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

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