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

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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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