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General Chemistry - Chemical Equilibria

  1. Dec 3, 2007 #1
    General Chemistry -- Chemical Equilibria

    Suppose 1.22 atm of CH4(g), 2.57 atm of C2H6(g), and 15.00 atm of O2(g) are placed in a flask at a given temperature. The reactions are given below.

    CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) KP = 1.0x10^4
    2 C2H6(g) + 7 O2(g) 4 CO2(g) + 6 H2O(g) KP = 1.0x10^8

    Calculate the equilibrium pressures of all gases.

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    There seems to one unknown that cannot be solved for! Any help is appreciated! (I'd rather not have you post the answer, but rather the method) Thanks.
     
    Last edited: Dec 3, 2007
  2. jcsd
  3. Dec 4, 2007 #2

    Gokul43201

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    There should only be 2 unknowns. Let x atm of CH4 and y atm of C2H6 be consumed. From the stoichiometry, you should be able to write the loss/gain in partial pressures of the other 3 components in terms of x and y.
     
  4. Dec 5, 2007 #3
    I could be completely wrong, since i have little experience with complicated equilibrium problems, but what I think you should do is subtract the equations, and then do the appropriate math with the K values. Remember adding equations multiplies the K's, subtracting equations divides the K's. In this case you will need to multiply one of the equations too. This means you raise the K value to what ur multiplying by, for example if your multiplying by 2 u raise the K^2, if you multiply by 1/2(or dividing by 2) you will raise K^(1/2). And then it becomes a simple ICE(initial, change, equilibrium) equation.

    You should have 1 equation now and 1 K value. The ICE will give you the correct input variables for all the pressures(you should be able to only use one variable 'x' but i'm not sure since i'm not doing this out) then just solve for x, and then plug in the value for x to find the partial pressures of all the gases.

    I'm assuming these equations occur in the same place and you don't solve them seperately. Stoichiometry won't work in this case because the products of one equation drive the other equation to the left, and it would just be too complicated for stoichiometry(i think...)
     
    Last edited: Dec 5, 2007
  5. Dec 7, 2007 #4

    Gokul43201

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    No, it's not too complicated. And the stoichiometry is essential.
     
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