0.100 mol of SO2 and 0.100 mol of O2 are introduced in a flask of 1.52 L. When in equilibrium, 203 is found to be 0.0916 mol. Determine kp.

2S03(g) <-> 2SO(g) + O2(g)

In order to solve this question 1st I made an ICE table and I got the values

2SO3 <-> 2SO + 02
I .100 .100
C .1- 0.0916 .1-.0458
E 0.0916 .0084 .0542

.0916/1.52 .0084/1.52 .0542/1.52
=.060M =.0055M =.0357M

Then I substitued each value into the kp expression and solved for the Kp. Is this correct?

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asked by K
  1. Sorry my ICE table came out wrong.

    It is actually suppose to be

    2SO3 <-> 2SO + 02
    I 0 .100 .100
    C 0 .1- 0.0916 .1-.0458
    E 0.0916 .0084 .0542

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    posted by K
  2. The boards don't recognize any space AFTER the first one but you can still space if you use periods; for example, as in
    ..........2SO3 ==> 2SO2 + O2.

    The equilibrium values look ok to me for SO3 and O2. For SO2, I think you have thrown away the last number which I wouldn't do. 0.1-0.0916 = 0.00840. (OK, so it's a zero but you shouldn't throw it away.)
    To that point you are ok. What you have calculated, and called Kp, actually is Kc. moles/L = M and you substituted into the Kc expression and solved for Kc.

    What you need to do next is to add all of the moles to find total moles, then find the mole fraction of each gas.
    XSO3 = moles SO3/total moles = ??
    XSO2 = moles SO2/total moles = ??
    XO2 = moles O2/total moles = ??
    Next you want the partial pressure of each. That is
    PSO3 = XSO3*Ptotal
    PSO2 = XSO2*Ptotal
    PO2 = XO2*Ptotal.
    Do you have a value for Ptotal in the problem, an equilibrium pressure given, or a temperature given in the problem/

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  3. No I wasn't, but can I just use the pressure formula of P() = nRT/V and then just add all of them together?

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    posted by K
  4. Okay I think I figured out how to do it all. Thanks for explaining.

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    posted by K
  5. You don't have a T; what will you substitute for T in P = nRT/V. If you have a total P you can calculate partial pressures of each then solve for Kp.

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