A 5.00 liter vessel contains carbon monoxide, carbon dioxide, hydrogen and water vapor at equilibrium at 980C. The equilibrium partial pressures are 300. torr, 300. torr, 90. torr, and 150 torr respectively. Carbon monoxide is then pumped into the system until the equilibrium partial pressures of hydrogen is 130. torr. Calculate the partial pressures of the other substances at the new equilibrium
Is this the equation?
........CO + H2O ==> H2 + CO2
E.......300..150.....90....300
Substitute into Kp expression and solve for Kp. I get 0.6.
Then x = pCO added.
.........CO + H2O ==> H2 + CO2
old E....300..150.....90....300
add......x.....................
new E.................130.....
So if pH2 - 90 and goes to 130 you must have added 40 torr. That makes added CO2 = 40 torr so new pCO2 = 300 + 40. If you added 40 torr to H2 and CO2 you must have taken away 40 from H2O the new pH2O must be 150-40 = 110 torr.
Substitute these new numbers into Keq expression and solve for pCO.
To calculate the partial pressures of the other substances at the new equilibrium, we can use the concept of the ideal gas law and the concept of equilibrium.
First, let's write down the balanced chemical equation for the reaction:
CO + H2O ↔ CO2 + H2
We are given the initial equilibrium partial pressures of the substances:
CO: 300. torr
CO2: 300. torr
H2: 90. torr
H2O: 150 torr
We also know that the reaction is at equilibrium, which means the forward and backward reaction rates are equal. From this information, we can construct an ICE (Initial, Change, Equilibrium) table as follows:
CO + H2O ↔ CO2 + H2
Initial 300. 150. 300. 90.
Change -x. +x.
Equilibrium. 300-x. 150-x. 300+x. 90+x.
As per the ICE table, the change in partial pressures of CO and H2O will be -x (since they are reactants), while the change in partial pressures of CO2 and H2 will be +x (since they are products).
Now, let's use the given information that the new equilibrium partial pressure of H2 is 130. torr.
Equilibrium partial pressures of substances:
CO: 300 - x
CO2: 300 + x
H2: 130. torr
H2O: 150 - x
From the reaction equation, we can see that the mole ratio between H2 and CO is 1:1. Therefore, the change in partial pressure of H2 is equal to the change in partial pressure of CO.
We can set up an equation using the partial pressures:
130 = 300 - x
To solve for x, we can rearrange the equation:
x = 300 - 130
x = 170. torr
Now that we know the value of x, we can calculate the partial pressures of the other substances at the new equilibrium:
CO: 300 - x = 300 - 170 = 130. torr
CO2: 300 + x = 300 + 170 = 470. torr
H2: 130. torr (already given)
H2O: 150 - x = 150 - 170 = -20. torr
After calculating, we find that the partial pressures of the substances at the new equilibrium are:
CO: 130. torr
CO2: 470. torr
H2: 130. torr
H2O: -20. torr
Note: The negative value for H2O partial pressure indicates that the water is mostly in the gaseous state and not condensed in the vessel.