Algebra for partial pressures....?
At 1100K, Kp = 0.25 for the rxn
2SO2(g) + O2(g) <---> 2SO3(g)
Calculate the equilibrium partial pressures of SO2, O2, and SO3 produced from an initial mixture in which Pso2 = Po2 = 0.50atm & Pso3 = 0.
So far I have set up...
2S02 = 0.5 - 2x
O2 = -.5 - x
2SO3 = 0.5 - 2x
I am very strong with math but my algebra has seemed to kill me on this specific problem. More-so these equilibrium problems where 2x is involved and the equilibrium constant is too big to leave out.
The answer in the book is
Pso2 = 0.38 atm
Po2 = 0.44 atm
Pso3 = 0.12 atm
Comments
Your expression for p(SO3) is not correct; it should be +2X (not 0.5-2X)
Kp = p(SO3)^2 / [p(SO2)^2 * p(O2)]
p(SO2) = 0.5 - 2X
p(O2) = 0.5 - X
p(SO3) = 0 + 2X = 2X
0.25 = (2X)^2 / [(0.5-2X)^2 * (0.5-X)]
I still end wirh a 3rd degree equation in X which I can only solve by approximation methods. I get X = 0.0621503
p(SO2) = 0.50 - 0.124 = 0.376 (0.38 atm)
p(O2) = 0.50-0.062 = 0.438 (0.44 atm)
p(SO3) = 0.00 + 0.124 = 0.124 (0.12 atm)
Use mole fraction extremely. finished moles = 6; finished stress = 10 atm. mole fraction of N2 is 3/6 = .5, so N2 is to blame for .5 X 10 atm = 5 atm.Mole fraction of O2 is two/6; mole fraction of CO2 is a million/6.. etc - crank it out all the partial pressures will upload as much as ten