Question

Is Iron's tendency to burn in pure oxygen because of the partial pressure of oxygen? Would it react the same if there was the same partial pressure of oxygen in a pressurized air environment?

Answer 1

@Frostbite

Answer 2

For a reaction whose equilibrium expression only contains gases,

∆G = ∆G° + RTlnKp

Kp for this oxidation rxn (4Fe + 3O2 <-> 2Fe2O3) is inversely proportional to the partial pressure of oxygen:

Kp = 1/(p_O2)^3

As you probably know sign of ∆G indicates the tendency for the reaction to proceed spontaneously.

Assuming temperature doesn't change drastically during the process, Kp is the only factor that affects ∆G.

For ∆G to be more negative (more spontaneous rxn), Kp must decrease;
for K to decrease, p_O2 must increase.

Therefore higher p_O2 leads to greater tendency for oxidation of Fe to occur.

As p_O2 only depends on quantity, temperature, and total volume encompassing the O2 molecules (p_O2=(n_O2)*RT/V), and not on other components in the gaseous solution, it doesn't matter whether it's pure or mixture of oxygen.

Same (partial) pressure of O2 at eq => same ∆G => same tendency for rxn to occur.