Answer:
We can solve this problem using the equilibrium expression and the stoichiometry of the reaction.
The equilibrium constant expression for the reaction is:
Keq = [Fe][CO2]/[FeO][CO]
Since the amount of Fe present is in excess, we can assume that its concentration remains essentially constant, and we can treat the reaction as if it were taking place between CO gas and solid FeO.
Let's start by writing the initial concentrations of CO and CO2 in the container before they reach equilibrium. We know that CO2 is injected at a pressure of 9.00 bar, so its initial concentration is:
[CO2] = 9.00 bar
Since there is no CO gas initially present, [CO] = 0.
The reaction stoichiometry tells us that for every mole of CO that reacts, one mole of CO2 is produced. Therefore, if x is the amount (in moles) of CO that reacts to reach equilibrium, the concentrations at equilibrium will be:
[CO2] = 9.00 bar + x bar
[CO] = x bar
At equilibrium, the equilibrium constant expression can be used to solve for x:
Keq = [Fe][CO2]/[FeO][CO]
0.90 = [Fe](9.00 bar + x bar)/[FeO](x bar)
We can simplify this expression by assuming that the amount of FeO converted to Fe and CO2 is small compared to the initial amount of FeO, and hence the amount of FeO left at equilibrium is approximately equal to the initial amount of FeO. Therefore, we can cancel out [FeO] from the above equation:
0.90 = [Fe](9.00 bar + x bar)/x
Solving for x, we get:
x = [Fe] * 9.00 bar / (0.90 + [Fe])
Now we can substitute the value of x into the equilibrium concentrations:
[CO2] = 9.00 bar + x bar
[CO] = x bar
We can also use the ideal gas law to relate the pressure of CO to its concentration:
P = nRT/V
where P is the pressure, n is the number of moles, R is the gas constant, T is the temperature, and V is the volume. Since the volume is not given, we can assume that it remains constant throughout the experiment and hence cancel it out from the equations. Therefore, the pressure of CO and CO2 at equilibrium will be:
[CO2] = 9.00 bar + x bar = 9.00 bar + ([Fe] * 9.00 bar / (0.90 + [Fe])) bar
[CO] = x bar = ([Fe] * 9.00 bar / (0.90 + [Fe])) bar
We can simplify this further by multiplying and dividing by 0.90:
[CO2] = (9.00 bar * (0.90 + [Fe]) + 9.00 bar * [Fe]) / (0.90 + [Fe]) = (8.10 bar + 9.00 bar * [Fe]) / (0.90 + [Fe])
[CO] = (9.00 bar * [Fe]) / (0.90 + [Fe])
Therefore, the partial pressure of CO at equilibrium is (9.00 bar * [Fe]) / (0.90 + [Fe]) bar, where [Fe] is the concentration of Fe in moles/liter. The units of the concentration depend on the volume of the container, which is not given.