Asked by Anonymous
                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
            
            
        Answers
                    Answered by
            DrBob222
            
    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.
    
........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.
                                                    There are no AI answers yet. The ability to request AI answers is coming soon!
                                            
                Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.