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One of the reaction that takes place in producing steel from iron ore is the reduction of iron (II) oxide by carbon monoxide to give iron metal and CO₂.

FeO(s)+CO(g)⇌Fe(s)+CO₂(g)
;Kp=0.265 atm at 1050 K.
What are the equilibrium partial pressures of CO and CO₂ respectively at 1050 K if the initial partial pressures are: PCO=1.4 atm and PCO₂=0.80 atm?

A. PCO=1.739 atm and PCO₂=0.461 atm
B. PCO=17.39 atm and PCO₂=0.461 atm
C. PCO=1.79 atm and PCO₂=0.46 atm
D. PCO=2.739 atm and PCO₂=1.461 atm

User Btwiuse
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Final answer:

The reaction in steel production FeO(s) + CO(g) ⇌ Fe(s) + CO₂(g) has equilibrium partial pressures of A: PCO = 1.739 atm and PCO₂ = 0.461 atm at 1050 K, given initial partial pressures and Kp = 0.265 atm. The correct option is A.

Step-by-step explanation:

In the context of steel production, the reaction FeO(s) + CO(g) ⇌ Fe(s) + CO₂(g) involves the reduction of iron (II) oxide by carbon monoxide to produce iron metal and carbon dioxide. Given the equilibrium constant Kp = 0.265 atm at 1050 K and initial partial pressures of PCO = 1.4 atm and PCO₂ = 0.80 atm, we can determine the equilibrium partial pressures by analyzing the changes in the reaction.

To solve this, we set up an ICE table (Initial, Change, Equilibrium) for the reaction. If x is the amount of FeO that reacts, then at equilibrium, the partial pressure of CO will decrease by x to become (1.4 - x) atm, and the partial pressure of CO₂ will increase by x to become (0.80 + x) atm. Plugging these values into the expression for Kp, we get:

Kp = PCO₂ / PCO = (0.80 + x) / (1.4 - x)

Solving for x, we get the change in partial pressures needed to reach equilibrium. Then we plug the value of x into the expressions for equilibrium partial pressures of CO and CO₂.

After calculating, we find the correct equilibrium partial pressures, which correspond to option A: PCO = 1.739 atm and PCO₂ = 0.461 atm.

User Amobiz
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