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. What are the equilibrium partial pressures of carbon monoxide and carbon dioxide when 1.00 atm of carbon monoxide and excess iron(II) oxide react in a sealed container at 1000°C.Answer: write Kp expression, do ICE chart with pressures, 0.287 atm CO2 and 0.713 atm CO

Answer 1

0.713atm for CO and 0.287atm for CO₂

Step-by-step explanation:

Based on the reaction:

FeO(s) + CO(g) ⇋ Fe(s) + CO₂(g)

Kp is defined as:

`Kp = (P_(CO_2))/(P_(CO))` = 0.403

When 1.00 atm of CO react with an excess of FeO, the pressures in equilibrium are:

PCO = 1.00atm - x

PCO₂ = x

Where x represents the reaction coordinate.

Replacing in Kp expression:

`0.403 = (x)/(1-x)`

0.403 - 0.403x = x

0.403 = 1.403x

0.287atm = x

Thus, pressures in equilibrium are:

PCO = 1.00atm - x = 0.713atm

PCO₂ = x = 0.287atm