Equilibrium Calculation for Fe3O4-CO Reaction at 600°C and 5.00 atm

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SUMMARY

The equilibrium calculation for the reaction Fe3O4(s) + CO(g) ↔ 3 FeO(s) + CO2(g) at 600°C and 5.00 atm reveals that the equilibrium constant Kp is 1.15. The initial moles present were 1 mole of Fe3O4, 2 moles of CO, 0.5 moles of FeO, and 0.3 moles of CO2. The correct calculation for the change in moles (x) at equilibrium is 0.93, indicating an error in the initial assumption that Δn = 0. The activities of solids do not affect the equilibrium expression, confirming that Kc equals Kp in this case.

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Homework Statement


The Reaction
Fe3O4(s) + CO(g) ↔ 3 FeO(s) + CO2(g)
took place at 600°C and at a constant total pressure of 5.00 atm. Determine the amount of each substance at equilibrium if there were 1 mole of Fe3O4, 2 moles of CO, .5 moles of FeO, and .3 moles of CO2 originally present in the mixture and knowing that Kp = 1.15.

Homework Equations


Kp = Kc* (R* T)^Δn

The Attempt at a Solution


So I've written out my RICE table. I assumed since Δn = 0 that Kc = Kp that I would be able to use Kc in my equation. Because the activities of solids is 1, I can assume they won't impact the equilibrium equation. My solution looked like
1.15 = (.3+x)/(2-x) which goes to x = .482. However the solution says that x should be .93. I'm not sure where I went wrong. Thank you in advance!
 
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