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Consider the following reaction: FeO(s) + CO(g) Fe (s) + CO2 (g) KP = 0.259 at 1000 K What are the equilibrium partial pressures at 1000 K if the initial partial pressures are PCO = 1.000 atm and PCO2 = 0.500 atm?

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Answer: 0.309 atm for CO2 and 1.191 atm for CO

Step-by-step explanation:

The given reaction is:

FeO(s) + CO(g) ↔ Fe(s) + CO2(g)

This is a reversible reaction, and the equilibrium constant expression (Kp) is given by the ratio of the partial pressures of the products to the reactants at equilibrium.

To solve for the equilibrium partial pressures, we can set up an ICE table:

FeO(s) CO(g) Fe(s) CO2(g)

Initial 1.000 1.000 0.000 0.500

Change -x x x -x

Equilibrium 1.000-x 1.000+x x 0.500-x

The equilibrium constant expression can be written as:

Kp = PCO2 / PCO

Note: Because FeO and Fe are solids, their concentrations do not appear in the equilibrium expression, as their activities are considered to be 1.

Substituting the given Kp value (0.259) and the equilibrium pressures from the ICE table:

0.259 = (0.5 + x) / (1.0 - x)

Solving for x:

0.259 * (1.0 - x) = 0.5 + x

0.259 - 0.259x = 0.5 + x

0.259 - 0.259x - x = 0.5

0.259 - 1.259x = 0.5

-1.259x = 0.5 - 0.259

-1.259x = 0.241

x = 0.241 / -1.259

x = -0.191

Substituting x back into the equilibrium expressions gives the equilibrium partial pressures:

PCO2 = 0.5 + x = 0.5 - 0.191 = 0.309 atm

PCO = 1.0 - x = 1.0 + 0.191 = 1.191 atm

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