kJ/mol - 298K( -.543Kj/mol) = -1490 kJ/mol


ΔG° = -RTlnK

ΔG° = -0.00831Kj/mol K*298K*K = -1490 kJ/mol

K=e^601.68

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