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Equilibrium Concentration

  1. Mar 23, 2006 #1
    A 1.00L vessel contains at equilibrium 0.300 mol of [itex]N_2[/itex], 0.400 mol [itex]H_2[/itex] and 0.100 mol [itex]NH_3[/itex]. If the temp is maintained constant, how many moles of [itex]H_2[/itex] must be introduced into the vessel in order to double the equilibrium concentration of [itex]NH_3[/itex]

    I said that if I let the change in conc. of [itex]H_2[/itex] be equal to x, then the change in conc. of [itex]\[NH_3\][/itex] would be 2/3x (from the chemical equation). Can't I just equation 2/3x with 0.100? It just seems too easy...
    Last edited: Mar 23, 2006
  2. jcsd
  3. Mar 23, 2006 #2
    The reaction is:

    [itex]N_2 + 3H_2[/itex] -> [itex]2NH_3[/itex]

    So, in equilibrium we have:
    Concentration of [itex]N_2[/itex]: 0.300 mol.dm-3
    Concentration of [itex]3H_2[/itex]: 0.400 mol.dm-3
    Concentration of [itex]2NH_3[/itex]: 0.100 mol.dm-3

    With this values you can determine the equilibrium constant Kc. I think you know the formula, if not1.

    But the problem asks how many moles of H2 must be introduced into the vessel in order to double the equilibrium concentration of NH3.

    Then in the new equilibrium the concentration of each substance must be:
    Concentration of [itex]N_2[/itex]: 0.300 + x mol.dm-3
    Concentration of [itex]3H_2[/itex]: 0.400 + 3x mol.dm-3
    Concentration of [itex]2NH_3[/itex]: 0.200 mol.dm-3

    Now you apply the equilibrium constant formula and determine x and the solution of the problem.

    1 http://en.wikipedia.org/wiki/Equilibrium_Constant
    Last edited by a moderator: Mar 23, 2006
  4. Mar 23, 2006 #3
    So there will be four solutions?
  5. Mar 24, 2006 #4
    Four solutions?
    The solution to the problem itself should be 3x. And you find x as I explained.
    If you don't understand I will resolve the problem entirely.
  6. Mar 24, 2006 #5
    I see now. Thanks for your help.
  7. Mar 24, 2006 #6
    You're welcome. I hope you check if this is correct with your teacher.
  8. Mar 25, 2006 #7


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    PPonte gave some very good points, except the stoichiometries should be modified

    Use factor labeling, and use x for the moles of hydrogen gas introduced.
  9. Mar 25, 2006 #8
    CGT, you are right. But I just used this stoichiometries in order to simplify the calculations since I prefer to use integral numbers than fractions. If x is the number of moles of hydrogen introduced, the final concentration of N2 is 0.300 + x/3 M.
    Last edited by a moderator: Mar 25, 2006
  10. Mar 23, 2007 #9
    Question: How the heck do you solve for x? I found that K = 0.52

    The equilibrium equation would be

    K = [[itex]2NH_3[/itex]]^2 / [[itex]H_2[/itex]]^3[[itex]N_2[/itex]]

    0.52 = (0.200)^2 / (0.400+3x)^3(0.300+x)

    Once you expand that, you get some ridiculous quartic equation. Is there an easier way to solve for x? :S
    Last edited: Mar 23, 2007
  11. Apr 26, 2008 #10
    hey... sorry if the thread is old, but I am having the same problem
    I can't solve
    0.52 = ((0.2)^2)/((0.3+x)*(0.4+3x)^3)
    I get a huge 4th degree polynomial
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