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