Question

A key step in the extraction of iron from its ore is feo(s) + co(g) fe(s) + co2(g) kp = 0.403 at 1000°c. this step occurs in the 700°c to 1200°c zone within a blast furnace. what are the equilibrium partial pressures of co(g) and co2(g) when 1.58 atm co(g) and excess feo(s) react in a sealed container at 1000°c?

Answer 1

The complete reaction of the problem, for better illustration, is

FeO(s) + CO(g) <--> Fe(s) + CO2(g)

The double-tailed arrow signifies that the reaction is in a dynamic chemical equilibrium. When the system is in equilibrium, the forward and the backward reaction rates have an equal ratio of Kp = 0.403 at 1000°C. The formula for Kp is

Kp = [partial pressure of products]/[partial pressure of reactants]

So, first, let's find the partial pressure of the compounds in the reaction.

                      FeO(s) + CO(g) <--> Fe(s) + CO2(g)
Initial                 x            1.58            0           0
Change           -1.58        -1.58       +1.58     +1.58
------------------------------------------------------------------
Equilbrium      x-1.58          0             1.58       1.58

Kp = [(1.58)(1.58)]/[(x-1.58)] = 0.403
x = 7.77 atm (this is the amount of excess FeO)

Therefore, the partial pressure of CO2 at equilibrium is 1.58 atm. There is no more CO because it has been consumed due to excess FeO.