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Equilibrium equations and coefficients

  1. Jul 11, 2007 #1
    Assignments given to me are drawn up in this manor:

    There is 1,34 mol of X, 1,69 mol of Y and 1,95 mol of Z at equilibrium.
    Calculate the equilibrium constant for the following reaction: [tex]2X + Y_2 \rightleftharpoons 2XY[/tex]


    or sometimes the coefficients are left out in the equilbrium equation.

    What I'm trying to say is that the coefficients are never the same as the number of moles given in the assignment.

    I want to use

    [tex]cC + dD \rightleftharpoons aA + bB[/tex]

    [tex]K=\frac{[A]^a ^b}{[C]^c[D]^d}[/tex]

    Should I use the coefficient in the equation or the exact number of moles from the assignment?
     
  2. jcsd
  3. Jul 11, 2007 #2

    chemisttree

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    The definition of the equilibrium constant is true regardless of how many moles of X or Y you (or anyone else) are given. Why would you want to change a definition? 2X + Y2 ---> 2XY means that under equilibrium there is a definite realtionship for each of these species.

    Start by writing the equlibrium equation for your example.
     
  4. Jul 11, 2007 #3
    I start with 1 mol H2 and 2 mol CO2.
    At one point, 85,5% of the hydrogen turned into water.

    They want the eq. constant of the following reaction:
    [tex]H_2(g) + CO_2(g) \rightleftharpoons H_2O(g) + CO(g)[/tex]

    Code (Text):
    ________________________________|H2_____|CO2____|H2O______|CO
    Amount of substance at start____|1______|2______|0________|0
    Change__________________________|-0,855_|-0,855_|+0,855___|+0,855
    Amount of substance at eq.______|0,145__|1,145__|0,855____|0,855
    Concentration at eq.____________|A______|B______|C________|D
    I didn't get my LaTeX table to work here, so I hope this is just as readable.

    If 85,5% of the 1 mol of hydrogen will turn into water, we will end up with 14,5% of the original amount; 0,145 mol of hydrogen. The other part of what was once hydrogen, is now water; an increase from 0 to 0,855.

    Facing the CO2, I'm not quite sure what happens. I'm guessing since the reaction uses as much H2 as it does CO2 (equivalence?), equal amounts are turned into products. Hence, 0,855 is turned into CO and we end up with 1,145 mol of CO2.

    Do correct me if I'm wrong.

    To be able to calculate the eq. constant, I need the concentrations at eq.
    Now, in a previous problem, I was handling liquids with a given volume. Calculating the concentration was simple.
    What do I do here?
     
  5. Jul 12, 2007 #4

    chemisttree

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    Excellent! You are most of the way there! Next, write out the expression for the equilibrium constant (Keq) using the above equation.


    Correct.

    No correction required...

    You can make up any volume you want to. It applies to each of the starting materials/products. Use something convenient... like a liter.
     
  6. Jul 12, 2007 #5
    So as long as I'm consistent, it doesn't matter how much I use?

    In that case, I'll go for a liter, as you mentioned. No division needed.

    Code (Text):
    ________________________________|H2_____|CO2____|H2O______|CO
    Amount of substance at start____|1______|2______|0________|0
    Change__________________________|-0,855_|-0,855_|+0,855___|+0,855
    Amount of substance at eq.______|0,145__|1,15___|0,855____|0,855
    Concentration at eq.____________|0,145__|1,15___|0,855____|0,855
    [tex]K=\frac{[H_2O][CO]}{[H_2][CO_2]}=\frac{0,855*0,855}{0,145*1,15}\approx 4,38[/tex]

    And there we go. :smile:
     
  7. Jul 12, 2007 #6

    chemisttree

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    Now try it with 23.7 liters and see what you get...
     
  8. Jul 13, 2007 #7
    I was thinking of the algebra behind it and...


    [tex]K=\frac{0,855*1*0,855*1}{0,145*1*1,15*1}[/tex]

    [tex]K=\frac{0,855*0,5*0,855*0,5}{0,145*0,5*1,15*0,5}[/tex]

    Since you apply the same volume to all the factors, they cancel each other out.

    It's the amount of substance that makes the eq. constant.

    Thanks CT!
     
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